Category: Class 11

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TN State Board 11th Physics Model Question Paper 5 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

PART – I

Answer all the questions: [15 × 1 = 15]

Question 1.
Two protons are travelling. along the same straight path but in opposite directions. The relative velocity between the two is ……………………
(a) c
(b) \(\frac{c}{2}\)
(c) 2c
(d) 0
Hint:
One of the velocity V₁ = V; Other velocity V₂ = -V
Relative velocity (V_rel) = \(\frac{V_{1}-V_{2}}{\left[1+\frac{V^{2}}{C^{2}}\right]}\) = \(\frac{V-(-V)}{\left[1+\frac{V^{2}}{C^{2}}\right]}\) = \(\frac{2v}{2}\)
V = C
V_relative = C
Answer:
(a) c

Question 2.
If the Earth stops rotating about its own axis, g remains unchanged at …………………..
(a) Equator
(b) Poles
(c) Latitude of 45°
(d) No where
Answer:
(b) Poles

Question 3.
When train stops, the passenger moves forward. It is due to ……………………
(a) Inertia of passenger
(b) Inertia of train
(c) Gravitational pull by Earth
(d) None of the above
Answer:
(a) Inertia of passenger

Question 4.
A particle of mass m moves in the xy plane with a velocity v along the straight line AB. If the angular momention of the particle with respect to origin O is L_A when it is at A and L_B when it is at B, then …………………….
(a) L_A = L_B
(b) L_A < L_B
(c) L_A > L_B
(d) The relationship between L_A and L_B depends uopn the slope of the line AB

Hint:
Magnitude of L is, L = mvr sin ϕ = mvd
d = r sin ϕ is the distance of closest approach of the particle so the origin, as ‘d’ is same for both particles.
So, L _A = L_B
Answer:
(a) L_A = L_B

Question 5.
A couple produces ……………………….
(a) Pure rotation
(b) Pure translation
(c) Rotation and translation
(d) No motion
Answer:
(a) Pure rotation

Question 6.
A body starting from rest has an acceleration of 20 ms~2 the distance travelled by it in the sixth second is ……………………..
(a) 110 m
(b) 130m
(c) 90m
(d) 50 m
Hint:
Distance travelled in n^th second, u = 0
S_n = u + \(\frac{1}{2}\)a(2n – 1)
S₆ = 0 + \(\frac{1}{2}\) × 20 × (2 × 6 – 1); S₆ = 110 m
Answer:
(a) 110 m

Question 7.
A lift of mass 1000 kg, which is moving with an acceleration of 1m/s² in upward direction has tension has developed in its string is ………………………
(a) 9800 N
(b) 10800 N
(c) 11000 N
(d) 10000 N
Hint:
Tension, T = mg + ma = m(g + a) = 1000 (10 + 1)
T = 11000 N
Answer:
(c) 11000 N

Question 8.
The relation between acceleration and displacement of four particles are given below ………………………..
(a) a_x = 2x
(b) a_x = + 2x²
(c) a_x = -2x²
(d) a_x = -2x
Answer:
(d) a_x = -2x

Question 9.
A sonometer wire is vibrating in the second overtone. In the wire there are, ……………………..
(a) Two nodes and two antinodes
(b) One node and two antinodes
(c) Four nodes and three antinodes
(d) Three nodes and three antinodes
Answer:
(d) Three nodes and three antinodes

Question 10.
Which of the following is the graph between the light (h) of a projectile and time (t), when it is projected from the ground ………………………..
(a)
(b)
(c)
(d)
Answer:
(c)

Question 11.
According to kinetic theory of gases, the rms velocity of the gas molecules is directly proportional to ………………………
(a) \(\sqrt{T}\)
(b) T³
(c) T
(d) T⁴
Hint:
The rms velocity, V_rms = \(\sqrt{3KT/m}\) ⇒ V_rms ∝ \(\sqrt{T}\)
Answer:
(a) \(\sqrt{T}\)

Question 12.
A body of mass m moving with velocity v collides head on with another body of mass 2m which is initially at rest. The ratio of K.E of colliding body before and after collision will be ……………………..
(a) 1 : 1
(b) 2 : 1
(c) 4 : 1
(d) 9 : 1
Hint:
KE of colliding bodies before collision = \(\frac{1}{2}\) mv²
After collision the mass = m + 2m = 3m
velocity becomes V’ = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)v = \(\frac{mv}{3m}\) = \(\frac{v}{3}\)
KE after collision = \(\frac{1}{2}\)m (\(\frac{V}{3}\)² = \(\frac{1}{9}\) (\(\frac{1}{2}\) mv²)
\(\frac{\mathrm{KE}_{\text {before }}}{\mathrm{KE}_{\text {after }}}=\frac{\frac{1}{2} \mathrm{mv}^{2}}{\frac{1}{9}\left(\frac{1}{2} \mathrm{mv}^{2}\right)}=9: 1\)
Answer:
(d) 9 : 1

Question 13.
Four particles have velocity 1, 0, 2 and 3ms^-1. The root mean square velocity of the particles is ……………………..
(a) 3.5 ms^-1
(b) \(\sqrt{3.5}\) ms^-1
(c) 1.5ms^-1
(d) Zero
Hint:

Answer:
(b) \(\sqrt{3.5}\) ms^-1

Question 14.
Two vibrating tuning forks produce progressive waves given by y₁ = 4 sin 500 πt and y₂ = 2 sin 506 πt where t is in seconds Number of beat produced per minute is ………………………
(a) 360
(b) 180
(c) 3
(d) 60
Hint:

f₂ – f₁ = 3 = beats per sec and 3 × 60 = 180 beats per min

Answer:
(b) 180

Question 15.
Workdone by a simple pendulum in one complete oscillation is …………………………..
(a) Zero
(b) Jmg
(c) mg cos θ
(d) mg sin θ
Answer:
(a) Zero

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
A girl is swinging in the sitting position. How will the period of the swing be changed if she stands up?
Answer:
This can be explained using the concept of a simple pendulum. We know that the time period of a simple pendulum is given by
T = 2π\(\sqrt{l/g}\) i.e; T ∝\(\sqrt{l}\)
When the girl stands up, the distance between the point of suspension and the centre of mass of the swinging body decreases i.e., l decreases, so T will also decrease.

Question 17.
A car starts to move from rest with uniform acceleration 10 ms^-2 then after 2 sec, what is its velocity?
Answer:
a =10 ms^-2;
t = 2s;
w = 0;
v = ?
v = u + at
v = 0 + 10 × 2
= 20 ms^-1

Question 18.
State Lami’s theorem?
Answer:
If a system of three concurrent and coplanar forces is in equilibrium, then Lami’s theorem states that the magnitude of each force of the system is proportional to sine of the angle between the other two forces. The constant of proportionality is same for all three forces.

Question 19.
Due to the action of constant torque, a wheel from rest makes n rotations in t seconds? Show that the angular acceleration of a wheel as \(\frac{4 \pi n}{t^{2}}\) rad s^-2
Answer:
Initial angluar velocity ω₀ = 0
Number of rotations in t seconds = n
angular displacement θ = 2πn
but, θ = ω₀t + \(\frac{1}{2}\)αt²
2πn = \(\frac{1}{2}\)αt²
α = \(\frac{4 \pi n}{t^{2}}\)

Question 20.
Why a given sound is louder in a hall than in the open?
Answer:
In a hall, repeated reflections of sound take place from the walls and the ceiling. These reflected sounds mix with original sound which results in increase the intensity of sound. But in open, no such a repeated reflection is possible. .’. sound will not be louder as in hall.

Question 21.
What are the differences between connection and conduction?
Answer:
Conduction:
Conduction is the process of direct transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the colder one. The objects which allow heat to travel easily through them are called conductors.

Convection:
Convection is the process in which heat transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.

Question 22.
Why two holes are made to empty an oil tin?
Answer:
When oil comes out through a tin with one hole, the pressure inside the tin becomes less than the atmospheric pressure, soon the oil stops flowing out. When two holes are made in the tin, air keeps on entering the tin through the other hole and maintains pressure inside.

Question 23.
If the length of the simple pendulum is increased by 44% from its original length, calculate the percentage increase in time period of the pendulum?
Answer:
Since T ∝ \(\sqrt{l}\) = Constant \(\sqrt{l}\)

∴ T_f = 1.2 T_i = T_i + 20% T_i

Question 24.
When do the real gases obey more correctly the gas equation PV = nRT?
Answer:
An ideal gas is one whose molecules have zero volume and no mutual force between them. At low pressure, the volume of a gas is large and so the volume occupied by the molecules is negligible in comparison to the volume of the gas.

At high temperature, the molecules have large velocities and so the intermolecular force has no influence on their motion. Hence at low pressure and high temperature, the behaviour of real gases approach the ideal gas behaviour.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
A stone is thrown upwards with a speed v from the top of a tower. It reaches the ground with a velocity 3v. What is the height of the tower?
Answer:
From equation of motion,
v’ = u + at ……………….. (1)
h = ut + \(\frac{1}{2}\) at²
here, v’ = av; u = v; a = +g
Using equ. (1)
3v = v + gt ⇒ 3v – v = gt
t = \(\frac{2v}{g}\)

Substitute ‘t’ value in equ. (2)
h = v(\(\frac{2v}{g}\)) + \(\frac{1}{2}\)g(\(\frac{2v}{g}\))² = \(\frac { 2v^{ 2 } }{ g } \) + \(\frac{1}{2}\)g (\(\frac{2v}{g}\))²
h = \(\frac { 2v^{ 2 } }{ g } \) + \(\frac { 2v^{ 2 } }{ g } \)
= \(\frac { 4v^{ 2 } }{ g } \) g = 10 ms^-2
= \(\frac { 4v^{ 2 } }{ 10 } \); h = \(\frac { 2v^{ 2 } }{ 5 } \)

Question 26.
An object is projected at an angle such that the horizontal range is 4 times of the maximum height. What is the angle of projection of the object?
Answer:
Given:
Horizontal range = 4 H_max
Horizontal range = \(\frac{u^{2} \sin 2 \theta}{g}\) = \(\frac{2 u^{2} \sin \theta \cos \theta}{g}\)
Maximum height = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
as given, \(\frac{2 u^{2} \sin \theta \cos \theta}{g}\) = \(\frac{4 u^{2} \sin ^{2} \theta}{2 g}\)
2 cos θ = 2 sin θ
tan θ = 1
∴ θ = 45°

Question 27.
A room contains oxygen and hydrogen molecules in the ratio 3 : 1. The temperature of the room is 27°C. The molar mass of 0₂ is 32 g mol^-1 and for H₂, 2 g mol^-1. The value of gas constant R is 8.32 J mol^-11 k^-1. Calculate rms speed of oxygen and hydrogen molecule?
Answer:
(a) Absolute Temperature T = 27°C = 27 + 273 = 300 K.
Gas constant R = 8.32 J mol^-1 K^-1
For Oxygen molecule: Molar mass

M = 32 gm/mol = 32 × 10^-3 kg mol^-1
rms speed v_rms = \(\sqrt { \frac { 3RT }{ M } } \) = \(\sqrt{\frac{3 \times 8.32 \times 300}{32 \times 10^{-3}}}\) = 483.73 ms^-1 ~ 484ms^-1

For Hydrogen molecule: Molar mass M = 2 × 10^-3 kg mol^-1
rms speed v_rms = \(\sqrt { \frac { 3RT }{ M } } \) = \(\sqrt{\frac{3 \times 8.32 \times 300}{2 \times 10^{-3}}}\) = 1934 ms^-1 = 1.93 K ms^-1

Note that the rms speed is inversely proportional to \(\sqrt{M}\) and the molar mass of oxygen is 16 times higher than molar mass of hydrogen. It implies that the rms speed of hydrogen is 4 times greater than rms speed of oxygen at the same temperature. \(\frac{1934}{484}\) ~ 4.

Question 28.
Explain about an angle of friction?
Answer:
The angle of friction is defined as the angle between the normal force (N) and the resultant force (R) of normal force and maximum friction force (\(f_{s}^{\max }\))

In the figure the resultant force is R = \(\sqrt{\left(f_{s}^{\max }\right)^{2}+\mathrm{N}^{2}}\)
tan θ = \(\frac{f_{s}^{\max }}{\mathrm{N}}\) ………………….. (1)

But from the frictional relation, the object begins to slide when \(f_{s}^{\max }=\mu_{\mathrm{s}} \mathrm{N}\)
or when \(\frac{f_{s}^{\max }}{\mathrm{N}}\) = µ_s ………………….. (2)

From equations (1) and (2) the coefficient of static friction is
µ_s = tan θ ……………………. (3)
The coefficient of static friction is equal to tangent of the angle of friction.

Question 29.
How does resolve a vector into its component? Explain?
Answer:
component of a resolve:
In the Cartesian coordinate system any vector \(\vec { A } \) can be resolved into three components along x, y and z directions. This is shown in figure. Consider a 3-dimensional coordinate system. With respect to this a vector can be written in component form as
\(\vec { A } \) = A_x \(\hat { i } \) + A_y\(\hat { j } \) + A_z\(\hat { k } \)
Components of a vector in 2 dimensions and 3 dimensions

Here Ax is the x-component of \(\vec { A } \), Ay is the y-component of \(\vec { A } \) and A_z is the z component of \(\vec { A } \).
In a 2-dimensional Cartesian coordinate system (which is shown in the figure) the vector \(\vec { A } \) is given by
\(\vec { A } \) = A_x \(\hat { i } \) + A_y \(\hat { j } \)

If \(\vec { A } \) makes an angle θ with x axis, and A_x and A_y are the components of A along x-axis and y-axis respectively, then as shown in figure,
A_x θ = A cos θ, A = A sin θ
where ‘A’ is the magnitude (length) of the vector \(\vec { A } \), A = \(\sqrt{\mathrm{A}_{x}^{2}+\mathrm{A}_{y}^{2}}\)

Question 30.
Derive an expression for energy of an orbiting satellite?
Answer:
The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is,
U = \(-\frac{\mathrm{GM}_{s} \mathrm{M}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\) ………………….. (1)

Here M_s -mass of the satellite, M_E -mass of the Earth, R_E – radius of the Earth.
The Kinetic energy of the satellite is
K.E = \(\frac{1}{2}\) M_sv² ………………….. (2)

Here v is the orbital speed of the satellite and is equal to
v = \(\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}}\)

Substituting the value of v in (2) the kinetic energy of the satellite becomes,
K.E = \(\frac{1}{2} \frac{\mathrm{GM}_{\mathrm{E}} \mathrm{M}_{s}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\)

Therefore the total energy of the satellite is
E = \(\frac{1}{2}\) \(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{M}_{s}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}-\frac{\mathrm{GM}_{s} \mathrm{M}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\)
E = \(-\frac{\mathrm{GM}_{s} \mathrm{M}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\)
The total energy implies that the satellite is bound to the Earth by means of the attractive gravitational force.

Note:
As h approaches ∞, the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth’s gravity and is not bound to Earth at large distance.

Question 31.
Explain in detail Newton’s law of cooling?
Answer:
Newton’s law of cooling:
Newton’s law of cooling body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac{dQ}{dT}\) ∝(T – T_s) …………………. (1)
The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,

T = Temperature of the object
T_s = Temperature of the surrounding

From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.

Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount dT in time dt, then the amount of heat lost is,
dQ = msdT …………………….. (2)

Dividing both sides of equation (2) by dt
\(\frac{dQ}{dT}\) = \(\frac{msdT}{dt}\) ………………….. (3)

From Newton’s law of cooling
\(\frac{dQ}{dT}\) ∝(T – T_s)
\(\frac{dQ}{dT}\) = -a(T – T_s) ………………………… (4)

Where a is some positive constant.
From equation (3) and (4)
-a (T – T_s) = ms\(\frac{dT}{dt}\)
\(\frac{d \mathrm{T}}{\mathrm{T}-\mathrm{T}_{s}}\) = -a\(\frac{a}{ms}\) dt …………………… (5)

Integrating equation (5) on both sides,

Where b₁ is the constant of integration. Taking exponential both sides we get,
T = T_s + \(b_{2} e^{\frac{-a}{m s} t}\) …………………….. (6)
Here b₂ = e^b1 Constant

Question 32.
Explain Laplace’s correction?
Answer:
Laplace’s correction: In 1816, Laplace satisfactorily corrected this discrepancy by assuming that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast.

Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a bad conductor of heat. Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the gas obeys Poisson’s law (not Boyle’s law as Newton assumed), which is
PV^γ = Constant …………………. (1)
where, γ = \(\frac { C_{ p } }{ C_{ v } } \) which is the ratio between specific heat at constant pressure and specific heat at constant volume. Differentiating equation (1) on both the sides, we get
\(V^{ \gamma }dP+P(\gamma ^{ V\gamma -1 }dV)=0\)

or

\(\gamma \mathrm{P}=-\mathrm{V} \frac{d p}{d \mathrm{V}}=\mathrm{B}_{\mathrm{A}}\) ………………….. (2)
where, B_A is the adiabatic bulk modulus of air. Now, substituting equation (2) in equation
V = \(\sqrt{\frac{B}{\rho}}\), the speed of sound in air is
v_A = \(\sqrt{\frac{\mathrm{B}_{\mathrm{A}}}{\rho}}=\sqrt{\frac{\gamma \mathrm{P}}{\rho}}=\sqrt{\gamma v_{\mathrm{T}}}\)
Since air contains mainly, nitrogen, oxygen, hydrogen etc, (diatomic gas), we take γ = 1.47. Hence, speed of sound in air is v_A = (\(\sqrt{1.4}\)) (280 m s^-1) = 331.30 ms^-1, which is very much closer to experimental data.

Question 33.
Explain types of equilibrium?
Answer:

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
What are the applications of dimensional analysis?
Verify s = ut + \(\frac{1}{2}\)at² by dimensional analysis?
Answer:

Given equation is dimensionally correct as the dimensions on the both side are same. Applications of dimensional analysis

1. Convert a physical quantity from one system of units to another.
2. Check the dimensional correctness of a given physical equation.
3. Establish relations among various physical quantities.

[OR]

(b) Explain the types of equilibrium with suitable examples?
Answer:
Translational motion – A book resting on a table.
Rotational equilibrium – A body moves in a circular path with constant velocity.
Static equilibrium – A wall-hanging, hanging on the wall.
Dynamic equilibrium – A ball decends down in a fluid with its terminal velocity.
Stable equilibrium – A table on the floor A pencil
Unstable equilibrium – standing on its tip.
Neutral equilibrium – A dice rolling on a game board.

Question 35 (a)
Explain the motion of block connected by a string in vertical motion?
Answer:
When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Vertical motion:
Consider two blocks of masses m₁ and m₂ (m₁ > m₂) connected by a light and inextensible string that passes over a pulley as shown in Figure 1.

Let the tension in the string be T and acceleration a.

When the system is released, both the blocks start, Two blocks connected by a string moving, m₂ vertically upward and m₁ downward with same acceleration a. The gravitational force m₁g on mass m₁ is used in lifting the mass m₂.

The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure 2.

Applying Newton’s second law for mass m₂
T\(\hat { j } \) – m₂\(\hat { j } \)g = m₂ a\(\hat { j } \)

The left hand side of the above equation is the total force that acts on m₂ and the right hand side is the product of mass and acceleration of m₂ in direction.

By comparing the components on both sides, we get
T -m₂g = m₂a ……………………. (1)

Similarly, applying Newton’s law second law of for mass m₁
T\(\hat { j } \) – m₁g\(\hat { j } \) = -m₁a\(\hat { j } \)

As mass m₁ moves downward (-\(\hat { j } \)), its accleration is along (-\(\hat { j } \))
By comparing the components on both sides, we get
T – m₁g = -m₁a
m₁g – T = m₁a …………………….. (2)

Adding equations (1) and (2), we get
m₁g – m₂g = m₁a + m₂a
(m₁g – m₂)g = (m₁ + m₂)a …………………….. (3)

From equation (3), the acceleration of both the masses is
a = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)g ……………………… (4)

If both the masses are equal (m₁ = m₂), from equation (4)
a = 0

This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest. To find the tension acting on the string, substitute the acceleration from the equation (4)
T – m₂g = m₂ \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)g
T = m₂g + m₂\(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)g ……………………… (5)
By taking m₂g common in the RHS of equation (5)

Equation (4) gives only magnitude of accleration.

For mass m₁ g the accleration vector is given by \(\vec{a}=-\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) g \hat{j}\)
For mass m₂ g the accleration vector is given by \(\vec{a}=-\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) g \hat{j}\)

[OR]

(b) Derive the kinematic equation of motion for constant acceleration?
Answer:
Consider an object moving in a straight line with uniform or constant acceleration ‘a’. Let u be the velocity of the object at time t = 0, and v be velocity of the body at a later time t.

Velocity – time relation:

(I) The acceleration of the body at any instant is given by the first derivative of the velocity with respect to time,
a = \(\frac{dv}{dt}\) or dv = a dt
Integrating both sides with the condition that as time changes from 0 to t, the velocity changes from u to v. For the constant acceleration,

Displacement – time relation:

(II) The velocity of the body is given by the first derivative of the displacement with respect to time.
v = \(\frac{ds}{dt}\) or ds = vdt
and since v = u + at,
We get ds = (u + at)dt
Assume that initially at time t = 0, the particle started from the origin. At a later time t, the particle displacement is s. Further assuming that acceleration is time-independent, we have
\(\int_{0}^{s} d s=\int_{0}^{t} u d t+\int_{0}^{t} a t d t \text { or } s=u t+\frac{1}{2} a t^{2}\) ……………………. (2)
Velocity – displacement relation

(III) The acceleration is given by the first derivative of velocity with respect to time.
a = \(\frac{dv}{dt}\) = \(\frac{dv}{ds}\) \(\frac{ds}{dt}\) = \(\frac{dv}{ds}\) v [since ds/dt = v] where s is displacement traversed.
This is rewritten as a = \(\frac{1}{2}\) \(\frac { dv^{ 2 } }{ s } \) or ds = \(\frac{1}{2a}\) d(v²)
Integrating the above equation, using the fact when the velocity changes from u² to v², displacement changes from u² to v², we get

We can also derive the displacement s in terms of initial velocity u and final velocity v.
From equation we can write,
at = v – u
Substitute this in equation, we get
s = ut + \(\frac{1}{2}\) (v -u)t
s = \(\frac{(u+v)t}{2}\)

Question 36 (a).
State and prove perpendicular axis theorem?
Answer:
Perpendicular axis theorem: This perpendicular axis theorem holds good only for plane laminar objects.

The theorem states that the moment of inertia of a plane laminar body about an axis perpendicular to its plane is equal to the sum of moments of inertia about two perpendicular axes lying in the plane of the body such that all the three axes are mutually perpendicular and have a common point.

Let the X and Y-axes lie in the plane and Z-axis perpendicular to the plane of the laminar object. If the moments of inertia of the body about X and Y-axes are Ix and IY respectively and Iz is the moment of inertia about Z-axis, then the perpendicular axis theorem could be expressed as,
I_z = I_x + I_y

To prove this theorem, let us consider a plane laminar object of negligible thickness on which lies the origin (O). The X and Y-axes lie on the plane and Z-axis is perpendicular to it as shown in figure. The lamina is considered to be made up of a large number of particles of mass m. Let us choose one such particle at a point P which has coordinates (x, y) at a distance r from O.

The moment of inertia of the particle about Z-axis is, mr².
The summation of the above expression gives the moment of inertia of the entire lamina about Z-axis as,
I_z = Σ mr²
Here, r² = x² + y²
Then, I_z = Σm(x² + y²)
I_z = Σmx² + Σmy²
In the above expression, the term Σmx² is the moment of inertia of the body about the Y-axis and similarly the term Σmy² is the moment of inertia about X-axis. Thus,
I_X = Σmy² and I_Y = Σmx²
Substituting in the equation for I_Z gives, I_Z = I_X + I_Y
Thus, the perpendicular axis theorem is proved.

[OR]

(b) Explain in detail the triangle law of addition?
Answer:
Let us consider two vectors \(\vec { A } \) and \(\vec { B } \) as shown in figure.

To find the resultant of the two vectors we apply the triangular law of addition as follows:

Represent the vectors \(\vec { A } \) and \(\vec { B } \) by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.

To explain further, the head of the first vector \(\vec { A } \) is connected to the tail of the second vector \(\vec { B } \). Let θ be the angle between A and B. Then R is the resultant vector connecting the tail of the first vector \(\vec { A } \) to the head of the second vector \(\vec { B } \).

The magnitude of \(\vec { R } \) (resultant) is given geometrically by the length of \(\vec { R } \)(OQ) and the direction of the resultant vector is the angle between \(\vec { R } \) and \(\vec { A } \). Thus we write \(\vec { R } \) = \(\vec { A } \) + \(\vec { B } \).
\(\overline { OQ } \) = \(\overline { OP } \) + \(\overline { PQ } \)

1. Magnitude of resultant vector:
The magnitude and angle of the resultant vector are determined by using triangle law of vectors as follows.

From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

From figure, let R is the magnitude of the resultant of \(\vec { A } \) and \(\vec { B } \).

For ∆OBN, we have OB² = ON² + BN²
⇒ R² = (A + B cos θ)² + (B sin θ)²
⇒ R² = A² + B² cos² θ + 2AB cos θ + B²sin² θ
⇒ R² = A² + B²(cos² θ + sin² θ) + 2AB cos θ
⇒R² = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

2. Direction of resultant vectors:
If θ is the angle between \(\vec { A } \) and \(\vec { B } \), then
|\(\vec { A } \) + \(\vec { B } \)| = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)
If \(\vec { R } \) makes an angle α with \(\vec { A } \), then in ∆OBN,
tan α = \(\frac{BN}{ON}\) = \(\frac{BN}{OA+AN}\)
tan α = \(\frac{B \sin \theta}{A+B \cos \theta}\) ⇒ α = tan^-1\(\left(\frac{B \sin \theta}{A+B \cos \theta}\right)\)

Question 37 (a).
Explain in detail the various types of errors?
Answer:
The uncertainty in a measurement is called an error. Random error, systematic error and gross error are the three possible errors.

(I) Systematic errors:
Systematic errors are reproducible inaccuracies that are consistently in the same direction. These occur often due to a problem that persists throughout the .experiment. Systematic errors can be classified as follows.

(1) Instrumental errors:
When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.

(2) Imperfections in experimental technique or procedure:
These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied

(3) Personal errors:
These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.

(4) Errors due to external causes:
The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity, or pressure during measurements may affect the result of the measurement.

(5) Least count error:
Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument’s resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.

(II) Random errors:
Random errors may arise due to random and unpredictable variations in experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer who performs the experiment. Random errors are sometimes called “chance error”.

When different readings are obtained by a person every time he repeats the experiment, personal error occurs. For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.
If n number of trial readings are taken in an experiment, and the readings are
a₁, a₂, a₃, ……………………….. a_n. The arithmetic mean is

[OR]

(b) To move an object, which one is easier, push or pull? Explain?
Answer:
When a body is pushed at an arbitrary angle θ (0 to \(\frac{π}{2}\)), the applied force F can be resolved into two components as F sin θ parallel to the surface and F cos θ perpendicular to the surface as shown in figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is no acceleration along the vertical direction the normal force N is equal to
N_push = mg + F cos θ …………………. (1)
As a result the maximal static friction also increases and is equal to
\(f_{s}^{\max }=\mu_{r} N_{\text {push }}=\mu_{s}(m g+F \cos \theta)\) …………………. (2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.

When an object is pulled at an angle 0, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is
N_pull = mg – F cos θ …………………….. (3)

Question 38 (a).
Describe the method of measuring angle of repose?
Answer:
Angle of Repose Consider an inclined plane on which an object is placed, as shown in figure. Let the angle which this plane makes with the horizontal be θ. For small angles of θ, the object may not slide down. As θ is increased, for a particular value of θ, the object begins to slide down. This value is called angle of repose. Hence, the angle of repose is the angle of inclined plane with the horizontal such that an object placed on it begins to slide.

Let us consider the various forces in action here. The gravitational force mg is resolved into components parallel (mg sin θ) and perpendicular (mg cos θ) to the inclined plane. The component of force parallel to the inclined plane (mg sin θ) tries to move the object down. The component of force perpendicular to the inclined plane (mg cos θ) is balanced by the Normal force (N).
N = mg cos θ ……………… (1)

When the object just begins to move, the static friction attains its maximum value
f_s = \(f_{s}^{\max }\)

This friction also satisfies the relation
\(f_{s}^{\max }\) = µ_s = mg sin θ …………………… (2)

Equating the right hand side of equations (1) and (2),
\(\left(f_{s}^{\max }\right) / \mathrm{N}=\sin \theta / \cos \theta\)

From the definition of angle of friction, we also know that
tan θ = µ_s ……………………… (3) in which θ is the angle of friction.

Thus the angle of repose is the same as angle of friction. But the difference is that the angle of repose refers to inclined surfaces and the angle of friction is applicable to any type of surface.

[OR]

(b) A block of mass m slides down the plane inclined at an angle 60° with an acceleration g/2. Find the co-efficient of kinetic friction?
Answer:
Kinetic friction comes to play as the block is moving on the surface.
The forces acting on the mass are the normal force perpendicular to surface, downward gravitational force and kinetic friction along the surface.
Along the x-direction

There is no motion along the y-direction as normal force is exactly balanced by the mg cos θ.
mg cos θ = N = mg/2

[OR]

(c) Write a note on triangulation method and radar method to measure larger distances?
Answer:
Triangulation method for the height of an accessible object:
Let AB = h be the height of the tree or tower to 6e measured. Let C be the point of observation at distance x from B. Place a range finder at C and measure the angle of elevation, ACB = θ as shown in figure.

From right angled triangle ABC,
tan θ = \(\frac{AB}{BC}\) = \(\frac{h}{x}\)
(or) height h = x tan θ
Knowing the distance x. the height h can be detennined.

RADAR method:
The word RADAR stands for radio detection and ranging. A radar can be used to measure accurately the distance of a nearby planet such as Mars. In this method, radio waves are sent from transmitters which, after reflection from the planet, are detected by the receiver.

By measuring, the time interval (t) between the instants the radio waves are sent and received, the distance of the planet can be determined as where v is the speed of the radio wave. As the time taken (t) is for the distance covered during the forward and backward path of the radio waves, it is divided by 2 to get the actual distance of the object. This method can also be used to determine the height, at which an aeroplane flies from the ground.

[OR]

(d) Jupiter is at a distance of 824.7 million km from the earth. Its angular diameter is measured to be 35.75”. Calculate the diameter of jupiter?
Answer:
Given,
Distance of Jupiter = 824.7 × 10⁶ km = 8.247 × 10¹¹ m
angular diameter = 35.72 × 4.85 × 10^-6 rad = 173.242 × 10^-6 rad
= 1.73 × 10^-4rad
∴ Diameter of Jupiter D = θ × d = 1.73 × 10^-4 rad × 8.247 × 10¹¹ m
14.267 × 10⁷ m = 1.427 × 10⁸ m (or) 1.427 × 10⁵ km

Students can Download Tamil Nadu 11th Maths Model Question Paper 4 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Maths Model Question Paper 4 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 90

PART – I

I. Choose the correct answer. Answer all the questions: [20 × 1 = 20]

Question 1.
If n(A) = 2 and n(B∪C) = 3 then n[(A × B) ∪ (A × C)] is ………………..
(a) 2³
(b) 3²
(c) 6
(d) 5
Answer:
(c) 6

Question 2.
For any two sets A and B, A∩(A∪B) = …………………….
(a) B
(b) ∅
(c) A
(d) none of these
Answer:
(c) A

Question 3.
cos 1° + cos 2° + cos 3° + cos 4° + cos 179° = …………………
(a) 0
(b) 1
(c) -1
(d) 89
Answer:
(a) 0

Question 4.
The value of log₉ 27 is ……………………
(a) \(\frac{2}{3}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{3}{4}\)
(d) \(\frac{4}{3}\)
Answer:
(b) \(\frac{3}{2}\)

Question 5.
The value of \(\frac{\sin 3 \theta+\sin 5 \theta+\sin 7 \theta+\sin 9 \theta}{\cos 3 \theta+\cos 5 \theta+\cos 7 \theta+\cos 9 \theta}\) = ………………..
(a) tan3θ
(b) tan6θ
(c) cot3θ
(d) cot6θ
Answer:
(b) tan6θ

Question 6.
In 3 fingers the number of ways 4 rings can be worn in ……………………. ways.
(a) 43 – 1
(b) 34
(c) 68
(d) 64
Answer:
(d) 64

Question 7.
Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is ……………….
(a) 11
(b) 12
(c) 10
(d) 6
Answer:
(b) 12

Question 8.
The H.M of two positive number whose AM and G.M. are 16, 8 respectively is ………………..
(a) 10
(b) 6
(c) 5
(d) 4
Answer:
(d) 4

Question 9.
The co-efficient of the term independent of x in the expansion of (2x+\(\frac{1}{3x}\))⁶ is …………………
(a) \(\frac{160}{27}\)
(b) \(\frac{160}{27}\)
(c) \(\frac{80}{3}\)
(d) \(\frac{80}{9}\)
Answer:
(a) \(\frac{160}{27}\)

Question 10.
The value of \(\left|\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right|^{2}\) is ………………..
(a) abc
(b) -abc
(c) 0
(d) a²b²c²
Answer:
(d) a²b²c²

Question 11.
The value of x for which the matrix A = \(\left[\begin{array}{cc}
e^{x-2} & e^{7+x} \\
e^{2+x} & e^{2 x+3}
\end{array}\right]\) is singular is …………………..
(a) 9
(b) 8
(c) 7
(d) 6
Answer:
(b) 8

Question 12.
If |\(\vec { a } \) + \(\vec { b } \)| = 60, |\(\vec { a } \) – \(\vec { b } \)| = 40 and |\(\vec { b } \)| = 46 then |\(\vec { a } \)| is …………………
(a) 42
(b) 12
(c) 22
(d) 32
Answer:
(c) 22

Question 13.
Given \(\vec { a } \) = 2\(\vec { i } \) + \(\vec { j } \) – 8\(\vec { k } \) and \(\vec { b } \) = \(\vec { i } \) + 3\(\vec { j } \) – 4\(\vec { k } \) then |\(\vec { a } \) + \(\vec { b } \)| = ………………….
(a) 13
(b) \(\frac{13}{3}\)
(c) \(\frac{4}{13}\)
(d) \(\frac{3}{13}\)
Answer:
(a) 13

Question 14.
If f(x) = \(\left\{\begin{array}{ccc}
k x & \text { for } & x \leq 2 \\
3 & \text { for } & 2
\end{array}\right.\) is continous at x = 2 then the value of k is ……………………
(a) \(\frac{3}{4}\)
(b) 0
(c) 1
(d) \(\frac{4}{3}\)
Answer:
(c) 1

Question 15.
If f: R→R is defined by f(x) = |x – 3| + |x – 4| for x∈R then \(\lim _{x \rightarrow 3^{-}}\) f(x) is equal to ………………..
(a) -2
(b) -1
(c) 0
(d) 1
Answer:
(c) 0

Question 16.
\(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+5 x+3}{x^{2}+x+3}\right)^{x}\) is ………………..
(a) e⁴
(b) e²
(c) e³
(d) 1
Answer:
(a) e⁴

Question 17.
\(\int \frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1}\) dx is ………………..
(a) e tan^-1(x + 1)
(b) tan^-1(e^x) + c
(c) e^x \(\frac{\left(\tan ^{-1} x\right)^{2}}{2}\) + c
(d) e^xtan^-1x + c
Answer:
(d) e^xtan^-1x + c

Question 18.
∫ \(\frac { secx }{ \sqrt { cos2x } } \) dx = …………………..
(a) tan^-1(sin x) + c
(b) 2 sin^-1(tan x) + c
(c) tan^-1(cos x) + c
(d) sin^-1 (tan x) + c
Answer:
(d) sin^-1 (tan x) + c

Question 19.
\(\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\) dx = ………………..
(a) x + c
(b) \(\frac { x^{ 3 } }{ 3 } \) + c
(c) \(\frac { 3 }{ x^{ 3 } } \) + c
(d) \(\frac { 1 }{ x^{ 2 } } \) + c
Answer:
(b) \(\frac { x^{ 3 } }{ 3 } \) + c

Question 20.
It is given that the events A and B are such that P(A) = \(\frac{1}{4}\), P(A/B) = \(\frac{1}{2}\), P(B/A) = \(\frac{2}{3}\) then P(B) = …………………
(a) \(\frac{1}{6}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{2}\)
Answer:
(b) \(\frac{1}{3}\)

PART – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
In the set Z of integers, define mRn if m – n is a multiple of 12. Prove that R is an equivalence relation?
Answer:
As m – m = 0 and 0 = 0 × 12, we have 0 is a multiple of 12; hence mRm proving that R is reflexive.
Let mRn. Then m – n = 12k for some integer k; thus n – m = 12(-k) and hence nRm.
This shows that R is symmetric.
Let mRn and nRp: then m – n = 12k and n – p = 12l for some integers k and l.
So m – p = 12(k + l) and hence mRp. This shows that R is transitive.
Thus R is an equivalence relation.

Question 22.
Simplify:
\(\frac { 1 }{ 2+\sqrt { 3 } } \) + \(\frac { 3 }{ 4-\sqrt { 5 } } \) + \(\frac { 6 }{ 7-\sqrt { 8 } } \)
Answer:

Question 23.
Find the value of sin 22 \(\frac{1}{2}\)°?
Answer:
We know that cos θ = 1 – 2 sin² \(\frac{θ}{2}\) ⇒ sin \(\frac{θ}{2}\) = ±\(\sqrt { \frac { 1-cos2\theta }{ 2 } } \)
Take θ = 45°, we get sin \(\frac{45°}{2}\) = ±\(\sqrt { \frac { 1-cos45°}{ 2 } } \), (taking positive sign only, since 22\(\frac{1}{2}\)° lies in the first quadrant)
Thus, sin 22\(\frac{1}{2}\)° = \(\sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{2}\).

Question 24.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th
hour and wth hour?
Answer:
No. of bacteria at the beginning = 30
No. of bacteria after 1 hour = 30 × 2 = 60
No. of bactena after 2 hours = 30 × 2² = 30 × 4 = 120
No. of bacteria after 4 hours = 30 × 2⁴ = 30 × 16 = 480
No. of bacteria after n^th hour = 30 × 2ⁿ

Question 25.
If the sum of the distance of a moving point in a plane from the axis is 1, then find the locus of the point?
Answer:
Let coordinates of a moving point P be (x, y).
Given that the sum of the distances from the axis to the point is always 1.
∴|x| + |y| = 1 ⇒ x + y = 1
⇒- x- y = 1 ⇒ x + y = 1
⇒ x – y = 1
Hence, these equations give us the locus of the point P which is a square.

Question 26.
If (a, a + b, a + b + c) is one set of direction ratios of the line joining (1, 0, 0) and (0, 1, 0) then find a set of values of a, b, c?
Answer:
Let A be the point (1, 0, 0) and B be the point (0, 1, 0) (i.e.,) \(\overline { OA } \) = \(\hat { i } \) and \(\overline { OB } \) = \(\hat { j } \).
Then \(\overline { AB } \) = \(\overline { OB } \) – \(\overline { OA } \) = \(\hat { j } \) – \(\hat { i } \) = –\(\hat { i } \) + \(\hat { j } \)
= (-1, 1, 0)
= (a, a + b, a + b + c)
⇒ a = -1, a + b = 1 and a + b + c = 0
⇒ a = -1, ⇒ -1 + b = 1; a + b + c = 0
⇒ b = 2; -1 + 2 + c = 0 ⇒ c + 1 = 0
⇒ c = -1
∴ a = -1, b = 2, c = -1.
Note: if we taken \(\overline { BA } \) then we get a = 1, b = -2 and c = 1.

Question 27.
Find \(\frac{dy}{dx}\) for y = (x² + 4x + 6)⁵
Answer:
Let u = x² + 4x + 6
⇒ \(\frac{du}{dx}\) = 2x + 4
Now y = u⁵ = \(\frac{dy}{du}\) = 5u⁴
∴ \(\frac{dy}{dx}\) = \(\frac{dy}{du}\) × \(\frac{du}{dx}\) = 5u⁴ (2x + 4)
= 5(x² + 4x + 6)⁴ (2x + 4)
= 5(2x + 4) (x² + 4x + 6)⁴

Question 28.
Evaluate ∫\(\sqrt { 25x^{ 2 }-9 } \) dx?
Answer:

Question 29.
A bag contains 5 white and 7 black balls. 3 balls are drawn at random. Find the probability that

1. all are white
2. one white and 2 black.

Answer:
Number of white balls = 5
Number of black balls = 7
Total number of balls = 12
Selecting 3 from 12 balls can be done in
¹²C₃ = \(\frac{12 \times 11 \times 10}{3 \times 2 \times 1}\) = 220 ways
∴n(S) = 220

1. Let A be the selecting 3 white balls.
∴n(A) = ⁵C₃ = ⁵C₂ = \(\frac{5×3}{2×1}\) = 10
∴P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{10}{220}\) = \(\frac{1}{22}\)

2. Let B be the event of selecting one white and 2 black balls.
∴n(B) = ⁵C₁ × ⁷C₂ = (5) (\(\frac{7×6}{2×1}\)) = 5(21) = 105
∴P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{105}{220}\) = \(\frac{21}{44}\).

Question 30.
If (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k?
Answer:
Area of ∆ with vertices (k, 2) (2, 4) and (3, 2) = \(\frac{1}{2}\) \(\left|\begin{array}{lll}
k & 2 & 1 \\
2 & 4 & 1 \\
3 & 2 & 1
\end{array}\right|\) = 4 given
⇒ \(\left|\begin{array}{lll}
k & 2 & 1 \\
2 & 4 & 1 \\
3 & 2 & 1
\end{array}\right|\) = 2(4) = 8
(i.e.,) k(4 – 2) – 2(2 – 3) + 1(4 – 12) ± 8
(i.e.,) 2k – 2(-1) + 1(-8) = ± 8
(i.e.,) 2k + 2 – 8 = 8
(i.e.,) 2k = 8 + 8 – 2 = 14
k = 14/2 = 7
∴k = 7
So k = 7 (or) k = -1.

2k + 2 – 8 = -8
⇒2k = – 8 + 8 – 2
2k = – 2
k = -1

PART – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
In the set Z of integers define mRn if m – n is a multiple of 12. Prove that R is an equivalence relation?

Question 32.
Prove that \(\frac{sin4x+sin2x}{cos4x+cos2x}\) = tan 3x?

Question 33.
A polygon has 90 diagonals. Find the number of its sides?

Question 34.
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)ⁿ are equal?

Question 35.
Find the equation of the line passing through the point (1, 5) and also divides the co-ordinate axes in the ratio 3 : 10?

Question 36.
Prove that \(\left|\begin{array}{ccc}
1 & x & x \\
x & 1 & x \\
x & x & 1
\end{array}\right|^{2}=\left|\begin{array}{ccc}
1-2 x^{2} & -x^{2} & -x^{2} \\
-x^{2} & -1 & x^{2}-2 x \\
-x^{2} & x^{2}-2 x & -1
\end{array}\right|\)

Question 37.
If G is the centroid of a traiangle ABC prove that \(\overline { GA } \) + \(\overline { GB } \) + \(\overline { GC } \) = 0

Question 38.
Find \(\frac{dy}{dx}\) for y = \(\sqrt { 1+tan2x } \)?

Question 39.
Evaluate \(\frac { \sqrt { x } }{ 1+\sqrt { x } } \) dx?

Question 40.
Find the relation between a and b if \(\underset { x\rightarrow 3 }{ lim } \) f(x) exists where f(x) = \(\left\{\begin{array}{cc}
a x+b & \text { if } x>3 \\
3 a x-4 b+1 \text { if } x<3
\end{array}\right.\)

PART – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41 (a).
From the curve y = |x|, draw(i) y = |x-1| + 1

1. y = |x + 1| + 1
2. y = |x + 2| – 3

[OR]

(b) Resolve into partial fraction \(\frac { x+4 }{ (x^{ 2 }-4)(x+1) } \)

Question 42 (a).
Find the number of positive integers greater than 6000 and less than 7000 which arc divisible by 5, provided that no digit is to be repeated?

[OR]

(b) If ⁿP_r = ⁿP_r+1 and ⁿC_r = ⁿC_r-1, find the values of n and r?

Question 43 (a).
In a ∆ABC, prove that b² sin 2C + c² sin 2B = 2bc sin A?

[OR]

(b) Differentiate the following s(t) = \(\sqrt[4]{\frac{t^{3}+1}{t^{3}-1}}\)?

Question 44 (a).
Find the equation of the lines make an angle 60° with the positive x axis and at a distance 5\(\sqrt{2}\) units measured from the point (4, 7) along the line x – y + 3 = 0

[OR]

(b) If y = A cos4x + B sin 4x, A and B are constants then Show that y₂ + 16y = 0

Question 45 (a).
Find the sum up to the 17^th term of the series \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}\) + …………..

[OR]

(b) A shopkeeper in a Nuts and Spices shop makes gifi packs of cashew nuts, raisins and almonds?

1. Pack I contains 100 gm of cashew nuts, 100 gm of raisins and 50 gm of almonds.
2. Pack-II contains 200 gm of cashew nuts, 100 gm of raisins and 100 gm of almonds.
3. Pack-III contains 250 gm of cashew nuts, 250 gm of raisins and 150 gm of almonds.
4. The cost of 50 gm of cashew nuts is ₹50, 50 gm of raisins is ₹10. and 50gm of almonds is ₹60. What is the cost of each gift pack?

Question 46 (a).
Find matrix C if A = \(\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\), B = \(\begin{bmatrix} -3 & 2 \\ 4 & -1 \end{bmatrix}\) and 5C + 28= A?

[OR]

(b) The probability that a new railway bridge will get an award for its design is 0.48, the probability that it will get an award for the efficient use of materials is 0.36, and that it will get both awards is 0.2. What is the probability, that

1. it will get at least one of the two awards
2. it will get only one of the awards.

Question 47 (a).
\(\underset { \alpha \rightarrow 0 }{ lim } \) \(\frac { sin(\alpha ^{ n }) }{ (sin\alpha )^{ m } } \)

[OR]

(b) Evaluate I = sin^-1 (\(\frac { 2x }{ (1+x)^{ 2 } } \)) dx?

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 1 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 1 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is …………………….
(a) 0%
(b) 4.4%
(c) 16%
(d) 8.4%
MgCO₃ → MgO + CO₂↑
MgCO₃: (1 × 24) + (1 × 12) + (3 × 16) = 84g
CO₂: (1 × 12) + (2 × 16) = 44g
100% pure 84 g MgCO₃ on heating gives 44 g CO₂
Given that 1 g of MgCO₃ on heating gives 0.44 g CO₂
Therefore, 84 g MgCO₃ sample on heating gives 36.96 g CO₂
Percentage of purity of the sample = \(\frac { 100% }{ 44gCO_{ 2 } } \) × 36.96 g CO₂ = 84%
Percentage of impurity = 16%
Answer:
(c) 16%

Question 2.
The electronic configuration of Eu (atomic no. 63) Gd (atomic no. 64) and Tb (atomic no. 65) are …………………….. [NEET – Phase II]
(a) [Xe] 4f⁶ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁸ 5d¹ 6s²
(b) [Xe] 4f⁷ 6s² [Xe] 4f⁸ 6s² and [Xe] 4f⁹ 6s²
(c) [Xe] 4f⁷ 6s² [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
(d) [Xe] 4f ⁶ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
Solution:
Eu : [Xe] 4f⁷, 5d⁰, 6s²
Gd : [Xe] 4f⁷, 5d¹, 6s²
Tb : [Xe] 4f⁹, 5d⁰, 6s²
Answer:
(b) [Xe] 4f⁷ 6s² [Xe] 4f⁸ 6s² and [Xe] 4f⁹ 6s²

Question 3.
Match the following:

Answer:

Question 4.
Various successive ionization enthalpies (in kJ mol^-1) of an element are given below.

The element is ……………………….
(a) Phosphorus
(b) Sodium
(c) Aluminium
(d) Silicon
Answer:
(c) Aluminium

Question 5.
Non-stoichiometric hydrides are formed by ……………………..
(a) Palladium, vanadium
(b) Carbon, nickel
(c) Manganese, lithium
(d) Nitrogen, chlorine
Answer:
(a) Palladium, vanadium

Question 6.
Which is the function of sodium – potassium pump?
(a) Maintenance of ion balance
(b) Used in nerve impulse conduction
(c) Transmitting nerve signals
(d) Regulates the blood level
Answer:
(c) Transmitting nerve signals

Question 7.
∆S is expected to be maximum for the reaction ………………………
(a) Ca_(s) + 1/2O_2(g) → CaO_(s)
(b) C_(s) + O_2(g) → CO_2(g)
(c) N_2(g) + O_2(g) → 2NO_(g)
(d) CaCO_3(s) → CaO_(s) + CO_2(g)
Solution:
In CaCO_3(s) → CaO_(s) + CO_2(g) entropy change is positive. In (a) and (b) entropy change is negative; in (c) entropy change is zero.
Answer:
(d) CaCO_3(s) → CaO_(s) + CO_2(g)

Question 8.
Which one of the following is a reversible reaction?
(a) Ripening of a banana
(b) Rusting of iron
(c) Tarnishing of silver
(d) Transport of oxygen by Hemoglobin in our body
Solution:
All the other three reactions are irreversible reactions. But the hemoglobin combines with O₂ in lungs to form oxyhemoglobin. The oxyhemoglobin has a tendency to form hemoglobin by releasing O₂. So it is a reversible reaction.
Answer:
(d) Transport of oxygen by Hemoglobin in our body

Question 9.
P₁ and P₂ are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if x₁ represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be ………………………….
Solution:
(a) P₁ + x₁ (P₂ – P₁)
(b) P₂ – x₁ (P₂ + P₁)
(c) P₁ – x₂ (P₁ – P₂)
(d) P₁ + x₂ (P₁ – P₂)
P_total = P₁ + P₂
= P₁x₁ + P₂ x₂
= P₁(1 – x₂) + P₂x₂ [∵x₁ + x₂ = 1, x₁ = 1 – x₂]
= P₁ – P₁x₂ + P₂x₂ = P₁ – x₂ = P₁ – x₂ (P₁ – P₂)
Answer:
(c) P₁ – x₂ (P₁ – P₂)

Question 10.
Which of the following molecule contain no n bond?
(a) SO₂

(b) NO₂

(c) CO₂

(d) H₂O

Solution:
Water (H₂O) contains only σ bonds and no π bonds.
Answer:
(d) H₂O

Question 11.
IUPAC name of  is …………………….
(a) Trimethylheptane
(b) 2 -Ethyl -3, 3- dimethyl heptane
(c) 3, 4, 4 – Trimethyloctane
(d) 2 – Butyl -2 -methyl – 3 – ethyl-butane.
Answer:
(c) 3, 4, 4 – Trimethyloctane

Question 12.
Which one of the following is an example for free radical initiators?
(a) Benzoyl peroxide
(b) Benzyl alcohol
(c) Benzyl acetate
(d) Benzaldehyde.
Answer:
(a) Benzoyl peroxide

Question 13.
Some meta-directing substituents in aromatic substitution are given. Which one is most – deactivating?
(a) – COOH
(b) – NO₂
(C) – C ≡ N
(d) – SO₃H
Answer:
(b) – NO₂

Question 14.
Consider the following statements.
(I) S_N² reaction is a bimolecular nucleophilic first order reaction.
(II) S_N² reaction take place in one step.
(III) S_N² reaction involves the formation of a carbocation. Which of the above statements is/are not correct?
(a) (II)
(b) (I) only
(c) (I) & (III)
(d) (III)
Answer:
(c) (I) & (III)

Question 15.
The questions given below consists of an assertion and the reason. Choose the correct option out of the choices given below each question.
Assertion (A): If BOD level of water in a reservoir is more than 5 ppm it is highly polluted.
Reason(R): High biological oxygen demand means high activity of bacteria in water.
(a) Both (A) and R are correct and (R) is the correct explanation of (A)
(b) Both (A) and R are correct and (R) is not the correct explanation of (A)
(c) Both (A) and R are not correct
(d) (A) is correct but( R) is not correct
Answer:
(d) (A) is correct but( R) is not correct

PART – II

Answer any six questions in which question No. 24 is compulsory. [6 × 2 = 12]

Question 16.
What is the actual configuration of copper (Z = 29)? Explain about its stability?
Answer:
Copper (Z = 29)
Expected configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s²
Actual configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
The reason is that fully filled orbitals have been found to have extra stability.

Copper has the electronic configuration [Ar] 3d¹⁰ 4s¹ and not [Ar] 3d⁹ 4s² due the symmetrical distribution and exchange energies of d electrons.

Symmetry leads to stability. The full filled configuration have symmetrical distribution of electrons and hence they are more stable than unsymmetrical configuration.

Question 17.
What is screening effect?
Answer:
Screening effect:
The repulsive force between inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus the inner shell elections act as a shield between the nucleus and the valence electrons. This effect is called shielding effect (or) screening effect.

Question 18.
Ice is less dense than water at 0°C. Justify this statement?
Answer:
In ice, each oxygen atom is surrounded tetrahedrally by through hydrogen bonds to four water molecules. That is, the presence of two hydrogen atoms and two lone electron pairs (on oxygen) in each water molecule results in a three-dimensional structure. The arrangements creates an open structure, which in turn accounts for the fact that ice is less dense than water at 0°C.

Question 19.
Aerosol cans carry clear warning of heating of the can. Why?
Answer:
Aerosol cans carry clear warning of heating of the can. As the temperature rises, pressure in the can will increase and ambient temperatures about 120°F may lead to explosions.

So aerosol cans should always be stored in dry areas where they will not be exposed to excessive temperatures. You should never throw an aerosol can onto a fire, or leave it in the direct sunlight, even it is empty. This is because the pressure will build up so much that the can will burst. It is due to 2 reasons.

1. The gas pressure increases.
2. More of the liquefied propellant turns into a gas.

Question 20.
One mole of PCl₅ is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant?
Answer:
Given that [PCl₅]_initial = \(\frac { 1mol }{ dm^{ 3 } } \)
[Cl₂]_eq = 0.6 mol dm^-3
PCl₅⇄ PCl₃ + Cl₂
[PCl₅]_eq = 0.6 mole dm^-3
[PCl5]_eq = 0.4 mole dm^-3
∴ K_C = \(\frac{0.6×0.6}{0.4}\)
K_C = 0.9

Question 21.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Mostly dissolution of gases in liquid is an exothermic process. It is because the fact that this process involves decrease of entropy.

Thus, increase of temperature tends to push the equilibrium towards backward direction as a result of which solubility of the gas decrease with rise in temperature.
(Gas + Solvent ⇄ Solution + Heat)

Question 22.
Give the general formula for the following classes of organic compounds?
(a) Aliphatic monohydric alcohol
(b) Aliphatic ketones
(c) Aliphatic amines.
Answer:
(a) Aliphatic monohydric alcohol

(b) Aliphatic ketones

(c) Aliphatic amines

Question 23.
Identify which of the following shows +1 and -I effect?

(I) -NO₂
(II) -SO₃H
(III) -I
(IV) -OH
(V) CH₃O^–
(VI) CH_3-

Answer:

Question 24.
Write the products A & B for the following reaction?

Answer:

PART – III

Answer any six questions in which question No. 33 is compulsory. [6 × 3 = 18]

Question 25.
Balance by oxidation number method: Mg + HNO₃ → Mg(N0₃)₂ + NO₂ + H₂O.
Answer:
Step 1:

Step 2:
Mg + 2HNO₃ → Mg(NO₃)₂ + NG₂ + H₂O

Step 3:
To balance the number of oxygen atoms and hydrogen atoms 2HNO₃ is multiplied by 2.
Mg + 4HNO₃ → Mg(NO₃)₂ + 2NO₂ + H₂O

Step 4:
To balance the number of hydrogen atoms, the H₂O molecule is multiplied by 2.
Mg + 4HNO₃ → Mg(NO₃)₂ + 2NO₂ + 2H₂O

Question 26.
For each of the following, give the sub level designation, the allowable m values and the number of orbitals?
Answer:

1. n = 4, l = 2
2. n = 5, l = 3
3. n = 7, l = 0

1. n = 4, l = 2
If l = 2, ‘m’values are -2,-1, 0, +1, +2. So, 5 orbitals such as d_xy, d_yz ,d_xz, \(d_{ x^{ 2 }-y^{ 2 } }\) and \(d_{ z^{ 2 } }\)

2. n = 5, l = 3
If l = 3, ‘m’ values are -3, -2, -1, 0, +1, +2, +3
So, 7 orbitals such as \(\mathrm{f}_{\mathrm{z}}^{3}, \mathrm{f}_{\mathrm{xz}}^{2}, \mathrm{f}_{\mathrm{yz}}^{2}, \mathrm{f}_{\mathrm{xyz}}, \mathrm{f}_{\mathrm{z}\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{x}\left(\mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{y}\left(3 \mathrm{x}^{2}-\mathrm{z}^{2}\right)}\)

3. n = 7, l = 0
If l = 0, ‘m’ values are 0. Only one value. So, 1 orbital such as 7s orbital.

Question 27.
Prove that ionization energy is a periodic property?
Answer:
Variation in a period:
On moving across a period from left to right, the ionization enthalpy value increases.
This is due to the following reasons:

1. Increase of nuclear charge in a period
2. Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus, thus ionization enthalpy increases. Hence, ionization energy is a periodic property.

Variation in a group:
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons:

1. A gradual increase in atomic size
2. Increase of screening effect on the outermost electrons due. to the increase of number of inner electrons.

Hence, ionization enthalpy is a periodic property.

Question 28.
Explain how heat absorbed at constant pressure is measured using coffee cup calorimeter with neat diagram?
Answer:

1. Measurement of heat change at constant pressure can be done in a coffee cup calorimeter.
2. We know that ∆H = qp (at constant P) and therefore, heat absorbed or evolved, q_p at constant pressure is also called the heat of reaction or enthalpy of reaction, ∆H_r
3. In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, q_p will be negative and ∆H_r will Reaction also be negative. Mixture
4. Similarly in an endothermic reaction, heat is absorbed, q_p is positive and ∆H_r will also be positive.

Question 29.
Consider the following reactions,
(a) H₂(g) + I₂(g) ⇄ 2 HI(g)
(b) CaCO₂(s) ⇄ CaO(s) + CO₂(g)
(c) S(s) + 3F₂(g) ⇄ SF₆(g)
In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product?
Answer:
(a) H₂(g) + I₂2(g) ⇄ 2HI(g)
In the above equilibrium reaction, volume of gaseous mdlecules is equal on both sides. So increase or decrease the volume will not affect the equilibrium and there will be no change in the yield of product.

(b)
Volume is greater in product side. By decreasing the pressure, volume will increase thus, to get more of product CO₂, the pressure should be decreased or volume should be increased.

(c)
Volume is lesser in product side. So by increasing the pressure, equilibrium shifts to the product side.

Question 30.
You are provided with a solid ‘A’ and three solutions of A dissolved in water one saturated, one unsaturated, and one super saturated. How would you determine each solution?
Answer:
(I) Saturated solution:
When maximum amount of solute is dissolved in a solvent, any more addition of solute will result in precipitation at a given temperature and pressure. Such a solution is called a saturated solution.

(II) Unsaturated solution:
When minimum amount of solute is dissolved in a solvent at a given temperature and pressure is called an unsaturated solution.

(III) Super saturated solution:
It is a solution that holds more solute than it normally could in its saturated form.

Example:

1. A saturated solution where the addition of more compound would not dissolve in the solution. 359 g of NaCl in 1 litre of water at 25°C.
2. An unsaturated solution has the capacity to dissolve more of the compound. 36 g of NaCl in 1 litre of water at 25°C.
3. A super saturated solution is the solution in which crystals can start growing. 500 g of NaCl in 1 litre of water at 25°C.

Question 31.
Explain about the salient features of molecular orbital theory?
Answer:

• When atoms combine to fonn molecules, their individual atomic orbitals lose their identity and form new orbitals called molecular orbitals.
• The shape of molecular orbitals depend upon the shapes of combining atomic orbitals.
• The number of molecular orbitals formed is the same as the number of combining atomic orbitals. Half the number of molecular orbitals formed will have lower energy and are called bonding orbitals, while the remaining half molecular orbitals will have higher energy and are called anti-bonding molecular orbitals.
• The bonding molecular orbitals are represented as σ (sigma), π (pi), δ (delta) and the corresponding anti-bonding orbitals are called C*, 7t* and 5*.
• The electrons in the molecule are accommodated in the newly formed molecular orbitals. The filling of electrons in these orbitals follow Aufbau’s Principle, Pauli’s exclusion principle and Hund’s rule as in the case of filling of electrons in the atomic orbitals.
• Bond order gives the number of covalent bonds between the two combining atoms.

The bond order of a molecule can be calculated using the following equation:
Bond order = \(\frac { N_{ b }-N_{ a } }{ 2 } \)
N_b = Number of electrons in bonding molecular orbitals.
N_a = Number of electrons in anti-bonding molecular orbitals.
(vii) A bond order of zero value indicates that the molecule does not exist.

Question 32.
Explain the types of addition reactions?
Answer:
Addition reactions are classified into three types they are,

1. Electrophilic addition reaction
2. Nucleophilic addition reaction
3. Free radical addition reaction

1. Electrophilic addition reaction:
An electrophilic addition reaction can be described as an addition reaction in whifth a reactant with multiple bonds as in a double or triple bond undergoes has its n bond broken and two new a bond are formed.

(ethane) Br Bf

2. Nucleophilic addition reaction:
A nucleophilic addition reaction is an addition reaction where a chemical compound with an electron deficient or electrophilic double or triple bond, a n bond, reacts with a nucleophilic which is an electron rich reactant with the disappearance of the double bond and creation of two new single or a bonds.

3. Free radical addition reaction:
It is an addition reaction in organic chemistry involving free radicals. The addition may occur between a radical and a non radical or between two radicals.

Question 33.
Complete the following reaction and identify A, B and C?

Answer:

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
(I) How many radial nodes for 2s, 4p, 5d and 4f orbitals exhibit? How many angular nodes?
(II) How many unpaired electrons are present in the ground state of

(a) Cr³⁺ (Z = 24)
(b) Ne (Z = 10)

[OR]

(b) (I) The electronic configuration of an atom is one of the important factor which affects the value of ionization potential and electron gain enthalpy. Explain?
(II) Explain why cation are smaller and anions are larger in radii than their parent atoms?
Answer:
(a) (I) Formula for total number of nodes = n – 1
For 2s orbital: Number of radial nodes = 1.
For 4p orbital: Number of radial nodes = n – l – 1.
= 4 – 1 – 1 = 2
Number of angular nodes = l
∴Number of angular nodes = l.
So, 4p orbital has 2 radial nodes and 1 angular node.

For 5d orbital:
Total number of nodes = n – 1
= 5 – 1 =4 nodes
Number of radial nodes = n – l – 1
= 5 – 2 – 1 = 2 radial nodes.
Number of angular nodes = l = 2
∴5d orbital have 2 radial nodes and 2 angular nodes.

For 4f orbital:
Total number of nodes = n – 1
= 4 – 1 = 3 nodes
Number of radial nodes = n – l – 1
= 4 – 3 – 1 = 0 node.
Number of angular nodes = l
= 3 nodes
∴ 5d orbital have 0 radial node and 3 angular nodes.

(II) (a) Cr (Z = 24) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
Cr³⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴.
It contains 4 unpaired electrons.
(b) Ne(Z=10) 1s² 2s² 2p⁶. No unpaired electrons in it.

[OR]

(b) (I)

• Electronic configuration of an atom affects the value of ionization potential and electron gain enthalpy.
• Half filled valence shell electronic configuration and completely filled valence shell electronic configuration are more stable than partially filled electronic configuration.
• For e.g. Beryllium (Z = 4) 1s² 2s² (completely filled electronic configuration)
  Nitrogen (Z = 7) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹ (half filled electronic configuration)
  Both beryllium and nitrogen have high ionization energy due to more stable nature.
• In the case of beryllium (1s² 2s²), nitrogen (1s² 2s² 2p³) the addition of extra electron will disturb their stable electronic configuration and they have almost zero electron affinity.
• Noble gases have stable ns² np⁶ configuration and the addition of further electron is unfavorable and they have zero electron affinity.

(II) A cation is smaller than the parent atom because it has fewer electrons while its nuclear charge remains the same. The size of anion will be larger than that of parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

Question 35 (a).
(I) Arrange NH₃, H₂O and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement?
(II) Can we use concentrated sulphuric acid and pure zinc in the preparation of dihydrogen?

[OR]

(b) (I) Beryllium halides are covalent whereas magnesium halides are ionic. Why?
(II) What happens when
Sodium metal is dropped in water?

1. Sodium metal is heated in free supply of air?
2. Sodium peroxide dissolves in water?

Answer:
(a) 1. Increasing magnitude of hydrogen bonding among NH₃, H₂O and HF is
HF > H₂O > NH₃

2. The extent of hydrogen bonding depends upon electronegativity and the number of hydrogne atoms available for bonding.

3. Among N, F and O the increasing order of their electronegativities are
N < O < F 4. Hence the expected order of the extent of hydrogen bonding is HF > H₂O > NH₃

(II) Conc. H₂SO₄ cannot be used because it acts as an oxidizing agent also and gets reduced to SO₂.
Zn + 2H₂SO₄ (Conc) → ZnSO₄ + 2H₂O + SO₂ Pure Zn is not used because it is non-porous and reaction will be slow. The impurities in Zn help in constitute of electrochemical couple and speed up reaction.

[OR]

(b)
(I) Beryllium ion (Be²⁺) is smaller in size and it is involved in equal sharing of electrons with halogens to form covalent bond, whereas magnesium ion (Mg²⁺) is bigger and it is involved in transfer of electrons to form ionic bond.

(II)

1. 2Na + 2H₂O → 2NaOH + H₂
2. 2Na + O₂ → Na₂O₂
3. Na₂O₂ + 2H₂O → 2NaOH + H₂O₂

Question 36 (a).
(I) Define Gibb’s free energy?
(II) You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below?
Answer:

[OR]

(b) (I) Define mole fraction.
(II) Differentiate between ideal solution and non-ideal solution.
Answer:
(a) (I) Gibbs free energy is defined as the part of total energy of a system that can be converted (or) available for conversion into work.
G = H – TS, where G = Gibb’s free energy
H – enthalpy;
T = temperature;
S = entropy

(II) For ethanol:
Given: T_b = 78.4°C = (78.4 + 273) = 351.4 K

[OR]

(b) (I) Mole fraction of a component is the ratio of number of moles of the component to the total number of moles of all components present in the solution.

(II)
Ideal Solution:

1. An ideal solution is a solution in which each component obeys the Raoult’s law over entire range of concentration.
2. For an ideal solution,
   ΔH_missing = 0, ΔV_mixing = 0
3. Example: Benzene and toulene n – Hexane and n – Heptane.

Non – Ideal Solution:

1. The solution which do not obey Raults’s law over entire range of concentrations are called non-ideal solution.
2. For an ideal solution,
   ΔH_mixing ≠ 0, ΔV_mixing ≠ 0
3. Example: Ethyl alcohol and Cyclo hexane, Benzene and aceton.

Question 37 (a).
(I) What is dipole moment?
(II) Describe Fajan’s rule?

[OR]

(b) (I) How does Huckel rule help to decide the aromatic character of a compound?
(II) Draw cis-trans isomers for the following compounds
(a) 2- chloro-2-butene
(b) CH₃-CCl=CH-CH₂CH₃
Answer:
(a)
(I)

1. The polarity of a covalent bond can be measured in terms of dipole moment which is defined as: m = q × 2d, where m is the dipole moment, q is the charge, 2d is the distance between the two charges.
2. The dipole moment is a vector quantity and the direction of the dipole moment points from the negative charge to positive charge.
3. The unit of dipole moment is Coulomb metre (C m). It is usually expressed in Debye unit (D).
4. 1 Debye = 3.336 × 10^-3o C m

(II)

1. The ability of a cation to polarise an anion is called its polarising ability and the tendency of ah anion to get polarised is called its polarisibility. The extent of polarisation in an ionic compound is given by the Fajans rule.

2. To show greater covalent character, both the cation and anion should have high charge on them. Higher the positive charge on the cation greater will be the attraction on the electron cloud of the anion. Similarly higher the magnitude of negative charge on anion, greater is its polarisability. For example, Na⁺ < Mg²⁺ < Al³⁺, the covalent character also follows the order:
NaCl < MgCl₂ < AlCl₃.

3. The smaller cation and larger anion show greater covalent character due to the greater extent of polarisation, e.g., LiCl is more covalent than NaCl.

4. Cation having ns² np⁶ nd⁰ configuration shows greater polarising power than the cations with ns² np⁶ configuration, e.g., CuCl is more covalent than NaCl.

[OR]

(b) (I) A compound is said to be aromatic, if it obeys the following rules:

1. The molecule must be cyclic.
2. The molecule must be co-planar.
3. Complete delocalisation of rc-electrons in the ring.
4. Presence of (4n + 2)π electrons in the ring where n is an integer (n = 0, 1, 2 …)

This is known as Huckel’s rule.
Example –  – Benzene

1. It is cyclic one.
2. It is a co-planar molecule.
3. It has six delocalised n electrons.
4. 4n + 2 = 6

4n = 6 – 2
4n = 4
⇒ n = 1
It obey Huckel’s rule, with n = 1, hence benzene is aromatic in nature.

(II) (a) 2-Chloro-2-butene:

(b)

Question 38 (a).
(I) Reagents and the conditions used in the reactions are given below. Complete the table by writing down the product and the name of the reaction?
Answer:

(II) What is the IUPAC name of the insecticide DDT? Why is their use banned in most of the countries?

[OR]

(b) (I) Explain about green chemistry in day-to-day life?
(II) How acetaldehyde is commercially prepared by green chemistry?
Answer:
(a)
(I)

(II)

•  The IUPAC name of the insecticide DDT is p, p’-dichloro-diphenyl trichloroethane.
• Even DDT is an effective insecticide. Now-adays it is banned because of its long term toxic effects.
• DDT is very persistent in the environment and it has a high affinity for fatty tissues. As a result, DDT gets accumulated in animal tissue fat, in particular that of birds of prey with subsequent thinning of their eggs shells and impacting their rate of reproduction. That is why DDT is banned in most of the countries.

[OR]

(b)
(I)
1. Dry cleaning of clothes:
Solvents like tetrachloroethylene used in dry cleaning of clothes, pollute the ground water and are carcinogenic. In place of tetrachloro ethylene, liquefied CO₂ with suitable detergent is an alternate solvent used. Liquefied CO₂ is not harmful to the ground water. Nowadays H₂O₂ is used for bleaching clothes in laundry, gives better result and utilises less water.

2. Bleaching of paper:
Conventional method of bleaching was done with chlorine. Nowadays H₂O₂ can be used for bleaching paper in the presence of catalyst.

1. Instead of petrol, methaftol is used as a fuel in automobiles.
2. Neem based pesticides have been synthesised, which are more safer than the chlorinated hydrocarbons.

(II) Acetaldehyde is commericially prepared by one step oxidation of ethene in the presence of ionic catalyst in aqueous medium with 90% yield.

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 2 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 2 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
An element X has the following isotopic composition ²⁰⁰X = 90 %, ¹⁰⁰X = 8 % and ²⁰²X = 2 %.
The weighted average atomic mass of the element X is closest to ………………………….
(a) 201 u
(b) 202 u
(c) 199 u
(d) 200 u
\(\frac { (200\times 90)+(199\times 18)+(202\times 2) }{ 100 } \)
= 199.96 = 200 u
Answer:
(d) 200 u

Question 2.
Which of the following is not used in writing electronic configuration of an atom?
(a) Aufbau principle
(b) Hund’s rule
(c) Pauli’s exclusion principle
(d) Heisenberg’s uncertainty principle
Answer:
(d) Heisenberg’s uncertainty principle

Question 3.
Assertion (A): Cr with electronic configuration [Ar]3d⁵ 4s¹ is more stable than [Ar] 3d⁴ 4s².
Reason (R): Half filled orbitals have been found to have extra stability than partially filled orbitals.
(a) A and R are correct and R is the correct explanation of A.
(b) A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) A and R are correct and R is the correct explanation of A.

Question 4.
The type of H-bonding present in ortho nitro phenol and p-nitro phenol are respectively ……………………..
(a) Inter molecular H-bonding and intra molecular H – bonding
(b) Intra molecular H-bonding and inter molecular H – bonding
(c) Intra molecular H – bonding and no H – bonding
(d) Intra molecular H – bonding and intra molecular H – bonding

Answer:
(b) Intra molecular H-bonding and inter molecular H – bonding

Question 5.
Match the following:

Answer:

Question 6.
Which of the following is the correct expression for the equation of state of van der Waals gas?
(a) \(\left(P+\frac{a}{n^{2} V^{2}}\right)\) (V – nb) = nRT
(b) \(\left(P+\frac{n a}{n^{2} V^{2}}\right)\) (V – nb) = nRT
(c) \(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT
(d) \(\left(P+\frac{n^{2} a^{2}}{V^{2}}\right)\) (V – nb) = nRT
Answer:
(c) \(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT

Question 7.
In a reversible process, the change in entropy of the universe is ………………………
(a) > 0
(b) >0
(c) <0
(d) = 0
Answer:
(d) = 0

Question 8.
Which of the following is not a general characteristic of equilibrium involving physical process?
(a) Equilibrium is possible only in a closed system at a given temperature
(b) The opposing processes occur at the same rate and there is a dynamic but stable condition
(c) All the physical processes stop at equilibrium
(d) All measurable properties of the system remains constant
Solution:
Correct statement: Physical processes occurs at the same rate at equilibrium.
Answer:
(c) All the physical processes stop at equilibrium

Question 9.
Stomach acid, a dilute solution of HCl can be neutralised by reaction with Aluminium hydroxide
Al(OH)₃ + 3HCl(aq) → AlCl₃ + 3H₂O
How many millilitres of 0.1 M Al(OH)₃ solution are needed to neutralise 21 mL of 0.1 M HCl?
(a) 14 mL
(b) 7 mL
(c) 21 mL
(d) None of these
Solution:
M₁ × V = M₂ × V₂
∵ 0.1 M Al(OH)₃ gives 3 × 0.1 = 0.3 M OH^– ions
0.3 × V₁ = 0.1 × 21
V₁ = \(\frac{0.1×21}{0.3}\) = 7 ml
Answer:
(b) 7 mL

Question 10.
Shape and hybridisation of IF₅ are ……………………..
(a) Trigonal bipyramidal, sp³d²
(b) Trigonal bipyramidal, sp³d
(c) Square pyramidal, sp³d²
(d) Octahedral, sp³d²
Solution:
IF₅ – 5 bond pair + 1 lone pair
∴ hybridisation sp³d²

Answer:
(c) Square pyramidal, sp³d²

Question 11.
Consider the following statements:

1. It is not possible for the carbon to form either C⁴⁺ (or) C^4- ions.
2. Carbon can form ionic bonds.
3. In compounds of carbon, it form covalent bonds.

Which of the above statement is/are not correct?
(a) (I) and (II)
(b) (III) only
(c) (I) only
(d) (II) only
Answer:
(d) (II) only

Question 12.
For the following reactions
(A) CH₃CH₂CH₂Br + KOH → CH₃-CH + KBr + H₂O
(B) (CH₃)₃CBr + KOH → (CH₃)₃ COH + KBr
(C)
Which of the following statement is correct?
(a) (A) is elimination, (B) and (G) are substitution
(b) (A) is substitution, (B) and (C) are elimination
(c) (A) and (B) are elimination and (C) is addition reaction
(d) (A) is elimination, B is substitution and (C) is addition reaction.
Answer:
(d) (A) is elimination, B is substitution and (C) is addition reaction.

Question 13.
Which of the following compound used for metal cleaning solvent?
(a) Methylene chloride
(b) Methyl chloride
(c) Chloroform
(d) Ethane
Answer:
(a) Methylene chloride

Question 14.
The number of possible isomers of C₆H₁₂ is ……………………..
(a) 2
(b) 3
(c) 5
(d) 6
Answer:
(c) 5

Question 15.
Ozone layer is depleted by the reactive ……………………..
(a) Hydrogen atom
(b) Oxygen atom
(c) Fluorine atom
(d) Chlorine atom
Answer:
(d) Chlorine atom

PART – II

Answer any six questions in which question No. 19 is compulsory. [6 × 2 = 12]

Question 16.
Calculate the average atomic mass of naturally occurring magnesium using the following data?

Answer:
Solution: Isotopes of Mg.
Atomic mass = Mg²⁴ = 23.99 × 78.99/100 = 18.95
Atomic mass = Mg²⁶ = 24.99 × 10/100 = 2.499
Atomic mass = Mg²⁵ = 25.98 × 11.01/100 = 2,860
Average Atomic mass = 24.309
Average atomic mass of Mg = 24.309

Question 17.
What are quantum numbers?
Answer:

1. The electron in an atom can be characterized by a set of four quantum numbers, namely – principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s).
2. When Schrodinger equation is solved for a wave function φ, the solution contains the first three quantum numbers n, 1 and m.
3. The fourth quantum number arises due to the spinning of the electron about its own axis.

Question 18.
How 2-ethylanthraquinone helps to prepare hydrogen peroxide?
Answer:
On an industrial scale, hydrogen peroxide is prepared by auto oxidation of 2-alkyl anthraquinol.

Question 19.
Calculate the pressure exerted by 2 moles of sulphur hexafluoride in a steel vessel of volume 6 dm³ at 70°C assuming it is an ideal gas?
Answer:
We will use the ideal gas equation for this calculation as below:
P = \(\frac{nRT}{V}\) = image 7 = 9.39 atm.

Question 20.
What are the important features of lattice enthalpy?
Answer:

1. Higher lattice energy shows greater electrostatic attraction and therefore a stronger bond in the solid.
2. The lattice enthalpy is greater for ions of higher charge and smaller radii.

Question 21.
What are aqueous and non-aqueous solution? Give example?
Answer:

1. If the solute is dissolved in the solvent water, the resultant solution is called as an aqueous solution, e.g., salt in water.
2. If the solute is dissolved in the solvent other than water such as benzene, ether, CCl₄ etc, the resultant solution is called a non aqueous solution, e.g., Br₂ in CCl₄.

Question 22.
What is bond enthalpy? How they relate with bond strength?
Answer:
The bond enthalpy is defined as the minimum amount of energy required to break one mole of a particular bond in molecules in their gaseous state. Larger the bond enthalpy stronger will be the bond.

Question 23.
What is triad system? Give example?
Answer:
(I) In this system hydrogen atom oscillates between three polyvalent atoms. It involves 1, 3 – migration of hydrogen atom from one polyvalent atom to other with in the molecule

(II) The most important type of triad system is keto-enol tautomerism and the two groups of tautomers are keto form and enol form.

(III)

Question 24.
What is stone leprosy? How is it formed?
Answer:

1. The attack on, the marble of buildings by acid rain is called stone leprosy.
2. Acid rain causes extensive damage to buildings made up of marble.

CaCO + H₂SO₄ → CaSO₄ + H₂O + CO₂↑

PART – III

Answer any six questions in which question No. 26 is compulsory. [6 × 3 = 18]

Question 25.
Calculate the equivalent mass of hydrated ferrous sulphate?
Answer:
Hydrated ferrous sulphate = FeSO₄.7H₂O
Ferrous sulphate – Reducing agent
Ferrous sulphate reacts with an oxidising agent in acid medium according to the equation.

16 parts by mass of oxygen oxidised 304 g of FeSO₄.
8 parts by mass of oxygen will oxidise \(\frac{304}{16}\) × 8 parts by mass of FeSO₄ = 152
Equivalent mass of Ferrous sulphate (Anhydrous) = 152
Equivalent mass of crystalline Ferrous sulphate FeSO₄.7H₂O = 152 + 126 = 278

Question 26.
Calculate the uncertainty in position of an electron, if ∆v = 0.1% and v = 2.2 × 10⁶ ms^-1.
Answer:
Mass of an electron = m = 9.1 × 10^-31 kg.
∆v = Uncertainty in velocity = \(\frac{0.1}{100}\) × 2.2 × 10⁶ ms^-1
∆v = 0.22 × 10⁴ = 2.2 × 10³ ms^-1
∆x. ∆v.m = \(\frac{h}{4π}\)

∆x = 2.635 × 10^-8
Uncertainty in position = 2.635 × 10^-8

Question 27.
Distinguish between diffusion and effusion?
Answer:
Diffusion:

1. Diffusion is the spreading of molecules of a substance throughout a space or a second subsance.
2. Diffusion refers to the ability of the gases to mix with each other.
3. E.g; Spreading of something such as brown tea liquid spreading through the water in a tea cup.

Effusion:

1. Effusion is the escape of gas molecules through a very small hole in a membrane into an evacuated area.
2. Effusion is a ability of a gas to travel through a small pin-hole.
3. E.g; Pouring out something like the soap studs bubbling out from a bucket of water.

Question 28.
For a chemical reaction the values of ∆H and ∆S at 300 K are – 10 kJ mol^-1 and – 20 JK^-1 mol^-1 respectively. What is the value of ∆G of the reaction? Calculate the ∆G of a reaction at 600K assuming ∆H and ∆S values are constant. Predict the nature of the reaction?
Answer:
Given:
∆H = -10 kJ mol^-1 = -10000 J mol^-1
∆S = – 20 JK^-1 mol^-1
T = 300 K

∆G?
∆G = ∆H – T∆S
∆G = -10 kJ mol^-1 – 300 K × (-20 × 10^-3) kJ K^-1 mol^-1
∆G = (-10 + 6) kJ mol^-1
∆G = – 4 kJ mol^-1

At 600 K,
∆G = – 10 kJ mol^-1 – 600 K × (-20 × 10^-3) k^-1 mol^-1
∆G = (-10 + 12) kJ mol^-1
∆G + 2 kJ mol^-1
The value of ∆G is negative at 300K and the reaction is spontaneous, but at 600K the value ∆G becomes positive and the reaction is non-spontaneous.

Question 29.
For the reaction: A₂(g) + B₂(g) ⇄ 2AB(g); H is -∆ve.
The following molecular scenes represent different reaction mixture (A – light grey, B-dark grey)

1. Calculate the equilibrium constant K_p and K_c.
2. For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions?
3. What is the effect of increase in pressure for the mixture at equilibrium?

Answer:

K_c > Q i.e; forward reaction is favoured.

(III) Since ∆n_g = 2 -2 = 0, thus, pressure has no effect. So by increasing the pressure, equilibrium will not be affected.

Question 30.
Derive the relationship between the relative lowering of vapour pressure and mole fraction of the solute?
Answer:

Question 31.
Differentiate between the principle of estimation of nitrogen in an organic compound by

1. Dumas method
2. Kjeldahl’s method

Answer:

1. Dumas method:
The organic compound is heated strongly with excess of CuO (Cupic Oxide) in an atmosphere of CO₂ where free nitrogen, CO₂ and H₂O are obtained.

2. Kjeldahl’s method:
A known mass of the organic compound is heated strongly with cone. H₂SO₄, a little amount of potassium sulphate arid a little amount of mercury (as catalyst). As a result of reaction, the nitrogen present in the organic compound is converted to ammonium sulphate.

Question 32.
In what way free radical affect the human body?
Answer:

1. Free radicals can disrupt cell membranes.
2. Increase the risk of many forms of cancer.
3. Damage the interior lining of blood vessels.
4. Eads to a high risk of heart disease and stroke.

Question 33.
Dissolved oxygen in water is responsible for aquatic life. What processes are responsible for the reduction in dissolved oxygen in water?
Answer:

1. Organic matter such as leaves, grass, trash can pollute water. Microorganisms present in water can decompose these organic matter and consume dissolved oxygen in water.
2. Eutrophication is a process by which water bodies receive excess nutrients that stimulates excessive plant growth.
3. This enhanced plant growth in water bodies is called algal bloom.
4. The growth of algae in extreme abundance covers the water surface and reduces the oxygen concentration in water. Thus, bloom-infeded water inhibits the growth of other living organisms in the water body.
5. This process in which the nutrient rich water support a dense plant population, kills animal life by depriving it of oxygen and results in loss of biodiversity is known as eutrophication.

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
(I) Calculate the equivalent mass of sulphuric acid?
(II) The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals.
(Atomic mass of Al = 27 u Atomic mass of O = 16 u)
2Al + Fe₂O₃ → Al₂O₃ + 2Fe; If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide.
(a) Calculate the mass of Al₂O₃ formed.
(b) How much of the excess reagent is left at the end of the reaction?

[OR]

(b) (I) Consider the following electronic arrangements for the d⁵ configuration?

(a)

(b)

(c)

1. Which of these represents the ground state?
2. Which configuration has the maximum exchange energy?

(II) An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion?
Answer:
(a) (I) Equivalent mass of sulphuric acid:
Sulphuric acid = H₂SO₄
Molar mass of Sulphuric acid = 2 + 32 + 64 = 96
Basicity of Sulphuric acid = 2

= \(\frac{96}{2}\) = 49 g eq^-1

(II) (a)

As per balanced equation 54 g Al is required for 112 g of iron and 102 g of Al₂O₃.
54 g of Al gives 102 g of Al₂O₃.
∴ 324 g of Al will give \(\frac{102}{54}\) × 324 = 612 g of AlcO₃

(b) 54 g of Al requires 160 g of Fe₂O₃ for welding reaction.
∴ 324 g of Al will require \(\frac{160}{54}\) × 324 = 960 g of Fe\(\frac{102}{54}\)O₃
∴ Excess Fe₂O₃ -Unreacted Fe₂O₃ = 1120 – 960 = 160 g 160 g of exces reagent is left at the end of the reaction.

[0R]

(b)
(I) 1.
2.

(II) Let the no. of electrons in the ion = x
∴ The no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + \(\frac{30.4x}{100}\) = x + 0.304 x
Now, mass number of ion = Number of protons + Number of neutrons
= (x + 3) + (x + 0.304 x)
∴ 56 = (x + 3) + (x + 0.304 x) or 2.304 x = 56 – 3 = 53
x = \(\frac{53}{2.304}\) = 23
Atomic number of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

Question 35 (a).
(I) State the Newland’s law of octaves?
(II) What are the two exceptions of block division in the periodic table?

[OR]

(b) (I) Complete the following reactions.
(a)

(b) 2 BeCl₂ + LiaH₄ →?

(II) What happens when quick lime reacts with
(a) H₂O and
(b) CO₂?

(I) The Law of octaves states that, “when elements are arranged in the order of increasing atomic weights, the properties of the eighth element are a repetition of the properties of the first element”.

(II)

1. Helium has two electrons. Its electronic configuration is 1s². As per the configuration, it is supposed to be placed in ‘s’ block, but actually placed in 18th group which belongs to ‘p’ block. Because it has a completely filled valence shell as the other elements present in 18th group. It also resembles with 18th group elements in other properties. Hence helium is placed with other noble gases.

2. The other exception is hydrogen. It has only one s-electron and hence can be placed in group 1. It can also gain an electron to achieve a noble gas arrangement and hence it can behave as halogens (17^th group elements). Because of these assumptions, position of hydrogen becomes a special case. Finally, it is placed separately at the top of the periodic table.

[OR]

(b) (I) (a) Beryllium oxide is heated with carbon and chloride to get BeCl₂

(b) Beryllium chloride is treated with LiAlH₄ to get beryllium hydride.
2BeCl₂ + LiAlH₄ → 2BeH₂ + Licl + Alcl₃

(II) (a) CaO + H₂O → Ca(OH)₂ (calcium hydroxide)
(b) CaO + CO₂ → CaO₃ (calcium carbonate)

Question 36 (a).
(I) State the first law of thermodynamics?
(II) Calculate the enthalpy of combustion of ethylene at 300 Kc at constant pressure, if its heat of combustion at constant volume (∆U) is -1406 kJ?

[OR]

(b)
(I) Explain how the equilibrium constant Kc predict the extent of a reaction? (3)
(II) Explain about the effect of catalyst in an equilibrium reaction? (2)
Answer:
(a) (I) The first law of thermodynamics states that “the total energy of an isolated system . remains constant though it may change from one form to another” (or) Energy can neither be created nor destroyed, but may be converted from one form to another.
(II) The complete ethylene combustion reaction can be written as,

C₂H_4(g) + 3O_2(g) → 2CO_2(g) + 2H₂O_(l)
ΔU = -1406 kJ
Δn = n_p(g) – n_r(g)
Δn = 2 – 4 – 2
ΔH = ΔU + RTΔn_g
ΔH = -1406 + (8.314 × 10^-3 × 300 × (-2)) ΔH = -1410.9 kJ

[OR]

(b) (I)

1. The value of equilibrium constant K_C tells us the extent of the reaction i.e., it indicatehow far the reaction has proceeded towards product formation at a given temperature.
2. A large value of K_C indicates that the reaction reaches equilibrium with high product yield on the other hand, lower value of K_C indicates that the reaction reaches equilibrium with low product yield.
3. If K_C > 10³, the reaction proceeds nearly to completion.
4. If K_C < 10^-3 the reaction rarely proceeds.
5. It the K_C is in the range 10^-3 to 10³, significant amount of both reactants and products are present at equilibrium.

(II) Addition of catalyst does not affect the state of equilibrium. The catalyst increases the rate of both the forward and reverse reactions to the same extent. Hence it does not change the equilibrium composition of the reaction mixture.

Question 37 (a).
(I) Briefly explain geometrical isomerism in alkenes by considering 2- butene as an example.
2 – butene: Geometrical isomerism: CH₃ – CH = CH – CH₃ 

(II) What is meant by condensed structure? Explain with an example.

[OR]

(b)
(I) Why cut apple turns a brown colour?
(II) Predict the product for the following reaction,

1.

2.

3.

Answer:
(a) (I)

• Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.
• In 2-butene, the carbon-carbon double bond is sp² hybridised. The carbon-carbon double bond consists of a s bond and a p bond. The presence of p bond lock the molecule in one position. Hence, rotation around C = C bond is not possible.
• 
• These two compounds are termed as geometrical isomers and are termed as cis and transform.
• The cis isomer is the one in which two similar groups are on the same side of the double bond. The trans isomer is that in which two similar groups are on the opposite side of the double bond. Hence, this type of isomerism is called cis-trans isomerism.

(II) The bond line structure can be further abbreviated by omittiilg all the these dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula.
e.g., 1,3 – butadiene.

[OR]

(b) (I)

1. Apples contains an enzyme called polyphenol oxidase (PPO) also known as tyrosinase.
2. Cutting an apple exposes its cells to the atmospheric oxygen and oxidizes the phenolic compounds present in apples. This is called the “enzymatic browning” that turns a cut apple brown.
3. In addition to apples, enzymatic browning is also evident in bananas, pears, avocados and even potatoes.

Question 38 (a).
Suggest the route for the preparation of the following from benzene?

1. 3 – chloro-nitrobenzene
2. 4 – chlorotoluene
3. Bromobenzene
4. m – dinitrobenzene

[OR]

(b) A hydrocarbon C₃H₆ (A) reacts with HBr to form compound (B). Compound (B) reacts with aqueous potassium hydroxide to give (C) of molecular formula C₃H₆O. What are (A) (B) and (C). Explain the reactions?
Answer:
(a)
1. Preparation of 3 – chloronitro – benzene from benzene: Benzene undergoes nitration and followed by chlorination and it leads to the formation of 3 – chloronitrobenzene.

2. Preparation 4-chlorotoulene from benzene: Benzene undergoes Friedal craft’s alkylation followed by chlorination and it leads to the formation of 4 – chlorotoulene.

3. Preparation of Bromobenzene from benzene: Benzene undergo bromination to give bromobenzene.

4. Preparation of m-dinitrobenzene from benzene: Benzene undergo twice the time nitration to give m-dinitrobenzene.

[OR]

(b)
1. The hydrocarbon with molecular formula C₃H₆ (A) is identified as propene,
CH – CH = CH₂
2. Propene reacts with HBr to form bromopropane CH₃ – CH₂ – CH₂Br as (B).

3. 1 – bromopropane react with aqueous potassium hydroxide to give 1 – propanol CH₃ – CH₂ – CH₂OH as (C).
4. 2 – bromo propane reacts with aqueous KOH to give 2-propanol as (C)

Students can Download Tamil Nadu 11th Physics Model Question Paper 3 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Physics Model Question Paper 3 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

PART – I

Answer all the questions: [15 × 1 = 15]

Question 1.
A new unit of length is chosen such that the speed of light in vacuum is unity. The distance between the Sun and the Earth in terms of the new unit, if light takes 8 minute and 20 sec to cover the distance is …………………..
(a) 100 new unit
(b) 300 new unit
(c) 500 new unit
(d) 700 new unit
Hint:
Speed is unity = 1 unit/sec
Time = 8 min and 20 sec = 500 sec
Distance b/w sun and earth = Speed × Time
= 1 × 500 = 500 unit
Answer:
(c) 500 new unit

Question 2.
For a satellite moving in an orbit around the Earth, the ratio of kinetic energy to potential energy is ………………….
(a) 2
(b) 1 : 2
(c) 1 : \(\sqrt{2}\)
(d) \(\sqrt{2}\)
Hint:
\(\frac { GMm }{ R^{ 2 } } \) = mω²R
K.E = \(\frac{1}{2}\)Iω² = \(\frac{1}{2}\)mR²ω² = \(\frac{GMm}{2R}\)
P.E = – \(\frac{GMm}{R}\) ⇒ So \(\frac{K.E}{|P.E|}\) = \(\frac{1}{2}\)
Answer:
(b) 1 : 2

Question 3.
In the equilibrium position a body has ……………….
(a) Maximum potential energy
(b) Minimum potential energy
(c) Minimum kinetic energy
(d) Neither maximum nor minimum potential energy
Answer:
(c) Minimum kinetic energy

Question 4.
The centrifugal force appears to exist ………………….
(a) Only in inertial frames
(b) Only in rotating frames
(c) In any accelerated frame
(d) Both in inertial and non-inertial frames
Answer:
(b) Only in rotating frames

Question 5.
A particle is moving with a constant velocity along a line parallel to positive x-axis. The magnitude of its angular momentum with respect to the origin is …………………..
(a) Zero
(b) Increasing with x
(c) Decreasing with x
(d) Remaining constant
Answer:
(d) Remaining constant

Question 6.
When 8 droplets of water of radius 0.5 mm combine to form a single droplet. The radius of it is …………………
(a) 4 mm
(b) 2 mm
(c) 1 mm
(d) 8 mm
Hint:
Volume of 8 droplets of water = 8 × \(\frac{4}{3}\) π(0.5)³
When each droplet combine to form one volume remains conserved
R³ = 8 × (0.5)³
R³ = (8 × (0.5)³)
R³ = 2 × 0.5 = 1 mm
Answer:
(c) 1 mm

Question 7.
Pressure head in Bernoulli’s equation is …………………
(a) \(\frac { P_{ \rho } }{ g } \)
(b) \(\frac { P }{ \rho g } \)
(c) ρg
(d) Pρg
Answer:
(b) \(\frac { P }{ \rho g } \)

Question 8.
The angle between particle velocity and wave velocity in a transverse wave is ……………………
(a) Zero
(b) π/4
(c) π/2
(d) π
Answer:
(c) π/2

Question 9.
If the masses of the Earth and Sun suddenly double, the gravitational force between them will …………………….
(a) Remains the same
(b) Increase two times
(c) Increase four times
(d) Decrease two times
Answer:
(c) Increase four times

Question 10.
A mobile phone tower transmits a wave signal of frequency 900 MHz, the length of the transmitted from the mobile phone tower ……………………
(a) 0.33 m
(b) 300 m
(c) 2700 × 10⁸m
(d) 1200 m
Hint:
f = 900 MHz = 900 × 10⁶ Hz
Speed of wave (c) = 3 × 10⁶ ms^-1
λ = \(\frac{v}{f}\) = \(\frac { 3\times 10^{ 8 } }{ 900\times 10^{ 6 } } \) = \(\frac{1}{3}\) = 0.33m
Answer:
(a) 0.33 m

Question 11.
The displacement y of a wave travelling in the x direction is given by
y = (2 × 10^-3) sin (300 t – 2x + \(\frac { \pi }{ 4 } \)), where x and y are measured in metres and t in second. The speed of the wave is …………………
(a) 150 ms^-1
(b) 300 ms^-1
(c) 450 ms^-1
(d) 600 ms^-1
Hint:
From standard equation of wave, Y = a sin (ωt – kx + ϕ)
ω = 300 ; k = 2
Speed of wave, V = \(\frac{ω}{k}\) = \(\frac{300}{2}\) = 150ms^-1
Answer:
(a) 150 ms^-1

Question 12.
The increase in internal energy of a system is equal to the workdone on the system. The process does the system undergoes is ……………………
(a) Isochoric
(b) Adiabatic
(c) Isobaric
(d) Isothermal
Answer:
(d) Isothermal

Question 13.
The minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop is ……………………..
(a) \(\sqrt{2gR}\)
(b) \(\sqrt{3gR}\)
(c) \(\sqrt{5gR}\)
(d) \(\sqrt{gR}\)
Answer:
(c) \(\sqrt{5gR}\)

Question 14.
If the rms velocity of the molecules of a gas in a container be doubled then the pressure of the gas will.
(a) Becomes 4 times of the previous value
(b) Becomes 2 times of its previous value
(c) Remains same
(d) Becomes \(\frac{1}{4}\) of its previous value
Hint:

Answer:
(a) Becomes 4 times of the previous value

Question 15.
Gravitational mass is proportional to gravitational ………………….
(a) Intensity
(b) Force
(c) Field
(d) All of these
Answer:
(b) Force

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
Write any four postulates of Kinetic theory of gases?
Answer:

1. A gas consists of a very large number of molecules. Each one is a perfectly identical elastic sphere.
2. The molecules of a gas are in a state of continuous and random motion. They move in all directions with all possible velocities.
3. The size of each molecule is very small as compared to the distance between them. Hence, the volume occupied by the molecule is negligible in comparison to the volume of the gas:
4. There is no force of attraction or repulsion between the molecules and the walls of the container.

Question 17.
Draw the free body diagram of the book at rest on the table?
Answer:

Question 18.
Three block are connected as shown in fig on a horizontal frictionless table. If m₁ = 1 kg, m₂ = 8 kg, m₃ = 27 kg and T₃ = 36 N then calculate tension T₂?
Answer:

Acceleration acquired by all blocks a = \(\frac { T_{ 3 } }{ m_{ 1 }+m_{ 2 }+m_{ 3 } } \) = \(\frac{36}{36}\) = 1ms^-2
∴ Tension T₂ = (m₁ + m₂) a
= (1 + 8) × 1 = 9N

Question 19.
What is power? Give its dimensional formula?
Answer:
The rate of work done is called power. Dimensional formula of power is ML² T^-3

Question 20.
What are geostationary and polar satellites?
Answer:
Geostationary Satellite: It is the satellite which appears at a fixed position and at a definite height to an observer on Earth.
Polar Satellite: It is the satellite which revolves in polar orbit around the Earth.

Question 21.
An iceberg of density 900 kg m^-3 is floating in water of density 1000 kg m^-3. What is the percentage of volume of iceberg outside the water?
Answer:
Fraction of volume inside water = Relative density of the body
\(\frac { V’ }{ V } \) = \(\frac { \rho }{ \rho ‘ } \) = \(\frac{900}{1000}\) = 0.9
Fraction of volume outside water = 1 – 0.9 = 0.1
Percentage of volume outside water = 0.1 × 100 = 10%

Question 22.
State stoke’s law and define terminal velocity?
Answer:
Stoke’s law:
When a body falls through a highly viscous liquid, it drags the layer of the liquid immediately in contact with it. This results in a relative motion between the different layers of the liquid. As a result of this, the falling body experiences a viscous force F.

Stoke performed many experiments on the motion of small spherical bodies in different fluids and concluded that the viscous force F acting on the spherical body depends on

1. Coefficient of viscosity q of the liquid
2. Radius a of the sphere and
3. Velocity v of the spherical body

Dimensionally it can be proved that ∴ F = k ηav
Experimentally Stoke found that k = 6π
This is Stoke’s law

Terminal velocity:
Terminal velocity of a body is defined as the constant velocity acquired by a body while falling through a viscous liquid.

Question 23.
The Earth without its atmosphere would be hospitably cold. Explain why?
Answer:
The lower layers of Earth’s atmosphere reflect infrared radiations from Earth back to the surface of Earth. Thus the heat radiation received by the earth from the Sun during the day are kept trapped by the atmosphere. If atmosphere of Earth were not there, its surface would become too cold to live.

Question 24.
A body A is projected upwards with velocity v₁ Another body B of same mass is proj eeted at an angle of 45°. Both reach the same height. Calculate the ratio of their initial kinetic energies?
Answer:
As A and B attain the same height therefore vertical component of initial velocity of B is equal to initial velocity of A
v₂ cos 45° = V₁ (or) \(\frac { v_{ 2 } }{ \sqrt { 2 } } \) = v₁

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
Explain the rules for counting significant figures with examples?
Rules for counting significant figures:
Answer:

Question 26.
Elastic headon collision, consider two particles one is moving and another one is stationary with their respective masses m and \(\frac { M }{ m } \). A moving particle meets collides elastically on stationary particle in the opposite direction. Find the kinetic energy of the stationary particle after a collision?
Answer:
mass of the moving particle m₁ = m (say)
mass of the stationary particle m₁ = \(\frac { 1 }{ m } \) M
Velocity of the moving particle before collision = v_1i (say)
Velocity of the stationary particle before collision = v_2i = 0
Velocity of the stationary particle after collision = v_2f (say)

Kinetic energy of the stationary particle after a collision

Question 27.
Calculate the angle for which a cyclist bends when he turns a circular path of length 34.3 m in \(\sqrt{22}\) s?
Answer:
Given Data:
l = 34.3 m, t = \(\sqrt{22}\) , g = 9.8 ms^-2, θ = ?
If r is radius of circular path, then length of path = 2πr = 34.3 m
r = \(\frac { 33.4 }{ 2\pi } \) and time taken t = \(\sqrt{22}\)s

As tan θ = \(\frac { v^{ 2 } }{ rg } \)
∴ tan θ = (\(\frac { 34.3 }{ \sqrt { 22 } } \))² × \(\frac { 2\pi }{ 34.3\times 9.8 } \)
tan θ = \(\frac { 34.3\times 34.3 }{ 22 } \) × \(\frac{2×22}{7×343×9.8}\) = \(\frac{34.3×2}{68.6}\) = 1 [∴θ = 45°]

Question 28.
Explain how density, moisture affect the velocity of sound in gases?
Answer:
Effect of density:
Let us consider two gases with different densities having same temperature and pressure. Then the speed of sound in the two gases are

Taking ratio of equation (1) and equation (2) we get

For gases having same value of γ,
\(\frac { v_{ 1 } }{ v_{ 2 } } \) = \(\sqrt { \frac { \rho _{ 2 } }{ \rho _{ 1 } } } \)
Thus the velocity of sound in a gas is inversely proportional to the square root of the density of the gas.

Effect of moisture (humidity):
We know that density of moist air is 0.625 of that of dry air, which means the presence of moisture in air (increase in humidity) decreases its density. Therefore, speed of sound increases with rise in humidity. From equation:
v = \(\sqrt { \frac { \gamma \rho }{ \rho } } \)

Let ρ₁, v₁ = and ρ₂, v₂ be the density and speeds of sound in dry air and moist air, respectively.

Since P is the atmospheric pressure, it can be shown that
\(\frac { \rho _{ 2 } }{ \rho _{ 1 } } \) = \(\frac { P }{ p_{ 1 }+0.625p_{ 2 } } \)
where p₁ and p₂ are the partial pressures of dry air and water vapour respectively. Then

Question 29.
Explain
(a) Why there are no lunar eclipse and solar eclipse every month?
(b) Why do we have seasons on earth?
Answer:
(a) If the orbits of the Moon and Earth lie on the same plane, during full Moon of every month, we can observe lunar eclipse. If this is so during new Moon we can observe solar eclipse.

But Moon’s orbit is tilted 5° with respect to Earth’s orbit. Due to this 5° tilt, only during certain periods of the year, the Sun, Earth and Moon align in straight line leading to either lunar eclipse or solar eclipse depending on the alignment.

(b) The common misconception is that ‘Earth revolves around the Sun, so when the Earth is very far away, it is winter and when the Earth is nearer, it is summer’.

Actually, the seasons in the Earth arise due to the rotation of Earth around the Sun with 23.5° tilt. Due to this 23.5° tilt, when the northern part of Earth is farther to the Sun, the southern part is nearer to the Sun. So when it is summer in the northern hemisphere, the southern hemisphere experience winter.

Question 30.
When a person breathes, his lungs can hold up to 5.5 1 of air at body temperature 37°C and atmospheric pressure (1 atm = 101 kPa). This air contains 21% oxygen, calculate the number of oxygen molecules in the lungs?
Answer:
We can treat the air inside the lungs as an ideal gas. To find the number of molecules, we can use the ideal gas law.
PV = NkT
Here volume is given in the Litre. 1 Litre is volume occupied by a cube of side 10 cm
1 Litre = 10 cm × 10 cm × 10 cm = 10^-3m^-3
N = \(\frac{PV}{kT}\) = \(\frac { 1.01\times 10^{ 5 }\times 5.5\times 10^{ -3 } }{ 1.38\times 10^{ -23 }\times 310 } \)
= 1.29 × 10²³ × \(\frac{21}{100}\)
Number of oxygen molecules = 2.7 × 10²² molecules

Question 31.
Give any five properties of vector product of two vectors?
Answer:
(I) The vector product of any two vectors is always another vector whose direction is perpendicular to the plane containing these two vectors, i.e., orthogonal to both the vectors \(\vec { A } \) and \(\vec { B } \), even though the vectors AandB may or may not be mutually orthogonal.

(II) The vector product of two vectors is not commutative, i.e., \(\vec { A } \) × \(\vec { B } \) ≠ \(\vec { B } \) × \(\vec { A } \) . But, \(\vec { A } \) × \(\vec { B } \) = –\(\vec { B } \) × \(\vec { A } \).

Here it is worthwhile to note that |\(\vec { A } \) × \(\vec { B } \)| = |\(\vec { B } \) × \(\vec { A } \)| = AB sin θ i.e., in the case of the product vectors \(\vec { A } \) × \(\vec { B } \) and \(\vec { B } \) × \(\vec { A } \), the magnitudes are equal but directions are opposite to each other.

(III) The vector product of two vectors will have maximum magnitude when sin 0 = 1, i.e., θ = 90° i.e., when the vectors \(\vec { A } \) and \(\vec { B } \) are orthogonal to each other.
(\(\vec { A } \) × \(\vec { B } \))_max = AB\(\hat { n } \)

(IV) The vector product of two non-zero vectors will be minimum when sin 0 = 0, i.e., 0 = 0° or 180°
(\(\vec { A } \) × \(\vec { B } \))_max = 0
i.e., the vector product of two non-zero vectors vanishes, if the vectors are either parallel or antiparallel.

(V) The self-cross product, i.e., product of a vector with itself is the null vector
\(\vec { A } \) × \(\vec { A } \) = AA sin 0°\(\hat { n } \) = \(\vec { 0 } \)
In physics the null vector \(\vec { 0 } \) is simply denoted as zero.

Question 32.
Why does a porter bend forward w hile carrying a sack of rice on his back?
Answer:
When a porter carries a sack of rice, the line of action of his centre of gravity will go away from the body. It affects the balance, to avoid this he bends. By which centre of gravity will realign within the body again. So balance is maintained.

Question 33.
A piece of wood of mass m is floating erect in a liquid whose density is p. If it is slightly pressed down and released, then executes simple harmonic motion. Show that its time period of oscillation is T = 2π\(\sqrt{m/Agρ}\)
Answer:
Spring factor of liquid(k) = Aρg
Inertia factor of wood = m
Time period T = 2π = image 12
T = 2π\(\sqrt{m/Aρg}\)

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
Describe the vertical oscillations of a spring?
Answer:
Vertical oscillations of a spring: Let us consider a massless spring with stiffness constant or force constant k attached to a ceiling as shown in figure. Let the length of the spring before loading mass m be L.

If the block of mass m is attached to the other end of spring, then the spring elongates by a length. Let F, be the restoring force due to stretching of spring. Due to mass m, the gravitational force acts vertically downward. We can draw free-body diagram for this system as shown in figure. When the system is under equilibrium,

F₁ + mg = 0 ……………… (1)
But the spring elongates by small displacement 1, therefore,
F₁ ∝ l ⇒ F₁ = -kl ……………….. (2)
Substituting equation (2) in equation (1) we get
-kl + mg = 0
mg = kl or
\(\frac{m}{k}\) = \(\frac{l}{g}\) ………………….. (3)

Suppose we apply a very small external force on the mass such that the mass further displaces downward by a displacement y, then it will oscillate up and down. Now, the restoring force due to this stretching of spring (total extension of spring is y + 1) is
F₂ ∝ (y + l)
F₂ = -k(y + l) = -ky – kl …………………… (4)

Since, the mass moves up and down with acceleration \(\frac{d^{2} y}{d t^{2}}\), by drawing the free body diagram for this case we get
-ky – kl + mg = m \(\frac{d^{2} y}{d t^{2}}\) ……………………. (5)
The net force acting on the mass due to this stretching is
F = F₂ + mg
F = -ky – kl + mg …………………… (6)
The gravitational force opposes the restoring force. Substituting equation (3) in equation (6), we get
F = – ky- kl + kl = -ky
Applying Newton’s law we get
m \(\frac{d^{2} y}{d t^{2}}\) = -ky
\(\frac{d^{2} y}{d t^{2}}\) = –\(\frac{k}{m}\)y ………………… (7)
The above equation is in the form of simple harmonic differential equation. Therefore, we get the time period as
T = 2π\(\sqrt{m/k}\) second ……………………. (8)
The time period can be rewritten using equation (3)
T = 2π\(\sqrt{m/k}\) = 2πl\(\frac{1}{g}\) second ……………………. (9)
The accleration due to gravity g can be computed by the formula
g = 4π²\((\frac { 1 }{ T } )^{ 2 }\)ms^-2 …………………….. (10)

[OR]

(b) Derive poiseuille’s formula for the volume of a liquid flowing per second through a pipe under streamlined flow?
Answer:
Consider a liquid flowing steadily through a horizontal capillary tube. Let v = (\(\frac{1}{g}\)) be the volume of the liquid flowing out per second through a capillary tube. It depends on (1) coefficient of viscosity (η) of the liquid, (2) radius of the tube (r), and (3) the pressure gradient (\(\frac{P}{l}\)) . Then,
v ∝η^ar^b(\(\frac{P}{l}\))^c
v = kη^ar^b(\(\frac{P}{l}\))^c …………………….. (1)

where, k is a dimensionless constant.
Therefore, [v] = \(\frac { Volume }{ Time } \) = [L³T^-1]; [ \(\frac{dP}{dX}\) ] = \(\frac { Pressure }{ Distance } \) = [ML^-2T^-2]
[η] = [Ml^-1T^-1] and [r] = [L]

Substituting in equation (1)
[L³T^-1] = [ML^-1T^-1]^a[L]^b [ML^-2T^-2]^c
M⁰L³T^-1 = M^a+bL^-a+b-2cT^-a-2c = -1

So, equating the powers of M, L and T on both sides, we get
a + c = 0, – a + b – 2c = 3, and – a – 2c = – 1

We have three unknowns a, b and c. We have three equations, on solving, we get
a = – 1, b = 4 and c = 1

Therefore, equation (1) becomes,
v = kη^-1r⁴(\(\frac{P}{l}\))¹

Experimentally, the value of k is shown to be , we have \(\frac{π}{8}\), we have
v = \(\frac{\pi r^{4} \mathrm{P}}{8 \eta /}\)

The above equation is known as Poiseuille’s equation for the flow of liquid through a narrow tube or a capillary tube. This relation holds good for the fluids whose velocities are lesser than the critical velocity (v_c).

Question 35 (a).
Describe briefly simple harmonic oscillation as a projection of uniform circular motion?
Answer:
Consider a particle of mass m moving with unifonn speed v along the circumference of a circle whose radius is r in anti-clockwise direction (as shown in figure). Let us assume that the origin of the coordinate system coincides with the center O of the circle.

If ω is the angular velocity of the particle and θ the angular displacement of the particle at any instant of time t, then θ = ωt. By projecting the uniform circular motion on its diameter gives a simple harmonic motion.

This means that we can associate a map (or a relationship) between uniform circular (or revolution) motion to vibratory motion. Conversely, any vibratory motion or revolution can be mapped to unifonn circular motion. In other words, these two motions are similar in nature.

Let us first project the position of a particle moving on a circle, on to its vertical diameter or on to a line parallel to vertical diameter as shown in figure. Similarly, we can do it for horizontal axis or a line parallel to horizontal axis.

The projection of uniform circular motion on a diameter of SHM:
As a specific example, consider a spring mass system (or oscillation of pendulum). When the spring moves up and down (or pendulum moves to and fro), the motion of the mass or bob is mapped to points on the circular motion.

Thus, if a particle undergoes uniform circular motion then the projection of the particle on the diameter of the circle (or on a line parallel to the diameter) traces straight line motion which is simple harmonic in nature. The circle is known as reference circle of the simple harmonic motion. The simple harmonic motion can also be defined as the motion of the projection of a particle on any diameter of a circle of reference.

[OR]

(b) State and prove Bernoulli’s theorem for a flow of incompressible non viscous and stream lined flow of fluid?
Answer:
Bernoulli’s theorem:
According to Bernoulli’s theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains a constant. Mathematically,
\(\frac{P}{ρ}\) + \(\frac{1}{2}\)v² + gh – constant
This is known as Bernoulli’s equation.
Proof:
Let us consider a flow of liquid through a pipe AB. Let V be the volume of the liquid when it enters A in a time t. Which is equal to the volume of the liquid leaving B in the same time. Let a_A, v_A and P_A be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively.

Let the force exerted by the liquid at A is
F_A = P_Aa_A

Distance travelled by the liquid in time t is d = v_At
Therefore, the work done is W = F_Ad = P_Aa_Av_At
But a_Av_At = a_Ad = V, volume of the liquid entering at A.

Thus, the work done is the pressure energy (at A), W = F_Ad = P_AV

Pressure energy per unit volume at
A = \($\frac{\text { Pressure energy }}{\text { Volume }}$\) = \(\frac { P_{ A }V }{ V } \) = P_A

Pressure energy per unit mass at
A = \($\frac{\text { Pressure energy }}{\text { Mass }}$\) = \(\frac { P_{ A }V }{ m } \) = \(\frac { P_{ A } }{ \frac { m }{ V } } \) = \(\frac { P_{ A } }{ \rho } \)

Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is
EP_A = P_AV = P_AV × (\(\frac{m}{m}\)) = m\(\frac { P_{ A } }{ \rho } \)

Potential energy of the liquid at A,
PE_A = mgh_A

Due to the flow of liquid, the kinetic energy of the liquid at A,
KE_A = \(\frac{1}{2}\)mv^2_A

Therefore, the total energy due to the flow of liquid at A,
E_A = EP_A + KE_A + PE_A
E_A = \(m \frac{P_{A}}{\rho}+\frac{1}{2} m v_{A}^{2}+m g h_{A}\)

Similarly, let a_B, V_B and P_B be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at B. Calculating the total energy at F_B, we get .
\(\mathrm{E}_{\mathrm{B}}=m \frac{\mathrm{P}_{\mathrm{B}}}{\rho}+\frac{1}{2} m v_{\mathrm{B}}^{2}+m g h_{\mathrm{B}}\)
From the law of conservation of energy.
E_A = E_B

Thus, the above equation can be written as
\(\frac { P }{ \rho g } \) + \(\frac{1}{2}\) \(\frac { v^{ 2 } }{ g } \) + h = Constant

The above equation is the consequence of the conservation of energy which is true until there is no loss of energy due to friction. But in practice, some energy is lost due to friction. This arises due to the fact that in a fluid flow, the layers flowing with different velocities exert frictional forces on each other. This loss of energy is generally converted into heat energy. Therefore, Bernoulli’s relation is strictly valid for fluids with zero viscosity or non-viscous liquids. Notice that when the liquid flows through a horizontal pipe, then
h = 0 ⇒ \(\frac { P }{ \rho g } \) + \(\frac{1}{2}\) \(\frac { v^{ 2 } }{ g } \) = Constant

Question 36 (a).
Explain perfect inelastic collision and derive an expression for loss of kinetic energy in perfect inelastic collision?
Answer:
In a perfectly inelastic or completely inelastic collision, the objects stick together permanently after collision such that they move with common velocity. Let the two bodies with masses m₁ and m₂ move with initial velocities u₁ and u₂ respectively before collision. Aft er perfect inelastic collision both the objects move together with a common velocity v as shown in figure.
Since, the linear momentum is conserved during collisions,

m₁u₁ + m₂u₂ = (m₁ + m₂) v

The common velocity can be computed by
v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\) ………………….. (1)

Loss of kinetic energy in perfect inelastic collision;
In perfectly inelastic collision, the loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, heat, light etc. Let KE_i be the total kinetic energy before collision and KE_f be the total kinetic energy after collision.
Total kinetic energy before collision,
KE_e = \(\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}\) …………………… (2)
Total kinetic energy after collision,
KE_f = \(\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}\) …………………….. (3)
Then the loss of kinetic energy is Loss of KE, ∆Q = KE_f – KE_i
= \(\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}-\frac{1}{2} m_{1} u_{1}^{2}-\frac{1}{2} m_{2} u_{2}^{2}\) ………………….. (4)
Substituting equation (1) in equation (4), and on simplifying (expand v by using the algebra (a + b)² = a² + b² + 2ab), we get
Loss of KE, ∆Q = \(\frac{1}{2}\) \(\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)\) (u₁ – u₂)²

[OR]

(b) Derive an expression for maximum height attained, time of flight, horizontal range for a projectile in oblique projection?
Answer:

Maximum height (h_max):
The maximum vertical distance travelled by the projectile during the journey is called maximum height. This is determined as follows:
For the vertical part of the motion
\(v_{y}^{2}=u_{y}^{2}+2 a_{y} s\)
Here, u_y= u sin θ, a = -g, s = h_max, and at the maximum height v = 0

Time of flight (T_f):
The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight. This time of flight is the time taken by the projectile to go from point O to B via point A as shown in figure.
We know that s_y = u_yt + \(\frac{1}{2}\)a_yt²
Here, s_y = y = 0 (net displacement in y-direction is zero), u_y = u sin θ, a_y = -g, t = T_f, Then
0 = u sin θ T_f – \(\frac{1}{2} g \mathrm{T}_{f}^{2}\)
T_f = 2u \(\frac{sin θ}{g}\) …………………….. (2)

Horizontal range (R):
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). This is found easily since the horizontal component of initial velocity remains the same. We can write.

Range R = Horizontal component of velocity % time of flight = u cos θ × T_f = \(\frac{u^{2} \sin 2 \theta}{g}\)
The horizontal range directly depends on the initial speed (u) and the sine of angle of projection (θ). It inversely depends on acceleration due to gravity ‘g’.

For a given initial speed u, the maximum possible range is reached when sin 2θ is
maximum, sin 2θ = 1. This implies 2θ = π/2 or θ = π/4
This means that if the particle is projected at 45 degrees with respect to horizontal, it attains maximum range, given by
R_max = \(\frac { u^{ 2 } }{ g } \).

Question 37 (a).
Explain the work-energy theorem in detail and also give three examples?
Answer:

1. If the work done by the force on the body is positive then its kinetic energy increases.
2. If the work done by the force on the body is negative then its kinetic energy decreases.
3. If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.
4. When a particle moves with constant speed in a circle, there is no change in the kinetic energy of the particle. So according to work energy principle, the work done by centripetal force is zero.

[OR]

(b) (i) Define molar specific heat capacity?
Answer:
Molar specific heat capacity is defined as heat energy required to increase the temperature of one mole of substance by IK or 1°C

(ii) Derive Mayer’s relation for an ideal gas?

Mayer’s relation: Consider p mole of an ideal gas in a container with volume V, pressure P and temperature T.

When the gas is heated at constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU.

If C_V is the molar specific heat capacity at constant volume, from equation.
C_V = \(\frac { 1 }{ \mu } \) \(\frac{dU}{dT}\) …………………… (1)
dU = µC_V dT ………………… (2)

Suppose the gas is heated at constant pressure so that the temperature increases by dT. If ‘Q’ is the heat supplied in this process and ‘dV’ the change in volume of the gas.
Q = pCpdT ……………. (3)

If W is the workdone by the gas in this process, then
W = P dV ………………….. (4)

But from the first law of thermodynamics,
Q = dU + W ………………… (5)

Substituting equations (2), (3) and (4) in (5), we get,
For mole of ideal gas, the equation of state is given by
\(\mu \mathrm{C}_{\mathrm{p}} d \mathrm{T}=\mu \mathrm{C}_{\mathrm{v}} d \mathrm{T}+\mathrm{P} d \mathrm{V}\)

Since the pressure is constant, dP = 0
C_pdT = C_VdT + PdV
∴ C_p = C_V + R (or) C_p – C_V = R …………………… (6)
This relation is called Mayer’s relation It implies that the molar specific heat capacity of an ideal gas at constant pressure is greater than molar specific heat capacity at constant volume.
The relation shows that specific heat at constant pressure (s_p) is always greater than specific heat at constant volume (s_v).

Question 38 (a).
Derive an expression of pressure exerted by the gas on the walls of the container?
Answer:
Expression for pressure exerted by a gas : Consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l.
The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.

The molecules of the gas exert pressure on the walls of the container due to collision on it. During each collision, the molecules impart certain momentum to the wall. Due to transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.

A molecule of mass m moving with a velocity \(\vec { v } \) having components (v_x, v_y, v_z) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with same speed and its x-component is reversed. This is shown in the figure. The components of velocity of the molecule after collision are (-v_x, v_y, v_z).

The x-component of momentum of the molecule before collision = mv_x
The x-component of momentum of the molecule after collision = – mv_x
The change in momentum of the molecule in x direction
= Final momentum – initial momentum = – mv_x – mv_x = – 2mv_x
According to law of conservation of linear momentum, the change in momentum of the wall = 2 mv_x
The number of molecules hitting the right side wall in a small interval of time ∆t.

The molecules within the distance of v_x∆t from the right side wall and moving towards the right will hit the wall in the time interval ∆t. The number of molecules that will hit the right side wall in a time interval ∆t is equal to the product of volume (Av_x∆t) and number density of the molecules (n).

Here A is area of the wall and n is number of molecules per \(\frac{N}{V}\) unit volume. We have assumed that the number density is the same throughout the cube.

Not all the n molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side.

The number of molecules that hit the right side wall in a time interval ∆t
= \(\frac{n}{2}\) Av_x∆t
In the same interval of time ∆t, the total momentum transferred by the molecules
\(\Delta \mathrm{P}=\frac{n}{2} \mathrm{A} v_{x} \Delta t \times 2 m v_{x}=\mathrm{A} v_{x}^{2} m n \Delta t\) ………………….. (2)
From Newton’s second law, the change in momentum in a small interval of time gives rise to force.

The force exerted by the molecules on the wall (in magnitude)
F = \(\frac{∆p}{∆t}\) = nmAv^2_x ……………………. (3)

Pressure, P = force divided by the area of the wall
P = \(\frac{F}{A}\) = nmAv^2_x ……………………….. (4)
p = \(nm\bar{v}_{x}^{2}\)

Since all the molecules are moving completely in random manner, they do not have same . speed. So we can replace the term vnmAv^2_x by the average \(\bar { v } \)^2_x in equation (4).
P = nm\(\bar { v } \)^2_x ……………………. (5)

Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, \(\bar{v}_{x}^{2}\) = \(\bar{v}_{y}^{2}\) = \(\bar{v}_{z}^{2}\). The mean square speed is written as
\(\bar{v}^{2}\) = \(\bar{v}_{x}^{2}\) + \(\bar{v}_{y}^{2}\) + \(\bar{v}_{z}^{2}\) = 3\(\bar{v}_{x}^{2}\)
\(\bar{v}_{x}^{2}\) = \(\frac{1}{3}\) \(\bar{v}^{2}\)
Using this in equation (5), we get
P = \(\frac{1}{3} n m \bar{v}^{2} \quad \text { or } P=\frac{1}{3} \frac{N}{V} m \bar{v}^{2}\) ………………….. (6)

[OR]

(b) Discuss the simple pendulum in detail?
Answer:
Simple pendulum

A pendulum is a mechanical system which exhibits periodic motion. It has a bob with mass m suspended by a long string (assumed to be massless and inextensible string) and the other end is fixed on a stand. At equilibrium, the pendulum does not oscillate and hangs vertically downward.

Such a position is known as mean position or equilibrium position. When a pendulum is displaced through a small displacement from its equilibrium position and released, the bob of the pendulum executes to and fro motion. Let l be the length of the pendulum which is taken as the distance between the point of suspension and the centre of gravity of the bob. Two forces act on the bob of the pendulum at any displaced position.

• The gravitational force acting on the body (\(\vec { F} \) = m\(\vec { g } \)) which acts vertically downwards.
• The tension in the string T which acts along the string to the point of suspension.

Resolving the gravitational force into its components:

1. Normal component: The component along the string but in opposition to the direction of tension, F_as = mg cos θ.
2. Tangential component: The component perpendicular to the string i.e., along tangential direction of arc of swing, F_ps = mg sin θ.

Therefore, The normal component of the force is, along the string,
\(\mathrm{T}-\mathrm{W}_{a s}=m \frac{v^{2}}{l}\)

Here v is speed of bob
T -mg cos θ = m \(\frac{v^{2}}{l}\)

From the figure, we can observe that the tangential component W_ps of the gravitational force always points towards the equilibrium position i.e., the direction in which it always points opposite to the direction of displacement of the bob from the mean position. Hence, in this case, the tangential force is nothing but the restoring force. Applying Newton’s second law along tangential direction, we have
\(m \frac{d^{2} s}{d t^{2}}+\mathrm{F}_{p s}=0 \Rightarrow m \frac{d^{2} s}{d t^{2}}=-\mathrm{F}_{p s}\)
\(m \frac{d^{2} s}{d t^{2}}=-m g \sin \theta\) …………………. (1)

where, s is the position of bob which is measured along the arc. Expressing arc length in terms of angular displacement i.e.,
s = lθ ………………… (2)
then its acceleration, \(\frac{d^{2} s}{d t^{2}}=l \frac{d^{2} \theta}{d t^{2}}\) …………………. (3)

Substituting equation (3) in equation (1), we get
\(\begin{aligned}
l \frac{d^{2} \theta}{d t^{2}} &=-g \sin \theta \\
\frac{d^{2} \theta}{d t^{2}} &=-\frac{g}{l} \sin \theta
\end{aligned}\) ………………….. (4)

Because of the presence of sin θ in the above differential equation, it is a non-linear differential equation (Here, homogeneous second order). Assume “the small oscillation approximation”, sin θ ~ 0, the above differential equation becomes linear differential equation.
\(\frac{d^{2} \theta}{d t^{2}}=-\frac{g}{l} \theta\) …………………… (5)

This is the well known oscillatory differential equation. Therefore, the angular frequency of this oscillator (natural frequency of this system) is
ω² = \(\frac{g}{l}\) …………………… (6)
∴ ω = \(\sqrt{g/l}\) in rad s^-1 ……………….. (7)

The frequency of oscillation is
f = \(f=\frac{1}{2 \pi} \sqrt{\frac{g}{l}} \text { in } \mathrm{Hz}\) ………………… (8)
and time period of sscillations is
T = 2π\(\sqrt{l/g}\) in second. ……………….. (9)

Students can Download Tamil Nadu 11th Maths Model Question Paper 3 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Maths Model Question Paper 3 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 90

PART – 1

I. Choose the correct answer. Answer all the questions: [20 × 1 = 20]

Question 1.
Let R be the universal relation on a set X with more than one element then R is ………………
(a) Not reflexive
(b) Not symmetric
(c) Transitive
(d) None of the above
Answer:
(c) Transitive

Question 2.
The value of log_ab log_bc log_ca is …………………..
(a) 2
(b) 1
(c) 3
(d) 4
Answer:
(b) 1

Question 3.
If log \(\log _{\sqrt{ }}\) 0.25 = 4 then the value of x is ………………..
(a) 0.5
(b) 2.5
(c) 1.5
(d) 1.25
Answer:
(a) 0.5

Question 4.
The product of r consecutive positive integers is divisible by ………………..
(a) r!
(b) (r-1)!
(c) (r+l)!
(d) r^r
Answer:
(a) r!

Question 5.
The value of tan75° – cot 75° is ……………….
(a) 1
(b) 2 + \(\sqrt{3}\)
(c) 2 – \(\sqrt{3}\)
(d) 2\(\sqrt{3}\)
Answer:
(d) 2\(\sqrt{3}\)

Question 6.
If (1 +x²)²(1 + x)² = a₀ + a₁ x + a₂x² …. + xⁿ⁺⁴ and if a₀, a₁, a₂, are in AP, then n is …………………..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 7.
If _nC₁₂ = _nC₅ then _nC₂ = …………………
(a) 72
(b) 306
(c) 152
(d) 153
Answer:
(d) 153

Question 8.
The line (p + 2q)x + (p- 3q)y =p – q for different values of p and q passes through the point …………………
(a) (\(\frac{3}{5}\), \(\frac{2}{5}\))
(b) (\(\frac{2}{5}\), \(\frac{2}{5}\))
(c) (\(\frac{3}{5}\), \(\frac{3}{5}\))
(d) (\(\frac{2}{5}\), \(\frac{3}{5}\))
Answer:
(d) (\(\frac{2}{5}\), \(\frac{3}{5}\))

Question 9.
The number of terms in the expansion of [(a + b)²]¹⁸ = ………………..
(a) 19
(b) 18
(c) 36
(d) 37
Answer:
(d) 37

Question 10.
A line passes through the point (2, 2) and is perpendicular to the line 3x + y = 5 then its y intercept is …………………
Answer:
(a) \(\frac{3}{4}\)
(b) \(\frac{4}{3}\)
(c) 5
(d) \(\frac{1}{3}\)

Question 11.
If a and b are the roots of the equation x² – kx + 16 = 0 satisfy a² + b² = 32, then the value of k is ………………..
(a) 10
(b) -8
(c) -8, 8
(d) 6
Answer:
(c) -8, 8

Question 12.
If A is a square matrix of order 3 then |kA| = ………………….
(a) k |A|
(b) k²|A|
(c) k³|A|
(d) k|A³|
Answer:
(c) k³|A|

Question 13.
If ABCD is a parallelogram then \(\bar { AB } \) + \(\bar { AD } \) + \(\bar { CD } \) + \(\bar { CD } \) = ………………..
(a) 2(\(\bar { AB } \) + \(\bar { AD } \))
(b) 4\(\bar { AC } \)
(c) 4\(\bar { BD } \)
(d) \(\bar { o } \)
Answer:
(d) \(\bar { o } \)

Question 14.
\(\lim _{x \rightarrow 0}\) x cot x = ………………….
(a) 0
(b) 1
(c) -1
(d) ∞
Answer:
(b) 1

Question 15.
If x = \(\frac { 1-t^{ 2 } }{ 1+t^{ 2 } } \) and y = \(\frac { 2t }{ 1+t^{ 2 } } \) then \(\frac{dy}{dx}\) = ………………..
(a) \(\frac{y}{x}\)
(b) \(\frac{-y}{x}\)
(c) –\(\frac{x}{y}\)
(d) \(\frac{x}{y}\)
Answer:
(c) –\(\frac{x}{y}\)

Question 16.
If y = \(\frac { (1-x)^{ 2 } }{ x^{ 2 } } \) then \(\frac{dy}{dx}\) is …………………..
(a) \(\frac { 2 }{ x^{ 2 } } \) + \(\frac { 2 }{ x^{ 3 } } \)
(b) –\(\frac { 2 }{ x^{ 2 } } \) + \(\frac { 2 }{ x^{ 3 } } \)
(c) –\(\frac { 2 }{ x^{ 2 } } \) – \(\frac { 2 }{ x^{ 3 } } \)
(d) –\(\frac { 2 }{ x^{ 3 } } \) + \(\frac { 2 }{ x^{ 2 } } \)
Answer:
(d) –\(\frac { 2 }{ x^{ 3 } } \) + \(\frac { 2 }{ x^{ 2 } } \)

Question 17.
If y = \(\frac{sinx+cosx}{sinx-cosx}\) then \(\frac{dy}{dx}\) at x = \(\frac { \pi }{ 2 } \) is ………………….
(a) 1
(b) 0
(c) -2
(d) 2
Answer:
(c) -2

Question 18.
\(\int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}\) dx is ……………………
(a) \(\frac{1}{2}\) sin2x + c
(b) –\(\frac{1}{2}\) sin2x + c
(c) \(\frac{1}{2}\) cos 2x + c
(d) – \(\frac{1}{2}\) cos 2x + c
Answer:
(b) –\(\frac{1}{2}\) sin2x + c

Question 19.
An urn contains 5 red and 5 black balls. A balls is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. The probability that the second ball drawn is red will be ………………
(a) \(\frac{5}{12}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{7}{12}\)
(d) \(\frac{1}{4}\)
Answer:
(b) \(\frac{1}{2}\)

Question 20.
Let A and B be two events such that P(\(\bar { AUB } \)) = \(\frac{1}{6}\) , Then the events A and B are P(A∩B) = 1/4 and P(\(\bar { A } \)) = 1/4 is ………………
(a) Equally likely but not independent
(b) Independent but not equally likely
(c) Independent and equally likely
(d) Mutually inclusive and dependent
Answer:
(b) Independent but not equally likely

PART – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
Let A and B are two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2) and (z, 1) are in A × B, find A and B where x, y, z are distinct elements?
Answer:
n(A) = 3 ⇒ set A contains 3 elements
n(B) = 2 ⇒ set B contains 2 elements
we are given (x, 1), (y, 2), (z, 1) are elements in A × B
⇒ A = {x, y, z} and B = {1, 2}

Question 22.
Solve |5x — 12| ← 2
Answer:
5x – 12 > -2 (or) 5x – 12 < 2 ⇒ 5x > -2 + 12 (= 10)
⇒ x > \(\frac{10}{5}\) = 2
x > 2
(or)
5x < 2 + 12 (= 14)
⇒ x < \(\frac{14}{5}\)
so 2 < x < \(\frac{14}{5}\)

Question 23.
If ¹⁰P_r-1 = 2 × 6 P_r, find r?
Answer:
¹⁰P_r-1 = 2 × 6P_r

⇒ (11 – r) (10 – r) (9 – r) (8 – r) (7 – r) = 10 × 9 × 4 × 7
= 5 × 2 × 3 × 3 × 2 × 2 × 7
= 7 × 6 × 5 × 4 × 3
⇒ 11 – r = 7
11 – 7 = r
r = 4

Question 24.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2^nd hour, 4^th hour and n^th hour?
Answer:
No. of bacteria at the beginning = 30
No. of bacteria after 1 hour = 30 × 2 = 60
No. of bacteria after 2 hours = 30 × 2² = 30 × 4 = 120
No. of bacteria after 4 hours = 30 × 2⁴ = 30 × 16 = 480
No. of bacteria after n^th hour = 30 × 2ⁿ

Question 25.
Find |A| if A = \(\left[\begin{array}{ccc}
0 & \sin \alpha & \cos \alpha \\
\sin \alpha & 0 & \sin \beta \\
\cos \alpha & -\sin \beta & 0
\end{array}\right]\)
Answer:
\(\left[\begin{array}{ccc}
0 & \sin \alpha & \cos \alpha \\
\sin \alpha & 0 & \sin \beta \\
\cos \alpha & -\sin \beta & 0
\end{array}\right]\)
= 0M₁₁ – sin αM₁₂ + cos αM₁₃
= 0 – sin α(0 – cos α sin β) + cos α(- sin α sin β – 0) = 0

Question 26.
Find the value of λ for which the vectors \(\vec { a } \) = 3\(\hat { i } \) + 2\(\hat { j } \) + 9\(\hat { k } \) and \(\vec { a } \) = \(\hat { i } \) + λ\(\hat { j } \) +3\(\hat { k } \) are parallel?
Answer:
Given \(\vec { a } \) and \(\vec { b } \) are parallel ⇒\(\vec { a } \) = t\(\vec { b } \) (where t is a scalar)
(i.e.,) 3\(\hat { i } \) + 2\(\hat { j } \) + 9\(\hat { k } \) = t\(\hat { i } \) + λ\(\hat { j } \) + 3\(\hat { k } \))
equating \(\hat { i } \) components we get 3 = t
equating \(\hat { j } \) components
(i.e); 2 = tλ
2 = 3λ ⇒λ = 2/3

Question 27.
Evaluate \(\underset { x\rightarrow \pi }{ lim } \) \(\frac{sin 3x}{sin 2x}\)
Answer:

Question 28.
Find the derivative of sinx² with respect to x²?
Answer:

Question 29.
Let the matrix M = \(\begin{bmatrix} x & y \\ z & 1 \end{bmatrix}\) if x, y and z are chosen at random from the set {1, 2, 3}, and repetition is allowed (i.e., x = y = z), what is the probability that the given matrix M is a singular matrix?
Answer:
If the given matnx M is singular, then = \(\begin{vmatrix} x & y \\ z & 1 \end{vmatrix}\) = 0
That is, x – yz = 0
Hence the possible ways of selecting (x, y, z) are
{(1, 1, 1), (2, 1, 2), (2, 2, 1), (3, 1, 3), (3, 3, 1)} = A (say)
The number of favourable cases n(A) = 5
The total number of cases are n(S) = 3³ = 27
The probability of the given matrix is a singular matrix is
P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{5}{27}\)

Question 30.
Evaluate \(\frac { x^{ 2 } }{ 1+x^{ 6 } } \)
Answer:

PART – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
If f, g, h are real valued functions defined on R, then prove that (f + g) o h = f o h + g o h. What can you say about fo(g + h)? Justify your answer?

Question 32.
Solve \(\frac{4}{x+1}\) ≤ 3 ≤ \(\frac{6}{x+1}\), x > 0?

Question 33.
Prove that cos^-1 \(\frac{4}{5}\) + tan^-1 \(\frac{3}{5}\) = tan^-1 \(\frac{27}{11}\)?

Question 34.
There are 15 candidates for an examination. 7 candidates are appearing for mathematics examination while the remaining 8 are appearing for different subjects. In how many ways
can they be seated in a row so that no two mathematics candidates are together?

Question 35.
Prove that if a, b, c are in H.P. if and only if \(\frac{a}{c}\) = \(\frac{a-b}{b-c}\)?

Question 36.
If (-4, 7) is one vertex of a rhombus and if the equation of one diagonal is 5x – y + 7 = 0, then find the equation of another diagonal?

Question 37.
Verify the existence of \(\underset { x\rightarrow 1 }{ lim } \) f(x), where f(x) = \(\left\{\begin{aligned}
\frac{|x-1|}{x-1}, & \text { for } x \neq 1 \\
0, & \text { for } x=1
\end{aligned}\right.\)

Question 38.
If y = sin^-1 x then find y?

Question 39.
Evaluate cot² x + tan² x?

Question 40.
Show that
\(\left|\begin{array}{ccc}
2 b c-a^{2} & c^{2} & b^{2} \\
c^{2} & 2 c a-b^{2} & a^{2} \\
b^{2} & a^{2} & 2 a b-c^{2}
\end{array}\right|=\left|\begin{array}{ccc}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|^{2}\)

PART – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41 (a).
The total cost of airfare on a given route is comprised of the base cost C and the fuel surcharge S in rupee. Both C and S are functions of the mileage m; C(m) = 0.4m + 50 and S(m) = 0.03m. Determine a function for the total cost of a ticket in terms of the mileage and find the airfare for flying ¡600 miles?

[OR]

(b) Evaluate \(\sqrt { x^{ 2 }+x+1 } \)?

Question 42 (a).
Determine the region in the plane determined by the inequalities y ≥ 2x and -2x + 3y ≤ 6?

[OR]

(b) If y(cos^-1 x)², prove that (1-x²) \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) – x \(\frac{dy}{dx}\) – 2 = 0. Hence find y₂ when x = 0?

Question 43(a).
Prove that ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r

[OR]

(b) If the binomial coefficients of three consecutive terms in the expansion of (a + x)ⁿ are in the ratio 1 : 7 : 42, then find n?

Question 44 (a).
(a) Prove that

1. sin A + sin( 120° + A) + sin (240° + A) = O
2. cos A+ cos (120° + A) + cos (120° – A) = O

[OR]

(b) A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

Question 45 (a).
Show that \(\left|\begin{array}{ccc}
a^{2}+x^{2} & a b & a c \\
a b & b^{2}+x^{2} & b c \\
a c & b c & c^{2}+x^{2}
\end{array}\right|\) is divisible by x⁴?

[OR]

(b) \(\left[\begin{array}{ccc}
0 & p & 3 \\
2 & q^{2} & -1 \\
r & 1 & 0
\end{array}\right]\) is skew-symmetric, find the values of p, q and r?

Question 46 (a).
In a shopping mall there is a hail of cuboid shape with dimension 800 × 800 × 720 units, which needs to be added the facility of an escalator in the path as shown by the dotted line in the figure. Find

1. The minimum total length of the escalator
2. The heights at which the escalator changes its direction
3. The slopes of the escalator at the turning points.

[OR]

(b) Evaluate \(\lim _{x \rightarrow a} \frac{\sqrt{x-b}-\sqrt{a-b}}{x^{2}-a^{2}}(a>b)\)

Question 47 (a).
Evaluate ∫\(\frac { 3x+5 }{ x^{ 2 }+4x+7 } \) dx

[OR]

(b) A factory has two Machines – I and II. Machine-I produces 60% of items and Machine-II produces 40% of the items of the total output. Further 2% of the items produced by Machine-I are defective whereas 4% produced by Machine-II are defective. If an itci is drawn at random what is the probability that it is defective?

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Let us look at these Government of Tamil Nadu State Board 11th Commerce Model Question Papers Tamil Medium with Answers 2020-21 Pdf. Students can view or download the Class 11th Commerce New Model Question Papers 2021 Tamil Nadu English Medium Pdf for their upcoming Tamil Nadu HSC Board Exams. Students can also read Tamilnadu Samcheer Kalvi 11th Commerce Guide.

TN State Board 11th Commerce Model Question Papers 2020 2021 English Tamil Medium

Tamil Nadu 11th Commerce Model Question Papers English Medium 2020-2021

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Tamil Nadu 11th Commerce Model Question Papers Tamil Medium 2020-2021

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11th Commerce Model Question Paper Design 2020-2021 Tamil Nadu

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It is necessary that students will understand the new pattern and style of Model Question Papers of 11th Standard Commerce Tamilnadu State Board Syllabus according to the latest exam pattern. These Tamil Nadu Plus One 11th Commerce Model Question Papers State Board Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN HSLC Board Exams and Score More marks.

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Students can Download Tamil Nadu 11th Maths Model Question Paper 2 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Maths Model Question Paper 2 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 90

PART – 1

I. Choose the correct answer. Answer all the questions: [20 × 1 = 20]

Question 1.
Let R be the set of all real numbers. Consider the following subsets of the plane R × R:
S= {(x, y) : y = x + 1 and 0 < x < 2} and T = {(x, y): x – y is an integer}
Then which of the following is true?
(a) T is an equivalence relation but S is not an equivalence relation.
(b) Neither S nor T is an equivalence relation
(c) Both S and T are equivalence relation
(d) S is an equivalence relation but T is not an equivalence relation.
Answer:
(a) T is an equivalence relation but S is not an equivalence relation.

Question 2.
If the set A has m elements the set B has n elements and the number of elements in A × B is …………………
(a) m + n
(b) mn
(c) \(\frac{m}{n}\)
(d) m²
Answer:
(b) mn

Question 3.
If \(\frac{ax}{(x+2)(2x-3)}\) = \(\frac{2}{x+2}\) + \(\frac{3}{2x-3}\) then a = ……………….
(a) 8
(b) 7
(c) 5
(d) 4
Answer:
(b) 7

Question 4.
The number of solutions of x² + |x – 1| = 1 is ………………….
(a) 1
(b) 0
(c) 2
(d) 3
Answer:
(c) 2

Question 5.
If a, 8, b are in A.P. a, 4, b are in G.P. and a, x, b are in H.P then x = ………………..
(a) 2
(b) 1
(c) 4
(d) 16
Answer:
(a) 2

Question 6.
If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then the total number of points of intersection are ……………….
(a) 45
(b) 40
(c) 10!
(d) 2¹⁰
Answer:
(a) 45

Question 7.
The value of e^2logx = …………………..
(a) 2x
(b) x²
(c) \(\sqrt{x}\)
(d) \(\frac{x}{2}\)
Answer:
(b) x²

Question 8.
The n^th term of the sequence 1, 2, 4, 7, 11 …. is …………………
(a) n³ + 3n² + 2n
(b) n³ – 3n² + 3n
(c) n\(\frac{(n+1)(n+2)}{3}\)
(d) \(\frac { n^{ 2 }-n+2 }{ 2 } \)
Answer:
(d) \(\frac { n^{ 2 }-n+2 }{ 2 } \)

Question 9.
The last term in the expansion (2+\(\sqrt{3}\))⁸ is ………………
(a) 81
(b) 27
(c) 9
(d) 3
Answer:
(a) 81

Question 10.
A line perpendicular to the line 5x -y = 0 forms a triangle with the coordinate axes. If the area of the triangle is 5sq.units, then its equation is …………………..
(a) x + 5y ± 5\(\sqrt{2}\) = 0
(b) x – 5y ± 5\(\sqrt{2}\) = 0
(c) 5x + y ± 5\(\sqrt{2}\) = 0
(d) 5x – y ± 5\(\sqrt{2}\) = 0
Answer:
(a) x + 5y ± 5\(\sqrt{2}\) = 0

Question 11.
A factor of the determinant \(\left|\begin{array}{ccc}
x & -6 & -1 \\
2 & -3 x & x-3 \\
-3 & 2 x & x+2
\end{array}\right|\) is ……………….
(a) x + 3
(b) 2x – 1
(c) x – 2
(d) x – 3
Answer:
(a) x + 3

Question 12.
If λ\(\vec { a } \) + 2λ\(\vec { j } \) + 2λ\(\vec { k } \) is a unit vector then the value of λ is ………………
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{9}\)
(d) \(\frac{1}{2}\)
Answer:
(a) \(\frac{1}{3}\)

Question 13.
One of the diagonals of parallelogram ABCD with \(\vec { a } \) and \(\vec { b } \) are adjacent sides is \(\vec { a } \) + \(\vec { b } \). The other diagonal BD is ………………….
(a) \(\vec { a } \) – \(\vec { b } \)
(b) \(\vec { a } \) – \(\vec { b } \)
(c) \(\vec { a } \) + \(\vec { b } \)
(d) \(\frac{\vec{a}+\vec{b}}{2}\)
Answer:
(b) \(\vec { a } \) – \(\vec { b } \)

Question 14.
If (1, 2, 4) and (2, -3λ, -3) are the initial and terminal points of the vector \(\vec { i } \) + 5\(\vec { j } \) – 7\(\vec { k } \) then the value of λ …………………..
(a) \(\frac{7}{3}\)
(b) –\(\frac{7}{3}\)
(c) \(\frac{5}{3}\)
(d) \(\frac{-5}{3}\)
Answer:
(b) –\(\frac{7}{3}\)

Question 15.
If y = mx + c and f(0) =f'(0) = 1 then f(2) = …………………..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 16.
The derivative of (x + \(\frac{1}{x}\))² w.r.to. x is ………………..
(a) 2x – \(\frac { 2 }{ x^{ 3 } } \)
(b) 2x + \(\frac { 2 }{ x^{ 3 } } \)
(c) 2(x + \(\frac{1}{x}\))
(d) 0
Answer:
(a) 2x – \(\frac { 2 }{ x^{ 3 } } \)

Question 17.
If f(x) is \(\left\{\begin{array}{cc}
a x^{2}-b, & -1<x<1 \\
\frac{1}{|x|}, & \text { elsewhere }
\end{array}\right.\) is differentiable at x = 1, then …………………
(a) a = \(\frac{1}{2}\), b = \(\frac{-3}{2}\)
(b) a = \(\frac{-1}{2}\), b = \(\frac{3}{2}\)
(c) a = \(\frac{-1}{2}\), b = \(\frac{-3}{2}\)
(d) a = \(\frac{1}{2}\), b = \(\frac{3}{2}\)
Answer:
(c) a = \(\frac{-1}{2}\), b = \(\frac{-3}{2}\)

Question 18.
∫\(\frac { \sqrt { tanx } }{ sin2x } \) dx is ………………
(a) \(\sqrt{tanx}\) + c
(b) 2\(\sqrt{tanx}\) + c
(c) \(\frac{1}{2}\) \(\sqrt{tanx}\) + c
(d) \(\frac{1}{4}\) \(\sqrt{tanx}\) + c
Answer:
(a) \(\sqrt{tanx}\) + c

Question 19.
An urn contains 5 red and 5 black balls. A balls is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. The probability that the second ball drawn is red will be ………………….
(a) \(\frac{5}{12}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{7}{12}\)
(d) \(\frac{1}{4}\)
Answer:
(b) \(\frac{1}{2}\)

Question 20.
It is given that the events A and B are such that P(A) = \(\frac{1}{4}\), P(A/B) = \(\frac{1}{2}\), and P(B/A) = \(\frac{2}{3}\) then
P(B) = ………………….
(a) \(\frac{1}{6}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{2}\)
Answer:
(b) \(\frac{1}{3}\)

PART – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
If n(P(A)) = 1024, n(A∪B) = 15 and n(P(B)) = 32 then find n(A∩B)
Answer:
n(P(A)) = 1024 = 2¹⁰ ⇒ n(A) = 10
n(A∪B) = 15
n(P(B)) = 32 = 2⁵ ⇒ n(B) = 5
We know n(A∪B) = n(A) + n(B) – n(A∩B)
(i.e) 15 = 10 + 5 – n(A∩B)
⇒ n(A∩B) = 15 – 15 = 0

Question 22.
Simplify (343)^2/3
Answer:
(343)^2/3 = (7³)^2/3 = 7^3×2/3 = 7² = 49

Question 23.
Show that cos36° cos 72° cos 108° cos 144° = \(\frac{1}{16}\)
Answer:
LHS = cos36° cos(90° – 18°) cos(90° – 18°) cos(90° + 18°) cos(180° – 36°)
= sin² 18° cos² 36°
= (\(\frac { \sqrt { 5-1 } }{ 4 } \))² (\(\frac { \sqrt { 5+1 } }{ 4 } \))² = \(\frac{1}{16}\) = RHS

Question 24.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour?
Answer:
There are 6 red balls, 5 white balls and 5 blue balls.
We have to select 3 balls of each colour.
∴Number of ways of selection = ⁶C₃ × ⁵C₃ × ⁵C₃
= \(\frac { 6! }{ 3!3! } \) × \(\frac { 5! }{ 3!2! } \) × \(\frac { 5! }{ 3!2! } \)
= 20 × 10 × 10 = 2000

Question 25.
Find |A| if A = \(\left[\begin{array}{ccc}
0 & \sin \alpha & \cos \alpha \\
\sin \alpha & 0 & \sin \beta \\
\cos \alpha & -\sin \beta & 0
\end{array}\right]\)
Answer:
\(\left[\begin{array}{ccc}
0 & \sin \alpha & \cos \alpha \\
\sin \alpha & 0 & \sin \beta \\
\cos \alpha & -\sin \beta & 0
\end{array}\right]\)
= 0M₁₁ – sin αM₁₂ + cos αM₁₃
= 0 – sin α(0 – cos α sin β) + cos α (-sin α sin β – 0) = 0

Question 26.
For any vector prove that \(\vec { r } \) = [\(\vec { r } \).\(\vec { i } \)) + (\(\vec { r } \).\(\vec { j } \))j + [\(\vec { r } \).\(\vec { k } \)}k
Answer:
Let \(\vec { r } \) = x\(\hat { i } \) + y\(\hat { j } \) + z\(\hat { k } \)

Question 27.
Calculate \(\lim _{x \rightarrow-2}\) (x³ – 3x + 6) (-x² + 15)
Answer:

Question 28.
Evaluate y = e^x sin x
Answer:

Question 29.
Integrate the following with respect to x
\(\frac{4}{3+4x}\) + (10x + 3)⁹ – 3 cosec(2x + 3) cot (2x + 3)
Answer:

Question 30.
P(A) = 0.6, P (B) = 0.5 and P(A∩B) = 0.2 find P(A/B)
Answer:
Given that P(A) = 0.6, P(B) = 0.5, and P(A∩B) = 0.2
P(A/B) = \(\frac { p(A∩B) }{ p(B) } \) = \(\frac{0.2}{0.5}\) = \(\frac{2}{5}\)

PART – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
A quadratic polynomial has one of its zeros 1 + \(\sqrt{5}\) and it satisfies p(1) = 2. Find the quadratic polynomial?

Question 32.
Prove that

1. tan^-1 (\(\frac{1}{7}\)) + tan^-1(\(\frac{1}{13}\)) = tan^-1(\(\frac{2}{9}\))
2. cos^-1\(\frac{4}{5}\) + tan^-1\(\frac{3}{5}\) = tan^-1\(\frac{27}{11}\)

Question 33.
The product of three increasing numbers in GP is 5832. If we add 6 to the second number and 9 to the third number, then resulting numbers form an AP. Find the numbers in GP?

Question 34.
Find the equation of the line passing through the point (5, 2) and perpendiular to the line joining the points (2, 3) and (3, -1)?

Question 35.
Find the area of the triangle whose vertices are (0,0), (1,2) and (4,3)?

Question 36.
If \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) are three vectors such that \(\vec { a } \) + 2\(\vec { b } \) + \(\vec { c } \) = 0 and |\(\vec { a } \)| = 3, |\(\vec { b } \)| = 4, |\(\vec { c } \)| = 7 fimd the angle between \(\vec { a } \) and \(\vec { b } \)

Question 37.
Evaluate: \({ \underset { x\rightarrow 0 }{ lim } }\) \(\frac { 3^{ x }-1 }{ \sqrt { 1+x-1 } } \)

Question 38.
Find \(\frac{dy}{dx}\) for y = tan^-1 \((\frac { cosx+sinx }{ cosx-sinx } )\)

Question 39.
Evaluate: ∫x⁵ e^x2

Question 40.
How many automobile license plates can be made, if each plate contains two different letters followed by three different digits?

PART – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41.
(a) If f : R – {-1, 1} → R is defined by f(x) = \(\frac { x }{ x^{ 2 }-1 } \), verify whether f is one-to-one or not?

[OR]

(b) Solve: log₂ x + log₄ x + log₈ x = 11

Question 42.
(a) Prove that \(\frac{\sin x+\sin 3 x+\sin 5 x+\sin 7 x}{\cos x+\cos 3 x+\cos 5 x+\cos 7 x}\) = tan 4x

[OR]

(b) If x + y + z = xyz, then prove that \(\frac { 2x }{ 1-x^{ 2 } } \) + \(\frac { 2y }{ 1-y^{ 2 } } \) + \(\frac { 2z }{ 1-z^{ 2 } } \) = \(\frac { 2x }{ 1-x^{ 2 } } \) \(\frac { 2y }{ 1-y^{ 2 } } \) \(\frac { 2z }{ 1-z^{ 2 } } \)

Question 43.
(a) If the letters of the word GARDEN are permuted in all possible ways and the strings thus formed are arranged in the dictionary order, then and the ranks of the words

1. GARDEN
2. DANGER

[OR]

(b) \(\underset { x\rightarrow a }{ lim } \) \(\frac{\sqrt{x-b}-\sqrt{a-b}}{x^{2}-a^{2}}\) (a>b)

Question 44.
(a) If the binomial coefficients of three consecutive terms in the expansion of (a + x)ⁿ are in the ratio 1 : 7 : 42 then find n?

[OR]

(b) Evalute \(\sqrt { x^{ 2 }+y^{ 2 } } \) = tan^-1(\(\frac{y}{x}\))

Question 45.
Let \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) be three vectors such that |\(\vec { a } \)| = 3, |\(\vec { b } \)| = 4, |\(\vec { c } \)| = 5 and each one of them being perpendicular to the sum of the other two, find |\(\vec { a } \) + \(\vec { b } \) + \(\vec { c } \)|.

[OR]

(b) Evaluate ∫sec³ 2xdx

Question 46.
(a) Find all the equations of the straight lines in the family of the lines y = mx – 3, for which m and the x-coordinate of the point of intersection of the lines with x – y = 6 are integers?

[OR]

(b) There are two identical boxes containing respectively 5 white and 3 red balls, 4 white and 6 red balls. A box is chosen at random and a ball is drawn from it

1. Find the probability that the ball is white
2. If the ball is white, what is the probability that it from the first box?

Question 47 (a).
If A_i B_i, C_i are the cofactors of a_i, b_i, c_i, respectively, i = 1 to 3 in

[OR]

(b) Express the matrix \(\left(\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right)\) as the sum of symmetric martix and a skew-symmetric martix?

Students can Download Tamil Nadu 11th Maths Model Question Paper 1 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Maths Model Question Paper 1 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 90

PART – 1

I. Choose the correct answer. Answer all the questions: [20 × 1 = 20]

Question 1.
For non empty sets A and B if A ⊂ B then (A × B) n (B × A) is equal to ………………….
(a) A ∩ B
(b) A × A
(c) B × B
(d) none of these
Answer:
(b) A × A

Question 2.
The solution set of the inequality |x – 1| ≥ |x – 3| is ………………..
(a) [0, 2]
(b) [2, ∞)
(c) (0, 2)
(d) (-∞, 2)
Answer:
(b) [2, ∞)

Question 3.
The numer of solutions of x² + |x – 1| = 1 is …………………
(a) 1
(b) 0
(c) 2
(d) 3
Answer:
(c) 2

Question 4.
Which of the following is not true?
(a) sin θ = – \(\frac{3}{4}\)
(b) cos θ = -1
(c) tan θ = 25
(d) sec θ = \(\frac{1}{4}\)
Answer:
(d) sec θ = \(\frac{1}{4}\)

Question 5.
Let f_k(x) = \(\frac{1}{k}\) [sin^k x + cos² x] where x ∈ R and k ≥ 1. Then f₄(x) – f₆(x) = …………………
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{12}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{1}{3}\)
Answer:
(b) \(\frac{1}{12}\)

Question 6.
If A and B are coefficients of xⁿ in the expansion of (1 + x)²ⁿ and (1 + x)^2n-1 respectively then \(\frac{A}{B}\) =
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{n}\)
(c) 1
(d) 2
Answer:
(d) 2

Question 7.
The value of 15C₈ + 15C₉ – 15C₆ – 15C₇ is …………………
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(a) 0

Question 8.
The slope of the line which makes an angle 45° with the line 3x -y = – 5 are …………………
(a) 1, -1
(b) \(\frac{1}{2}\), -2
(c) 1, \(\frac{1}{2}\)
(d) 2, \(\frac{-1}{2}\)
Answer:
(b) \(\frac{1}{2}\), -2

Question 9.
The sum of the binomial co-efficients is ………………….
(a) 2n
(b) 2ⁿ
(c) n²
(d) 1
Answer:
(b) 2ⁿ

Question 10.
If the square of the matrix \(\begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix}\) satisfy the relation
(a) 1 + α² + βγ = 0
(b) 1 – α² – βγ = 0
(c) 1 – α² + βγ = 0
(d) 1 + α² – βγ = 0
Answer:
(b) 1 – α² – βγ = 0

Question 11.
The value of x for which the matrix A = \(\begin{bmatrix} e^{ x-2 } & e^{ 7+x } \\ e^{ 2+x } & e^{ 2x+3 } \end{bmatrix}\) is singular is ………………….
(a) 9
(b) 8
(c) 7
(d) 6
Answer:
(b) 8

Question 12.
If A = \(\begin{pmatrix} \lambda & 1 \\ -1 & -\lambda \end{pmatrix}\) then for what value of λ, A² = 0?
(a) 0
(b) ±1
(c) -1
(d) 1
Answer:
(b) ±1

Question 13.
lim_x→1 \(\frac { xe^{ x }-sinx }{ x } \) is ……………….
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(d) 0

Question 14.
If the points whose position vectors 10\({ \vec { i } }\) + 3\({ \vec { j } }\) + 12\({ \vec { i } }\) – 5\({ \vec { j } }\) and a\({ \vec { i } }\) + 11\({ \vec { j } }\) are collinear then the value of a is ……………….
(a) 3
(b) 5
(c) 6
(d) 8
Answer:
(d) 8

Question 15.
If y = mx + c and f(0) = f'(0) = 1 then f(2) = ………………..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 16.
lim_x→1 \(\frac { e^{ x }-e }{ x-1 } \) = ……………….
(a) 1
(b) e
(c) ∞
(d) 0
Answer:
(b) e

Question 17.
∫2^3x+5 dx is ………………..
(a) \(\frac { 3(2^{ 3x+5 }) }{ log2 } \) + c
(b) \(\frac { 2^{ 3x+5 } }{ 2log(3x+5) } \) + c
(c) \(\frac { 2^{ 3x+5 } }{ 2log3 } \) + c
(d) \(\frac { 2^{ 3x+5 } }{ 3log2 } \) + c
Answer:
(d) \(\frac { 2^{ 3x+5 } }{ 3log2 } \) + c

Question 18.
∫x² cos xdx is ………………..
(a) x²sinx + 2xcosx – 2 sinx + c
(b) x² – 2xcosx – 2sinx + c
(c) -x² + 2xcosx + 2sinx + c
(d) -x² – 2xcosx + 2 sinx + c
Answer:
(a) x²sinx + 2xcosx – 2 sinx + c

Question 19.
∫\(\frac { x }{ 1+x^{ 2 } } \) dx = ………………
(a) tan^-1 x + c
(b) log (1 + x²) + c
(c) log x + c
(d) \(\frac{1}{2}\) log (1 + x²) + c
Answer:
(d) \(\frac{1}{2}\) log (1 + x²) + c

Question 20.
Ten coins are tossed. The probability of getting atleast 8 heads is …………….
(a) \(\frac{7}{64}\)
(b) \(\frac{7}{32}\)
(c) \(\frac{7}{128}\)
(d) \(\frac{7}{16}\)
Answer:
(c) \(\frac{7}{128}\)

PART – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
To secure an A grade one must obtain an average of 90 marks or more in 5 subjects each of maximum 100 marks. If A scored 84, 87, 95, 91 in first four subjects, what is the minimum mark be scored in the fifth subject to get an A grade in the course?
Answer:
Required marks = 5 × 90 = 450
Total marks obtained in 4 subjects = 84 + 87 + 95 + 91 = 357
So required marks in the fifth subject = 450 – 357 = 93.

Question 22.
Prove that (1 + tan1°) (1 + tan2°) (1 + tan 3°) …. (1 + tan 44°) is a multiple of 4?
Answer:
45°= 1 + 44 (or) 2 + 43 (or) 3 + 42 (or) 22 + 23
So we have 22 possible pairs
sa the product is (2) (22) = 44
which is ÷ by 4

Question 23.
Find the distinct permutations of the letters of the word MISSISSIPPI?
Answer:
MISSISSIPPI
Number of letters = 11
Here M-1 time, I-4timcs, S-4times, P-2timcs
So total number of arrangement is of this word = \(\frac { 11! }{ 4!4!2! } \)
\(\frac { 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4! }{ 4\times 3\times 2\times 1\times 2\times 1\times 4! } \) = 34650

Question 24.
Compare |A| using Sarrus rule if A = \(\left[\begin{array}{ccc}
3 & 4 & 1 \\
0 & -1 & 2 \\
5 & -2 & 6
\end{array}\right]\)
Answer:

|A| = [3(-1) (6) + 4(2)(5) + 1(0)(-2)] -[5(-1)(1) + (-2)(2)3 + 6(0)(4)] = [-18 + 40 + 0] – [-5 – 12 + 0] = 22 + 17 = 39.

Question 25.
If |\({ \vec { a } }\)|, |\({ \vec { b } }\)| = 6, |\({ \vec { c } }\)| =7 and \({ \vec { a } }\) + \({ \vec { b } }\) + \({ \vec { c } }\) = 0 find \({ \vec { a } }\).\({ \vec { b } }\) + \({ \vec { b } }.\) \({ \vec { c } }\) + \({ \vec { c } }\).\({ \vec { a } }\)
Answer:
Given \(\bar { a } \)+\(\bar { b } \)+\(\bar { c } \) = 0
⇒ (\(\bar { a } \)+\(\bar { b } \)+\(\bar { c } \))² = 0
(i.e;) |\(\bar { a } \)|² + |\(\bar { b } \)|² + |\(\bar { c } \)|² + 2 [\(\bar { a } \).\(\bar { b } \)+\(\bar { b } \).\(\bar { c } \)+\(\bar { c } \).\(\bar { a } \)) = 0
⇒ 5²+6²+7²+2(\(\bar { a } \).\(\bar { b } \)+\(\bar { b } \).\(\bar { c } \)+\(\bar { c } \).\(\bar { a } \)) = 0
⇒2(\(\bar { a } \).\(\bar { b } \)+\(\bar { b } \).\(\bar { c } \)+\(\bar { c } \).\(\bar { a } \)) = – 25 – 36 – 49 = -110
⇒\(\bar { a } \).\(\bar { b } \)+\(\bar { b } \).\(\bar { c } \)+\(\bar { c } \).\(\bar { a } \) = \(\frac{-110}{2}\) = -55

Question 26.
Find \(\lim _{ t\rightarrow 0 }{ \frac { \sqrt { t^{ 2 }+9-3 } }{ t^{ 2 } } } \)
Answer:
We can’t apply the quotient Iieorem it.:iiediatcly. Use the algebra technique of rationalising the numerator.

Question 27.
Evaluate y = e^-x.log x?
Answer:
y = e^-x log x = uv(say)
Here u = e^-x and v = log x
⇒ u’ = -e^-x and v’ = uv’ + vu’
Now y = uv ⇒y’ = uv’ + vu’
(i.e). \(\frac{dy}{dx}\) = e^-x(\(\frac{1}{x}\)) + log x(- e^-x)
= e^-x(\(\frac{1}{x}\) – log x)

Question 28.
Evaluate ∫\(\frac { 1 }{ \sqrt { 1+(4x)^{ 2 } } } \) dx
Answer:
Let I = ∫\(\frac { 1 }{ \sqrt { 1+4x^{ 2 } } } \)dx = ∫\(\frac { 1 }{ \sqrt { 1+(2x^{ 2 }) } } \) dx
Putting 2x = t ⇒ 2 dx = dt ⇒ dx = \(\frac{1}{2}\) dt
Thus, I = \(\frac{1}{2}\)∫\({ \frac { 1 }{ \sqrt { 1^{ 2 }+t^{ 2 } } } }\) dt
I = \(\frac{1}{2}\) log |t+\(\sqrt { t^{ 2 }+1 } \) + c = \(\frac{1}{2}\) log |2x + \(\sqrt { (2x)^{ 2 }+1 } \)| + c
I = \(\frac{1}{2}\) log |2x+\(\sqrt { 4^{ 2 }+1 } \)| + c.

Question 29.
Nine coins are tossed once. Find the probablity to get atleast 2 heads?
Answer:
Let S be the sample and A be the event of getting at least two heads.
Therefore, the event \(\bar { A } \) denotes, getting at most one head.
n(S) = 2⁹ = 521, n(\(\bar { A } \)) = 9C₀ + 9C₁ = 1 + 9 = 10
P(\(\bar { A } \)) = \(\frac{10}{512}\) = \(\frac{5}{256}\)
P(A) = 1- P(\(\bar { A } \)) = 1 – \(\frac{5}{256}\) = \(\frac{251}{256}\)

Question 30.
If P(A) denotes the power set of A, then find n(P(P(ϕ))))?
Answer:
Since P(∅) contains 1 element, P(P(∅)) contains 2¹ elements and hence P(P(P(∅))) contains 2² elements. That is, 4 elements.

PART – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
A simple cipher takes a number and codes it, using the function f(x) = 3x – 4. Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line y = x (by drawing the lines)?

Question 32.
Show that 4 sin A sin (60° + A). sin(60° – A) = sin 3A?

Question 33.
How many strings of length 6 can be formed using letters of the word FLOWER if (i) either starts with F or ends with R?

Question 34.
Find the distance of the line 4x – y = 0 from the point P(4, 1) measured along the line making an angle 135° with the positive x- axis?

Question 35.
Using factor theorem prove that \(\left|\begin{array}{lll}
x+1 & 3 & 5 \\
2 & x+2 & 5 \\
2 & 3 & x+4
\end{array}\right|\) = (x – 1)² + ( x+ 9)

Question 36.
Find the unit vectors perpendicular to each of the vectors \({ \vec { a } }\) + \({ \vec { b } }\) and \({ \vec { a } }\) – \({ \vec { b } }\). Where \({ \vec { a } }\) = \({ \vec { i } }\) + \({ \vec { j } }\) + \({ \vec { k } }\) and b = \({ \vec { i } }\) + 2\({ \vec { j } }\) + 3\({ \vec { k } }\)?

Question 37.
If y = tan^-1 \(\left(\frac{1+x}{1-x}\right)\), find y’?

Question 38.
Evaluate ∫\(\frac { x }{ \sqrt { 1+x^{ 2 } } } \) dx?

Question 39.
If A and B are two events such that P(A∪B) = 0.7, P(A∩B) = 0.2 and P(B) = 0.5 then show that A and B are independent?

Question 40.
Resolve into partial fractions \(\frac { 2x^{ 2 }+5x-11 }{ x^{ 2 }+2x-3 } \)

PART – IV

IV. Answer all the questIons [7 × 5 = 35]

Question 41.
(a) A photocopy store charges Rs. 1.50 per copy for the first 10 copies and Rs. 1.00 per copy after the 10th copy. Let x be the number of copies, and let y be the total cost of photocopying.
(a) Draw graph of the cost as x goes from 0 to 50 copies?
(b) Find the cost of making 40 copies

[OR]

(b) If the difference of the roots of the equation 2x² – (a + 1)x + a – 1 = 0 is equal to their product then prove that a =2?

Question 42.
(a) Prove that tan 70° – tan 20° – 2 tan 40° = 4 tan 10°?

[OR]

(b) In a ∆ABC, if a = 2\(\sqrt{3}\), b = 2\(\sqrt{2}\) and C = 75° find the other side and the angles?

Question 43.
(a) Use induction to prove that n³ – 7n + 3, is divisible by 3, for all natural numbers n?

[OR]

(b) Evaluate y = (x² + 1) \(\sqrt [ 3 ]{ x^{ 2 }+2 } \)?

Question 44.
(a) Show that f(x) f(y) = f(x + y), where f(x) = \(\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\)

[OR]

(b) The chances of X, Y and Z becoming managers of a certain company are 4 : 2 : 3. The probabilities that bonus scheme will be introduced if X, Y and Z become managers are 0.3, 0.5 and 0.4 respectively. 1f the bonus scheme has been introduced, what is the probability that Z was appointed as the manager?

Question 45.
(a) If the equation λx² – 10xy + 12y² + 5x – 16y – 3 = O represents a pair of straight lines, find

1. The value of λ and the separate equations of the lines
2. Angle between the lines

[OR]

(b) Show that

1. \(\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{3 n^{2}+7 n+2}=\frac{1}{6}\)
2. \(\lim _{n \rightarrow \infty} \frac{1^{2}+2^{2}+\ldots+(3 n)^{2}}{(1+2+\ldots+5 n)(2 n+3)}=\frac{9}{25}\)
3. \(\lim _{n \rightarrow \infty} \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots .+\frac{1}{n(n+1)}=1\)

Question 46.
(a) A man repays an amount of ₹3250 by paying ₹20 in the first month and then increases the payment by ₹15 per month. How long will it take him to clear the amount?

[OR]

(b) Find the area of the triangle whose vertices are A (3, -1, 2), B (1, -1, -3) and C (4, -3, 1)?

Question 47.
(a) Evaluate (x + 1) \(\sqrt { 2x+3 } \)?
[OR]

(b) Evaluate ∫cosec²x dx?

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 1 Basic Concepts of Chemistry and Chemical Calculations Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

11th Chemistry Guide Basic Concepts of Chemistry and Chemical Calculations Text Book Back Questions and Answers

Textual Questions:

I. Choose the best answer:

Question 1.
40 ml of methane is completely burnt using 80 ml of oxygen at room temperature. The volume of gas left after cooling to room temperature
(a) 40 ml CO₂
(b) 40 ml CO₂ gas and 80 ml H₂o gas
(c) 60 ml CO₂ gas and 60 ml H₂o gas
(d) 120 ml CO₂ gas
Answer:
(a) 40 ml CO₂
Solution:
CH₄(g) + 2O₂ → CO₂(g) + 2H₂O (1)

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct.

Question 2.
An element X has the following isotopic composition ²⁰⁰X = 90%, ¹⁹⁹X = 8% and ²⁰²X = 2%. The weighted average atomic mass of the element X is closest to
(a) 201 u
(b) 202 u
(c) 199 u
(d) 200 u
Answer:
(d) 200 u
Solution:
X = \(\frac{(200 \times 90)+(199 \times 8)+(202 \times 2)}{100}\) = 199.96 = 200 u

Question 3.
Assertion:
Two mole of glucose contains 12.044 × 10²³ molecules of glucose
Reason:
Total number of entities present in one mole of any substance is equal to 6.02 × 10²²
(a) both assertion and reason are true and the reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(c) assertion is true but reason is false
Solution:
Correct reason:
Total number of entities present in one mole of any substance is equal to 6.022 × 10²³.

Question 4.
Carbon forms two oxides, namely carbon monoxide and carbon dioxide. The equivalent mass of which element remains constant?
(a) Carbon
(b) Oxygen
(c) both carbon and oxygen
(d) neither carbon nor oxygen
Answer:
(b) Oxygen
Solution:
Reaction 1:
2C + O₂ → 2CO
2 × 12g carbon combines with 32g of oxygen. Hence, Equivalent mass of carbon
\(\frac{2 \times 12}{32}\) × 8 = 6
Reaction 2:
C + O₂ → CO₂
12 g carbon combines with 32 g of oxygen. Hence, Equivalent mass of carbon
= \(\frac{12}{32}\) × 8 = 6

Question 5.
The equivalent mass of a trivalent metal element is 9 g eq^-1 the molar mass of its anhydrous oxide is
(a) 102 g
(b) 27 g
(c) 270 g
(d) 78 g
Answer:
(a) 102 g
Solution:
Let the trivalent metal be M³
Equivalent mass = mass of the metal / 3 eq
9 g eq^-1 = mass of the metal / 3 eq
Mass of the metal = 27 g
Oxide formed M₂O₃ ;
Mass of the oxide = (2 × 27) + (3 × 16) = 102 g

Question 6.
The number of water molecules in a drop of water weighing 0.018 g is
(a) 6.022 × 10²⁶
(b) 6.022 × 10²³
(c) 6.022 × 10²⁰
(d) 9.9 × 10²²
Answer:
(c) 6.022 × 10²⁰
Solution:
Weight of the water drop 0.018 g
No. of moles of water in the drop = Mass of water / molar mass = 0.018/18 = 10^-3 mole
No of water molecules present in 1 mole of water = 6.022 × 10²³
No. water molecules in one drop of water(10^-3 mole)
= 6.022 × 10²³ × 10^-3
= 6.022 × 10²⁰

Question 7.
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is
(a) 0 %
(b) 4.4 %
(c) 16 %
(d) 8.4 %
Answer:
(c) 16 %
Solution:
MgCO₃ → MgO + CO₂↑
MgCO₃:
(1 × 24) + (1 × 12) + (3 × 16) = 84 g
CO₂: (1 × 12) + (2 × 16) = 44g
100 % pure 84 g MgCO₃ on heating gives 44 g CO₂
Given that 1 g of MgCO₃ on heating gives 44 g of CO₂
Therefore, 84 g MgCO₃ sample on heating gives 36.96 g CO₂
Percentage of purity of the sample = \(\frac{100 \%}{44 \mathrm{~g} \mathrm{CO}_{2}}\) × 36.96 g of CO₂ = 84 %
Percentage of impurity = 16 %

Question Question 8.
When 6.3 g of sodium bicarbonate is added to 30 g of acetic acid solution, the residual solution is found to weigh 33 g. The number of moles of carbon dioxide released in the reaction is
(a) 3
(b) 0.75
(c) 0.075
(d) 0.3
Answer:
(c) 0.075
solution:
NaHCO₃ + CH₃COOH → CH₃OONa + H₂O + CO₂
6.3 g + 30 g → 33 g + x
The amount of CO₂ released, x = 3.3 g
No. of moles of CO₂ released = 3.3/44 = 0.075 mol.

Question 9.
When 22.4 litres of H₂ (g) is mixed with 11.2 litres of Cl₂ (g), each at 273 K at 1 atm the moles of HCl (g), formed is equal to
(a) 2 moles of HCl (g)
(b) 0.5 moles of HCl (g)
(c) 1.5 moles of HCl (g)
(d) 1 moles of HCl (g)
Answer:
(d) 1 moles of HCl (g)
Solution:
H_2(g) + Cl_2(g) → 2HCl_(g)

Amount of HCl formed = 1 mol

Question 10.
Hot concentrated sulphuric acid is a moderately strong oxidising agent. Which of the following reactions does not show oxidising behaviour?
(a) Cu + 2H₂ → CuSO₄ + SO₂ +2 H₂O
(b) C + 2 H₂SO₄ → CO₂ + 2 SO₂ +2 H₂O
(c) BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
(d) none of the above
Answer:
(c) BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
Solution:

Question 11.
Choose the disproportionation reaction among the following redox reactions.
(a) 3Mg_(s) + N_2(g) → Mg₃N_2(s)
(b) P_4(s) + 3NaOH + 3H₂O → PH_3(g) + 3NaH₂PO_2(aq)
(c) Cl_2(g) + 2KI_(aq) → 2KCl_(aq) + I_2(s)
(d) Cr₂O_3(s) + 2Al_(s) → Al₂O_3(s) + 2Cr_(s)
Answer:
b) P_4(s) + 3NaOH + 3H₂O → PH_3(g) + 3NaH₂PO_2(aq)
Solution:

Question 12.
The equivalent mass of potassium permanganate in an alkaline medium is
MnO₄ + 2H₂O + 3e → MnO₂ + 4OH^–
(a) 31.6
(b) 52.7
(c) 79
(d) None of these
Answer:
(b) 52.7
Solution:
The reduction reaction of the oxidising agent (Mn0₄) involves the gain of 3 electrons.
Hence the equivalent mass = (Molar mass of KMnO₄)/3 = 158.1/3 = 52.7

Question 13.
Which one of the following represents 180g of water?
(a) 5 Moles of water
(b) 90 moles of water
(c) \(\frac{6.022 \times 10^{23}}{180}\)

(d) \(\frac{6.022 \times 10^{23}}{1.7}\)
Answer:
(d) \(\frac{6.022 \times 10^{23}}{1.7}\)
Solution:
No. of moles of water present in 180 g = Mass of water / Molar mass of water = 180 g/18 gmol^-1 = 10 moles
One mole of water contains = 6.022 × 10²³ water molecules
10 mole of water contains = 6.022 × 10²³ × 10 × 6.022 × 10²⁴ water molecules

Question 14.
7.5 g of a gas occupies a volume of 5.6 litres at 0° C and 1 atm pressure. The gas is
(a) NO
(b ) N₂O
(c) CO
(d) CO₂
Answer:
(a) NO
Solution:
7.5 g of gas occupies a volume of 5.6 liters at 273 K and 1 atm pressure Therefore, the mass of gas that occupies a volume of 22.4 liters
= \(\frac{7.5 g}{5.61}\) × 22.41 = 30 g
Molar mass of NO (14 + 16) = 30 g

Question 15.
Total number of electrons present in 1.7 g of ammonia is
(a) 6.022 × 10²³
(b) \(\frac{6.022 \times 10^{22}}{1.7}\)
(c) \(\frac{6.022 \times 10^{24}}{1.7}\)
(d) \(\frac{6.022 \times 10^{23}}{1.7}\)
Answer:
(a) 6.022 × 10²³
Solution:
No. of electrons present in one ammonia (NH₃) molecule (7 + 3) = 10
No. of moles of ammonia = \(\frac{\text { Mass }}{\text { Molar mass }}\)
= \(\frac{1.7 \mathrm{~g}}{17 \mathrm{~mol}^{-1}}\) = 0.1 mol
= 0.1 × 6.022 × 10²³ = 6.022 × 10²²
= No. of electrons present in 0.1 mol of ammonia

Question 16.
The correct increasing order of the oxidation state of sulphur in the anions
SO₄^2-, SO₃^2-, S₂O₄^2-, S₂O₆^2- is

(a) SO₃^2- < SO₄^2- < S₂O₄^2- < S₂O₆^2-

(b) SO₄^2- < S₂O₄^2- < S₂O₆^2- < SO₃^2-

(c) S₂O₄^2- < SO₃^2- < S₂O₆^2- < SO₄^2-

(d) S₂O₆^2- < S₂O₄^2- < SO₄^2- < SO₃^2-
Answer:
(c) S₂O₄^2- < SO₃^2- < S₂O₆^2- < SO₄^2-
Solution:

Question 17.
The equivalent mass of ferrous oxalate is
(a) \(\frac{\text { molar mass of ferrous oxalate }}{1}\)

(b) \(\frac{\text { molar mass of ferrous oxalate }}{2}\)

(c) \(\frac{\text { molar mass of ferrous oxalate }}{3}\)
(d) none of these
Answer:
(c) \(\frac{\text { molar mass of ferrous oxalate }}{3}\)
Solution:

Question 18.
If Avogadro number were changed from 6.022 × 10²³ to 6.022 × 10²⁰, this would change
(a) the ratio of chemical species to each other in a balanced equation
(b) the ratio of elements to each other in a compound
(c) the definition of mass in units of grams
(d) the mass of one mole of carbon
Answer:
(d) the mass of one mole of carbon

Question 19.
Two 22.4 liter containers A and B contains 8 g of O₂ and 8 g of SO₂ respectively at 273 K and 1 atm pressure, then
(a) Number of molecules in A and B are the same
(b) Number of molecules in B is more than that in A.
(c) The ratio between the number of molecules in A to number of molecules in B is 2 : 1
(d) Number of molecules in B is three times greater than the number of molecules in A.
Answer:
(c) The ratio between the number of molecules in A to number of molecules in B is 2 : 1
Solution:
No. of moles of oxygen = 8 g/32 g = 0.25 moles of oxygen
No. of moles of sulphur dioxide = 8 g/64 g
= 0.125 moles of sulphur dioxide
Ratio between the no. of molecules = 0.25 : 0.125 = 2 : 1

Question 20.
What is the mass of precipitate formed when 50 ml of 8.5 % solution of AgNO₃ is mixed with 100 ml of 1.865 % potassium chloride solution?
(a) 3.59 g
(b) 7 g
(c) 14 g
(d) 28 g
Answer:
(a) 3.59 g
Solution:
AgNO₃ + KCl → KNO₃ + AgCl
50 mL of 8.5 % solution contains 4.25 g of AgNO₃
No. of moles of AgNO₃ present in 50 mL of 8.5 % AgNO₃ solution
= Mass / Molar mass = 4.25 / 170 = 0.025 moles
Similarly, No of moles of KCl present in 100 mL of 1.865 % KCl solution = 1.865 / 74.5 = 0.025 moles
So total amount of AgCl formed is 0.025 moles (based on the stoichiometry)
Amount of AgCl present in 0.025 moles of AgCl
= No. of moles × molar mass = 0.025 × 143.5 = 3.59 g

Question 21.
The mass of a gas that occupies a volume of 612.5 ml at room temperature and pressure (25°C and 1 atm pressure) is 1.1 g. The molar mass of the gas is
(a) 66.25 g mol^-1
(b) 44 g mol^-1
(c) 24.5 g mol^-1
(d) 662.5 g mol^-1
Answer:
(b) 44 g mol^-1
Solution:
No. of moles of a gas that occupies a volume of 612.5 mL at room temperature and pressure (25° C and 1 atm pressure)
= 612.5 × 10^-3 L/24.5 L mol^-1
= 0.025 moles
We know that,
Molar mass = Mass / no. of moles
= 1.1 g / 0.025 mol
= 44 g mol^-1

Question 22.
Which of the following contain the same number of carbon atoms as in 6 g of carbon -12.
(a) 7.5 g ethane
(b) 8 g methane
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)
Solution:
No. of moles of carbon present in 6 g of C – 12 = Mass/Molar mass
= 6/12 = 0.5 moles
= 0.5 × 6.022 × 10²³ carbon atoms.
No. of moles in 8 g of methane
= 8/16 = 0.5 moles
= 0.5 × 6.022 × 10²³ carbon atoms.
No. of moles in 7.5 g of ethane = 7.5 / 16 = 0.25 moles
= 2 × 0.25 × 6.022 × 10²³ carbon atoms.

Question 23.
Which of the following compound(s) has/have a percentage of carbon same as that in ethylene (C₂H₄)
(a) Propene
(b) ethyne
(c) benzene
(d) ethane
Answer:
(a) Propene
Solution:
Percentage of carbon in ethylene (C₂H₄) = \(\frac{\text { mass of carbon }}{\times 100 \text { Molar mass }}\)
= \(\frac{24}{28}\) × 100 = 85.71 %
Percentage of carbon in propene (C₃H₆) = \(\frac{24}{28}\) × 100 = 85.71 %

Question 24.
Which of the following is/are true with respect to carbon -12.
(a) relative atomic mass is 12 u
(b) oxidation number of carbon is +4 in all its compounds.
(c) 1 mole of carbon-12 contains 6.022 × 10²² carbon atoms.
(d) all of these
Answer:
(a) relative atomic mass is 12 u

Question 25.
Which one of the following is used as a standard for atomic mass.
(a) ₆C¹²
(b) ₇C¹²
(c) ₆C¹³
(d) ₆C¹⁴
Answer:
(a) ₆C¹²

II. Write brief answer to the following questions:

Question 26.
Define relative atomic mass.
Answer:
Relative atomic mass is defined as the ratio of the average atomic mass factor to the unified atomic mass unit.
Relative atomic mass (A_r) = \(\frac{\text { Average mass of the atom }}{\text { Unified atomic mass }}\)

Question 27.
What do you understand by the term mole?
Answer:
The mole is defined as the amount of a substance which contains 6.023 x 10²³ particles such as atoms, molecules, or ions. It is represented by the symbol.

Question 28.
Define equivalent mass.
Answer:
Gram equivalent mass of an element, compound or ion is the mass that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine. Equivalent mass has no unit but gram equivalent mass has the unit g eq-1.
Gram equivalent mass = \(\frac{\text { Molar mass }\left(\mathrm{g} \mathrm{mol}^{-1}\right)}{\text {Equivalence factor }\left(\mathrm{eq} \mathrm{mol}^{-1}\right)}\)

Question 29.
Distinguish between oxidation and reduction.
Answer:

Question 30.
What do you understand by the term oxidation number.
Answer:
Oxidation:
According to the classical concept, oxidation is a process of addition of oxygen or removal of hydrogen.
Removal of hydrogen
2H₂S + O₂ → H₂O + 2S
Addition of oxygen
C + O₂ → CO₂
According to the electronic concept, loss of electrons is called oxidation reaction.
Ca → Ca²⁺ + 2e^–
During oxidation, oxidation number increases.
Dining oxidation, reducing agent gets oxidised.

Reduction:
Reduction is a process of removal of oxygen or addition of hydrogen.
Addition of hydrogen
Ca + H₂ → CaH₂
Removal of oxygen
Zn O + C → Zn + CO
According to the electronic concept, gain of electrons is called a reduction reaction.
Zn²⁺ + 2e^– → Zn
During reduction, oxidation number decreases.
During reduction, oxidising agent gets reduced.

Question 31.
Calculate the molar mass of the following compounds.
Answer:
(i) urea [CO(NH₂)₂]:
Molar mass of urea = (4 × Atomic mass of hydrogen) + ( 1 × Atomic mass of carbon) + ( 2 × Atomic mass of nitrogen) + ( 1 × Atomic mass of oxygen)
= (4 × 1.008) +(1 × 12) + (2 × 14)+ ( 1 × 16)
= 4.032 + 12 + 28 + 16 = 60.032 g mol^-1

(ii) acetone[CH₃COCH₃]
Molar mass of Acetone = (6 × Atomic mass of hydrogen) + ( 3 × Atomic mass of carbon) + (1 × Atomic mass of oxygen) = (6 × 1.008) + (3 × 12) + ( 1 × 16)
= 6.048 + 36 + 16 = 52.024 g mol^-1.

(iii) boric acid [H₃BO₃]:
Molar mass of Boric acid = (3 × Atomic mass of hydrogen) + ( 1 × Atomic mass of boron) + ( 3 × Atomic mass of oxygen)
= (3 × 1.008) + (3 × 11) + ( 1 × 16)
= 3.024 + 33 + 16 = 52.024 g mol^-1.

(iv) sulphuric acid [H₂SO₄] = (2 × Atomic mass of hydrogen) + ( 1 × Atomic mass of sulphur) + ( 4 × Atomic mass of oxygen)
= (2 × 1.008) +(1 × 32) + (4 × 16)
= 2.016 + 32 +64 = 98.016 g mol^-1.

Question 32.
The density of carbon dioxide is equal to 1.965 kgm^-3 at 273 K and 1 atm pressure. Calculate the molar mass of CO₂.
Answer:
Given:
The density of C0₂ at 273 K and 1 atm pressure = 1.965 kgm^-3
Molar mass of CO₂ =?
At 273 K and 1 atm pressure, 1 mole of CO₂ occupies a volume of 22.4 L
Mass of 1 mole of CO₂ = \(\frac{1.965 \mathrm{Kg}}{1 \mathrm{~m}^{3}}\) × 22.4 L
= \(\frac{1.965 \times 10^{3} \mathrm{~g} \times 22.4 \times 10^{-3} \mathrm{~m}^{3}}{1 \mathrm{~m}^{3}}\)
= 44.01 g
Molar mass of CO₂ = 44 gmol^-1

Question 33.
Which contains the greatest number of moles of oxygen atoms.
1 mol of ethanol
1 mol of formic acid
1 mol of H₂O
Answer:

Answer: Formic acid

Question 34.
Calculate the average atomic mass of naturally occurring magnesium using the following data:

Answer:
Average atomic mass
= \(\frac{(78.9923 .99)(1024.99)(11.0125 .98)}{100}\)
= \(\frac{2430.9}{100}\)
= 24.31 u

Question 35.
In a reaction x + y + xyz. identify the Limiting reagent if any, in the following reaction mixtures.
(a) 200 atoms of x + 200 atoms of y + 50 molecules of z₂.
(b) 1 mol of x + 1 mol ofy + 3 mol of z₂.
(c) 50 atoms of x + 25 atoms of y + 50 molecules of z₂.
(d) 2.5 mol ofx + 5 mol ofy + 5 mol of z₂.
Reaction:
x + y + z₂ → xyz₂

(a) 200 atoms of x + 200 atoms of y + 50 molecules of z₂ According to the reaction, 1 atom of x reacts with one atom of y and one molecule of z to give product. In the case (a) 200 atoms of x, 200 atoms of y react with 50 molecules of z₂ (4 part) i.e. 50 molecules of z₂ react with 50 atoms of x and 50 atoms of y. Hence z is the limiting reagent.

(b) 1 mol of x + 1 mol of y + 3 mol of z₂
According to the equation 1 mole of z₂ only react with one mole of x and one mole of y. If 3 moles of z₂ are there, z is limiting reagent.

(c) 50 atoms of x + 25 atoms of y + 50 molecules of z₂
25 atoms of y react with 25 atoms of x and 25 molecules of z₂. So y is the limiting reagent.

(d) 2.5 mol of x + 5 mol of y + 5 mol of z₂
2.5 mol of x react with 2.5 mole of y and 2.5 mole of z₂. So x is the limiting reagent.

Question 36.
Mass of one atom of an element is 6.645 × 10²³ g How many moles of element are there in 0.320 kg.
Answer:

there in 0.320 kg
Given:
mass of 1 atom = 6.645 × 10^-23 g
∴ mass of 1 mole of atom = 6.645 × 10^-2323 g × 6.022 × 10²³ = 40 g
∴ Number of moles of element in 0.320 kg = \(\frac{1 \mathrm{~mol}}{40 \mathrm{~g}}\) × 0.320 kg
= \(\frac{1 \mathrm{~mol} \times 320 \mathrm{~g}}{40 \mathrm{~g}}\) = 8 mol

Question 37.
What is the difference between molecular mass and molar mass? Calculate the molecular mass and molar mass for carbon monoxide.
Answer:
Molecular mass:

• Relative molecular mass is defined as the ratio of the mass of the molecule to the unified atomic mass unit.
•  It can be calculated by adding the relative atomic masses of its constituent atoms.
• For carbon monoxide (CO) Molecular mass = Atomic mass of carbon + Atomic mass of oxygen 12 + 16 = 28 u.

Molar mass:

• It is defined as the mass of one mole of a substance.
• The molar mass of a compound is equal to the sum of the relative atomic masses of its constituent expressed in g mol-¹.
• For carbon monoxide (CO) 12 + 16 = 28 g mol^-1 Both molecular mass and molar mass are numerically the same but the units are different.

Question 38.
What is the empirical formula of the following?
(i) Fructose (C₆H₁₂O₆) found in honey
(ii) Caffeine (C₈H₁₀N₄O₂ )a substance found in tea and coffee.
Answer:

Question 39.
The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of AC = 27 u Atomic mass of O = 16 u)
2Al + Fe₂O₃ → Al₂O₃ + 2Fe; If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide
(i) Calculate the mass of Al₂O₃ formed.
(ii) How much of the excess reagent is left at the end of the reaction?
Answer:
Given:
2Al + Fe₂O₃ → Al₂O₃ + 2Fe

Molar mass of Al₂O₃ formed = 6mol × 102 g mol^-1 = 612 g
[Al₂O₃ = (2 × 27) + (3 × 16) = 54 + 48 = 102]
Excess reagent = Fe₂O₃.
Amount of excess reagent left at the end of the reaction = 1 mol × 160 g mol^-1 = 160 g
[Fe₂O₃ = (2 × 56) + (3 × 16) = 112 + 48 = 160 g] = 160 g

Question 40.
How many moles of ethane is requircd to produce 44 g of CO₂ (g) after combustion. Balanced equation for the combustion of ethane.
Answer:
C₂H₆ + \(\frac{7}{2}\)O₂ → 2CO₂ + 3H₂O
⇒ 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
∴ To produce 4 moles of CO₂, 2 moles of ethane is required
To produce 1 mole (44g) of CO₂ required number of moles of ethane
= \(\frac{1}{2}\) mole of ethane
= 0.5 mole of ethane

Question 41.
Hydrogen peroxide is an oxidising agent. It oxidises ferrous ion to ferric ion and reduced itself to water. Write a balanced equation.
Answer:

Question 42.
Calculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47.
Answer:

Empirical formula = C₆H₆O
η = Molar mass / Calculated empirical formula mass = \(\frac{2 \times \text { vapour density }}{94}\)
= \(\frac{2 \times 47}{94}\) = 1
∴ Molecular formula(C₆H₆O) × 1 = C₆H₆O

Question 43.
A Compound on analysis gave Na = 14.31% S = 9.97% H= 6.22% and 0= 69.5% calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as water of crystallization, (molecular mass of the compound is 322).
Answer:

∴ Empirical formula = Na₂SH₂₀O₁₄
η = Molar mass / Calculated empirical formula mass
= \(\frac{322}{322}\) = 1
[Na₂SH₂₀O₁₄ = (2 × 23) + (1 × 32) + (20 × 1) + (14 × 16)
= 46 + 32 + 20 + 224 = 322]
Molecular formula = Na₂SH₂₀O₁₄
Since all the hydrogen in the compound present as water
∴ Molecular formula is Na₂SH₂₀O₁₄

Question 44.
Balance the following equations by oxidation number method
(i) K₂Cr₂O₇ + KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + I₂ + H₂O
Answer:

K₂Cr₂O₇ + 6KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + I₂ + H₂O
K₂Cr₂O₇ + 6KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + H₂O
K₂Cr₂O₇ + 6KI + 7H₂SO₄ → 4K₂SO₄ + Cr₂(SO₄) + I₂ + 7H₂O

(ii) KMnO₄ + Na₂SO₃ → MnO₂ + Na₂SO₄ + KOH
Answer:

⇒ 2KMnO₄ + 3Na₂SO₃ → MnO₂ + Na₂SO₄ + KOH
⇒ 2KMnO₄ + 3Na₂SO₃ → 2MnO₂ + 3Na₂SO₄ + KOH
⇒ 2KMnO₄ + 3Na₂SO₃ → MnO₂ + Na₂SO₄ + 2KOH

(iii) Cu + HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O
Answer:

Cu + 2HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O
Cu + 2HNO₃ + 2HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

(iv) KMnO₄ + H₂C₂O₄ + H₂SO₄ → K₂SO + MnSO₄ + CO₂ + H₂O
Answer:

KMnO₄ + 5H₂C₂O₄ + H₂SO₄ → K₂SO + MnSO₄ + CO₂ + H₂O
2KMnO₄ + 5H₂C₂O₄ + H₂SO₄ → 2MnSO₄ + 10CO₂ + H₂O
2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → K₂SO + 2MnSO₄ + 10CO₂ + 8H₂O

Question 45.
Balance the following equations by ion electron method.
(i) KMnO₄ + SnCl₂ + HCl → MnCl₂ SnCl₄ + H₂O + KCl
Answer:

(ii) C₂O₄^2- + Cr₂O₇^2- →
Cr³⁺ + CO₂ (in acid medium)
Answer:

(iii) Na₂S₂O₃ + I₂ → Na₂S₄O₆ + NaI ₂ (in acid medium)
Answer:
S₂O₃^2- → S₄O₆^2- ………….(1)
Half reaction ⇒ I₂ → I_– …………….(2)

(iv) Zn + NO₃^– → Zn⁺² + NO
Half reactions are

11th Chemistry Guide Basic Concepts of Chemistry and Chemical Calculations Additional Questions and Answers

I. Choose the best answer:

Question 1.
_____ consists of more than one chemical entity present without any chemical interactions.
(a) Mixtures
(b) Pure substances
(c) Compounds
(d) Elements
Answer:
(a) Mixtures

Question 2.
_______ are made up of molecules which contain two or more atoms of different elements.
(a) Mixtures
(b) Compounds
(c) Pure substances
(d) Elements
Answer:
(b) Compounds

Question 3.
Match the correct pair:

(a) A – ii, B – i, C – iii
(b) A – i, B – ii, C – iii
(c) A – ii, B – iii, C – i
(d) A – iii, B – ii, C – i
Answer:
(c) A – ii, B – iii, C – i

Question 4.
_______ atom is considered as standard by the IUPAC for calculating atomic masses.
(a) C – 13
(b) C – 15
(c) C – 14
(d) C – 12
Answer:
(d) C – 12

Question 5.
The value of unified mass is equal to
(a) 1.6605 × 10^-27 kg
(b) 1.6736 × 10^-27 kg
(c) 1.6605 × 10²⁷ kg
(d) 1.6736 × 10^-29 kg
Answer:
(a) 1.6605 × 10^-27 kg

Question 6.
Chlorine consists of two naturally occurring isotopes ₁₇Cl³⁵ and ₁₇Cl³⁷ in the ratio 77 : 23. The average relative atomic mass of chlorine is
(a) 36.45 u
(b) 35.56 u
(c) 35.46 u
(d) 35.65 u
Answer:
(c) 35.46 u

Question 7.
The relative atomic masses of hydrogen, oxygen, and carbon are 1.008 u, 16 u, and 12 u respectively. The relative molecular mass of glucose (C₆ H₁₂ O₆ ) is
(a) 170.096 u
(b) 189.096 u
(c) 180.096 u
(d) 190.086 u
Answer:
(c) 180.096 u

Question 8.
The specific amount of a substance is represented in SI unit is
(a) amu
(b) mole
(c) atomic mass
(d) equivalent mass
Answer:
(b) mole

Question 9.
Commonly used medicines for treating heart bum and acidity are called
(a) antipyretics
(b) analgesics
(c) antiseptics
(d) antacids
Answer:
(d) antacids

Question 10.
The typical concentration of hydrochloric acid in gastric acid is
(a) 0.1 M
(b) 0.01 M
(c) 0.082 M
(d) 0.82 M
Answer:
(c) 0.082 M

Question 11.
Antacids used to treat acidity contain mostly
(a) Al(OH)₃
(b) Mg(OH)₂
(c) (a) or (b)
(d) Ca(OH)₂
Answer:
(c) (a) or (b)

Question 12.
The value of Avogadro number is
(a) 6.022 × 10²³
(b) 6.022 × 10²²
(e) 6.022 × 10^-25
(d) 6.022 × 10^-23
Answer:
(a) 6.022 × 10²³

Question 13.
The volume occupied by one mole of any substance in the gaseous state st 273 K and 1 atm pressure is _______ (in litres)
(a) 24.5
(b) 22.4
(e) 22.71
(d) 21.18
Answer:
(b) 22.4

Question 14.
The equivalent mass of KMnO₄(Molar mass 158 g mol ^-1 based on the equation
MnO₄ + 8H⁺ + 5e → Mn²⁺ + 4H₂O
(a) 158
(b) 52.66
(c) 31.6
(d) 40
Answer:
(c) 31.6

Question 15.
The molecular formula of acetic acid is C₂H₄O₂ Its empirical formula is
(a) C₂H₄O₂
(b) C₂H₂O
(c) CH₂O₂
(d) CH₂O
Answer:
(d) CH₂O

Question 16.
The number of moles of hydrogen required to produce 20 moles of ammonia is
(a) 10
(b) 15
(c) 30
(d) 20
Answer:
(c) 30

Question 17.
Which of the following is/are redox reactions?
(i) 4Fe + 3O₂ → 2Fe₂O₃
(ii) H₂S + Cl₂ → 2HCl + S
(iii) CuO + C → Cu + CO
(iv) S +H₂ → H₂S
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
Answer:
(d) (iii) and (iv)

Question 18.
The oxidation number of sulphur on H₂SO₄ is
(a) -2
(b) -6
(c) +2
(d) +6
Answer:
(d) +6

Question 19.
Choose the correct pair:

(a) A – i, B – iv, C- iii, D – ii
(b) A – iii, B – i, C – iv, D – ii
(c) A – i, B – ii, C – iv, D – iii
(d) A – iii, B – iv, C – i, D – ii
Answer:
(b) A – iii, B – i, C – iv, D – ii

Question 20.
The change in the oxidation number of Manganese in the following reaction 2 KMnO₄ + 10 FeSO₄ + 8 H₂SO₄ → K₂SO₄ + 5 Fe₂(SO₄)₃ + 8 H₂O
(a) +2 to +7
(b) +2 to +5
(c) +7 to +2
(d) +5 to +2
Answer:
(c) +7 to +2

Question 21.
Which of the following statements are correct about oxidation?
(i) Removal of oxygen
(ii) Addition of hydrogen
(iii) Loss of electron
(iv) Gain of electron
(v) Addition of oxygen
(vi) Removal of hydrogen
(a) i, ii, iv
(b) iii, v, vi
(c) i, iii, v
(d) iii, iv, v
Answer:
(b) iii, v, vi

Question 22.
Choose the correct oxidation number of oxygen in the following compounds.

(a) A -iii, B- iv, C – i, D – ii.
(b) A – iv, B – i, C – iii, D – ii
(c) A – iii, B – i, C – iv, D – ii
(d) A – iv, B – iii, C – i, D – ii
Answer:
(c) A – iii, B – i, C – iv, D – ii

Question 23.
The correct order of electron releasing tendency of the following elements is
(a) Zn > Cu > Ag
(b) Cu > Zn > Ag
(c) Ag > Zn > Cu
(d) Ag > Cu > Zn
Answer:
(a) Zn > Cu > Ag

Question 24.
H₂O₂ → 2 H₂O + O₂ is a
(a) Displacement reaction
(b) Combinationreaction
(c) Decomposition reaction
(d) Disproportionate reaction
Answer:
(d) Disproportionate reaction

Question 25.
The molar mass and empirical formula mass of a compound are 78 and 13 respectively. The molecular formula of the compound is (Empirical formula is CH)
(a) C₂H₂O₂
(b) C₂H₄
(c) C₆H₆
(d) C₃H₈
Answer:
(c) C₆H₆

II. Very Short Question and Answers (2 Marks):

Question 1.
State Avogadro’s Hypothesis.
Answer:
It states that ‘Equal volume of all gases under the same conditions of temperature and pressure contain the same number of molecules’.

Question 2.
How is matter classified physically?
Answer:
Matter can be classified as solids, liquids, and gases based on their physical state. The physical state of matter can be converted into one another by modifying the temperature and pressure suitably.

Question 3.
How is matter classified chemically?
Answer:
Matter can be classified into mixtures and pure substances based on chemical compositions.

Question 4.
Calculate a number of moles of carbon atoms ¡n three moles of ethane.
Answer:
Ethane – Molecular formula = C₂H₆
1 mole of ethane contains 2 atoms of carbon (6.023 x 10²³ C)
∴ 3 moles of ethane contains 6 atoms of Carbon.
∴ No. of moles of Carbon atoms = 3 x 6.023 x 10²³ Carbon atoms.
= 18.069 x 10²³ Carbon atoms.

Question 5.
What are pure substances? How are they classified?
Answer:
Pure substances are composed of simple atoms or molecules. They are further classified as elements and compounds.

Question 6.
Mass of one atom of an element ¡s 6.66 x 10²³ g. How many moles of the element are there in 0.320 kg?
Answer:
Mass of one atom of an element = 6.66 x 10²³g
No. of moles = \(\frac {Mass}{Molecular mass}\) 3
Molecular mass = Mass of 1 atom x Avogadro number
6.66 x 10²³ x 6.023 x 10²³
= 6.66 x 6.023 = 40.11318
Number of moles = \(\frac {Mass}{Molecular mass}\) = \(\frac{0.320 \mathrm{kg} \times 10^{3}}{40}\) = 8 moles.

Question 7.
What are compounds? Give examples.
Answer:
Compounds are made up of molecules which contain two or more atoms of different elements.
Example: Carbon dioxide (CO₂), Sodium Chloride (NaCl).

Question 8.
Calculate the weight of 0.2 moles of sodium carbonate.
Answer:
Sodium carbonate = Na₂CO₃
Molecular mass of Na₂CO₃ = (23 x 2)+(12 x 1)+(16 x 3)
= 46 + 12 + 48 = 106 g
Mass of 1 mole of Na₂CO₃ = \(\frac{106 \times 0.2}{1}\) = 21.2 g

Question 9.
Define relative atomic mass.
Answer:
The relative atomic mass is defined as the ratio of the average atomic mass factor to the unified atomic mass unit.
Relative atomic mass
(A_r) = Average mass of the atom / Unified atomic mass

Question 10.
What is the average atomic mass?
Answer:
Average atomic mass is defined as the average of the atomic masses of all atoms in their naturally occurring isotopes.

Question 11.
Calculate the equivalent mass of barium hydroxide.
Answer:
Barium hydroxide = Ba(OH)₂
Molecular mass of Ba(OH)₂ = 137 + (16 x 2) + (1 x 2) = 171.0 g / mol.
Acidity = 2
Equivalent mass of Ba(OH)₂ = \(\frac {17 1.0}{2}\) = 85.5

Question 12.
What is a mole?
Answer:
One mole is the amount of substance of a system, which contains as many elementary particles as there are atoms in 12 g of carbon – 12 isotope. The elementary particles can be molecules, atoms, ions, electrons or any other specified particles.

Question 13.
What do you understand by the terms empirical formula and molecular formula?
Answer:
Empirical Formula:

• It is the simplest formula.
• It shows the ratio of the number of atoms of different elements in one molecule of the compound.

Molecular Formula:

• It is the actual formula.
• It shows the actual number of different types of atoms present in one molecule of the compound.

Question 14.
What is Avogadro’s number?
Answer:
The total number of entities present in one mole of any substance is equal to 6.022 × 10²³. This number is called the Avogadro number.

Question 15.
State Avogadro hypothesis.
Answer:
Equal volume of all gases under the same conditions of temperature and pressure contain equal number of molecules.

Question 16.
Calculate the mass of sodium (in kg) present in 95 kg of a crude sample of sodium nitrate whose percentage purity is 70%.
Answer:
Sodium Nitrate = NaNO₃
Molecular mass of Sodium Nitrate = 23 + 14 + 48 = 85
100% pure 85 g of NaNO₃ contains 23 g of Sodium.
100% pure 95 x 103 g of NaNO₃ will contains \(\frac {23}{85}\) x 95 x 10³
= 25.70 x 10³ g of Sodium.
100% pure NaNO₃  contains 25.70 x 10³ g of Sodium.
∴ 70% pure NaNO₃  will contains = 17990 g (or) 17.99 Kg of Na.

Question 17.
Define molar volume.
Answer:
The volume occupied by one mole of any substance in the gaseous state at a given temperature and pressure is called molar volume.

Question 18.
What is gram equivalent mass?
Answer:
Gram equivalent mass of an element, compound or ion is the mass that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.
Gram equivalent mass = (Molar mass(g mol^-1)) / Equivalence factor (eq mol^-1)

Question 19.
What is meant by Plasma state? Give an example.
Answer:
The gaseous state of matter at a very high temperature containing gaseous ions and free-electron is referred to as the Plasma state. e.g. Lightning.

Question 20.
What is the acidity of a base? Give an example.
Answer:
The acidity of a base is the number of moles of ionizable OH- ions present in 1 mole of the base. The acidity of potassium hydroxide (KOH) is 1.

Question 21.
What is empirical formula ola compound?
Answer:
The empirical formula of a compound is the formula written with the simplest ratio of the number of different atoms present in one molecule of the compound as a subscript to the atomic symbol.

Question 22.
Chlorine has a fractional average atomic mass. Justify this statement.
Answer:
Chlorine molecule has two isotopes as in ₁₇Cl³⁵, ₁₇ Cl³⁷ in the ratio of 77 : 23, so when we are calculating the average atomic mass, it becomes fractional.
The average relative atomic mass of Chlorine = \(\frac {(35 x 77) + (37 x 23)}{100}\) = 35.46 amu

Question 23.
What is meant by Stoichiometry?
Answer:
Stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation in moles. The quantity of reactants and products can be expressed in moles or in terms of mass unit or as volume.

Question 24.
What are limiting and excess reagents?
Answer:
When a reaction is carried out using non-stoichiometric quantifies of the reactants, the product yield will be determined by the reactant that is completely consumed and is called the limiting reagent. It limits the further reaction to take place. The other reagent which is in excess is called the excess reagent.

Question 25.
Define Avogadro Number.
Answer:
Avogadro number is the number of atoms present in one mole of an element or number of molecules present in one mole of a compound. The value of Avogadro number (N) = 6.023 x 10²³

Question 26.
Define oxidation number.
Answer:
The oxidation number is defined as the imaginary charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to a set of rules.

Question 27.
Calculate the equivalent mass of Copper. (Atomic mass of copper = 63.5)
Answer:
Equivalent mass = \(\frac {Atomic mass}{Valency}\)
Equivalent mass of Copper = \(\frac {63.5}{2}\) = 31.75 g eq^-1.

Question 28.
Mention the types of redox reactions?
Answer:
The types of redox reactions are combination reaction, decomposition reaction, displacement reaction, disproportionate reaction and competitive electron transfer reactions.

III. Short Question and Answers (3 Marks):

Question 1.
Describe the chemical classification of matter.
Answer:
Matter can be classified into mixtures and pure substances based on chemical compositions. Mixtures consists of more than one chemical entity present without any chemical interactions. They can be further classified as homogeneous or heterogeneous mixtures based on their physical appearance. Pure substances are composed of simple atoms or molecules. They are further classified as elements and compounds.

Question 2.
Distinguish between element and compound.
Answer:
An element consists of only one type of atom. Element can exists as monoatomic or polyatomic units. The polyatomic elements are called molecules.
Example: Monoatomic unit – Gold (Au), Copper (Cu);
Polyatomic unit: Hydrogen (H₂), Phosphorous (P₄).

Compounds are made up of molecules which contain two or more atoms of different elements.
Example: Carbon dioxide (CO₂), Glucose (C₆ H₁₂ O₆)

Properties of compounds are different from those of their constituent elements. For example, sodium is a shiny metal, and chlorine is an irritating gas. But the compound formed from these two elements, sodium chloride shows different characteristics as it is crystalline solid, vital for biological functions.

Question 3.
What is average atomic mass? How is average atomic mass of chlorine calculated?
Answer:
Average atomic mass is defined as the average of the atomic masses of all atoms in their naturally occurring isotopes. Chlorine consists of two naturally occurring isotopes ₁₇Cl³⁷ and ₁₇Cl³⁵ in the ratio 77 : 23, the average relative atomic mass of chlorine is
= ((35 × 77) + (37 × 23)) / 100
= 35.46 u

Question 4.
What will be the mass of one ¹²C atom in g?
Answer:
Molar mass of ¹²C = 12.00 g mol^-1.
∴ Mass of 6.023 x 10²³ carbon atom = 12.0 g
∴ Mass of 1 carbon atom = \(\frac{12}{6.023 \times 10^{23}}\) = 1.992 x 10 g.

Question 5.
Discuss the role of antacids.
Answer:
Gastric acid is a digestive fluid formed in the stomach and it contains hydrochloric acid. The typical concentration of the acid in gastric acid is 0.082 M. When the concentration exceeds 0.1M it causes heartburn and acidity. Antacids used to treat acidity contain mostly magnesium hydroxide or aluminium hydroxide that neutralizes the excess acid. The chemical reactions are as follows.

3 HCl + Al(OH)₃ → AlCl₃ + 3H₂O
2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O.

From the above reactions, we know that 1 mole of aluminium hydroxide neutralizes 3 moles of HCl while 1 mole of magnesium hydroxide neutralizes 2 moles of HCl.

Question 6.
Explain gram equivalent mass.
Answer:
Gram equivalent mass of an element, compound, or ion is the mass that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.
Consider the following reaction:
Zn + H₂ SO₄ → ZnSO₄ + H₂
In this reaction, 1 mole of zinc (65.38 g) displaces one mole of hydrogen molecule (2.016 g). Mass of zinc required to displace 1.008 g hydrogen is
= \(\frac{65.38}{2.016}\) × 1.008
= \(\frac{65.38}{2}\)
= 32.69
The equivalent mass of zinc = 32.69
The gram equivalent mass of zinc = 32.69 g eq^-1.
The expression used to calculate gram equivalent mass is
Gram equivalent mass = Molar mass(g mol^-1) / Equivalence factor (eq mol^-1)

Question 7.
Calculate the gram equivalent mass of sulphuric acid.
Answer:
Basicity of sulphuric acid (H₂SO₄) = 2eq mol^-1
Molar mass of H₂SO₄ = (2 × 1) +(1 × 32) +(4 × 16) = 56 g mol^-1
Gram equivalent mass of H₂SO₄ = \(\frac{98}{2}\)
= 49 g eq^-1

Question 8.
Calculate the gram equivalent mass of potassium hydroxide.
Answer:
Acidity of potassium hydroxide (KOH) = 1 eq mol^-1
Molar mass of KOH = (1 × 39) + (1 × 16) + (1 × 1) = 56 g mol^-1
Gram equivalent mass of KOH = \(\frac{56}{1}\) = 56 g eq^-1.

Question 9.
Calculate the gram equivalent mass of Potassium permanganate.
Answer:
Potassium permanganate is an oxidizing agent. Molar mass of
KMnO₄ = (1 × 39)+ (1 × 55)+ (4 × 16)
= 158 g mol^-1

In an acidic medium, permanganate is reduced during oxidation and is given by the following equation,
MnO₄^– + 8H⁺ + 5e → Mn² + 4H₂O
Therefore, n = 5.
Gram equivalent mass of KMnO₄ = \(\frac{158}{5}\) = 31.6 g mol^-1

Question 10.
An acid found in tamarind on analysis shows the following percentage composition: 32 % Carbon; 4 % Hydrogen; 64 % Oxygen. Find the empirical formula of the compound.
Answer:

The ratio of C : H : O is 2 : 3 : 3 and hence, the empirical formula of the compound is CH₂O

Question 11.
An organic compound present in vinegar has 40 % Carbon, 6.6 % Hydrogen: and 53.4 % Oxygen. Find the empirical formula of the compound.
Answer:

The ratio of C: H : O is 1 : 2 : 1 and hence, the empirical formula of the compound is CH₂O.

Question 12.
How much copper can be obtained from 100 g of anhydrous copper sulphate?
Answer:
Anhydrous copper sulphate = CuSO₄
Molecular mass of CuSO₄ = 63.5 + 32 + (16 x 4)
= 63.5 + 32 + 64
= 159.5 g
159.5 g of CuSO₄ contains 63.5 g of copper.
∴ 100 g of CuSO₄ contains \(\frac{6.35}{159.5}\) x 100 = 0.39811 x 100 = 39.81 g of Copper.

Question 13.
Explain the classical concept of oxidation and reduction.
Answer:
According to classical concept, the addition of oxygen or removal of hydrogen is called oxidation.
Consider the following reactions,
4Fe + 3O₂ → 2Fe₂O₃
H₂S +Cl₂ → 2 HCl + S

In the first reaction, which is responsible for the rusting of iron, the oxygen adds on to the metal, iron. In the second reaction, hydrogen is removed from Hydrogen sulphide.
According to classical concept, addition of hydrogen or removal of oxygen is called reduction.
Consider the following reactions,
CuO + C → Cu + CO
S + H₂ → H₂S
In the first reaction, oxygen is removed from cupric oxide and in the second reaction, hydrogen is added to sulphur.

Question 14.
Describe the electron concept of oxidation and reduction.
Answer:
The reaction involving loss of electron is termed as oxidation and gain of electron is termed as reduction.
Fe²⁺ → Fe³⁺ + e^– (loss of electron – oxidation)
Cu²⁺ + 2e^– → Cu ( gain of electron – reduction)

Question 15.
Describe the oxidation number concept of oxidation and reduction.
Answer:
During redox reactions, the oxidation number of elements changes. A reaction in which oxidation number of the element increases is called oxidation whereas the oxidation number of the element 3 decreases is called reduction.

In this reaction, manganese in potassium permanganate favours the oxidation of ferrous sulphate into ferric sulphate by gets reduced.

Question 16.
Write notes on displacement reaction.
Answer:
Redox reactions in which an ion or an atom in a compound is replaced by an ion or an atom of another element are called displacement reactions. They are further classified into
(i) metal displacement reactions
(ii) non-metal displacement reactions.

(i) Metal displacement reactions:
Place a zinc metal strip in an aqueous copper sulfate solution taken in a beaker. The intensity of blue colour of the solution slowly reduced and finally disappeared. The zinc metal strip became coated with brownish metallic copper. This is due to the following metal displacement reaction.
CuSO₄(aq)+ Zn(s) → Cu + ZnSO₄.

(ii) Non-metal displacement reaction:
Zn + 2HCl → ZnCl₂ + H₂

IV. Long Question and Answers:

Question 1.
What is a matter? Explain its classification.
Answer:
Matter is defined as anything that has mass and occupies space. All matter is composed of atoms.

Physical Classification:
Matter can be classified as solids, liquids and gases based on their physical state. The physical state of matter can be converted into one another by modifying the temperature and pressure suitably.

Chemical Classification:

Classification of Matter

Matter can be classified into mixtures and pure substances based on chemical compositions. Mixtures consists of more than one chemical entity present without any chemical interactions. They can be further classified as homogeneous or heterogeneous mixtures based on their physical appearance. Pure substances are composed of simple atoms or molecules. They are further classified as elements and compounds.

An element consists of only one type of atom. Element can exists as monoatomic or polyatomic units. The polyatomic elements are called molecules.
Example: Monoatomic unit – Gold (Au), Copper (Cu);
Polyatomic unit: Hydrogen (H₂), Phosphorous (P₄).

Compounds are made up of molecules which contain two or more atoms of different elements.
Example: Carbon dioxide (CO₂), Glucose (C₆H₁₂O₆)
Properties of compounds are different from those of their constituent elements. For example, sodium is a shiny metal, and chlorine is an irritating gas. But the compound formed from these two elements, sodium chloride shows different characteristics as it is a crystalline solid, vital for biological function.

Question 2.
How is empirical formula of a compound determined from the elemental analysis?
Answer:
The empirical formula of a compound determined from the elemental analysis by the following steps.
(i) Since the composition is expressed in percentage, we can consider the total mass of the compound as 100 g and the percentage values of individual elements as a mass in grams.
(ii) Divide the mass of each element by its atomic mass. This gives the relative number of moles of various elements in the compound.
(iii) Divide the value of a relative number of moles obtained in the step – (ii) by the smallest number of them to get the simplest ratio.
(iv) In case the simplest ratios obtained in step – (iii) are not whole numbers then they may be converted into the whole numbers by multiplying a suitable smallest number.

Question 3.
An organic compound present in vinegar has 40% carbon, 6.6 % hydrogen, and 53.4 % oxygen, Find the empirical formula and molecular formula of the compound. (Given, Molar mass: 60 g mol^-1).
Answer:

The ratio of C: H : O is 1 : 2 : 1 and hence, the empirical formula of the compound is CH₂O.
Empirical Formula mass = (1 × 12 + 1 × 2 + 1 × 16) = 12 + 2 + 16 = 30.
Whole number η =  Molar mass / Empirical formula mas = \(\frac{60}{30}\) = 2
Therefore, Molecular formula = (CH₂O)₂ = C₂H₄O₂

Question 4.
An organic compound present in vinegar has 40% carbon, 6.6 % hydrogen, and 53.4 % oxygen, Find the empirical formula and molecular formula of the compound. (Given, Molar mass: 90 g mol^-1).
Answer:

The ratio of C : H : O is 1 : 2 : 1 and hence, the empirical formula of the compound is CH₂O.
Empirical Formula mass = (1 × 12 + 1 × 2 × 16) = 12 + 2 + 16 = 30.
Whole number, n = \(\frac{\text { Molar mass }}{\text { Empirical formula mass }}\)
= \(\frac{90}{30}\) = 3
Therefore, Molecular formula = (CH₂O)₃ = C₃H₆O₃

Question 5.
What is a redox reaction? Explain the different concepts of redox reaction.
Answer:
The reaction involving loss of electron is oxidation and gain of electron is reduction. Both these reactions take place simultaneously and are called as redox reactions.

Classical concept of oxidation and reduction:
According to classical concept, addition of oxygen or removal of hydrogen is called oxidation.
Consider the following reactions,

4Fe + 3O₂ → 2Fe₂O₃
H₂S + Cl₂ → 2HCl + S

In the first reaction, which is responsible for the rusting of iron, the oxygen adds on to the metal, iron. In the second reaction, hydrogen is removed from Hydrogen sulphide.
According to classical concept, addition of hydrogen or removal of oxygen is called reduction.
Consider the following reactions,
CuO + C → Cu + CO
S + H₂ → H₂S

In the first reaction, oxygen is removed from cupric oxide, and in the second reaction, hydrogen is added to sulphur.

Electron concept of oxidation and reduction.
The reaction involving loss of electron is termed as oxidation and gain of an electron is termed as reduction.
Fe²⁺ → Fe³⁺ + e^– (loss of electron – oxidation)
Cu²⁺ + 2e^– → Cu (gain of electron – reduction)

Oxidation number concept of oxidation and reduction:
During redox reactions, the oxidation number of elements changes. A reaction in which oxidation number of the element increases is called oxidation whereas the oxidation number of the element decreases is called reduction.

In this reaction, manganese in potassium permanganate favours the oxidation of ferrous sulphate into ferric sulphate by gets reduced.

Question 6.
What is an oxidation number? State the rules to find the oxidation number.
Answer:
Oxidation number is defined as the imaginary charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to set of rules. A term that is often used interchangeably with oxidation number is oxidation state.

1. The oxidation state of a free element (i.e., uncombined state) is zero.
   Example: H₂ Cl₂, Na, and S8 have the oxidation number of zero.
2. For a monoatomic ion, the oxidation state is equal to the net charge on the ion.
   Example: The oxidation number of sodium in Na⁺ is +1.
   The oxidation number of chlorine in Cl^– is -1.
3. The algebraic sum of oxidation states of all atoms in a molecule is equal to zero, while in ions, it is equal to the net charge on the ion.
   Example: In H₂SO₄, 2 × (oxidation number of hydrogen) + 1 × (oxidation number of sulphur) + 4 × (oxidation number of oxygen) = 0.
4. Hydrogen has an oxidation number of +1 in all its compounds except in metal hydrides where it has -1 value.
   Example: Oxidation number of hydrogen in hydrogen chloride (HCl) is + 1.
   Oxidation number of hydrogen in sodium hydride (NaH) is -1.
5. Fluorine has an oxidation state of -1 in all its compounds.
6. The oxidation state of oxygen in most compounds is -2. Exceptions are peroxides, superoxides, and compounds with fluorine.
   Example: Oxidation number of oxygen
   (i) in water is -2,
   (ii) in hydrogen peroxide is -1,
   (iii) in superoxides such as KO₂ is 4, and
   (iv) in oxygen difluoride (OF₂) is +2.
7. Alkali metals have an oxidation state of +1 and alkaline earth metals have an oxidation state of +2 in all their compounds.

Question 7.
Balance the following chemical equation by oxidation number method.
KMnO₄ + FeSO₄ + H₂SO₄ → K₂SO₄ + Fe(SO₄)₃ + 8H₂O
Answer:
Using oxidation number concept, the reactants which undergoes oxidation and reduction are as follows:

The oxidation number of Mn in KMnO₄ changes from +7 to +2 by gaining five electrons and the oxidation number of Fe FeSO₄ changes from +2 to +3 by loosing one electron.
Since the total number of electrons lost is equal to the total number of electrons gained, the number of electrons, by cross multiplication of the respective formula with suitable integers on reactant side.

Based on the reactant side, the products are balanced.

2KMnO₄ + 10 FeSO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + Fe( SO₄)₃ + H₂O

Balance the other elements except H and O atoms. K and S are balanced as follows

2KMnO₄ + 10 FeSO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + Fe( SO₄)₃ + H₂O
The difference of 8 – S atoms in reactant side, has to be balanced by multiplying H₂SO₄ by ‘8’. The equation now becomes,

2KMnO₄ + 10 FeSO₄ + 8 H₂SO₄ → K₂SO₄ + MnSO₄ + Fe( SO₄)₃ + H₂O

‘H’ and ‘O’ atoms are balanced by multiplying H₂O molecules in the product side by ‘8’.

2KMnO₄ + 10 FeSO₄ + 8 H₂SO₄ → K₂SO₄ + 2 MnSO₄ + 5 Fe( SO₄)₃ + 8 H₂O
The above equation is a balanced equation.

Question 8.
How is the following equation ¡s balanced by Ion electron method?
MnO₄^– + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O.
Answer:
Using the oxidation number concept, the reactants which undergoes oxidation and reduction are as follows;

The two half reactions are,
Fe²⁺ → Fe³⁺ + 1e ……….(1)
MnO₄^– + 5e + 8H⁺ → Mn²⁺ + 4H₂O ……….(2)
Balancing the atoms and charges on both sides of the half reactions.
There is no change in the equation (1) whereas in equation (2), there are four ‘O’ atoms on the reractant side .
Therefore, four H₂ is added on the product side, to balance ‘H’ – add, 8 H+ in the reactant side.
MnO₄^– + 5e + 8H⁺ → Mn²⁺ + 4H₂O ………..(3)
The two half reactions are equated in such a way that the number of electrons lost is equal to number of electrons gained.
Adding the two half reactions as follows:
(1) × 5 5 Fe²⁺ → 5 Fe³⁺ + 5e ……………(4)
(3) × 1 MnO₄^– + 5e + 8H⁺ → Mn²⁺ + 4H₂O ………(5)

(4) + (5)
MnO₄^– + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5 Fe³⁺ + 4H₂O ………..(6)
The equation (6) is a balanced equation.

Question 9.
Calculate the percentage composition of the elements present in lead nitrate. How many Kg of 02 can be obtained from 50 kg of 70% pure lead nitrate?
Answer:
Lead nitrate = Pb (NO₃)₂
Molecular mass of lead nitrate = 207 + (14 x 2) + (16 x 6)
= 207 + 28 + 96 = 331 g / mol.
331 g of lead nitrate contains 96 g of oxygen.
∴ 50 x 10³ g of lead nitrate will contain \(\frac {96}{331}\) x 50 x 10³
= 14501.5 g
= 14.501 Kg of oxygen.
100 % pure lead nitrate contains 14.501 Kg of oxygen.
70 % pure lead nitrate will contain = \(\frac {14.501}{100}\) x 70 = 10.15 Kg of oxygen.
.’. 70 % pure lead nitrate will contain 10.15 Kg of oxygen.

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