Students must start practicing the questions from CBSE Sample Papers for Class 11 Physics with Solutions Set 7 are designed as per the revised syllabus.
CBSE Sample Papers for Class 11 Physics Set 7 with Solutions
Time Allowed : 3 hours
Maximum Marks : 70
General Instructions:
- All questions are compulsory. There are 40 questions.
- This Question paper has four sections : Section A, Section B, Section C, Section D
- Section A contains twenty five questions of one mark each, Section B contains five questions of two marks each, Section C contains seven questions of three marks each, Section D contains three question of five marks each.
- There is no overall choice. However, internal choices have been provided in seven questions of one mark, two questions of two marks, two questions of three marks and two questions of five marks weightage. You have to attempt only one of the choices in such questions.
- You may use the following values of physical constants wherever necessary:
c = 3 × 108 m/s,
h = 6.63 × 10-34 Js
e = 1.6 × 10-19 C,
Radius of Earth, Re = 6.4 × 106 m
Universal Gravitational constant. G = 6.67 × 10-11 Nm2kg-2
mass of electron, me = 9.1 × 10-31 kg,
mass of neutron, mn = 1.675 × 10-27 kg
mass of proton, mp = 1.673 × 10-27 kg
Avogadro’s number = 6.023 × 1023 atom per gram
Boltzmann constant = 1.38 × 10-23 JK-1
Section – A
Question numbers 1 to 25 carry 1 mark each.
Question 1.
The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give
(A) 2.75 and 2.74
(B) 2.74 and 2.73
(C) 2.75 and 2.73
(D) 2.74 and 2.74 [1]
Answer:
Option (D) is correct.
Question 2.
A bicyclist comes to a skidding stop in 10m. During this process, the force on the bicycle due to the road is 200N and is directly opposed to the motion. The work done by the cycle on the road is [1]
(A) + 2000J
(B) -200J
(C) zero
(D) -20,000J
Answer:
Option (C) is correct.
Explanation: Just because road does not move at all so the work done by the cycle on the road must be zero.
Question 3.
Conservation of momentum in a collision between particles can be understood from
(A) Conservation of energy.
(B) Newton’s first law only.
(C) Newton’s second law only
(D) Both Newton’s second and third law. [1]
OR
The horizontal range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be
(A) 60 m
(B) 71 m
(C) 100 m
(D) 141 m
Answer:
Option (D) is correct.
OR
Option (C) is correct
Explanation:
R = \(\frac{u^2 \sin 2 \theta}{g}\) ; R ∝ sin 2θ
∴ \(\frac{\mathrm{R}_1}{\mathrm{R}_2}\) = \(\frac{\sin 2 \times 15^{\circ}}{\sin 2 \times 45^{\circ}}\) = \(\frac{\sin 30^{\circ}}{\sin 90^{\circ}}\) = \(\frac{1}{2}\)
R2 = 2R1 = 2 × 50m = 100m
Question 4.
A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall? [1]
(A) Kinetic energy.
(B) Potential energy.
(C) Total mechanical energy.
(D) Total linear momentum.
Answer:
Option (C) is correct
Question 5.
The density of a non-uniform rod of length 1 m is given by ρ(x) = a(1 + bx2)
where a and b are constants and 0 ≤ x ≤ 1.
The centre of mass of the rod will be at
(A) \(\frac{3(2+b)}{4(3+b)}\)
(B) \(\frac{4(2+b)}{3(3+b)}\)
(C) \(\frac{3(3+b)}{4(2+b)}\)
(D) \(\frac{4(3+b)}{3(2+b)}\) [1]
Answer:
Option (A) is correct.
Explanation: Density, ρ(x) = a(1 + bx2), at b = 0, ρ(x)= a= constant centre of mass, at x = 0.5 m
(A) \(\frac{3}{4}\) × \(\frac{3}{4}\) = \(\frac{1}{2}\) = 0.5
(B) \(\frac{4}{3}\) × \(\frac{2}{3}\) ≠ 0.5
(C) \(\frac{3}{4}\) × \(\frac{3}{2}\) ≠ 0.5
(D) \(\frac{4}{3}\) × \(\frac{3}{2}\) ≠ 0.5
Question 6.
Both earth and moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon
(A) will be elliptical
(B) will not be strictly elliptical because the total gravitational force on it is not central.
(C) is not elliptical but will necessarily be a closed curve.
(D) deviates considerably from being elliptical due to influence of planets other than earth. [1]
OR
During inelastic collision between two bodies, which of the following quantities always remain conserved ?
(A) Total kinetic energy.
(B) Total mechanical energy.
(C) Total linear momentum.
(D) Speed of each body.
Answer:
Option (B) is correct
OR
Option (C) is correct
Question 7.
A spring is stretched by applying a load to its free end. The strain produced in the spring is :
(A) volumetric.
(B) shear.
(C) longitudinal and shear.
(D) longitudinal. [1]
OR
An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2,5 cm and 3.75 cm. The ratio of the velocities in two pipes is
(A) 9 : 4
(B) 3 : 2
(C) \(\sqrt{3}\) : \(\sqrt{2}\)
(D) \(\sqrt{2}\) : \(\sqrt{3}\)
Answer:
Option (C) is correct
OR
Option (A) is correct
Explanation:
or \(\frac{v_1}{v_2}\) = \(\frac{a_1}{a_2}\)
\(\left(\frac{d_2}{d_1}\right)^2\) = \(\left(\frac{3.75}{2.5}\right)^2\) = \(\frac{9}{4}\)
Question 8.
Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is
(A) 2γ-1
(B) \(\left(\frac{1}{2}\right)^{\gamma-1}\)
(C) \(\left(\frac{1}{1-\gamma}\right)^2\)
(D) \(\left(\frac{1}{\gamma-1}\right)^2\) [1]
Answer:
Option (A) is correct
Explanation: Gas in container A is compressed isothermally.
P1V1 = P2V2
P0(2V0) = P2(V0)
⇒ P0(2V0) = P2(V0)
Gas in container B is compressed adiabatically
Hence, ratio of final pressure
= \(\frac{\left(P_2\right)_B}{\left(P_2\right)_A}\) = \(\frac{P_o(2)^\gamma}{2 P_o}\) = 2γ-1
Question 9.
Volume versus temperature graphs for a given mass of an ideal gas at two different values of constant pressure. [1]
What can be inferred about relation between P1 & P2 ?
(A) P1 > P2
(B) P1 = P2
(C) P1 < P2
(D) data is insufficient
Answer:
Option (A) is correct
Explanation:
V ∝ T
\(\frac{\mathrm{V}}{\mathrm{T}}\) = constant = \(\frac{1}{\mathrm{P}}\)
In graph, slope at P2 is more than slope at, P1,
∴ P1 > P2
Question 10.
The displacement of a particle varies with according to the relation y = a sin ωt + b cos ωt.
(A) The motion is oscillatory but not S.H.M.
(B) The motion is S.H.M. with amplitude a + b.
(C) The motion is S.H.M. with amplitude a2 + b2.
(D) The motion is S.H.M. with amplitude \(\sqrt{\left(a^2+b^2\right)}\) [1]
Answer:
Option (D) is correct.
Explanation: Let
a = Acos ϕ …(i)
b = A sin ϕ …(ii)
Squaring and adding (i) & (ii)
a2 + b2 = A2
y = a sin ωt + b cos ωt …(iii)
‘y’ can be written as
y = A cos ϕ sin ωt + A sin ϕ cos ϕt
= A sin(ωt + ϕ)
This is an equation of SHM with amplitude
A = \(\sqrt{a^2+b^2}\)
Commonly Made Error
Most of the students do not able to find correct answer.
Answering Tip
Students should learn about different types of motion, like as S.H.M. and periodic Motion.
Question 11.
What does the speedometer records—the average speed or the instantaneous speed ? [1]
Answer:
It records (or measures) the instantaneous speed at any instant of time.
Question 12.
Is human body a rigid body ? [1]
Answer:
It is deformable. Different parts of the body show some relative displacement so human body is not strictly a rigid body.
Question 13.
Give a relation for impulse in scalar form.
OR
Is force of friction independent of path ? [1]
Answer:
I = F × f = p2 – p1
where notations have their usual meanings.
OR
No, the frictional force is a non-conservative force, i.e., the work done by it depends upon the actual path followed by a body.
Question 14.
State the Principle of conservation of total mechanical energy. [1]
Answer:
The total mechanical energy of a system remains same, if the working forces on it are of conservative nature.
Question 15.
What is the rotational analogue of force ? [1]
OR
What is moment of inertia of a solid sphere about its diameter ?
Answer:
Rotational analogue of force is torque.
OR
I = \(\frac{2}{5}\)MR2, where M is the mass and R is radius of the solid sphere.
Question 16.
What is the celestial sphere ? [1]
Answer:
The huge hemispherical surface of sky in which stars and planets etc. appear to be struck is called celestial sphere.
Question 17.
Name two classes of solids. [1]
OR
How does the viscosity of gases depend upon temperature ?
Answer:
Solids are of two types :
- Crystalline solids
- Glassy or amorphous solids.
OR
For gases η × T1/2.
Question 18.
What is pyrometer ? [1]
Answer:
Pyrometer is a radiation thermometer. It can be used to measure the temperature of a furnace or sun, etc.
Question 19.
Why motion of earth about its axis is periodic but not S.H.M. ? [1]
Answer:
Earth is simply rotating about its axis. There is no to and fro motion but time of rotation of earth is fixed. Thus, motion of earth about its axis is periodic but not simple harmonic.
Commonly Made Error
Confusion between periodic motion’ and ‘Simple Harmonic Motion’.
Answering Tip
Students should be able to distinguish between periodic, oscillatory and simple harmonic motion.
Question 20.
What is an epoch ? Name the unit in which it is measured. [1]
Answer:
The initial phase of the oscillating particle is called epoch. It is expressed in radians.
Read the following passage and answer any four of five questions:
Galileo’s Isochronous Pendulum: Galileo observed a lamp swinging from ceiling at Pisa cathedral ceiling. He was the first scientist to observe how long it took any object suspended from a rope or chain (a pendulum) to swing back and forth. There were no wrist watches at that time, so Galileo used his own pulse as a time measurement. Galileo observed that no matter how big the swings were, as in when the lamp was first swung, to how small the swings were as the lamp returned to a standstill, the time it took for each swing to complete was exactly the same. So, he concluded that the oscillations are isochronous. Anyhow, it was proved by advanced experimentation that the isochronism of simple pendulum is correct within 1% under the 30° of amplitude. It is confirmed that the period changes less than 1 % until the amplitude is 30°. On the other hand, the period increased by more than 10% when the amplitude became larger than 80°. At the end of his life he devised a scheme for using a pendulum to regulate a mechanical clock. However, the first reliable pendulum clock was only demonstrated by Huygens 15 years after Galileo’s death.
Question 21.
Who first observed the time period of a pendulum?
(A) Aristotle
(B) Pythagoras
(C) Galileo
(D) Democritus [1]
Answer:
Option (C) is correct
Explanation: Galileo observed a lamp swinging from ceiling at Pisa cathedral ceiling. He was the first scientist to observe how long it took any object suspended from a rope or chain (a pendulum) to swing back and forth.
Question 22.
How the time period of a pendulum was measured in the beginning of 16th century?
(A) By a stop watch
(B) By using pulse beats
(C) By using vibration of a stringed instrument
(D) By using a sand clock [1]
Answer:
Option (B) is correct.
Explanation: Since no watches were available at that time, Galileo used his own pulse as a time measurement.
Question 23.
Galileo observed, that the time period of the pendulum:
(A) Does not vary with the size of swing
(B) Increases when the size of swing increases
(C) Decreases when the size of swing increases
(D) Unpredictable [1]
Answer:
Option (A) is correct
Explanation: Galileo observed that no matter how big the swings were, as in when the lamp was first swung, to how small the swings were as the lamp returned to a standstill, the time it took for each swing to complete was exactly the same.
Question 24.
First reliable pendulum clock was only demonstrated by:
(A) Vincenzo Viviani
(B) Huygens
(C) Maffeo Barberini
(D) Newton [1]
Answer:
Option (B) is correct
Explanation: The first reliable pendulum clock was only demonstrated by Huygens 15 years after Galileo’s death.
Question 25.
Isochronism of simple pendulum is within acceptable limit as long as the amplitude is under
(A) 90°
(B) 60°
(C) 30°
(D) 80° [1]
Answer:
Option (C) is correct.
Explanation: Anyhow, it was proved by advanced experimentation that the isochronism of simple pendulum is correct within 1% under the 30° of amplitude. It is confirmed that the period changes less than 1% until the amplitude is 30°. On the other hand, the period increased by more than 10% when the amplitude became larger than 80°.
Section – B
Question numbers 26 to 30 carry 2 marks each.
Question 26.
How is Newton’s second law of motion consistent with the Newton’s first law ? [2]
Answer:
From Newton’s, force F = m × a. If F = 0, then a = 0, i.e., of motion no external force means no acceleration. It implies no change in state of rest or of uniform motion in a straight line which is nothing but first law of motion.
Question 27.
Why Newton’s second law of motion is not applicable to the motion of a rocket ? [2]
Answer:
Newton’s second law of motion, i.e., \(\vec{F}\) = m × \(\vec{a}\) is applicable only if the mass (m) of the body remains constant. In case of the rocket, the mass continuously decreases and hence \(\vec{F}\) = m × \(\vec{a}\) is not applicable.
Commonly Made Error
Most of the students do not able to write correct answer of this question.
Answering Tip
Students should learn about Newton’s second law of motion and its applications.
Question 28.
Prove that the maximum horizontal range is four times the maximum height attained by the projectile, when fired at an inclination so as to have maximum horizontal range. [2]
OR
Find the direction for an umbrella when rain falls vertically with speed 20 ms-1 and wind blows from east to west with a speed of 15 ms-1.
Answer:
The horizontal range is maximum for θ = 45° and it is given by
Rmax = \(\frac{u^2}{g}\) … (i)
The maximum height attained,
H = \(\frac{u^2 \sin ^2 \theta}{2 g}\)
Therefore, for θ = 45°
Hmax = \(\frac{u^2 \sin ^2 45^{\circ}}{2 g}\) = \(\frac{u^2}{4 g}\) … (ii)
From the equation (i) and (ii),
Rmax = 4Hmax
Commonly Made Error
Many student do not able to solve this problem.
Answering Tip
Students should learn tormula of maximum height and maximum range.
OR
From the figure
tan θ = \(\frac{v_w}{v_r}\) = \(\frac{15}{20}\) = \(\frac{3}{4}\)
tan θ = 0.75
θ ≈ 37° east from vertical
Question 29.
What is an ideal fluid ? [2]
Answer:
An ideal fluid should be non-viscous and incompressible.
(A) It should be non-viscous, i.e., dragging force or other opposing force should be zero.
(B) It should have steady flow.
(C) It should be incompressible.
(D) The flow of such fluid should be stream line.
Question 30.
Milk is poured into a cup of tea and is mixed with a spoon. What happens to the work done to mix the tea and milk? [2]
OR
What is calorie ? Define one calorie.
Answer:
The work done in mixing the tea and milk gets converted into heat energy. In principle, the temperature of the tea and milk will rise.
OR
Calories is C.G.S. unit of heat. One calorie is the amount of heat required to raise the temperature of 1 g of water through 1°C.
Section – C
Question numbers 31 to 37 carry 3 marks each.
Question 31.
Draw
(A) acceleration – time
(B) velocity – time
(C) position – time graphs representing motion of an object under free fall. Neglect air resistance. [3]
Answer:
Question 32.
A solid sphere of mass m and radius r is rolling on a horizontal surface.
What fraction of total energy of the sphere is :
(i) Kinetic energy of rotation ?
(ii) Kinetic energy of translation ? [3]
Answer:
Mass of sphere = m
Radius of sphere = r
Moment of inertia = (2,/5)(mr2)
Total energy = KR + KT
Total energy = KTOTAL = (1/2)Iω2 + (1/2)mυ2
= (1/2)(2/5)(mυ2)(υ2/r2) + (1/2) mυ2
= (1/2) (7/5)(mυ2)
The ratio KR/KTOTAL = \(\frac{(1 / 2)(2 / 5)(m v)^2}{(1 / 2)(7 / 5)(m v)^2}\) = \(\frac{2}{7}\)
The ratio Kr/KTOTAL = \(\frac{(1 / 2)(m v)^2}{(1 / 2)(7 / 5)(m v)^2}\) = \(\frac{5}{7}\)
Question 33.
What is coefficient of restitution ? Give its values for
(i) elastic collision,
(ii) inelastic collision and
(iii) perfectly inelastic collisions. [3]
OR
An engine draws a train up an incline of 1 in 100 at the rate of 36 kmh-1. If the resistance due to friction is 5 kg wt. per ton, find out the power of the engine. Mass of train and engine is 100 metric ton.
Answer:
The ratio of relative velocity of separation after collision to the relative velocity of approach before collision.
Coefficient of restitution,
i.e., e = \(\frac{v_2-v_1}{u_1-u_2}\)
where u1 and u2 are initial velocities of the two colliding bodies and υ1, υ2 are their final velocities after collision.
(i) For elastic collision, velocity of separation is equal to the velocity of approach.
∴ e = 1
(ii) For inelastic collision, velocity of separation is not zero but always less than the velocity of approach.
∴ 0 < e < 1
(iii) For perfectly inelastic collision, the colliding bodies stick to each other and move with same velocity.
e = 0
OR
Given : m = 100 metric ton
= 100 × 1000 kg
Total force of friction,
f1 = 100 × 5 = 500 kg wt
= 500 × 9.8 N = 4900 N
sin θ = \(\frac{1}{10}\)
Suppose, f2 = Downward force on the train = component of its weight acting in downward direction parallel to the inclined plane = mg sin θ
= 100 × 1000 × 9.8 × \(\frac{1}{100}\)
= 9800 N
When F be the total force against which engine has to work, then
F = f1 + f2
= 4900 + 9800 = 14700 N
υ = velocity of train
= 36 km h-1
= 36 × \(\frac{5}{18}\)ms-1
= 10 ms-1
∴ Power of the engine, is given by
P = F × υ = 14700 × 10
= 147000 watt
= 147 kW
Question 34.
We cannot move even a finger without disturbing all the stars. Explain. [3]
Answer:
According to Newton’s law of gravitation, everybody in this universe attracts every other body with a force which is inversely proportional to the square of the distance between them. When we move our finger, the distance of the objects with respect to finger changes, disturbing the entire universe including stars.
Question 35.
State and prove Stefan-Boltzmann law. [3]
OR
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear ?
Answer:
It states that the power (E) emitted per unit area by a body at absolute temperature, T is given by
E = \(\varepsilon \sigma T^4\)
where
E = Energy being emitted per second per unit area of surface
ε = Emissivity or emissive power of that body (which is equal to 1 for black body)
σ = Stefan’s constant
= 5.67 × 10-8 Jm-2 KT-4 s-1.
If temperature of the surrounding is to be taken into consideration then the net rate of loss of energy as heat per unit area per second is given by
E = εσ(T4 – \(\mathrm{T}_0^4\))
If the total area or surface is A, then the total energy radiated in time,
E = Aεσt(T4 – \(\mathrm{T}_0{ }^4\))
OR
Given : The maximum force which the bigger piston can bear,
F = 3000 kgf
= 3000 × 9.8 N
Area of cross-section of piston,
A = 425 cm2
= 425 × 10-4 m2.
When P = maximum pressure on the bigger piston
Then P = \(\frac{F}{A}\) = \(\frac{3000 \times 9.8}{425 \times 10^{-4}}\)
= 6.92 × 105 Pa
As the liquid transmits pressure equally in all directions, hence the maximum pressure the smaller piston can bear is 6.92 × 105 Pa
Question 36.
Calculate
(i) r.m.s velocity, and
(ii) mean kinetic energy of one gram molecule of hydrogen at S.T.P.
(Given, density of hydrogen at S.T.P. is 0.09 kg/m3.) [3]
Answer:
Here, ρ = 0.09 kg/m3
At S.T.P. ρ = 1.01 × 105Pa
(i) According to kinetic theory of gases
P = \(\frac{1}{3} \rho c^2\)
or C = \(\sqrt{\frac{3 P}{\rho}}\)
= \(\sqrt{\frac{3 \times 1.01 \times 10^5}{0.09}}\)
= 1837.5 m/s
(ii) Volume occupied by one mole of hydrogen at S.T.P.
= 22.4 litres = 22.4 × 10-3 m3
Mass of hydrogen,
M = Volume × Density
= 22.4 × 10-3 × 0.09 kg
= 2.016 × 10-3 kg
mean K.E. of one gram molecule of hydrogen at S.T.P.
= \(\frac{1}{2} \mathrm{MV}_{\mathrm{rms}}^2\)
= \(\frac{1}{2}\) × 2.016 × 10-3 × (1837.5)2
= 3.4 × 103J
Question 37.
Give a relation for velocity of a wave. [3]
Answer:
Section – D
Question numbers 38 to 40 carry 5 marks each.
Question 38.
A cricket bowler releases the ball in two different ways
(A) giving it only horizontal velocity, and
(B) giving it horizontal velocity and a small downward velocity.
The speed υs at the time of release is the same. Both are released at a height H from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance. [5]
Answer:
For (A) = \(\frac{1}{2} v_z^2\) = gH or υz = \(\sqrt{2 g H}\)
Speed at ground = \(\sqrt{v_s^2+v_z^2}\) = \(\sqrt{v_s^2+2 g H}\)
For(B) also [\(\frac{1}{2} m v_{\mathrm{s}}^2\) + mgH] is the total energy of the ball when it hits the ground.
So the speed would be the same for both (A) and (B).
Question 39.
Derive the relation between torque and moment of inertia. [5]
OR
Find the components along the x, y, z-axes of the angular momentum of a particle, whose position vector is \(\vec{r}\) with
components x, y, z and momentum is \(\vec{p}\) with components px, py, pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
Answer:
Relation between torque and Moment of inertia:
Consider a rigid body rotating about a given axis with a uniform angular acceleration α, under the action of a torque.
Let the body consist of particles of masses m1, m2, m3, …, mn, at perpendicular distance r1, r2, r3 …, rn, respectively from the axis of rotation. (as shown in figure)
As the body is rigid, angular acceleration α of all the particles of the body is the same. However, linear accelerations of the particles depend on their distance from the axis. If a1, a2, a3, …, an are the respective linear accelerations of the particles, then,
a1 = r1α, a2 = r2α, a3 = … 1
Force on particle of mass m is
f1 = m1a1 = m1r1α
Moment of this force about the axis of rotation
f1 × r1 = (m1r1α) × r1 = m1\(r_1^2 \alpha\)
Similarly, moment of forces on other particles about the axis of rotation are \(m_2 r_2^2 \alpha, m_3 r_3^2 \alpha\) ……
\(m_n r_n{ }^2 \alpha\).
∴ Torque acting on the body,
where, \(m_1 r_1^2\) = moment of inertia of the body about the given axis of rotation.
If α = 1, \(\tau\) = I × 1 or I = \(\tau\)
\(\vec{\tau}\) = \(I \vec{\alpha}\)
OR
Angular momentum ‘l of a particle-
If given particle moves only in x-y plane,
Then z = 0, pz = 0,
hence, lx = 0, ly = 0, lz = (xpy – ypx)\(\hat{k}\)
(which is only z-component of l)
Therefore, particle moves only in x-y plane, the angular momentum has only a z-component.
Question 40.
A steel rod of length 2l, cross sectional area A and mass M is set rotating in a horizontal plane about an axis passing through the centre. If Y is the Young’s modulus for steel, find the extension in the length of the rod. (Assume the rod is uniform.) [5]
OR
Derive an expression to find the specific heat with the help of calorimeter.
Answer:
Consider an element at r of width dr. Let T(r) and T (r + dr) be the tensions at the two edges, respectively.
Net centrifugal force in element = ω2rdm [ω = angular velocity of rod]
= ω2r µ dr [∴ µ = mass/length]
-T(r + dr) + T(r) = µω2rdr
–\(\frac{d \mathrm{~T}}{d r}\)dr = µω2rdr
-dT = µω2rdr
∵ Tènsion and centrifugal force are opposite.
Let the increase in length of the clement dr be dδ therefore, Young’s Modulus,
OR
Let
W = Water equivalent of calorimeter.
T1 = Initial temperature of water and calorimeter
m1 = Mass of water.
m2 = Mass of substance
C = Specific heat of the substance
T2 = Temperature of the substance
Rise in temperature of water and calorimeter
= (T – T1)
Fall in temperature of substance = (T2 – T)
Heat gained by water and calorimeter
= (m1 + W) (T – T1) ..(i)
Heat lost by the substance
= Cm2(T2 – T) …(ii)
It we assume that there is no stray loss of heat Ihen
Heat lost = Heat gained
Cm2. (T2 – T) = (m1 + W)(T – T1)
C = \(\frac{\left(m_1+\mathrm{W}\right)\left(\mathrm{T}-\mathrm{T}_1\right)}{m_2\left(\mathrm{~T}_2-\mathrm{T}\right)}\)