Students must start practicing the questions from CBSE Sample Papers for Class 11 Physics with Solutions Set 6 are designed as per the revised syllabus.
CBSE Sample Papers for Class 11 Physics Set 6 with Solutions
Section – A
Question numbers 1 to 25 carry 1 mark each.
Question 1.
The mean length of an object is 5 cm. Which of the following measurements is most accurate?
(A) 4.9 cm
(B) 4.805 cm
(C) 5.25 cm
(D) 5.4 cm [1]
Answer:
Option (A) is correct.
Question 2.
A vehicle travels half the distance L with speed υ1 and the other half with speed υ2 then its average speed is
(A) \(\frac{v_1+v_2}{2}\)
(B) \(\frac{2 v_1+v_2}{v_1+v_2}\)
(C) \(\frac{2 v_1 v_2}{v_1+v_2}\)
(D) \(\frac{\left(v_1+v_2\right)}{v_1 v_2}\) [1]
Answer:
Option (C) is correct.
Explanation: Time taken to travel first half distance, t1 = \(\frac{\frac{\mathrm{L}}{2}}{v_1}\) = \(\frac{\mathrm{L}}{2 v_1}\)
Time taken to travel second half distance, t2 = \(\frac{\frac{\mathrm{L}}{2}}{v_2}\) = \(\frac{\mathrm{L}}{2 v_2}\)
Question 3.
A body of mass 2 kg travels according to the law x(t) = pt + qt2 + rt3 where p = 3 ms-1, q = 4 ms -2, and r = 5 m s-3. The force acting on the body at t = 2 seconds is : [1]
(A) 136 N
(B) 134 N
(C) 158 N
(D) 68 N
OR
Consider the quantities-pressure, power, energy, impulse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are :
(A) Impulse, pressure and area
(B) Impulse and area
(C) Area and gravitational potential
(D) Impulse and pressure
Answer:
Option (A) is correct.
Explanation: Here, x(t) = pt + qt2 + rt3 where, p = 3 m/s, q = 4 m/s2 and r = 5 m/s3 m = 2 kg
Velocity, v = \(\frac{d x}{d t}\) = \(\frac{d}{d t}\)(pt + qt2 + rt3)
= p + 2qt + 3rt2
Acceleration, a = \(\frac{d v}{d t}\) = 2q + 6rt
\(\left(\frac{d v}{d t}\right)_{t=2}\) = a = 2(4 m/s2) + 6(5 m/s3) × (2 s)
\(\left(\frac{d v}{d t}\right)_{t=2}\) = 8 m/s2 + 60 m/s2 = 68 m/s2
∴ Force acting on the body of mass 2 kg is F = ma = (2 kg)(68 m/s2)
= 136 N
OR
Option (B) is correct
Explanation: Impulse is a vector quantity. Area of a surface is a vector which is along normal to the surface in the outward direction.
Question 4.
The potential energy function for a particle executing linear SHM is given by V(x) = \(\frac{1}{2} k x^2\) where k is the force
constant of the oscillator (Fig.). The graph of V (x) versus x is shown in the figure. A particle of total energy E turns back when it reaches x = ±xm, then which of the following is correct ? [1]
(A) V = 0, k = E
(B) V = E, k = 0
(C) V < E, k = 0
(D) V = 0, k < E.
Answer:
Option (C) is correct.
Explanation: At any instant of time, the total energy of an oscillator is the sum of kinetic energy and potential energy.
Total energy E = U + K
E = \(\frac{1}{2}\)mυ2 + \(\frac{1}{2}\)kx2
Question 5.
A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed ω. A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is : [1]
(A) 2ω
(B) ω
(C) \(\frac{\omega}{2}\)
(D) 0
Answer:
Option (A) is correct.
Explanation: As no external torque acts on the system, angular momentum should be constant
i.e. Iω = constant … (i)
from equation (i),
I1ω1 = I2ω2
[ω1 & ω2 are angular velocities before and after jumping]
We have, (Moment of inertia)
I = mr2
m1 = 2M
m2 = M
r1 = r1 = R
ω = Angular velocity
I1ω1 = I2ω2
2MR2ω1 = MR2ω2
ω2 = 2ω1
Question 6.
In our solar system, the inter-planetary region has chunks of matter (much smaller in size compared to planets) called asteroids. They
(A) will not move around the sun since they have very small masses compared to the sun.
(B) will move in an irregular way because of their small masses and will drift away into outer space.
(C) will move around the sun in closed orbits but not obey Kepler’s laws.
(D) will move in orbits like planets and obey Kepler’s laws. [1]
OR
A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 revolutions per minute, its kinetic energy would be
(A) 250π2
(B) 100 π2
(C) 5 π2
(D) 0
Answer:
Option (D) is correct.
OR
Option (A) is correct.
Explanation:
Given mass, in = 5 kg,
Radius, R = 1 m
v = 300rpm = \(\frac{300}{60}\)rps = 5 rps
The angular speed,
ω = 2πυ = 2π × 5 = 10× rad/s
The linear speed is
y = ωR = (10π)(rad/s)(1 m)
= 10π m/s
K.E. = K = \(\frac{1}{2}\)mv2
= 250π2 J
Question 7.
A rigid bar of mass M is supported symmetrically by three wires each of length 1. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to [1]
(A) Ycopper/Yiron
(B) \(\sqrt{\frac{Y_{\text {iron }}}{Y_{\text {cupper }}}}\)
(C) \(\frac{Y_{\text {iron }}^2}{Y_{\text {copper }}^2}\)
(D) \(\frac{Y_{\text {iron }}}{Y_{\text {copper }}}\)
OR
The angle of contact at the interface of water-glass is 0°, Ethyl alcohol- glass is 0°, Mercury-glass is 140° and methyl iodide-glass is 30°. A glass capillary is put in a trough containing one of these four liquids. It is observed that the meniscus is convex. The liquid in the trough is
(A) water
(B) ethyl alcohol
(C) mercury
(D) methyl iodide.
Answer:
Option (B) is correct.
Explanation:
Let T be tension in each wire. As the bar is supported symmetrically by three wires,
∴ Extension in each wire is same.
As Y = \(\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}\)
If D is the diameter of the wire, then
As per the conditions ot the problem, tension is same for each wire.
∴ Y ∝ \(\frac{1}{D^2}\) or D ∝ \(\sqrt{\frac{1}{Y}}\)
F (tension) since Length L, and extension Δ L, are same
∴ \(\frac{D_{\text {copper }}}{D_{\text {iron }}}\) = \(\sqrt{\frac{Y_{\text {iron }}}{Y_{\text {copper }}}}\)
OR
Option (C) is correct.
Question 8.
Three copper blocks of masses M1, M2 and M3 kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at T1, T2, T3 (T1 > T2 > T3). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (S is specific heat of copper) [1]
Answer:
Option (B) is correct.
Question 9.
A particle is acted simultaneously by mutually perpendicular simple harmonic motions x = acos ωt and y = asin ωt. The trajectory of motion of the particle will be [1]
(A) an ellipse.
(B) a parabola.
(C) a circle.
(D) a straight line.
Answer:
Option (C) is correct.
Explanation:
x = acos ωt ……(i)
y = a sin ωt ..(ii)
squaring and adding (i) & (ii), we get
x2 + y2 = a2 cos2 ωt + a2 sin2 ωt
= a2
The trajectory of motion will be a circle.
Question 10.
The displacement of a particle varies with time 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 \(\sqrt{a^2+b^2}\) [1]
Answer:
Option (D) is correct
Explanation:
Let a = A cos ϕ …. (i)
b = A sin ϕ …. (ii)
Squaring and adding (i) & (ii)
a2 + b2 = A2
y = a sin ωt + b cosωt
‘y’ can be written as
y = A cos ϕ sinωt + A sin ϕ cos ωt
= A sin(ωt + ϕ)
This is an equation of SHM with amplitude A = \(\sqrt{a^2+b^2}\)
Question 11.
Do A.U. (Astronomical Unit) and A represent the same magnitude of distance ? [1]
OR
Write the dimensional formula of wave-length and frequency of a wave ?
Answer:
No.
1 A.U. = 1.496 × 1011 m. (mean distance of earth from the sun)
1 A = 10-10 m. (atomic size)
OR
Dimensional formula of wavelength
= [M0L1T0]
Dimensional formula of frequency
= [M0L0T-1]
Question 12.
What is de-acceleration or retardation ? [1]
Answer:
Negative acceleration’s called retardation or de acceleration, i.e., when velocity of an object is steadily decreasing then the object is said to have retardation.
Question 13.
Why rockets are given conical shapes ?
Answer:
The rockets are given conical shapes so as to reduce atmospheric friction.
Question 14.
Does the work done in raising a load on to a platform depend upon how fast is it raised ? [1]
Answer:
No, because the work done is independent of time.
Question 15.
What is rotatory motion ? [1]
OR
What is couple ?
Answer:
When a body rotates about a fixed axis, the motion of the body is said to be rotatory motion.
OR
Two equal and opposite forces with some finite perpendicular distance between them constitute a couple of forces.
Question 16.
Where does a body weigh more at the surface of earth or in a mine ? [1]
Answer:
At the surface of earth, a body weighs more.
Question 17.
Which is more elastic : Water or air ? Give reason.
OR
When wax is rubbed on cloth, the cloth becomes water proof. Why ? [1]
Answer:
Elasticity is reciprocal of compressibility. Air is more compressive than water. Hence water is more elastic than air.
OR
The capillaries formed in threads disappear when wax is rubbed on cloth.
Question 18.
Does specific heat of a gas possess a unique value ? [1]
Answer:
No. It is of two types for gases-specific heat at constant pressure CP and specific heat at constant volume CV.
Question 19.
How much volume does one mole of a gas occupy at N.T.P. ? [1]
Answer:
22.4 litre.
Question 20.
What is the basic condition for the motion of a particle to be S.H.M. ? [1]
Answer:
The necessary and sufficient condition for motion to be simple harmonic is that the restoring force must be linear, i.e., F = -ky
where k is known as force constant. Here negative sign indicate that the restoring force (F) is always directed towards the mean position.
Read the following text and answer any 4 of the following question on the basis of the same.
Sagging of a bridge
A bridge is designed such that it can withstand the load of the flowing traffic, the force of winds and its own weight.
Let us consider the case of a beam loaded at the centre and supported near its ends as shown in Figure.
A beam of length I, breadth b, and depth d when loaded at the centre by a load W sags by an amount given by
From the equation, we see that to reduce the bending for a given load, one should use a material with a large Young’s modulus Y. For a given material, increasing the depth d rather than the breadth b is more effective in reducing the bending, since δ is proportional to d-3 and to b-1 (of course the length l of the span should be as small as possible).
Amongst bridge materials steel has the highest and most favourable strength qualities, and it is therefore suitable for the most daring bridges with the longest spans. Normal building steel has compressive and tensile strengths of 370 N/sq. mm, about ten times the compressive strength of a medium concrete and a hundred times its tensile strength. A special merit of steel is its ductility due to which it deforms considerably before it breaks, because it begins to yield above a certain stress level.
Question 21.
To reduce bending of a beam
(A) For a given length and material, depth should be greater than breadth.
(B) For a given length and material, breadth should be greater than depth.
(C) For a given length and material, depth should be equal to breadth.
(D) Breadth and depth has no effect. [1]
Answer:
Option (A) is correct.
Explanation: For a beam of length l, breadth b, and depth d when loaded at the centre by a load W sags by an amount given by
d = \(\frac{W l^3}{4 b d^3 Y}\)
From the equation, we see that to reduce the bending for a given load, one should increase the depth d rather than the breadth b, since d is proportional to d-3 and to b-1.
Question 22.
Compressive strength of normal building steel is about …… times of the compressive strength of medium concrete. [1]
(A) 2
(B) 10
(C) 100
(D) 1000
Answer:
Option (B) is correct
Explanation: Amongst bridge materials, steel has the highest and most strength qualities, and it is therefore suitable for the most daring bridges with the longest spans. Normal building steel has compressive and tensile strengths about ten times the compressive strength of a medium concrete and a hundred times of its tensile strength.
Question 23.
What is the special merit of steel over concrete is its
(A) Malleability
(B) Brittleness
(C) Conductivity
(D) Ductility [1]
Answer:
Option (D) is correct.
Explanation: A special merit of steel is its ductility due to which it deforms considerably before it breaks, because it begins to yield above a certain stress level.
Question 24.
A bar of length l, breadth b, and depth d, supported at two ends when loaded at the centre by a load W sags by an amount given by
(A) δ = \(\frac{W l^3}{4 b d^3 \mathrm{Y}}\)
(B) δ = \(\frac{\mathrm{Wl}}{4 b d^3 \mathrm{Y}}\)
(C) δ = \(\frac{W l^3}{4 b^3 d Y}\)
(D) δ = \(\frac{W Y l^3}{4 b d^3}\) [1]
Answer:
Option (A) is correct.
Explanation: Let us consider the case of a beam loaded at the centre and supported near its ends. A beam of length l, breadth b, and depth d when loaded at the centre by a load W sags by an amount given by
δ = \(\frac{W l^3}{4 b d^3 \mathrm{Y}}\)
Question 25.
Why ductility is the special merit of steel?
(A) Ductility allows structures to bend and deform to some extent without rupturing.
(B) Ductility offers the structure a high rigidity
(C) Ductility prevents the structure to sag while overloaded
(D) Ductility offers less corrosion [1]
Answer:
Option (A) is correct.
Explanation: In overload situation, to prevent sudden rupture ductility allows the structure to bend and deform to some extent without rupturing. High ductility is critical in applications such as metal cables and structural beams.
Section – B
Question numbers 26 to 30 carry 2 marks each.
Question 26.
What are the different methods to derive the equation for uniform acceleration motion?
OR
What is pseudo force ? [2]
Answer:
- From definitions of average velocity and acceleration.
- From velocity time graphs.
- By using calculus.
OR
The fictitious or imaginary force used to balance an actual or material force is called pseudo force,
e.g., centrifugal force, F = \(-\frac{m v^2}{r}\) is a pseudo force. Pseudo forces do not have material sources, they are simply produced because of acceleration or rotation of the frame itself.
Question 27.
When the horse suddenly stops, the rider falls in forward direction. Why ? Explain.
OR
How does Newton’s second law of motion give the concept of inertial mass ? [2]
Answer:
When the horse suddenly stops, the rider falls in the forward direction due to the inertia of motion.
Explanation : The lower portion of the rider comes to rest along with the horse while the upper portion of the rider still continues to move forward. Hence, he falls forward.
OR
Mass is given by the equation a = \(\frac{F}{M}\), i.e., M = \(\frac{F}{a}\) is called inertial mass. Clearly, higher the mass means lesser the acceleration when applied force is constant. Thus mass of a body resists the acceleration, i.e., rate of change in velocity coming up due to external force or in other words it is the measure of inertia of a body.
Question 28.
A thief jumps from the upper storey of a house with a load on his back. What is the force of the load on his back when the thief is in air ? [2]
Answer:
When the thief is in air, he is in the state of free fall, and hence in the state of weightlessness. So, the force of the load on his back is zero.
Question 29.
A simple pendulum performs S.H.M. about x = 0 with an amplitude a and time period T. What is the speed of the pendulum at x = a/2 ? [2]
Answer:
Speed of pendulum, υ = ω\(\sqrt{a^2-x^2}\)
υ = \(\frac{2 \pi}{\mathrm{T}} \sqrt{a^2-\frac{a^2}{4}}\)
= \(\frac{\sqrt{3} a \pi}{T}\)
Question 30.
Two exactly identical pendulums are oscillating with amplitude 2 cm and 6 cm. Calculate the ratio of their energies of oscillations. [2]
Answer:
Total energy of the bob of simple pendulum is given by, E = mω2r2, i.e., E ∝ r2,
So, \(\frac{E_1}{E_2}\) = \(\frac{r_1^2}{r_2^2}\) = \(\left(\frac{2}{6}\right)^2\) = \(\frac{1}{9}\)
Section – C
Question numbers 31 to 37 carry 3 marks each.
Question 31.
Give position-time graphs for one object moving with (a) negative velocity, (b) moving with positive velocity and (c) at rest. [3]
Answer:
Question 32.
A force of 100 N’ gives a mass m1, an acceleration of 10 ms-2 and of 20 ms-2 to a mass m2. What acceleration must be given to it if both the masses are tied together ? [3]
Answer:
Suppose, a = acceleration produced if m1 and m2 are tied together.
F = 100 N.
Let a1 and a2 be the acceleration produced in m1 and m2 respectively.
∴ a1 = 10 ms-2, a2 = 20 ms-2 (given)
Question 33.
A light body and a heavy body have same linear momentum. Which one has greater K.E. ? [3]
Answer:
Here, p1 = p2, m1υ1 = m2υ2
If m1 < m2, E2 < E1
⇒ E1 > E2
i.e., lighter body has more K.E.
Question 34.
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]
OR
What would happen if gravity suddenly disappear ?
Answer:
Mass of sphere = m, radius = r
Moment of Inertia = \(\frac{2}{5} m r^2\)
Total energy = KR + Kr
(i) Fraction of K.E. of rotation = \(=\frac{\mathrm{K} \cdot \mathrm{E}_{\cdot \mathrm{R}}}{\mathrm{K} \cdot \mathrm{E}_{\cdot \mathrm{Total}}}\)
= \(\frac{\frac{1}{2}\left(\frac{2}{5}\right) m v^2}{\frac{1}{2}\left(\frac{7}{5}\right) m v^2}\) = \(\frac{2}{7}\)
(ii) Fraction of K.E. of translation
= \(\frac{\mathrm{K}_{\mathrm{T}}}{\mathrm{K}_{\text {Total }}}\) = \(\frac{\frac{1}{2} m v^2}{\frac{1}{2}\left(\frac{7}{5} m v^2\right)}\) = \(\frac{5}{7}\)
OR
If gravity suddenly disappears,
- All bodies will lose their weights.
- We shall be thrown away from the surface of earth due to centrifugal force.
- The motion of planets around the sun will cease because centripetal force shall not be provided.
- Motion of the satellite around earth will also be not possible as no centripetal force will be provided. (1 mark each for any 3 points)
Question 35.
(a) What is conduction ?
(b) Give an expression for thermal conductivity.
(c) Define coefficient of thermal conductivity. [3]
Answer:
(a) Conductance is the process of flow of heat from one point to another through a substance without the actual transfer of particle.
(b) \(\frac{d \mathrm{Q}}{d \mathrm{~T}}\) = \(\frac{\mathrm{KA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{d}\)
where K is coefficient of conductivity. A is the area of cross-section of the face, \(\frac{\mathrm{T}_1-\mathrm{T}_2}{d}\) is temperature gradient.
(c) Coefficient of thermal conductivity (K) is defined as the amount of heat flowing in a second at steady slate between two opposite faces of a cube of side 1 m kept at a temperature difference of 1 K. S.I. unit of K is Js-1m-1K-1
Question 36.
Two capillaries of same length and radii in the ratio 1 : 2 are connected in series. A liquid flows through them in a stream lined condition if the pressure across the two extreme ends of the combination is 1 m of water, find the pressure difference across first capillary. [3]
Answer:
Here, \(\frac{\pi \mathrm{P}_1 r^4}{8 \eta 1}\) = \(\frac{\pi \mathrm{P}_2(2 r)^4}{8 \mathrm{\eta} l}\)
i.e., P1 = 16P2
Given that P1 + P2 = 1 m
P1 + \(\frac{\mathrm{P}_1}{16}\) = 1
or P1 = \(\frac{16}{17}\) = 0.94 m
Commonly Made Error
Students could not evaluate the pressure difference across the capillary in combination.
Answering Tip
Students should familiarize with capillary action in series combination which is similar to resistance in series.
Question 37.
The frequency of oscillations of a mass m suspended by a spring is f1. If the length of the spring is cut to one-half, the same mass oscillates with frequency f2. Determine the value f2/f1. [3]
OR
A particle moves with S.H.M. in a straight line. In the first second after starting from rest, it travels a distance x1 cm and in the next second it travels a distance x2 cm in the same direction. Prove that the amplitude of oscillation is \(\frac{2 x_1^2}{\left(3 x_1-x_2\right)}\)
Answer:
Let the full spring be the combination of two springs in series, each of force constant k. Then in case (1) : the effective spring constant (k1) is given by
k1 = \(\frac{k \times k}{k+k}\) = \(\frac{k}{2}\)
Frequency of oscillation,
In case (2) : when the spring is cul to one-half, the effective spring constant k2 = k.
Frequency of oscillation,
OR
As the particle starts from rest, it must start from the extreme position. Hence, when t = 0, x = r, where r is the required amplitude.
Using the relation,
x = rcos ωt
or r – x1 = rcosω × 1
= rcosω
and r – (x1 + x2) = rcosω × 2
= rcos 2ω
or r – x1 = r(2cos2ω – 1) …(ii)
Solving (i) and (ii),
r = \(\frac{2 x_1^2}{3 x_1-x_2}\)
Section – D
Question numbers 38 to 40 carry 5 marks each.
Question 38.
A racing car travels on a track (without banking) ABCDEFA. ABC is a circular arc of radius 2R. CD and FA are straight paths of length R and DEF is a circular arc of radius R = 100 m. The coefficient of friction on the road is µ = 0.1. The maximum speed of the car is 50 ms-1. Find the minimum time for completing one round. [5]
Answer:
The centripetal force to keep car in circular motion is provided by frictional force (inward to centre).
For DEF,
Question 39.
A rocket accelerates straight up by ejecting gas downwards. In a small time interval Δt, it ejects a gas of mass Δm at a relative speed u. Calculate KE of the entire system at t + Δt and t and show that the device that ejects gas work = \(\left(\frac{1}{2}\right) \Delta m u^2\) in this time interval (neglect gravity). [5]
OR
A uniform disc of radius R, is resting on a table on its rim. The coefficient of friction between disc and table is µ. Now the disc is pulled with a force F as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?
Answer:
Let mass of rocket at any time t = m
Velocity of rocket at any time t = υ
Δm = mass of gas ejected in time interval Δt.
By Newton’s third law,
Reaction force on rocket (upward) = Action force by burnt gases (downward)
\(\frac{m d v}{d t}\) = \(\frac{d m}{d t}|u|\) (∵ F = mu)
or mΔv = Δmu ⇒ mΔυ – uΔm = 0
Substitute this value in (i),
K = \(\frac{1}{2} \Delta m u^2\)
OR
Let us consider the given diagram, frictional force (f) is acting in the opposite direction of F.
Now, let ‘a’ and α be the linear and angular acceleration respectively.
Let the acceleration of the centre of mass of disc be ‘a’, then
F — f = Ma …(i)
where M is mass of disc.
Force of friction applies torque about centre.
The angular acceleration of the disc is α = a/R. (pure rolling)
or Ma = 2f
F – f = 2f or F = 3f …(ii)
from (i) and (ii), we get
f ≤ µN
Thus, f = µN.
Since there is no sliding.
or f ≤ µg
or F ≤ 3µMg.
Question 40.
Derive the expression for heat flow through a compound wall. [5]
OR
Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of 57°C is drunk. You can take body (tooth) temperature to be 37°C and α = 1.7 × 10-5/°C, bulk modulus for copper = 140 × 109 N/m2.
Answer:
Consider a compound wall (or a slab) made of two materials A and B of thickness d1 and d2.
Let K1 and K2 be the coefficients of thermal conductivity and θ1 and θ2 are the temperatures of the end faces (θ1 > θ2) and θ is the temperature of the surface in contact.
For material A,
Q1 = \(\frac{\mathrm{K}_1 \mathrm{~A}_1\left(\theta_1-\theta\right)}{d_1}\) …. (i)
For material B,
Q2 = \(\frac{\mathrm{K}_2 \mathrm{~A}_2\left(\theta-\theta_2\right)}{d_2}\) …. (ii)
At steady state Q1 = Q2
Substituting the value of θ in eqn. (i)
In general for any no. of walls
OR
Given:
Change in temperature
ΔT = 57°C – 37°C = 20°C
Linear expansion, α = 1.7 × 10-5°C-1
Cubical expansion, r = 3α
= 3 × 1.7 × 10-5
= 5.1 × 10-5 K-1
Let V be the volume of cavity, due to increase in temperature if, volume increased by ΔV,
ΔV = γVΔT
or \(\frac{\Delta \mathrm{V}}{\mathrm{V}}\) = γΔT
Thermal stress produced = B × volumetric strain
= B × \(\frac{\Delta \mathrm{V}}{\mathrm{V}}\)
= B × γΔT
Thermal stress = 140 × 109 × 1.7 × 10-5 × 20
= 14280 × 1o4
= 1.428 × 108 N/m2
Thermal stress is about 103 times of atmospheric pressure i.e., 1.01 × 105 N/m2.