Practice questions on Thermal Expansion and Thermal stress
Question 1 A pendulum clock gives correct time at 20°C at a place where g=9.8m/ s2. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where of = 9.788m/s2. At what temperature will it give correct time? Coefficient of linear expansion of steel = 12X10-6°C-1.
Time period of pendulum clock at temperature θ is T=2π√(lθ/g)
at temperature θ
lθ= l0 (1+αθ)
T = 2π√[l0(1+αθ)/g]
Now at temperature 20°C,
T20=(2π√l0) [(1 + 12 × 10-6X20)/9.8]1/2
= 2.007√l0
Now at some other place where acceleration due to gravity is g = 9.788 time period at temperature θ is
T = (2π√l0) [(1 + 12 × 10-6Xθ)/9.788] 1/2
but T = 2.007 as calculated earlier so
(2.007√l0) = (2π√l0)[(1+12 × 10-6Xθ)/9.788]1/2
calculating for value of θ we get
θ=-82.06°C.
Question 2 A metal rod 30.0 cm long expands by 0.075 cm when its temperature is raised from 0°C to 100°C.A rod of a different metal and of same length expands by 0.045 cm for the same raise in temperature. A third rod also 30.0 cm long is made up of the pieces of each of the above metals placed and to end. and expends 0.065 cm between 0°C and 100°C. Find the length of each portion of the composite bar.
we would first calculate the coefficient of linear expansion of metals 1 and 2.For first metal given l = 30cm at temp 0°C and change in length when temperature changes to 100°C
Δl = 0.075
using Δl/l = αΔθ
αm1 = (0.075/30 )×(1/100)
= 2.5 x 10-3 °C-1
for second metal at 0°C l= 30cm
at 100ºC change in length due to thermal expansion Δl= 0.045.
αm2= Δl/lΔθ
= (0.045/30) ×( 1/100)
= 1.5 × 10-5 °C-1.
Composite rod made up of both the metals has l1 length of first metal and l2 of second metal such that at 0°C . l1+l2 = 30 cm. (i)
In an increase in temperature from 0°C to 100°C change in length
Δl = Δl1+Δl2= 0.065 cm
but ,
Δl1=αm1l1Δθ
Δl2=αm2l2Δθ
thus,
αm1l1Δθ+αm2l2Δθ= 0.065
since, Δθ= 100°C
αm1l1+αm2l2 = 0.065/100
substituting the values of αm1 and αm2 we get
2.5l1+1.5l2 = 65 (ii)
using equations (i) and (ii) we can find the values of l1=20cm and l2=10cm.
Question 3 Suppose that a steel hoop could be constructed around the earth's equator, just fitting it at a temperature 20°C. What would be the thickness of space between hoop and the earth if the temperature of the hoop were increased by 1 °C.
αsteel = 1.2 × 10-5 °C-1
hoop made of steel is constructed around earth's equator. So at temperature 20°C
RHoop = radius of earth
= 6.38 × 106 m
with one degree increase in temperature expansion in the hoop
ΔR = αRHoopΔθ
= 1.2 × 10-5 × 6.38 × 106 m
= 76.5m.
This would be thickness of space between hoop and earth.
Question 4 Two brass rods and one steel rod of equal length L0 and equal cross-sectional are joined rigidly at their ends .All rods are in the state of zero tension at 0° C. Find the length of the system when the temperature is increased to θ.
Coefficient of linear expansion of steel=αs
Coefficient of linear expansion of brass=αb
Young modulus of steel=Ys
Young modulus of brass=Yb
Suppose αs > αb
It means steel rod will expand more that brass rod.
Since they are all joined at their ends. Steel rod will try to expand brass when heated to θ and it will resisted by the brass rod. It means a tensile force F will develop in brass and while a compressive force of 2F will develop in steel
So final strain in each case
For steel
Thermal strain-Longitudinal strain due to compression=Final strain
(αsθ)-(2F/AYs)=ΔL/L0---(1)
For brass
Thermal strain+Longitudinal strain=Final strain
(αbθ)+(F/AYb)=ΔL/L0---(2)
Multiplying Equation (2) by 2 and then Adding to Equation (1),it can be calculated that
$\Delta L= \frac {L_0( 2\alpha _b Y_b + \alpha _s Y_s) \theta}{2Y_b + Y_s}$
Now length of the system when temperature is changed to θ
L=L0+ΔL where ΔL is calculated as above
Question 5 What will the length of steel and copper rod .So that length of steel rod is 5cm longer then the copper rod at all the temperature (αcopper=1.7X10-5 /C,αsteel=1.1X10-5 /C)
Let αc and αs are the coefficient of linear expansion of copper and steel. Suppose the steel rod is of length L1 and copper rod of length L2,Such that al all the temperatures
L1-L2=5----(1)
This would happen when increase in length of steel rod ΔL1 is of equal to increase in length of copper rod ΔL2 at all change in temperature ΔT
or
L1αsΔT=L2αcΔT
or
L2=L1αs/αc
putting the values of α
L2=11/17L1
Putting this in equation ---(1)
L1=14.17cm
and also L2=9.17 cm
Question 6 A bimetallic strip is formed out of two identical strips .One of the copper and other of brass. The coefficient of linear expansion of the two metals are αc and αB.On heating the strip through the temperature change ΔT,the strip bends into an arc of the circle. Find the radius of the curvature of the strip
Let d be the thickness of the strip and L0 be the length of the two identical strips of copper and brass. On heating the bimetallic strip through the temperature change
ΔT,length of copper strip becomes lc and length of brass strip becomes lb and as given in question, strip bends to a circular arc.
Let R be the mean radius of curvature of the bi-metallic strip.
RB --> radius of curvature of the brass strip
RC --> radius of curvature of the copper strip
Then RB-RC=d the thickness of the bimetallic strip
Now
lc=l0(1+αcΔT)
lb=l0(1+αbΔT)
then
lb-lc=l0(αb-αc)ΔT---(1)
If θ is the angle subtended by the bend strip at the center of curvature O,
then
lb=θRB
lc=θRC
or
lb-lc=θ(RB-RC)
=θd--(2)
from equation 1 and 2
θd=l0(αb-αc)ΔT ---(3)
The mean radius of curvature of bimetallic strip is approximately given by
Rθ=l0 ---(4)
So from equation (3) and (4)
(l0d/R)=l0(αb-αc)ΔT
Thus R=d/(αb-αc)ΔT
Question 7 Show that the moment of inertia of the solid body of any shape changes with the temperature as I=I0(1+2αθ) .Where I0 is the moment of inertia at 0° C and α coefficient of linear expansion of the solid
At 0° C, the Moment of Inertia
of the solid is given by
I0=mR02
m --mass of the body
R0--- radius at 0° C
At temperature θ° C
ΔR=R0αθ
Radius of the body now becomes
R=R0(1+αθ)
Now I=mR2
=mR02(1+2αθ+α2θ2)
=I0(1+2αθ+α2θ2)
As α2θ2 is negligible as comparison to αθ
Thus ,at temperature moment of inertia of body is I=I0(1+2αθ)
Question 8
A bimetallic strip is made of aluminium and steel ($\alpha _{Al} > \alpha _{Steel}$) on heating, the strip will
(a) remain straight
(b) get twisted
(c) will bend with aluminium on concave side
(d) will bend with steel on concave side. Solution
Both strips of Al and steel are fixed together initially in bimetallic strip. When both are heated then expansion in steel will be smaller than aluminium. So Al
strip will be convex side and steel on concave side verifies the option (d).
Question 9
A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly
(a) its speed of rotation increases.
(b) its speed of rotation decreases.
(c) its speed of rotation remains same.
(d) its speed increases because its moment of inertia increases Solution
(b): On heating a uniform metallic rod its length will increase so Moment of Inertia of rod increased from $I_1$ to $I_2$
Now from law of conservation of angular momentum
$I_1 \omega_1 = I_2 \omega_2$
As $I_1 < I_2$ , $\omega _1 > \omega _2$, so Angular Speed decreased.
Verifies option (b)
Question 10
Two rods one of aluminium and other made of steel having initial length $L_1$ and $L_2$ are connected together to form a single rod of length $L_1 + L_2$ . The coefficient of linear expansion for aluminium and steel are $\alpha _a$ and $\alpha _S$ respectively. If the length of each road increases by the same amount when their temperature are raised by T degree Celsius, the what is the ratio between $ \frac {L_1}{L_1 + L_2}$ ?
a.$ \frac {\alpha _s}{\alpha _a}$
b.$ \frac {\alpha _a}{\alpha _s}$
c.$ \frac {\alpha _s}{\alpha _a + \alpha _s}$
d.$ \frac {\alpha _a}{\alpha _a + \alpha _s}$ Solution
Question 11
A pendulum clock loses 12s a day if the temperature is 40 °C and gains 4s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion ($\alpha$) of the metal of the pendulum shaft are respectively :
(a)25°C ; $\alpha =1.85 \times 10^{-5} $/°C
(b)60°C ;$\alpha =1.85 \times 10^{-4}$ /°C
(c)30°C ;$\alpha=1.85 \times 10^{-3}$/°C
(d)55°C ;$\alpha=1.85 \times 10^{-2}$/°C Solution
Time loss or gain per sec is given by
$ \Delta t=\frac {1}{2} \alpha ? \Delta T$
$12=12 \alpha (40-T) \times 86400$ ----(1) ( as 1 day=84400 sec)
$4=12 \alpha (T-20) \times 86400$ ----(2)
From (1) and (2)
$3=\frac {40-T}{T-20}$
Solving T=25°C and putting in (2)
$\alpha =\frac {8}{5 \times 86400}$
$=1.85 \times 10^{-5}$ /°C
Question 12
A cubical block is floating inside a bath. The temperature of system is increased by small temperature $\Delta T$. It was found that the depth of submerged portion of cube does not change. Find the relation between coefficient of linear expansion ($\alpha$) of the cube and volume expansion of liquid ($\gamma$).
a. $\gamma =2\alpha$
b. $\gamma =\alpha$
c. $\gamma =3\alpha$
d. $\gamma =\frac {1}{2} \alpha$ Solution
Let x be the depth submerged.At initial temperature for the equilibrium of the block
$AL \rho _b g = Ax \rho _l g$
$L\rho_b = x \rho _l$ . . . (i)
At final temperature
$A^{'} = A(1 + 2 \alpha \Delta T)$
$\rho _{l}^{'}= \rho _l (1 - \gamma \Delta T)$
For the equilibrium of the block
$A(1 + 2 \alpha \Delta T)(x) (1 - \gamma \Delta T) = AL \rho _b = A x \rho _l$
$ 1 + 2\alpha \Delta T - \gamma \Delta T= 1$
$\gamma =2\alpha$
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