Practice questions on Thermal Expansion and Thermal stress

Time period of pendulum clock at temperature θ is T=2π√(l_θ/g)
at temperature θ
l_θ= 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.

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
α_m1 = (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.
α_m2= 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=α_m1l₁Δθ
Δl₂=α_m2l₂Δθ
thus,
α_m1l₁Δθ+α_m2l₂Δθ= 0.065
since, Δθ= 100°C
α_m1l₁+α_m2l₂ = 0.065/100
substituting the values of α_m1 and α_m2 we get
2.5l₁+1.5l₂ = 65 (ii)
using equations (i) and (ii) we can find the values of l₁=20cm and l₂=10cm.

hoop made of steel is constructed around earth's equator. So at temperature 20°C
R_Hoop = radius of earth
= 6.38 × 10⁶ m
with one degree increase in temperature expansion in the hoop
ΔR = αR_HoopΔθ
= 1.2 × 10-5 × 6.38 × 106 m
= 76.5m.
This would be thickness of space between hoop and earth.

Suppose α_s > α_b
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
(α_sθ)-(2F/AY_s)=ΔL/L₀---(1)

For brass
Thermal strain+Longitudinal strain=Final strain
(α_bθ)+(F/AY_b)=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

Let α_c and α_s 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
L₁α_sΔT=L₂α_cΔT

or
L₂=L₁α_s/α_c

putting the values of α
L₂=11/17L₁
Putting this in equation ---(1)
L₁=14.17cm
and also L₂=9.17 cm

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 l_c and length of brass strip becomes l_b and as given in question, strip bends to a circular arc.
Let R be the mean radius of curvature of the bi-metallic strip.
R_B --> radius of curvature of the brass strip
R_C --> radius of curvature of the copper strip
Then R_B-R_C=d the thickness of the bimetallic strip
Now
l_c=l₀(1+α_cΔT)
l_b=l₀(1+α_bΔT)
then
l_b-l_c=l₀(α_b-α_c)ΔT---(1)
If θ is the angle subtended by the bend strip at the center of curvature O,
then
l_b=θR_B
l_c=θR_C
or
l_b-l_c=θ(R_B-R_C)
=θd--(2)
from equation 1 and 2
θd=l₀(α_b-α_c)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(α_b-α_c)ΔT
Thus R=d/(α_b-α_c)ΔT

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)

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).

(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)

Now
$L_1 \alpha _a t=L_2 \alpha _s t$
$L_1 \alpha _a =L_2 \alpha _s $
$\frac {L_1}{L_2} = {\alpha _s}{\alpha _a}$
$\frac {L_1}{L_1 + L_2}= \frac {\alpha _s}{\alpha _a + \alpha _s}$

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

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$
$\gamma =2\alpha$