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Numerical on Thermal Expansion

Q1. A metal rod A of length 25 cm expands by 0.05 cm, when its temperature is raised from 0°C to 100°C. Another rod B of a different metal of length 40 cm expands by 0.04 cm for the same rise in temperature. A third rod C of length 50 cm made up of pieces of rods A and B placed end to end expands by 0.03 cm on heating from 0°C to 50°C. Find the length of each portion of the composite rod.

Solution. For Rod A :-

0.05 = 25 α_A (100)

⇒ α_A= 2×10^-5 ºC^-1 ……….(1)

Also for rod B :-

0.04 = 40 α_B (100)

⇒ α_B= 10^-5 ºC^-1 ……….(2)

Now for 3rd composite rod C :-

l_A + l_B = 50cm ………..(3)

⇒ Δl_A + Δl_B = 0.03

l_Aα_A (50) + l_Bα_B (50) = 0.03

l_A× 2 × 10^-5 × (50) + l_B× 10^-5 × (50) = 0.03

⇒ 2l_A + l_B = 60 ………..(4)

from (3) and (4), we get

l_A = 10 cm & l_B = 40 cm

Q2. If a non-isotropic solid has coefficients of linear expansion $α_{x}, α_{y} and α_{z}$ for three mutually perpendicular directions in the solid, then what is the coefficient of volume expansion for the solid?

Solution. Consider a cube of that material with edges parallel to x, y & z axis having side l₀ at T = 0ºC.

At a temperature T, the new dimensions of the cube will be :-

l_x = l₀[1 + α_xT]

l_y = l₀[1 + α_yT]

l_z = l₀[1 + α_zT]

So volume of solid,

V = l₀[1 + α_xT] × l₀[1 + α_yT] × l₀[1 + α_zT]

⇒ V = (l₀)³ [1 + α_xT][1 + α_yT][1 + α_zT]

As V₀ = (l₀)³

⇒ V = V₀ [1 + α_xT][1 + α_yT][1 + α_zT]

⇒ V = V₀ [1 + (α_x + α_y + α_z) T] …..(1)

[neglecting higher powers of α_x, α_y and α_z]

now finally if γ is coefficient of volume expansion of the solid then

V = V₀ [1 + γ T] …..(2)

comparing equation (1) and (2), we get

γ = α_x + α_y + α_z

Q3. An isosceles triangle is formed with a rod of length $l_{1}$ and coefficient of linear expansion $α_{1}$ for the base and two thin rods each of length $l_{2}$ and coefficient of linear expansion $α_{2}$ for the two pieces, if the distance between the apex and the midpoint of the base remain unchanged as the temperatures varied, show that

$\displaystyle \frac{{{l}_{1}}}{{{l}_{2}}}=2\sqrt{\frac{{{\alpha }_{2}}}{{{\alpha }_{1}}}}$

Solution. First Method :-

The height of the triangle, $\displaystyle h=\sqrt{l_{2}^{2}-\frac{l_{1}^{2}}{4}}$

As the temperature increases, the height remains the same as the old length.

$\displaystyle \Rightarrow \sqrt{l_{2}^{2}{{(1+{{\alpha }_{2}}t)}^{2}}-\frac{l_{1}^{2}}{4}{{(1+{{\alpha }_{1}}t)}^{2}}}=\sqrt{l_{2}^{2}-\frac{l_{1}^{2}}{4}}$

$\displaystyle \Rightarrow \frac{{{l}_{1}}}{{{l}_{2}}}=2\sqrt{\frac{{{\alpha }_{2}}}{{{\alpha }_{1}}}}$

Second Method :-

As with change in temperature T , h remains constant, hence

$\displaystyle \frac{dh}{dT}=0$

$\displaystyle \Rightarrow \frac{d({{h}^{2}})}{dT}=0$

$\displaystyle \Rightarrow \frac{d}{dT}(l_{2}^{2}-\frac{l_{1}^{2}}{4})=0$

$\displaystyle 2{{l}_{2}}\frac{d{{l}_{2}}}{dT}-\frac{2}{4}{{l}_{1}}\frac{d{{l}_{1}}}{dT}=0$

$\displaystyle 2{{l}_{2}}\frac{d{{l}_{2}}}{dT}=\frac{2}{4}{{l}_{1}}\frac{d{{l}_{1}}}{dT}$

$\displaystyle \Rightarrow {{l}_{2}}\frac{{{l}_{2}}{{\alpha }_{2}}dT}{dT}=\frac{{{l}_{1}}}{4}\frac{{{l}_{1}}{{\alpha }_{1}}dT}{dT}$

$\displaystyle 4{{\alpha }_{2}}l_{2}^{2}={{\alpha }_{1}}l_{1}^{2}$

$\displaystyle \Rightarrow \frac{{{l}_{1}}}{{{l}_{2}}}=2\sqrt{\frac{{{\alpha }_{2}}}{{{\alpha }_{1}}}}$

Q4. A circular hole in an aluminum plate is 2.54 cm in diameter at 0ºC. What is the diameter when the temperature of the plate is raised to 100ºC. Given α_Al = 2.3×10^-5ºC^-1.

Solution. Change in diameter is a linear change, hence

D_T = D₀ (1 + α_AlΔT) = 2.54 (1 + 2.3×10^-5×100) = 2.5458 cm

Q5. A glass beaker holds exactly 1 liter at 0ºC.

What is its volume at 50ºC?
If the beaker is filled with mercury at 0ºC, what volume of mercury overflows when the temperature is 50ºC?

Given α_g = 8.3×10^-6 ºC^-1 & γ_Hg = 1.82×10^-4 ºC^-1

Solution.

i. New volume of beaker at 50ºC

V _beaker = V₀(1 + 3α_gΔT) = (1)[1 + 3×8.3×10^-6×50] = 1.001 liter

ii. New volume of mercury at 50ºC is

V _Hg = V₀(1 + γ_HgΔT) = (1)[1 + 1.82×10^-4×50] = 1.009 liter

the volume of the mercury which overflows ΔV = (1.009 – 1.001) liter = 0.008 liter = 8 ml

Q6. A 1 meter steel scale is to be prepared such that the millimeter intervals are to be accurate within 5×10^-4 mm at a certain temperature. Determine maximum permissible temperature variation during the ruling of the mm mark. Given α_steel = 13.22×10^-6 ºC^-1

Solution. Here

Δl = 5×10^-4mm , l = 1mm , α_steel = 13.22×10^-6 ºC^-1

Δl = α_steel × l × ΔT

5×10^-4 = 13.22×10^-6 ×1×ΔT

∴ ΔT = 500/13.22 = 37.82 ºC

Numerical on Thermal Expansion

Numerical on Thermal Expansion

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