JEE Advanced Thermometer & Thermal Expansion PYQ with Solutions

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Topic Question Set

Q 1 :

When a block of iron floats in mercury at $0 ° C$ , fraction $k_{1}$ of its volume is submerged, while at the temperature $60 ° C,$ a fraction $k_{2}$ is seen to be submerged. If the coefficient of volume expansion of iron is $γ_{Fe}$ and that of mercury is $γ_{Hg},$ then the ratio $\frac{k_{1}}{k_{2}}$ can be expressed as [2001]

$\frac{1 + 60 γ_{Fe}}{1 + 60 γ_{Hg}}$
$\frac{1 - 60 γ_{Fe}}{1 + 60 γ_{Hg}}$
$\frac{1 + 60 γ_{Fe}}{1 - 60 γ_{Hg}}$
$\frac{1 + 60 γ_{Hg}}{1 + 60 γ_{Fe}}$

1

2

3

4

(1)

For equilibrium when temperature $0 ° C$ fig (a)

$Upthrust = Wt. of body$

$∴$ $K_{1} V d_{2} g = V d_{1} g$

$\Rightarrow K_{1} = \frac{d_{1}}{d_{2}} \dots (i)$

For equilibrium when temperature increases to $60 °$ fig. (b)

When the temperature is increased the density will decrease.

$∴$ $d_{1}' = d_{1} (1 + γ_{Fe} \times 60)$

and $d_{2}' = d_{2} (1 + γ_{Hg} \times 60)$

Again upthrust = Wt. of body

$∴$ $K_{2} V' d_{2}' g = V' d_{1}' g$

$∴$ $K_{2} [\frac{d_{2}}{1 + γ_{Hg} \times 60}] = \frac{d_{1}}{1 + γ_{Fe} \times 60}$

$∴$ $K_{2} [\frac{1 + γ_{Fe} \times 60}{1 + γ_{Hg} \times 60}] = \frac{d_{1}}{d_{2}}$

$\Rightarrow \frac{K_{1}}{K_{2}} = \frac{1 + γ_{Fe} \times 60}{1 + γ_{Hg} \times 60}$

Q 2 :

Steel wire of length 'L' at $40 °$ C is suspended from the ceiling and then a mass 'm' is hung from its free end. The wire is cooled down from $40 °$ C to $30 °$ C to regain its original length 'L'. The coefficient of linear thermal expansion of the steel is $10^{- 5} / °$ C, Young's modulus of steel is $10^{11} {N/m}^{2}$ and radius of the wire is 1 mm. Assume that $L ≫$ diameter of the wire. Then the value of $' m'$ in kg is nearly [2011]

(3)

$We know that, Y = \frac{F / A}{Δ l / l}$

$∴$ $Y = \frac{m g / A}{Δ l / l} = \frac{m g l}{A Δ l} \dots (i)$

Also, $Δ l = l α Δ T \dots (i i)$

From (i) and (ii)

$Y = \frac{m g l}{A l α Δ T} = \frac{m g}{A α Δ T}$

$∴$ $m = \frac{Y A α Δ T}{g} = \frac{10^{11} \times π {(10^{- 3})}^{2} \times 10^{- 5} \times 10}{10} = π \approx 3$

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