Question

You have with you, a mug of 500 mL, and a spoon of length 12 cm, both made of pure aluminum. When the AC in the room brings down the room temperature from ${34}^{\circ }C$, to ${16}^{\circ }C$ you find upon precise measurement, that the spoon got shorter by 0.00499 cms. What will be the new volume of the mug?

• A. 501.12 mL
• B. 502.3 mL
• C. 499.38 mL
• D. 498.54 mL

Answer 1

The correct option is C

499.38 mL

Since the room is getting cooled, objects will shrink slightly. We finally need to know, by how much the volume of the mug shrinks - this tells us, we need to find out the coefficient of volume expansion, $\gamma$, of aluminium.

The only useful information we have is - the spoon, also made of aluminium, has shrunk in length by 0.00499 cms. So, let's try to find ${\alpha }_{Al}$ from this, and use the $\gamma$ = $3\alpha$ relation to obtain ${\gamma }_{Al}$.

The new length of the spoon will be-

$L^{\prime }$ = ($L$ - 0.00499) cms

= 11.995 cms

$\therefore$ From the relation for expansion,

$L^{\prime }$ = $L$ $(1+\alpha \Delta T)$

$\Rightarrow$ $11.995$ = $12[1+\alpha \times ({16}^{\circ }C-{34}^{\circ }C\left)\right]$

$\Rightarrow$ $11.995$ = $12+12\times \alpha \times (-18)$

$\Rightarrow$ ${\alpha }_{Al}$ = $\left[\frac{12-11.995}{12\times 18}\right]$${/}^{\circ }C$

$\Rightarrow$ ${\alpha }_{Al}$ = $23.1\times {10}^{-6}{/}^{\circ }C$.

$\therefore$${\gamma }_{Al}$ = $3{\alpha }_{Al}$ = $69.3\times {10}^{-6}$/^{\circ}C\).

$\therefore$The new volume $V^{\prime }$ will be -

$V^{\prime }$ = $V(1+{\gamma }_{Al}\Delta T)$ = $500[1+69.3\times {10}^{-6}\times (-18\left)\right]c{m}^3$

= $499.376c{m}^3$.

Thus, the capacity of the mug shrinks by a small amount of $0.624mL$.