Question

A Bakelite beaker has volume capacity of $500cc$ at $30°C$. When it is partially filled with$V_m$ volume ($30°C$) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If ${\gamma }_{\left(beaker\right)}=6\times {10}^{-6}°C^{-1}$ and ${\gamma }_{\left(mercury\right)}=1.5\times {10}^{-4°}{c}^{-1}$¹, where $\gamma$ is the coefficient of volume expansion, then $V_m\left(incc\right)$ is close to

Answer 1

Step 1. Given data

The volume of the beaker,$V_b=500cc$

Coefficient of volume expansion for the beaker, ${\gamma }_b=6\times {10}^{-6}°C^{-1}$

Coefficient of volume expansion for the mercury, ${\gamma }_m=1.5\times {10}^{-4}°C^{-1}$

Step 2. Finding the volume of the mercury, $V_m$

The volume of the empty part $=$ Volume of beaker $-$ Volume of mercury

Also, it is given that, there is no change in the unfilled volume of the beaker with the varying temperature.

So, a change in the volume of the beaker $=$ change in volume of mercury

$V_b{\gamma }_b∆T=V_m{\gamma }_m∆T$ [$∆T$ is change in temperature]

$V_b{\gamma }_b=V_m{\gamma }_m$

$V_m=\frac{V_b{\gamma }_b}{{\gamma }_m}$

$V_m=\frac{500\times 6\times {10}^{-6}}{1.5\times {10}^{-4}}$

$V_m=20cc$

Hence, the volume of the mercury is $20cc$