Question

A container of capacity 700 ml is filled with two different liquids of volume 200 ml and 500 ml with respective volume expansivities as $1.4\times {10}^{-5}{/}^{0}C$ and $2.1\times {10}^{-5}{/}^{0}C$. During the heating of the vessel, it is observed that neither liquid overflows nor empty space is created. The volume expansitivity of the container is

• A. $1.9\times {10}^{-5}{/}^{0}C$
• B. $1.9\times {10}^{-6}{/}^{0}C$
• C. $1.6\times {10}^{-5}{/}^{0}C$
• D. $1.6\times {10}^{-6}{/}^{0}C$

Answer 1

The correct option is A $1.9\times {10}^{-5}{/}^{0}C$

Let the increase in temperature be $T^{\circ }C$

Increase in volume of liquid 1 = $200\ast T\ast 1.4\ast {10}^{-5}=280T\ast {10}^{-5}ml$

Increase in volume of liquid 1 = $500\ast T\ast 2.1\ast {10}^{-5}=1050T\ast {10}^{-5}ml$

Total increase in volume = $1330T\ast {10}^{-5}ml$

Let the volume expansitivity of container be $\gamma {/}^{\circ }C$

As it is observed that neither liquid overflows nor empty space is created

Increase of volume of container = Increase in volume of liquid

=> $700T\gamma =1330T\ast {10}^{-5}$

=> $\gamma =1.9\ast {10}^{-5}{/}^{\circ }C$

So volume expansitivity of container $=1.9\ast {10}^{-5}{/}^{\circ }C$

Answer. A