Q. From a steel wire of density $\rho$ is suspended a brass block of density$\rho_{B}$. The extension of steel wire comes to $l$. If the brass block is now fully immersed in a liquid of density $\rho_{L}$, the extension becomes $l'$. The ratio $l/l'$ will be
A) $\frac{\rho_{B} - \rho}{\rho_{L} - \rho}$
B) $\frac{\rho_{L}}{\rho_{B} - \rho_{L}}$
C) $\frac{\rho_{B} - \rho_{L}}{\rho_{B}}$
D) $\frac{\rho_{B}}{\rho_{B} - \rho_{L}}$
Explanation
(d) Let $V$ be the the volume of the brass block weight of brass block $= V\rho_{B}g$
Weight of brass block when immersed in liquid $= V\ \rho_{B}g - V\rho_{L}g$. If $A$ is the area of cross-section of steel wire, then
$Y = \frac{V\rho_{B}g}{A} \times \frac{L}{l} = \frac{(V\rho_{B}g - V\rho_{L}g)}{A} \times \frac{L}{l^{'}}$
Or $\frac{l}{l^{'}} = \frac{\rho_{B}}{\rho_{B} - \rho_{L}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.