Question

A streamlined body falls through air from a height h on the surface of a liquid. If d and D(D > d) represents the densities of the material of the body and liquid respectively, then the time after which the body will be instantaneously at rest, is

• A. $\sqrt { \frac { 2 h } { g } }$
• B. $\sqrt { \frac { 2 h } { g } \cdot \frac { D } { d } }$
• C. $\sqrt { \frac { 2 h } { g } \cdot \frac { d } { D } }$
• D. $\sqrt { \frac { 2 h } { g } } \left( \frac { d } { D - d } \right)$

Answer 1

Answer: (D)

$\sqrt { \frac { 2 h } { g } } \left( \frac { d } { D - d } \right)$

Answer 2

Upthrust – weight of body = apparent weight

$V D g - V d g = V d a$

Where a = retardation of body ∴ $a = \left( \frac { D - d } { d } \right) g$

The velocity gained after fall from h height in air, $v = \sqrt { 2 g h }$

Hence, time to come in rest,

$t = \frac { v } { a } = \frac { \sqrt { 2 g h } \times d } { ( D - d ) g } = \sqrt { \frac { 2 h } { g } } \times \frac { d } { ( D - d ) }$