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. \[\sqrt {\dfrac{{2h}}{g}} \]
B. \[\sqrt {\dfrac{2h}{g}\,\dfrac{D}{d}} \]
C. \[\sqrt {\dfrac{2h}{g}\,\dfrac{d}{D}} \]
D. \[\sqrt {\dfrac{{2h}}{g}} \left( {\dfrac{d}{{D - d}}} \right)\]

Answer 1

Hint: When the body is submerged in a fluid, then the buoyant force acting on the body is constant and also the force gravity acting on the body is constant. So, the net force acting on the body is constant.

Formula used:
\[{F_B} = \rho Vg\]
Where \[{F_B}\] is the magnitude of buoyant force acting on a body of volume V by the fluid of density\[\rho \] and g is the acceleration due to gravity.
\[v = u + at\], this is called the first equation of motion, where v is the final velocity after time t when acceleration a is acting on the body having initial velocity u.
\[F = ma\]
where F is the force acting on a body of mass m and a is the acceleration produced by the force applied.

Complete step by step solution:
When the body starts falling, then the initial velocity is zero and the force acting on the body is due to the weight of the body. If the velocity of the body at the surface of water is \[{v_s}\] then using equation of motion,
\[{v_s} = \sqrt {{u^2} + 2gh} \]
\[\Rightarrow {v_s} = \sqrt {{0^2} + 2gh} \]
\[\Rightarrow {v_s} = \sqrt {2gh} \]

Let the volume of the body is V, then the mass of the body is \[m = dV\]. The force of gravity acts downward. Taking the direction downward as negative then the direction upward will be positive.
\[{F_g} = - mg\]
\[\Rightarrow {F_g} = - dVg\]

The upward buoyant force acting on the body is,
\[{F_B} = DVg\]
So, the net force acting on the body is,
\[F = {F_g} + {F_B}\]
\[\Rightarrow F = \left( {D - d} \right)Vg\]
Using Newton’s second law of motion, the acceleration in the body is,
\[a = \dfrac{F}{m} \\ \]
\[\Rightarrow a = \dfrac{{\left( {D - d} \right)Vg}}{{dV}}\]

Now, using the first equation of motion, the final velocity of the body is zero at rest, the initial velocity is equal to the velocity at the surface of water.
If time taken to come to rest is t, then
\[0 = - \sqrt {2gh} + \left( {\dfrac{{\left( {D - d} \right)g}}{d}} \right)t \\ \]
\[\Rightarrow t = \sqrt {\dfrac{{2gh}}{{{g^2}}}} \left( {\dfrac{d}{{\left( {D - d} \right)}}} \right) \\ \]
\[\therefore t = \sqrt {\dfrac{{2h}}{g}} \left( {\dfrac{d}{{D - d}}} \right)\]
Hence, time taken to come to rest is \[t = \sqrt {\dfrac{{2h}}{g}} \left( {\dfrac{d}{{D - d}}} \right)\].

Therefore, the correct option is D.

Note: We should be careful about using equations of motion because all the physical quantities are vector and hence the direction should be determined prior to putting the values in the equation of motion.