First slide

Fluid statics

Question

A stream-lined body falls through air from a height h on the surface of a liquid. Let d and D denote the densities of the material of the body and the liquid respectively. If D > d, then the time after which the body will be instantaneously at rest is:

Moderate

A

$\sqrt{\frac{2 h}{g}}$
B

$\sqrt{\frac{2 h D}{g d}}$
C

$\sqrt{\frac{2 h d}{g D}}$
D

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

Solution

Velocity u of the body when it enters the liquid is given by mgh = $\frac{1}{2} {mu}^{2} or u = \sqrt{2 gh}$
Let Volume of the body = V
Mass of the body = Vd
Weight of the body = Vdg
Mass of liquid displaced = VD
Net upward force = VDg-VDg
= ${Vg}_{} (D - d)$
Retardation = $\frac{net weight}{mass}$
= $\frac{V (D - d) g}{Vd} = (\frac{D - d}{d}) g$
Acceleration a = - $(\frac{D - d}{d}) g$
Final velocity, v in the liquid when the body is instantaneosuly at rest is zero. Let the time taken be t.
v = u+at
0 = $\sqrt{2 gh} - (\frac{D - d}{d}) gt . (\frac{D - d}{d}) gt = \sqrt{2 gh}$
$t = [\frac{d}{D - d}] \sqrt{\frac{2 h}{g}}$

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