First slide

Fluid statics

Question

A rectangular container having length of 0.5 m, breadth of 0.3 m and height of 0.4 m is containing water to a depth of 30 cm. A wooden block of mass 2 kg is gently released on the surface of water and it floats in water. Then hydrostatic pressure on the bottom of the container is

Moderate

A

3.133 $\frac{K N}{m^{2}}$
B

4.124 $\frac{K N}{m^{2}}$
C

8.625 $\frac{K N}{m^{2}}$
D

5.15 $\frac{K N}{m^{2}}$

Solution

Total force exerted on the bottom of the container is given by
F = Weight of water + Weight of block
$= (0.5 \times 0.3 \times 0.3) \times 10^{3} \times 10 + 2 \times 10 N$
$= 470 N$
$∴$ Hydrostatic pressure at the bottom of the container $= \frac{470}{0.5 \times 0.3} \frac{N}{m^{2}} = 3.133 \frac{K N}{m^{2}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App