First slide

Fluid statics

Question

A cubical block of wood 10 cm on a side floats at the interface between oil and water, as in Fig. with its lower face 2 cm below the interface. The density of the oil is 0.6 g ${cm}^{- 3}$ . The gauge pressure at the lower face of the block is

Easy

A

84 Pa
B

384 Pa
C

484 Pa
D

784 Pa

Solution

P = Pressure due to l0 cm of oil column + Pressure due to 2 cm of water column
= 10x0.6g+2x1g
= 8g = 8 x 980 dyne ${cm}^{- 2}$
= $\frac{8 \times 980}{10} {Nm}^{- 2} = 784 Pa$

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