# A small circular membrane is 10cm below the surface of a pool of mercury...

Question 163
A
1.17 x 107 Nm-2
B
6.80 x 105Nm-2
C
1.17x105Nm-2
D
1.03 x 105Nm-2
E
1.36 x104 Nm-2
##### Explanation

Pressure at a depth for fluid with constant density such as mercury is given as

$$p = p_{0} + \rho hg$$

where $$p_{0}$$ = atmospheric pressure.

$$p_{0} = \rho hg =13600 \times \frac{76}{100} \times 10$$

= $$103,360 Nm^{-2}$$

$$p = 103,360 + (13600 \times \frac{10}{100} \times 10)$$

$$p = 103,360 + 13,600$$

$$p = 116,960 Nm^{-2} \approxeq 1.17 \times 10^{5} Nm^{-2}$$

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