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

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Question 163
1.17 x 107 Nm-2
6.80 x 105Nm-2
1.03 x 105Nm-2
1.36 x104 Nm-2

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.

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

= \(103,360 Nm^{-2}\)

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

\(p = 103,360 + 13,600\)

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

Correct option: C

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