Mercury is filled in a tube of radius $$2$$ cm up to a height of $$30$$ cm. The force exerted by mercury on the bottom of the tube is _____ N. (Given, atmospheric pressure $$= 10^5$$ Nm$$^{-2}$$, density of mercury $$= 1.36 \times 10^4$$ kg m$$^{-3}$$, $$g = 10$$ m s$$^{-2}$$, $$\pi = \frac{22}{7}$$)

Mercury fills a tube of radius 2 cm to height 30 cm. Find the force on the bottom.

The total pressure at the bottom is atmospheric pressure plus the pressure due to the mercury column: $$P = P_0 + \rho g h$$. Substituting gives $$P = 10^5 + (1.36 \times 10^4)(10)(0.3) = 10^5 + 40800 = 140800 \text{ N/m}^2$$.

Using $$r = 2$$ cm = 0.02 m, the area of the bottom is $$A = \pi r^2 = \frac{22}{7} \times (0.02)^2 = \frac{22}{7} \times 4 \times 10^{-4} = \frac{88}{7} \times 10^{-4} \text{ m}^2$$.

The force on the bottom is $$F = P \times A = 140800 \times \frac{88}{7} \times 10^{-4}$$, which simplifies as $$F = 140800 \times \frac{88}{70000} = 140800 \times 1.2571 \times 10^{-3}$$, and further $$F = \frac{140800 \times 88}{70000} = \frac{12390400}{70000} = 177.006 \approx 177 \text{ N}$$.

The correct answer is 177 N.