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 = 10⁵ N m⁻², g = 10 m s⁻², π = 22/7) pressure, density of mercury = 1.36 × 10⁴ kg m⁻³ Numerical answer type. Work it out, then open the answer.

Answer: 177. Pressure at the bottom: P = P₀ + ρ g h = 10⁵ + 1.36×10⁴×10×0.30 = 1.408×10⁵ Pa. Area: A = π r² = 22/7×(0.02)² = 1.257×10⁻³ m². F = PA = 1.408×10⁵×1.257×10⁻³ ≈ 177 N So the force is 177 N.

Mercury is filled in a tube of radius 2 cm up to a height of 30 cm… | JEE Main 2024 (4 Apr, Shift 2)

Q9JEE Main 2024 · 4 APR · SHIFT 2Medium

Pressure in a Fluid

Mercury is filled in a tube of radius $2\text{ cm}$ up to a height of $30\text{ cm}$ . The force exerted by mercury on the bottom of the tube is _________ N.

(Given, atmospheric $= 10^{5}\text{ N m}^{-2},\ g = 10\text{ m s}^{-2},\ \pi = \frac{22}{7})$ pressure, density of mercury

$= 1.36 \times 10^{4}\text{ kg m}^{-3}$

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 177

Pressure at the bottom: $P = P_0 + \rho g h = 10^5 + 1.36\times10^4\times10\times0.30 = 1.408\times10^5\text{ Pa}$ .
Area: $A = \pi r^2 = \dfrac{22}{7}\times(0.02)^2 = 1.257\times10^{-3}\text{ m}^2$ .
$F = PA = 1.408\times10^5\times1.257\times10^{-3} \approx 177\text{ N}$
So the force is 177 N.

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