Question

Water from a tap of cross-sectional area \(1 \, \text{cm}^2\), falls vertically downwards at \(2 \, \text{m/s}\). The cross-sectional area of the stream, 20 cm below the tap is (assume that pressure is constant throughout and the flow is streamlined; \(g = 10 \, \text{m/s}^2\)).

• A. \(7.07 \, \text{cm}^2\)
• B. \(1 \, \text{cm}^2\)
• C. \(0.707 \, \text{cm}^2\)
• D. \(2 \, \text{cm}^2\)

Hint:

In fluid dynamics, the continuity equation \(A_1 v_1 = A_2 v_2\) and Bernoulli's principle are useful to solve problems related to the flow of fluids, where the mass flow rate remains constant.

Answer 1

The Correct Option is C

The problem involves the principle of conservation of mass and Bernoulli’s equation. As the water falls, the velocity increases, but the mass flow rate (product of the cross-sectional area and velocity) remains constant. Let the area of cross-section at the tap be \(A_1 = 1 \, \text{cm}^2\) and the velocity at the tap is \(v_1 = 2 \, \text{m/s}\). At a point 20 cm below, let the cross-sectional area be \(A_2\) and the velocity be \(v_2\). The principle of conservation of mass gives: \[ A_1 v_1 = A_2 v_2 \] Using the energy conservation principle (Bernoulli’s equation), we have: \[ \frac{v_2^2}{2} = g h + \frac{v_1^2}{2} \] where \(h = 0.2 \, \text{m}\) is the height difference. Now solving for \(v_2\): \[ v_2 = \sqrt{v_1^2 + 2gh} \] Substitute the given values: \[ v_2 = \sqrt{(B)^2 + 2 \times 10 \times 0.2} = \sqrt{4 + 4} = \sqrt{8} \approx 2.828 \, \text{m/s} \] Now, applying the continuity equation: \[ A_1 v_1 = A_2 v_2 \quad \Rightarrow \quad A_2 = \frac{A_1 v_1}{v_2} = \frac{1 \times 2}{2.828} \approx 0.707 \, \text{cm}^2 \] Thus, the cross-sectional area at 20 cm below the tap is approximately \(0.707 \, \text{cm}^2\).