Question

Water flows through a hose pipe whose internal diameter is 4 cm at a speed of 1 ms⁻¹. If water has to emerge at a speed of 4 ms⁻¹, the diameter of the nozzle should be

• A. 1 cm
• B. 2 cm
• C. 4 cm
• D. 0.5 cm

Answer 1

Answer: (B)

2 cm

Answer 2

The continuity equation for an incompressible fluid states that the volume flow rate is constant: $A_1 v_1 = A_2 v_2$. The cross-sectional area of a pipe is related to its diameter by $A = \frac{\pi d^2}{4}$. Substituting this into the continuity equation gives $\frac{\pi d_1^2}{4} v_1 = \frac{\pi d_2^2}{4} v_2$, which simplifies to $d_1^2 v_1 = d_2^2 v_2$. Given: $d_1 = 4$ cm, $v_1 = 1$ m/s, and $v_2 = 4$ m/s. We need to find $d_2$. $(4 \text{ cm})^2 \times (1 \text{ m/s}) = d_2^2 \times (4 \text{ m/s})$ $16 \text{ cm}^2 \times 1 \text{ m/s} = d_2^2 \times 4 \text{ m/s}$ $d_2^2 = \frac{16 \text{ cm}^2 \times 1 \text{ m/s}}{4 \text{ m/s}}$ $d_2^2 = 4 \text{ cm}^2$ Taking the square root, $d_2 = 2$ cm.