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CGP EDU Academic Team
Published on: September 12, 2026
A man crosses a 320 m wide river perpendicular to the current in 4 minutes. If in still water he can swim with a speed 5/3 times that of the current, then the speed of the current, in m/min is
Text Solution
Verified by ExpertsThe correct answer is:
D
\(\mathbf{v}_r^2 = \mathbf{v}_m^2 - \mathbf{v}^2\)
\(v = \frac{320}{4} \text{ m/min} = 80 \text{ m/min}\)
\(\mathbf{v}_{m} = \frac{5}{3} \mathbf{v}_{r}\)
\(v_{r}^{2} = t \frac{5}{3} (v_{r})^{2} - (80)^{2}\)
\(\frac{16}{9} v_r^2 = (80)^2\)
\(v_r = 60 \frac{m}{min}\)
u^{2}, u_{0} form canours linear
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