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(C) Let $L$ be the length of the lake and $V$ be the velocity of the boat relative to the water.On a quiet day,the velocity of the water is $0$. The t

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(C) Let $L$ be the length of the lake and $V$ be the velocity of the boat relative to the water.On a quiet day,the velocity of the water is $0$. The time taken for the round trip is:$t_{Q} = \frac{L}{V} + \frac{L}{V} = \frac{2L}{V} \quad ...(i)$On a rough day,let $v$ be the velocity of the uniform current. During the onward journey,the effective speed is $(V + v)$,and during the return journey,the effective speed is $(V - v)$.The time taken for the round trip on a rough day is:$t_{R} = \frac{L}{V + v} + \frac{L}{V - v} = \frac{L(V - v) + L(V + v)}{V^{2} - v^{2}} = \frac{2LV}{V^{2} - v^{2}} = \frac{2L}{V(1 - (v/V)^{2})} \quad ...(ii)$Comparing equations $(i)$ and $(ii)$:$t_{R} = t_{Q} 	imes \frac{1}{1 - (v/V)^{2}}$Since $(v/V)^{2} > 0$,the denominator $1 - (v/V)^{2}  1$.Therefore,$t_{R} > t_{Q}$. The time required on a rough day is more.

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