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A steam boat goes across a lake and come...

A steam boat goes across a lake and comes back : (a) on a quiet day when the water is still and (b) on a rough day when there is a uniform current so as to help the journey onward and to impede the journey backward. If the speed of launch on both days same, in which case will it complete the journey in lesser time?

A

case (A)

B

case (B)

C

same in both

D

Nothing can be predicted

Text Solution

Verified by Experts

If the breadth of the lake is L and velocity of boat is V. Time in going and coming back on a quite day
` t_(Q) = (L)/(V)+ (L)/(V) = (2L)/(V)` ….(A)
Now if v is the velocity of air-current then time taken in going across the lake,
`t_(1) = (L) /(V + v)`
[as current helps the motion]
and time taken in coming back
`t_(2) = (L)/(V + v) `
[as current opposes the motion]
`t_(R) = t_(1) + t_(2) = (2L)/(V[1-(v//V)^(2)])` ....(B)
From eq. (A) & (B) ` (t_(R))/(t_(Q)) = (1)/([1-(v//V)^(2)]) gt 1 `
[ as ` 1 - (v^(2))/(V^(2)) lt ] " " i.e. t_(R) gt t_(Q)`
i.e. time taken to complete the journey on quite day
is lesser than that on rough day
Hence correct answer is (B)
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