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A ball of mass 5.0 gm and relative densi...

A ball of mass 5.0 gm and relative density 0.5 strikes the surface of the water with a velocity of 20 m/sec. It comes to rest at a depth of 2m. Find the work done by the resisting force in water: (take `g = 10 m//s^(2)`)

A

`6 J`

B

` + 7. 5 J`

C

`9 J`

D

`-10 J`

Text Solution

Verified by Experts

The correct Answer is:
C
By work energy theorem,
`Delta KE` = work done by (gravity + buoyant force + resistance force)
` 0 - 1/2 mv^2 = mg xx 2 - 2 mg xx 2 + w, w = -1/2 mv^2 + mg xx 2`
`= - 1/2 xx 0.05 xx 20^2 + 0.05 xx 10 xx 2 = -9 J`.
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