A body of mass M is dropped from a height h on a sand floor. If the body penetrates xm into the sand, the average resistance offered by the sand to the body is
A
Mg(hx)
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B
Mg(1+hx)
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C
Mgh+Mgx
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D
Mg(1−hx)
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Solution
The correct option is BMg(1+hx) If the body strikes the sand floor with a velocity v, then Mgh=12Mv2
Let F be the resisting force acting on the body. Then, the resultant force = F−mg
Using work-energy theorem (F−Mg)x=12Mv2 (F−Mg)x=Mgh⇒F=Mg(1+hx) ∴(b)