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Question

# A small block is freely sliding down from the top of a rough inclined plane whose angle of inclination is 450. The block reaches bottom then it completes a smooth vertical circle. If the coefficient of friction is 0.5, the ratio of minimum vertical height of inclined plane to radius of vertical circle is:

A
3:1
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B
5:1
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C
5:2
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D
7:3
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Solution

## The correct option is B 5:1μ=0.5.Wo by friction=μNa=(0.5)(mg√2)(√2H)=mgH2minimum velocity required for completely velocity lively=√5gR work energy therefore on workmgH=mV2o2+mgH2.mgH2=m2(√5gR)2gH=5gR.HR=51

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