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Question

A copper block of mass 1 kg slides down on a rough inclined plane of inclination 37 at a constant speed. Find the increase in the temperature of the block as it slides down through 60cm assuming that the loss in mechanical energy goes into the copper block as thermal energy. (Specific heat of copper = 420Jkg1K1,g=10ms2) .

A
6.6×103C
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B
7.6×103C
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C
8.6×103C
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D
9.6×103C
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Solution

The correct option is C 8.6×103C

Height lost =60sin370=60×35=36m
Loss in PE=1×10×0.36=3.6J
Vi=0m/s
Vf=2gh=2×10×36×102=7.20
KEi=0
KEf=12mV0f=12×1×7.20=3.6J
Loss in Mechanical energy =3.6J
We know,
Q=msΔT
3.6=420×1×ΔT
ΔT=3.6420=3600420×103=8.57×1030c [C]

1220746_1167030_ans_3907a9f7fd57486699b692d41ae742f7.jpg

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