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Q.

A steel ball of mass 0.1 kg falls freely from a height of 10 m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is (specific heat of steel = 460 JKg-1K-1) (g = 10 ms-2).

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a

0.01oC

b

0.1oC

c

1oC

d

1.1oC

answer is B.

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Detailed Solution

loss in Potential Energy = Heat Energy produced = Q and Q = ms∆t .Loss in Potential Energy =mg(H−h)=msΔt10(10−5.6)=460×Δt10×4.6=460×ΔtΔt=0.1∘C

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A steel ball of mass 0.1 kg falls freely from a height of 10 m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is (specific heat of steel = 460 JKg-1K-1) (g = 10 ms-2).