A steel ball of mass 0.1 kg falls freely from a height of l0 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 = 460K/kg/∘C,g=10m/s2
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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 energy =mgh−h′=0.1×10×(10−5.4)=4.6J Now, 4.6J=msΔθ=0.1×460×Δθ∴Δθ=0.1∘C