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A piece of lead is dropped from a height of 210 m. If 50 % of its striking energy is converted into heat calculate the rise in temperature (Specific heat of lead = 0.03 KCal / kg, J = 4200J/KCal and g = 10 m/s2)

a
30°C
b
21°C
c
2530c
d
5530C

detailed solution

Correct option is C

(m×10×210)×50100=m×0.03×4200×Δθ⇒Δθ=253∘C

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