Question

The weight of a hammer is 0.5 kg. It falls at the ratio of 10 strokes per minute on a piece of lead of mass 0.03 kg at 25ºC with a velocity 1.0 m/s. The number of strokes of hammer required to melt the lead piece will be (melting point of lead = 325ºC, specific heat of lead = 0.03 cal/gm-ºC and latent heat = 6 kcal/kg) –

• A. 2590
• B. 3710
• C. 7560
• D. 9350

Answer 1

Answer: (C)

7560

Answer 2

KE imparted in one stroke = $\frac{1}{2}$×$\frac{1}{2}$×(1)²

̃ $\frac{1}{4}$ Joule

Heat required to melt lead

=[30 × 3 × 10^–2 × 300 ]+[ 30 × 6 ]

̃ 270 + 180 ̃ 450 cal

̃ 1890 J

So no. of strokes = $\frac{1890}{0.25}$=7560