A bullet of mass 50g and specific heat capacity 800kg−1K−1is initially at a temperature 20∘C. It is fired vertically upwards with a speed of 200ms−1and on returning to the starting point strikes a lump of ice at 0∘C and gets embedded in it. Assume that all the energy of the bullet is used up in melting. Neglect the friction of air. Latent heat of fusion of ice =3.36×105Jkg−1
A bullet of mass 50g and specific heat capacity 800kg−1K−1is initially at a temperature 20∘C. It is fired vertically upwards with a speed of 200ms−1and on returning to the starting point strikes a lump of ice at 0∘C and gets embedded in it. Assume that all the energy of the bullet is used up in melting. Neglect the friction of air. Latent heat of fusion of ice =3.36×105Jkg−1
The correct options are
A kinetic Energy of bullet used in melting is 1000 J.
C The mass of ice melted is slightly greater than 5 g.
If the friction offered by air is neglected, the speed of the bullet on returning to be staring point will be equal to its initial speed. The kinetic energy of the bullet is.
K.E.=12mv2=12(50×10−3)×(200)2=1000J
Heat lost by bullet for its temperature to fall from 20∘C to 0∘C=(50×10−3)×800×20=800J
If x kg is the mass of ice melted, then
⇒x=5.3×10−3kg=5.3g
Hence, the correct choices are (1) and (3)
A
kinetic Energy of bullet used in melting is 1000 J.
C
The mass of ice melted is slightly greater than 5 g.
If the friction offered by air is neglected, the speed of the bullet on returning to be staring point will be equal to its initial speed. The kinetic energy of the bullet is.
K.E.=12mv2=12(50×10−3)×(200)2=1000J
Heat lost by bullet for its temperature to fall from 20∘C to 0∘C=(50×10−3)×800×20=800J
If x kg is the mass of ice melted, then
⇒x=5.3×10−3kg=5.3g
Hence, the correct choices are (1) and (3)
