JEE PHYSICS MCQ: THERMODYNAMICS

1 Medium

1 of 1

Application +4 | -1

A bullet of mass

10 \times 10^{-3} \mathrm{~kg}

moving with a speed of

20 \mathrm{~ms}^{-1}

hits an ice block

\left(0^{\circ} \mathrm{C}\right)

990 \mathrm{~g}

kept at rest on a frictionless floor and gets embedded in it. If ice takes

50 \%

of KE lost by the system, the amount of ice melted (in grams) approximately is:

(J=4.2 \mathrm{~J} / \mathrm{cal}

; latent heat of ice

=80 \mathrm{~cal} / \mathrm{g}

)

View answer and explanation

Correct option: D

3 \times 10^{-3}

Option D (

3 \times 10^{-3}

) is correct because it is the calculated amount of ice melted based on the kinetic energy lost by the bullet and the latent heat of fusion.

More Questions from THERMODYNAMICS