A lead bullet at $27\ ^oC$ just melts when stooped by an obstacle. Assuming that $25\%$ of heat is absorbed by the obstacle, then the velocity of the bullet at the time of striking  ....... $m/s$  ( $M.P.$ of lead $= 327\,^oC$ , specific heat of lead $= 0.03\,cal/g\,^oC$ , latent heat of fusion of lead $= 6\,cal/g$ and $J = 4.2\,joule/cal$ )

If the earth suddenly stops revolving and all its rotational $KE$ is used up in raising its temperature and if $'s'$ is taken to be the specific heat of the earth's material, the rise of temperature of the earth will be : ( $R -$ radius of the earth and $\omega  =$ its angular velocity, $J =\,Joule$ constant)

A bullet moving with a uniform velocity v, stops suddenly after hitting the target and the whole mass melts be $m$, specific heat $S,$ initial temperature $25°C$ melting point $ 475°C$ and the latent heat $L.$ Then $v$ is given by

On a new scale of temperature (which is linear) and called the $W\, scale$, the freezing and boiling points of water are $39\,^oW$ and $239\,^oW$ respectively. What will be th temperature on the new scale, corresponding to a temperature of $39\,^oC$ on the Celsius scale? ............ $^\circ \mathrm{W}$

A lead bullet at $27\,^oC$ just melts when stopped by an obstacle. Assuming that $25\%$ of heat is absorbed by the obstacle, then the velocity of the bullet at the time of striking ........ $m/s$ ( $M. P.$ of lead $= 327\,^oC,$ specific heat of lead $= 0.03\,cal/g\,^oC,$ latent heat of fusion of lead $= 6\,cal/g$ and $J = 4.2\,joule/cal$ )

The graph $AB$ shown in figure is a plot of temperature of a body in degree Celsius and degree Fahrenheit. Then