Question

A drilling machine of power $P$ watts is used to drill a hole in copper block of mass $M \,kg$. If the specific heat of copper is $s \,J \,kg ^{-1}{ }^{\circ} C ^{-1}$ and $40 \%$ of the power is lost due to heating of the machine, the rise in the temperature of the block in $T$ seconds will be $\left(\right.$ in $\left.{ }^{\circ} C \right)$ :

• A. $\frac{0.6\, P T}{M s}$
• B. $\frac{0.6\, P}{M s T}$
• C. $\frac{0.4\, P T}{M s}$
• D. $\frac{0.4\, P}{M s T}$

Answer 1

Answer: (A)

Useful power available $=60 \%$ of $P$
$=0.6\, P$
Energy consumed in $T$ seconds,
$E =0.6\, P T$
$M s \Delta T =0.6\, P T$
$\Delta T =\frac{0.6\, P T}{M s}$