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A drilling machine of power $P w a t t s$ is used to drill a hole in copper block of mass $M k g$ . If the specific heat of copper is $s J k g^{- 1^{°}} C^{- 1}$ 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 (in ${}^{o}C$ )

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$\frac{0.6 P T}{M s}$

$\frac{0.6 P}{M s T}$

$\frac{0.4 P T}{M s}$

$\frac{0.4 P}{M s T}$

answer is A.

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Detailed Solution

Useful power available $= 60 %$ of $P = 0.6 P$ watts. Therefore, energy consumed in T seconds $= 0.6 P T$ joules. This energy is used up in heating block. If the rise in temperature $Δ θ$ , then
$0.6 P T = M s Δ θ$

or $Δ θ = \frac{0.6 P T}{M s}$

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