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Two bodies of equal masses are heated at a uniform rate under identical conditions. The change in temperature $(θ)$ in the two cases is shown graphically. The ratio of their specific heats in the solid state of I and II is

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3 : 1

2 : 1

1 : 3

1 : 2

answer is D.

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Detailed Solution

$Q = m s Δ θ$
$P t = m s Δ θ$
$\Rightarrow \frac{Δ θ}{t} = \frac{P}{m s} = s l o p e \Rightarrow S \propto \frac{1}{s l o p e}$
$\Rightarrow \frac{S_{I}}{S_{I I}} = \frac{(40 / 4)}{(60 / 2)} = \frac{1}{3}$

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