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(B) In the steady state, the power dissipated by the rod is equal to the heat lost to the surroundings. Asbestos is a poor conductor of heat compared

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In case-$1$, power lost as well as the temperature of the rod is higher.

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(B) In the steady state, the power dissipated by the rod is equal to the heat lost to the surroundings. Asbestos is a poor conductor of heat compared to air, so the rod covered with asbestos (case-$1$) will lose heat less efficiently than the rod exposed to the atmosphere (case-$2$). Therefore, the temperature of the rod in case-$1$ will be higher. Since the resistance of the graphite rod decreases with an increase in temperature $(R \propto 1/T)$, the resistance of the rod in case-$1$ will be lower than in case-$2$. Given that the potential difference $V$ is constant, the power dissipated is given by $P = V^2/R$. Since $R$ is lower in case-$1$, the power dissipated $P$ will be higher in case-$1$.

$A$ constant potential difference is applied to the ends of a graphite rod, whose resistance decreases with a rise in temperature. The rod can be $(1)$ covered with asbestos or $(2)$ left open to the atmosphere. Answer for the steady state.

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