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(A, B, C) $1$. Energy radiated by the body per unit time is given by $P = \sigma A (T^4 - T_0^4)$. Given $T = T_0 + 10$, so $T^4 = (T_0 + 10)^4 = T_0^

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(A, B, C) $1$. Energy radiated by the body per unit time is given by $P = \sigma A (T^4 - T_0^4)$. Given $T = T_0 + 10$, so $T^4 = (T_0 + 10)^4 = T_0^4(1 + 10/T_0)^4 \approx T_0^4(1 + 40/T_0) = T_0^4 + 40 T_0^3$.$2$. Thus, $P = \sigma A (T_0^4 + 40 T_0^3 - T_0^4) = 40 \sigma A T_0^3 = 40 \sigma A T_0^4 / T_0 = 40 	imes 460 / 300 \approx 61.3 \,W$. Thus, option $A$ is correct.$3$. For option $B$, $P = \sigma A (T^4 - T_0^4)$. Differentiating with respect to $T_0$ (keeping $T$ constant), $dP/dT_0 = -4 \sigma A T_0^3$. The change in power $\Delta P = 4 \sigma A T_0^3 \Delta T_0$. Since $A = 1 \,m^2$, $\Delta W = 4 \sigma T_0^3 \Delta T_0$. Thus, option $B$ is correct.$4$. For option $C$, $P \propto A$. Reducing $A$ reduces $P$, helping maintain temperature. Thus, option $C$ is correct.$5$. For option $D$, according to Wien's displacement law, $\lambda_m T = b$. If $T$ increases, $\lambda_m$ decreases (shifts to shorter wavelengths). Thus, option $D$ is incorrect.

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