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(A) According to Newton's law of cooling,the rate of heat loss is given by: $-ms\frac{dT}{dt} = e\sigma A(T^4 - T_0^4)$For small temperature differenc

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(A) According to Newton's law of cooling,the rate of heat loss is given by: $-ms\frac{dT}{dt} = e\sigma A(T^4 - T_0^4)$For small temperature differences,this simplifies to:$-\frac{dT}{dt} = \frac{4e\sigma AT_0^3}{ms}(T - T_0)$Let $k = \frac{4e\sigma AT_0^3}{ms}$. Since $m = ho V$,we have $k = \frac{4e\sigma AT_0^3}{ho Vs}$.Thus,the rate of cooling $|\frac{dT}{dt}|$ is inversely proportional to the product of density and specific heat $(ho s)$:$|\frac{dT}{dt}| \propto \frac{1}{ho s}$Calculating $(ho s)$ for both liquids:$(ho s)_A = (8 	imes 10^2) 	imes 2000 = 16 	imes 10^5 = 1.6 	imes 10^6\, Jm^{-3}K^{-1}$$(ho s)_B = 10^3 	imes 4000 = 4 	imes 10^6\, Jm^{-3}K^{-1}$Since $(ho s)_A  |\frac{dT}{dt}|_B)$.Therefore,liquid $A$ cools down faster than liquid $B$,which corresponds to graph $A$.

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