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(B) According to Newton's law of cooling,the rate of heat loss is $\frac{dQ}{dt} = 4 \sigma A 	heta_0^3 (\Delta 	heta)$.Since $\frac{dQ}{dt} = ms \l

Vedclass · Accepted Answer

$\frac{12 \sigma 	heta_0^3 \Delta 	heta}{rds}$

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(B) According to Newton's law of cooling,the rate of heat loss is $\frac{dQ}{dt} = 4 \sigma A 	heta_0^3 (\Delta 	heta)$.Since $\frac{dQ}{dt} = ms \left( -\frac{d	heta}{dt} ight)$,the rate of cooling is $-\frac{d	heta}{dt} = \frac{4 \sigma A 	heta_0^3 (\Delta 	heta)}{ms}$.Here,mass $m = 	ext{Volume} 	imes 	ext{density} = (\frac{4}{3} \pi r^3) d$ and surface area $A = 4 \pi r^2$.Substituting these values: $-\frac{d	heta}{dt} = \frac{4 \sigma (4 \pi r^2) 	heta_0^3 (\Delta 	heta)}{(\frac{4}{3} \pi r^3) d s}$.Simplifying the expression: $-\frac{d	heta}{dt} = \frac{16 \pi r^2 \sigma 	heta_0^3 (\Delta 	heta)}{\frac{4}{3} \pi r^3 d s} = \frac{12 \sigma 	heta_0^3 \Delta 	heta}{rds}$.

$A$ suspended sphere has density $d$ and specific heat $s$. The radius of the sphere is $r$. The temperature difference between the sphere and the surroundings $(\Delta \theta)$ is very small. If the temperature of the surroundings is $\theta_0$,then the rate of cooling of the sphere will be .......

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