Two metallic spheres S_1 and S_2 are made of | vedclass

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(B) The rate of cooling $R$ is given by the formula $R = \frac{d	heta}{dt} = \frac{A \epsilon \sigma (T^4 - T_0^4)}{mc}$.Since both spheres are made

Vedclass · Accepted Answer

$(1/3)^{1/3}$

Vedclass · Answer

(B) The rate of cooling $R$ is given by the formula $R = \frac{d	heta}{dt} = \frac{A \epsilon \sigma (T^4 - T_0^4)}{mc}$.Since both spheres are made of the same material and have the same surface finish,$\epsilon$,$\sigma$,$T$,$T_0$,and the density $ho$ are constant.Thus,$R \propto \frac{A}{m}$.Since $A = 4\pi r^2$ and $m = ho \cdot \frac{4}{3}\pi r^3$,we have $A \propto r^2$ and $m \propto r^3$,which implies $r \propto m^{1/3}$.Substituting this into the proportionality,$R \propto \frac{r^2}{r^3} \propto \frac{1}{r} \propto \frac{1}{m^{1/3}}$.Therefore,the ratio of the rates of cooling is $\frac{R_1}{R_2} = \left(\frac{m_2}{m_1}ight)^{1/3}$.Given $m_1 = 3m_2$,we get $\frac{R_1}{R_2} = \left(\frac{m_2}{3m_2}ight)^{1/3} = \left(\frac{1}{3}ight)^{1/3}$.

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