A solid copper cube of edges 1 cm is suspended in an evacuated enclosure. Its temperature is found to fill from $100{}^\circ C$ to $99{}^\circ C$ in 100 s. Another solid copper cube of edges 2 cm, with similar surface nature, is suspended in a similar manner. The time required for this cube to cool from $100{}^\circ C$ to $99{}^\circ C$will be approximately.
A. 25 s
B. 50 s
C. 200 s
D. 400 s

Hint:Here a solid copper cube is suspended in an evacuated enclosure. We have another solid copper cube suspended in a similar manner. We have to find the time taken by the second cube to cool so that we can apply an equation for the rate of cooling. Remember the factors affecting the rate of cooling.




Formula used:
We have rate of cooling of a body, $R=\frac{\Delta \theta }{t}=\frac{A\varepsilon \sigma ({{T}^{4}}-T_{0}^{4})}{mc}$
Where A is the area, m is the mass, $\varepsilon $is the emissivity and $\sigma $is the Stefan-Boltzmann constant.



Complete answer:
We have a solid copper cube of edges 1cm which is suspended in an evacuated chamber and it takes 100 s to cools down from $100{}^\circ C$ to $99{}^\circ C$. Now we have another solid copper cube of edge 2 cm. It is given that it is similar in nature with the first cube. We have to find the time taken by the second cube to cool down $100{}^\circ C$ to $99{}^\circ C$.
We have equation for rate of cooling as:
$R=\frac{\Delta \theta }{t}=\frac{A\varepsilon \sigma ({{T}^{4}}-T_{0}^{4})}{mc}$
From the question, we can say that $\sigma $, $({{T}^{4}}-T_{0}^{4})$, $\Delta \theta $are constants.
Therefore, we can connect the relation between edge and time taken for cooling as:
$t\propto \frac{m}{A}$
That is,
$t\propto \frac{m}{A}\propto \frac{volume}{A}\propto a$
Where a is the edge length.
So, we can say that time taken to cool is directly proportional to the edge length of the cube.
Therefore, we can write:
$\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{a}_{1}}}{{{a}_{2}}}$
Where ${{t}_{1}}$ is the time taken to cooling for first cube of edge length 1 cm and ${{t}_{2}}$be the time taken by the second cube to cool down.
Now time taken for second cube to cool down is ${{t}_{2}}=2\times 100=200s$
Therefore, the answer is option (C)



Note: In this we should be crisp about our knowledge in heat and thermodynamics. Answer to this question is simple but you may get confused with a big equation if your concepts are not clear.