Question

If a sphere and a cube of the same material having the same volume are heated up to the same temperature and allowed to cool in the same surrounding then the ratio of amounts of radiations emitted by them will be
(A) $\dfrac{4\pi }{3}:1$
(B) $1:1$
(C) $\dfrac{1}{2}{{\left( \dfrac{4\pi }{3} \right)}^{2/3}}:1$
(D) ${{\left( \dfrac{\pi }{6} \right)}^{1/3}}:1$

Answer 1

Hint In this question, we find the ratio of amounts of radiation emitted by a sphere and a cube. We find the radiation emitted by the sphere then find radiation emitted by the cube. And then find their ratio.

Complete Step by Step Answer:
The radiation emitted,
 $Q=\sigma At({{T}^{4}}-T_{0}^{4})$
Here,
 $\sigma $ = Density of the material
 $A$ = Area of the object
$T$= Temperature of the body
 ${{T}_{0}}$= Temperature of surrounding
 $Q$= Total radiation emitted by an object

Here for sphere and cube density, the temperature of the body and temperature of the surrounding is the same.
 $\therefore \dfrac{{{Q}_{sphere}}}{{{Q}_{cube}}}=\dfrac{{{A}_{sphere}}}{{{A}_{cube}}}$
Area of sphere = $4\pi {{r}^{2}}$
Area of cube = $6{{a}^{2}}$
Now we substitute the expression in the given formula then we get,
 $\dfrac{{{Q}_{sphere}}}{{{Q}_{cube}}}=\dfrac{4\pi {{r}^{2}}}{6{{a}^{2}}}.............................(1)$
The volume of sphere = $\dfrac{4}{3}\pi {{r}^{3}}$
The volume of cube = ${{a}^{3}}$
Now we compare both volumes of sphere and volume of a cube
 $\dfrac{4}{3}\pi {{r}^{3}}={{a}^{3}}$
On simplifying we get,
 $a={{\left( \dfrac{4}{3}\pi \right)}^{1/3}}r......................(2)$
 Substituting the value of equation (2) in equation (1) we get,
 $\dfrac{{{Q}_{sphere}}}{{{Q}_{cube}}}=\dfrac{4\pi {{r}^{2}}}{6\left\{ {{\left( \dfrac{4}{3}\pi \right)}^{1/3}}r \right\}}$
On further solving this equation we get,
 $\dfrac{{{Q}_{sphere}}}{{{Q}_{cube}}}=\dfrac{{{\left( \dfrac{\pi }{6} \right)}^{1/3}}}{1}$
 Here is the ratio of amounts of radiations emitted by a sphere and a cube is ${{\left( \dfrac{\pi }{6} \right)}^{1/3}}:1$

So, the option D is correct

Note: To solve this type of question we have to know about the fundamentals of radiation generation. And what amount of temperature is needed for the material to generate a high amount of radiation. And the knowledge about the calculation of the volume and area of the object. After learning these fundamentals you can easily solve these types of questions.