Question

The radius of a metal sphere at room temperature T is R, and the
coefficient of linear expansion of the metal is . The sphere is heated a little by a temperature $\Delta T$ so that its new temperature is $T+\Delta T$. The increase in the volume of the sphere is approximately

• A. $\pi R^2\alpha \Delta T$
• B. $2\pi R\alpha \Delta T$
  
• C. $4\pi R^3\alpha \Delta T$
  
• D. $4\pi R^3\alpha \Delta T/3$
  

Answer 1

The correct option is C $4\pi R^3\alpha \Delta T$


Given,

radius of the sphere at room temperature is 𝑅.

Coefficient of linear expansion = $\alpha$

Initial volume of the sphere, 𝑉=$\frac{4}{3}\pi R^3$

As the temperature increases $\Delta T$ its radius increases, so its volume also increases.


So, coefficient of volume expansion,

$\gamma =3\alpha$

Increase in volume $\Delta V=V\gamma \Delta T$

$\Delta V=\frac{4}{3}\pi R^3\left(3\alpha \right)\Delta T$

$\Delta V=4\pi R^3\alpha \Delta T$

Final answer: Option (d)