Question

A sphere of mass $M$ and diameter $D$ is heated by temperature $\Delta T$. If coefficient of linear expansion is $\alpha$, then find the change in the surface area of sphere.

• A. $\pi D^2\alpha \Delta T(\alpha \Delta T+2)$
• B. $\pi D^2\alpha \Delta T(\alpha \Delta T-2)$
• C. $\pi D^2\alpha \Delta T(\alpha \Delta T+4)$
• D. $\pi D^2\alpha \Delta T(\alpha \Delta T-4)$

Answer 1

The correct option is A $\pi D^2\alpha \Delta T(\alpha \Delta T+2)$

Given that a sphere of mass $M$ and diameter $D$ is heated by temperature $\Delta T$.
Initial surface area of sphere $A=4\pi R^2$
$=4\pi {\left(\frac{D}{2}\right)}^2$
$=\frac{4\pi D^2}{4}=\pi D^2..........\left(1\right)$
Surface area of sphere after heating by a temperature $\Delta T$
$A^{\prime }=4\pi {\left(\frac{D^{\prime }}{2}\right)}^2=\pi (D^{\prime }{)}^2........(2)$
wher $D^{\prime }$ is the new diameter.

Now, we know that
$D^{\prime }=D(1+\alpha \Delta T).........\left(3\right)$
Substituting $\left(3\right)$ in $\left(2\right)$, we can write that,
$A^{\prime }=\left(\pi D^2\right){(1+\alpha \Delta T)}^2$
$=\pi D^2(1+{\alpha }^2\Delta T^2+2\alpha \Delta T).........\left(4\right)$
To find the change in surface area, we subtract $\left(1\right)$ from $\left(4\right)$
$\therefore$ we get,
$A^{\prime }-A=\pi D^2({\alpha }^2\Delta T^2+2\alpha \Delta T)$
$\Rightarrow \pi D^2\alpha \Delta T(\alpha \Delta T+2)$
Hence, option (a) is the correct answer.