91Ó°ÊÓ

Volume of the ball at${60}^0\mathrm{C}$is${\mathrm{V}}_0=50{\mathrm{cm}}^3$

Final temperature,${\mathrm{T}}_{\mathrm{F}}=30.{00}^0\mathrm{C}$

Thermal expansion can occur due to an increase in temperature. In the given problem, we have to find the volume of a lead ball at. For that, we can use the concept of volume expansion.

Formula:

The linear expansion of a body,$∆\mathrm{L}=\mathrm{³¢Î\pm }∆\mathrm{T}$ …(¾±)

where$\mathrm{Î\pm }$ is the coefficient of linear expansion of body.

The volume change in expansion of a body,$∆\mathrm{V}={\mathrm{V}}_0{\mathrm{β}}_{\mathrm{Pb}}∆\mathrm{T}$ …(¾±¾±)

where,${\mathrm{β}}_{\mathrm{Pb}}$ is the volume expansion coefficient of body.

Since the volume expansion coefficient for lead relation to the linear expansion coefficient is given as:${\mathrm{β}}_{\mathrm{Pb}}=3{\mathrm{Î\pm }}_{\mathrm{Pb}}$

The coefficient of linear expansion of lead is given as${\mathrm{Î\pm }}_{\mathrm{Pb}}=29×{10}^{-6}/{}^0\mathrm{C}$

Thus, the volume coefficient is given as:

$\begin{array}{rcl}{\mathrm{β}}_{\mathrm{Pb}}& =& 3×29×{10}^{-6}/{}^0\mathrm{C}\\ & =& 87×{10}^{-6}/{}^0\mathrm{C}\end{array}$

Also, the temperature difference is given as:

$\begin{array}{rcl}∆\mathrm{T}& =& {30}^0\mathrm{C}-{60}^0\mathrm{C}\\ & =& -{30}^0\mathrm{C}\end{array}$

So, change in volume using equation (ii) is given by:

$\begin{array}{rcl}∆\mathrm{V}& =& 50{\mathrm{cm}}^3×87×{10}^{-6}/{}^0\mathrm{C}\left(-{30}^{\mathrm{o}}\mathrm{C}\right)\\ & =& -0.1305{\mathrm{cm}}^3\end{array}$

The final volume is given by the sum of the original volume and the change in volume expansion, hence

$\begin{array}{rcl}\mathrm{V}& =& {\mathrm{V}}_0+∆\mathrm{V}\\ & =& 50{\mathrm{cm}}^3-0.1305{\mathrm{cm}}^3\\ & =& 49.8695{\mathrm{cm}}^3\\ & ≈& 49.87{\mathrm{cm}}^3\end{array}$

Hence, the volume of the lead ball is$49.87{\mathrm{cm}}^3$.