91Ó°ÊÓ

1. Distance,$\text{L}=2.00\text{ c³¾or}0.02\text{″¾}$
2. Speed of the source,$\text{v}=100×{10}^{−9}\text{m}/\text{s}$
3. Thermal expansion coefficient for aluminum rod$\mathrm{Î\pm }=23×{10}^{−6}/°\text{ C}$

When an object's temperature changes, it expands and grows larger, a process known as thermal expansion. We know that the speed is the ratio of distance to time. In this experiment, the radioactive source is moving with a constant speed of 100nm/s. The source is radioactive, and the rod is heated as shown in figure. Therefore, thermal expansion takes place, and using the equation for linear expansion, we can find the rate of change of the temperature of the rod with respect to time.

Formula:

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

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

Dividing the equation (i) by time increment$∆\mathrm{t}$ and equating it to the speed,$\text{v}=100×{10}^{−9}\text{m}/\text{s}$we get

$\begin{array}{c}\frac{\mathrm{Δ}\text{L}}{\mathrm{Δ}\text{t}}=\frac{\text{L}\mathrm{Î\pm Δ}\text{T}}{\mathrm{Δ}\text{t}}\\ \text{v}=\mathrm{Î\pm }\text{L}\frac{\mathrm{Δ}\text{T}}{\mathrm{Δ}\text{t}}\text{(}âˆ\mu \text{speedofthebody,v}=\frac{\mathrm{Δ}\text{L}}{\mathrm{Δ}\text{t}}\text{)}\\ \frac{\text{v}}{\mathrm{Î\pm }\text{L}}=\frac{\mathrm{Δ}\text{T}}{\mathrm{Δ}\text{t}}\\ \frac{\mathrm{Δ}\text{T}}{\mathrm{Δ}\text{t}}=\frac{100×{10}^{−9}\text{″¾}/\text{s}}{23×{10}^{−6}\text{ }/°\text{ C}×0.02\text{″¾}}\text{ (}âˆ\mu \text{substitutingthegivenvalues)}\\ =0.2173°\text{ C}/\text{s}\end{array}$

Hence, the constant rate of temperature is$0.2173°\text{ C}/\text{s}$