Question

A rod of length l and radius r is bent into a ring of radius R with a cut which makes an angle of at the centre. If coefficient of thermal expansion is positive then on heating

Answer 1

Answer:

r and R Increases but teta remains same

Explanation:

INTERATOMIC SEPARATION INCREASES IRRESPECTIVE OF ITS POSITION ON THE BODY

Answer 2

Answer:

If coefficient of thermal expansion is positive then on heating is$r^{2}$

Explanation:

Given: A rod of length l and radius r is bent into a ring of radius R with a cut which makes an angle of at the centre.

To find: If coefficient of thermal expansion is positive then on heating

Solution:

Due to thermal expansion,

$\Delta L=L a \Delta T \text {, }$

where,

$L \rightarrow original length\\ a \rightarrow coefficient of linear expansion\\ \Delta T \rightarrow change in temperature$

Now, Young's modulus,

$\mathrm{Y}=\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}$

where,

$F \rightarrow stress force\\ A \rightarrow area of cross-section of rod\\ \Rightarrow F=\frac{Y A \Delta L}{L} \\ \Rightarrow F=\frac{Y A L a \Delta T}{L}[ from (1) ] \\ \Rightarrow \mathrm{F}=\mathrm{Y} A a \Delta \mathrm{T}$

$\begin{aligned}&\text { So, } F \propto A \\&\Rightarrow F \propto r^{2}\left[A=\pi r^{2}\right]\end{aligned}$