⇒Thermal Property of Matter

Correct Option is (C)

The increase in volume is $\frac{\mathrm{\Delta }\mathrm{V}}{\mathrm{V}}\times 100=0.225$

Change in temperature is $\mathrm{\Delta }\mathrm{T}={30}^{\circ }\mathrm{C}$

We know, $\gamma =3\alpha$

$\therefore$ The change in volume is $\mathrm{\Delta }\mathrm{V}=\mathrm{V}\gamma \mathrm{\Delta }\mathrm{T}$

$\begin{array}{rl}& \therefore \frac{\mathrm{\Delta }V}{V}\times 100=3\alpha \mathrm{\Delta }\mathrm{T}\times 100\\ & \therefore 0.225=3\alpha \times 30\times 100\\ & \therefore \alpha =\frac{0.225}{3\times 30\times 100}\\ & \therefore \alpha =2.5\times {10}^{-5}/{}^{\circ }\mathrm{C}\end{array}$

Correct Option is (B)

Given: $\mathrm{\Delta }l=60\mathrm{cm},{\alpha }_{\mathrm{A}}=18\times {10}^{-6}/{}^{\circ }\mathrm{C}$,

${\alpha }_B=27\times {10}^{-6}/{}^{\circ }\mathrm{C}$

$\mathrm{\Delta }l$ is constant at all temperatures.

We know $\mathrm{\Delta }l=l\alpha \mathrm{\Delta }t$

Let the length of the rods at a temperature $0^{\circ }\mathrm{C}$ be $l_{\mathrm{A}}$ and $l_{\mathrm{B}}$

$\therefore$ At temperature ${\mathrm{t}}^{\circ }\mathrm{C}$

$\begin{array}{ll}& l_{\mathrm{A}}{\alpha }_{\mathrm{A}}{\mathrm{t}}_{\mathrm{A}}=l_{\mathrm{B}}{\alpha }_{\mathrm{B}}{\mathrm{t}}_{\mathrm{B}}\\ & l_{\mathrm{A}}(18)\times {10}^{-6}=l_{\mathrm{B}}(27)\times {10}^{-6}\text{.... (i)}\\ & \mathrm{\Delta }l=l_{\mathrm{A}}-l_{\mathrm{B}}\\ & \mathrm{\Delta }l=\frac{3}{2}{l}_{\mathrm{B}}-l_{\mathrm{B}}\text{.... from (i)}\\ & \mathrm{\Delta }l=\frac{1}{2}{l}_{\mathrm{B}}\\ \therefore & l_{\mathrm{B}}=2\mathrm{\Delta }l\\ \therefore & l_{\mathrm{B}}=2\times 60=120\mathrm{cm}\\ \therefore & l_{\mathrm{A}}=\frac{3}{2}\times 120=180\mathrm{cm}\end{array}$

Correct Option is (A)

We know,

Coefficient of Linear expansion $\alpha =\frac{{\mathrm{L}}_2-{\mathrm{L}}_1}{{\mathrm{L}}_1\mathrm{\Delta }\mathrm{T}}$ ...... (i)

Given: $\mathrm{\Delta }\mathrm{T}=60-0={60}^{\circ }\mathrm{C}$

${\mathrm{L}}_1=2\mathrm{m}$ and ${\mathrm{L}}_2=2.0016$

Substituting the given values into (i),

$\alpha =\frac{.0016}{120}=1.33\times {10}^{-5}/{}^{\circ }\mathrm{C}$

Correct Option is (C)

Let the original temperature be $0^{\circ }\mathrm{C}$

Volume of $\mathrm{A}={\mathrm{V}}_1=l\times \pi (2\mathrm{r}{)}^2$

After heating, volume of $A$ will become

$\begin{array}{rl}& V_1^{\prime }=V_1(1+\gamma \mathrm{\Delta }T)\\ & \frac{(V_1^{\prime }-V_1)}{V_1}=\gamma \mathrm{\Delta }T\Rightarrow V^{\prime }{}_1-V_1\propto V_1\end{array}$

Similarly for rod $\mathrm{B}$,

$\begin{array}{rl}& \frac{(V_2^{\prime }-V_2)}{V_2}=\gamma \mathrm{\Delta }T\Rightarrow V_2^{\prime }-V_2\propto V_2\\ \therefore & \frac{\mathrm{\Delta }{V}_1}{\mathrm{\Delta }{V}_2}=\frac{(2r{)}^2}{2/r^2}=\frac{2}{1}\end{array}$