(2)

$Y=0.5\times {10}^{11}{\text{N/m}}^2, \alpha ={10}^{-5}/°\text{C}$

$l=1\text{m}, A={10}^{-3}{\text{m}}^2, \Delta T=100°\text{C}$

$F=YA\alpha \Delta T$

$F=0.5\times {10}^{11}\times {10}^{-3}\times {10}^{-5}\times 100$

$F=50\times {10}^3\text{N}$

(1)

As per question, $\Delta l_{\text{Cu}}=\Delta l_{\text{Al}}$

or,   $l_{\text{Cu}}{\alpha }_{\text{Cu}}\Delta T=l_{\text{Al}}{\alpha }_{\text{Al}}\Delta T$

$l_{\text{Al}}=\frac{l_{\text{Cu}}{\alpha }_{\text{Cu}}}{{\alpha }_{\text{Al}}}=\frac{88\times 1.7\times {10}^{-5}}{2.2\times {10}^{-5}}=68$ $\mathrm{cm}$

(2)

Linear expansion of brass = ${\alpha }_1$

Linear expansion of steel = ${\alpha }_2$

Length of brass rod = $l_1,$ Length of steel rod = $l_2$

On increasing the temperature of the rods by $\Delta T$, new lengths would be

         $l_1\text{'}=l_1(1+{\alpha }_1\Delta T)$                                    ...(i)

         $l_2\text{'}=l_2(1+{\alpha }_2\Delta T)$                                    ...(ii)

Subtracting eqn. (i) from eqn. (ii), we get

      $l_2\text{'}-l_1\text{'}=(l_2-l_1)+(l_2{\alpha }_2-l_1{\alpha }_1)\Delta T$

According to the question,

     $l_2\text{'}-l_1\text{'}=l_2-l_1$               (for all temperatures)

 $\therefore$ $l_2{\alpha }_2-l_1{\alpha }_1=0 \text{or} l_1{\alpha }_1=l_2{\alpha }_2$

(4)

Let ${\rho }_0$ and ${\rho }_T$ be densities of glycerin at $0°C$ and $T°C$ respectively. Then,

        ${\rho }_T={\rho }_0(1-\gamma \Delta T)$

where $\gamma$ is the coefficient of volume expansion of glycerine and $\Delta T$ is rise in temperature.

        $\frac{{\rho }_T}{{\rho }_0}=1-\gamma \Delta T \text{or} \gamma \Delta T=1-\frac{{\rho }_T}{{\rho }_0}$

Thus,  $\frac{{\rho }_0-{\rho }_T}{{\rho }_0}=\gamma \Delta T$

Here, $\gamma =5\times {10}^{-4}{\text{K}}^{-1}$ and $\Delta T=40°C=40\text{K}$

$\therefore$   The fractional change in the density of glycerin

$=\frac{{\rho }_0-{\rho }_T}{{\rho }_0}=\gamma \Delta T=(5\times {10}^{-4}{\text{K}}^{-1})\left(40\text{K}\right)=0.020$