$${\gamma _r} = {\gamma _a} + {\gamma _v}$$   where $${\gamma _r} = $$ coefficient of real expansion, $${\gamma _a} = $$  coefficient of apparent expansion and $${\gamma _v} = $$  coefficient of expansion of vessel.
For copper: $${\gamma _r} = C + 3{\alpha _{Cu}} = C + 3A$$
For silver, $${\gamma _r} = S + 3{\alpha _{Ag}}$$
$$ \Rightarrow C + 3A = S + 3{\alpha _{Ag}} \Rightarrow {\alpha _{Ag}} = \frac{{C - S + 3A}}{3}$$

The area of circular hole increases when we heat the metal sheet & expansion of metal sheet will be independent of shape & size of the hole.

Let $${A_0}$$ and $${A_t}$$ be the areas of cross-section of the tube at temperature 0 and $$t$$ respectively,
$$l = $$  length of the liquid column (constant)
$${V_0}$$ and $${V_t}$$ be the volumes of the liquid at expansion. temperature 0 and $$t$$ respectively,
$$\eqalign{ & {V_0} = \ell {A_0} \cr & {V_t} = \ell {A_t} \cr & {V_t} = {V_0}\left( {1 + \gamma t} \right){A_t} = {A_0}\left( {1 + 2\alpha t} \right) \cr & \therefore {V_t} = \ell {A_0}\left( {1 + 2\alpha t} \right) = {V_0}\left( {1 + \gamma t} \right) \cr & = \ell {A_0}\left( {1 + \gamma t} \right)\,\,{\text{or}}\,\,\gamma = 2\alpha . \cr} $$

$$\eqalign{ & \rho = \frac{{{\rho _0}}}{{1 + \gamma \Delta \theta }} \cong {\rho _0}\left( {1 - \gamma \Delta \theta } \right);\frac{{\Delta \rho }}{\rho } = - \gamma \Delta \theta \cr & \frac{1}{{100}} = 3 \times {10^{ - 5}}\Delta \theta ;\,\Delta \theta = \frac{{100}}{3} = {33.3^ \circ }C \cr} $$

$$\eqalign{ & {W_1} = mg - V{d_a}g \cr & {W_2} = mg - V'd{'_a}g \cr & = mg - V\left( {1 + 50{\gamma _b}} \right)\frac{{{d_a}g}}{{\left( {1 + 50{\gamma _a}} \right)}} \cr & = mg - V{d_a}g\left[ {\frac{{1 + 50{\gamma _b}}}{{1 + 50{\gamma _a}}}} \right] \cr & {\text{Given }}{\gamma _b} < {\gamma _a} \cr & \therefore \,\,1 + 50{\gamma _b} < 1 + 50{\gamma _a} \cr & \therefore \,\,\frac{{1 + 50{\gamma _b}}}{{1 + 50{\gamma _a}}} < 1 \cr & \therefore \,\,{W_2} > {W_1}\,\,{\text{or }}{W_1} < {W_2} \cr} $$

$$\eqalign{ & V + \Delta V = {\left( {L + \Delta L} \right)^3} = {\left( {L + \alpha L\Delta T} \right)^3} \cr & = {L^3} + \left( {1 + 3\alpha \Delta T + 3{\alpha ^2}\Delta {T^2} + {\alpha ^3}\Delta {T^3}} \right) \cr & \Rightarrow {\alpha ^2}\,{\text{and}}\,{\alpha ^3}\,{\text{terms}}\,{\text{are}}\,{\text{neglected}}{\text{.}} \cr & \therefore V\left( {1 + \gamma \Delta T} \right) = V\left( {1 + 3\alpha \Delta T} \right) \cr & 1 + \gamma \Delta T = 1 + 3\alpha \Delta T \cr & \therefore \gamma = 3\alpha . \cr} $$

The volume of cavity inside the solid ball increases when it is heated.

$$\eqalign{ & {V_1} = \frac{{\left( {45 - 25} \right)g}}{{{d_1}}}\,{\text{and}}\,{V_2} = \frac{{\left( {45 - 25} \right)g}}{{{d_2}}} \cr & \therefore \frac{{{V_1}}}{{{V_2}}} = \frac{{20}}{{18}} \times \frac{{{d_2}}}{{{d_1}}} \cr & {\text{or}}\,\,\frac{{{V_1}}}{{{V_1}\left( {1 + {r_{{\text{metal}}}} \times 10} \right)}} = \frac{{20}}{{18}} \times \left[ {\frac{{{d_1}}}{{\left( {1 + {r_\ell } \times 10} \right){d_1}}}} \right] \cr} $$
After simplifying, we get
$$\alpha = \frac{{{\gamma _{{\text{metal}}}}}}{3} = 2.6 \times {10^{ - 3}}{/^ \circ }C$$

Initial diameter of tyre $$ = \left( {1000 - 6} \right)\,mm = 994\,mm,$$      so initial radius of tyre $$R = \frac{{994}}{2} = 497\,mm$$
and change in diameter $$\Delta D = 6\,mm$$
so $$\Delta R = \frac{6}{2} = 3\,mm$$
After increasing temperature by $$\Delta \theta $$  tyre will fit onto wheel
Increment in the length (circumference) of the iron tyre
$$\eqalign{ & \Delta L = L \times \alpha \times \Delta \theta = L \times \frac{\gamma }{3} \times \Delta \theta \,\,\left[ {{\text{As}}\,\alpha = \frac{\gamma }{3}} \right] \cr & 2\pi \Delta R = 2\pi R\left( {\frac{\gamma }{3}} \right)\Delta \theta \Rightarrow \Delta \theta \cr & = \frac{3}{\gamma }\frac{{\Delta R}}{R} = \frac{{3 \times 3}}{{3.6 \times {{10}^{ - 5}} \times 497}} \Rightarrow \Delta \theta \simeq {500^ \circ }C \cr} $$

Given, the value of coefficient of volume expansion of glycerin is $$5 \times {10^{ - 4}}{K^{ - 1}}.$$
As, original density of glycerin, $$\rho = {\rho _0}\left( {1 + Y\Delta T} \right)$$
$$ \Rightarrow \rho - {\rho _0} = {\rho _0}Y\Delta T\,$$
Thus, fractional change in the density of glycerine for a rise of $${40^ \circ }C$$  in its temperature,
$$\eqalign{ & \frac{{\rho - {\rho _0}}}{{{\rho _0}}} = Y\Delta T = 5 \times {10^{ - 4}} \times 40 \cr & = 200 \times {10^{ - 4}} = 0.020 \cr} $$