Thermal Expansion MCQ Questions & Answers in Heat and Thermodynamics | Physics

$$\eqalign{ & {\gamma _{{\text{real }}}} = {\gamma _{{\text{app}}{\text{.}}}} + {\gamma _{{\text{vessel}}}} \cr & {\text{So}}\,\,{\left( {{\gamma _{{\text{app}}{\text{.}}}} + {\gamma _{{\text{vessel}}}}} \right)_{{\text{glass}}}} = {\left( {{\gamma _{{\text{app}}}} + {\gamma _{{\text{vessel}}}}} \right)_{{\text{steel}}}} \cr & \Rightarrow 153 \times {10^{ - 6}} + {\left( {{\gamma _{{\text{vessel}}}}} \right)_{{\text{glass}}}} \cr & = 144 \times {10^{ - 6}} + {\left( {{\gamma _{{\text{vessel}}}}} \right)_{{\text{steel}}}} \cr & {\text{Further,}}\,\,{\left( {{\gamma _{{\text{vessel}}}}} \right)_{{\text{steel}}}} = 3\alpha = 3 \times \left( {12 \times {{10}^{ - 6}}} \right) \cr & = 36 \times {10^{ - 6}}{/^ \circ }C \cr & \Rightarrow 153 \times {10^{ - 6}} + {\left( {{\gamma _{{\text{vessel}}}}} \right)_{{\text{glass}}}} \cr & = 144 \times {10^{ - 6}} + 36 \times {10^{ - 6}} \cr & \Rightarrow {\left( {{\gamma _{{\text{vessel}}}}} \right)_{{\text{glass}}}} = 3\alpha = 27 \times {10^{ - 6}}{/^ \circ }C \cr & \Rightarrow \alpha = 9 \times {10^{ - 6}}{/^ \circ }C \cr} $$

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
$$\eqalign{ & \Delta L = L \times \alpha \times \Delta \theta = L \times \frac{\gamma }{3} \times \Delta \theta \,\,\left[ {{\text{As}}\,\alpha = \frac{\gamma }{3}} \right] \cr & 2\pi \Delta R = 2\pi R\left( {\frac{\gamma }{3}} \right)\Delta \theta \Rightarrow \Delta \theta \cr & = \frac{3}{\gamma }\frac{{\Delta R}}{R} = \frac{{3 \times 3}}{{3.6 \times {{10}^{ - 5}} \times 497}} \Rightarrow \Delta \theta \simeq {500^ \circ }C \cr} $$

$$\frac{{{\text{Time lost}}}}{{{\text{gained per day}}}} = \frac{1}{2} \propto \,\Delta \theta \times 86400\,{\text{second}}$$
$$\eqalign{ & 12 = \frac{1}{2}\alpha \left( {40 - \theta } \right) \times 86400\,\,\,.....\left( {\text{i}} \right) \cr & 4 = \frac{1}{2}\alpha \left( {\theta - 20} \right) \times 86400\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
On dividing we get, $$3 = \frac{{40 - \theta }}{{\theta - 20}}$$
$$\eqalign{ & 3\theta - 60 = 40 - \theta \cr & 4\theta = 100 \cr & \Rightarrow \,\,\theta = {25^ \circ }C \cr} $$