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

$$\eqalign{ & {\text{Coefficient of linear expansion}} = \frac{{{\text{Change in length}}}}{{{\text{Original length}} \times {\text{rise in temperature}}}} \cr & {\text{i}}{\text{.e}}{\text{.}}\,\alpha = \frac{{\Delta l}}{{lt}} \cr & {\text{or}}\,\Delta l = l\,\alpha t \cr & {\text{For brass rod,}}\;\,\Delta {l_1} = {l_1}{\alpha _1}t \cr & {\text{For steel rod,}}\,\Delta {l_2} = {l_2}{\alpha _2}t \cr & {\text{Since,}}\,{l_2} - {l_1} = {\text{constant}}\,\,\left( {{\text{given}}} \right) \cr & {\text{So,}}\,\Delta {l_2} - \Delta {l_1} = 0 \cr & {\text{or}}\,\Delta {l_2} = \Delta {l_1} \cr & \therefore {l_2}{\alpha _2}t = {l_1}{\alpha _1}t \cr & {\text{As}}\,t \ne 0,\,{\text{hence}}\,{l_2}{\alpha _2} = {l_1}{\alpha _1} \cr} $$

$$V = {V_0}\left( {1 + \alpha \Delta \theta } \right)$$    where $${\Delta \theta }$$  is the difference in temperature, $$\alpha $$ is the coefficient of volume expansion.
i.e., $$\frac{{V - {V_0}}}{{{V_0}}}\frac{1}{{\Delta \theta }} = \alpha $$
$$\eqalign{ & \alpha = \frac{{\frac{m}{{{d_2}}} - \frac{m}{{{d_1}}}}}{{\frac{m}{{{d_1}}}}}.\frac{1}{{\Delta \theta }} = \frac{{\left( {{d_1} - {d_2}} \right)}}{{\left( {{d_1}{d_2}} \right)}} \times \frac{{{d_1}}}{{\Delta \theta }} \cr & = \frac{{\left( {{d_1} - {d_2}} \right)}}{{{d_2}}}.\frac{1}{{\Delta \theta }} = \frac{{1000 - 958.4}}{{958.4 \times 96}} \cr & = 4.5 \times {10^{ - 4}}\,{\text{per}}\,{\text{degree}}\,K. \cr} $$

Suppose, height of liquid in each arm before rising the temperature is $$l.$$

With temperature rise height of liquid in each arm increases i.e., $${l_1} > l$$   and $${l_2} > l$$
Also $$l = \frac{{{l_1}}}{{1 + \gamma {t_1}}} = \frac{{{l_2}}}{{1 + \gamma {t_2}}}$$
$$ \Rightarrow {l_1} + \gamma \,{l_1}{t_2} = {l_2} + \gamma \,{l_2}{t_1} \Rightarrow \gamma = \frac{{{l_1} - {l_2}}}{{{l_2}{t_1} - {l_1}{t_2}}}.$$

Among glass, wood and metals, metals expand more for same rise in temperature.

The lengths of each rod increases by the same amount
$$\eqalign{ & \therefore \,\,\Delta {\ell _a} = \Delta {\ell _s} \cr & \Rightarrow \,\,{\ell _1}{\alpha _a}t = {\ell _2}{\alpha _s}t \cr & \Rightarrow \,\,\frac{{{\ell _2}}}{{{\ell _1}}} = \frac{{{\alpha _a}}}{{{\alpha _s}}} \cr & \Rightarrow \,\,\frac{{{\ell _2}}}{{{\ell _1}}} + 1 = \frac{{{\alpha _a}}}{{{\alpha _s}}} + 1 \cr & \Rightarrow \,\,\frac{{{\ell _2} + {\ell _1}}}{{{\ell _1}}} = \frac{{{\alpha _a} + {\alpha _s}}}{{{\alpha _s}}} \cr & \Rightarrow \,\,\frac{{{\ell _1}}}{{{\ell _1} + {\ell _2}}} = \frac{{{\alpha _s}}}{{{\alpha _a} + {\alpha _s}}} \cr} $$

As we know, Bulk modulus

$$\eqalign{ & K = \frac{{\Delta P}}{{\left( {\frac{{ - \Delta V}}{V}} \right)}} \cr & \Rightarrow \,\,\frac{{\Delta V}}{V} = \frac{P}{K} \cr & V = {V_0}\left( {1 + \gamma \,\Delta t} \right) \cr & \frac{{\Delta V}}{{{V_0}}} = \gamma \,\Delta t \cr & \therefore \,\,\frac{P}{K} = \gamma \,\Delta t \cr & \Rightarrow \,\,\Delta t = \frac{P}{{\gamma K}} \cr & = \frac{P}{{3\,\alpha K}} \cr} $$

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

$$\eqalign{ & {T_1} = {0^ \circ }C,{T_2} = {100^ \circ }C \cr & \gamma = \frac{{{V_2} - {V_1}}}{{100\;{V_1}}};\alpha = \frac{{{\ell _2} - {\ell _1}}}{{100{\ell _1}}} \cr & {\ell _1} = 10\,cm;{\ell _2} - {\ell _1} = 0.08\,mm = 0.008\,cm \cr & \alpha = \frac{{0.008}}{{10 \times 100}} = 8 \times {10^{ - 6}}{/^ \circ }C \cr & \gamma = 3\alpha = 24 \times {10^{ - 6}}{/^ \circ }C \cr & \therefore 24 \times {10^{ - 6}} = \frac{{{V_2} - 1000}}{{1000 \times 100}} \cr & {V_2} - 1000 = 24 \times {10^{ - 6}} \times {10^5} = 2.4 \cr & \therefore {V_2} = 1002.4\,cc \cr} $$

Due to volume expansion of both mercury and flask, the change in volume of mercury relative to flask is given by
$$\eqalign{ & \Delta V = {V_0}\left[ {{\gamma _L} - {\gamma _g}} \right]\Delta \theta = V\left[ {{\gamma _L} - 3{\alpha _g}} \right]\Delta \theta \cr & = 50\left[ {180 \times {{10}^{ - 6}} - 3 \times 9 \times {{10}^{ - 6}}} \right]\left( {38 - 18} \right) \cr & = 0.153\,cc \cr} $$

Strain developed
$$\varepsilon = \alpha \Delta T = \left( {12 \times {{10}^{ - 6}}} \right)\left( {50} \right) = 6 \times {10^{ - 4}}$$
Strain will be negative, as the rod is in a compressed state.