Conduction MCQ Questions & Answers in Heat and Thermodynamics | Physics

$$\eqalign{ & {q_1} = \frac{{K\,2\,A\left( {100} \right)}}{\ell } \cr & {q_2} = \frac{{A\left( {100} \right)}}{{\frac{\ell }{K} + \frac{\ell }{K}}} \cr & = \frac{{K\,A\left( {100} \right)}}{2} \cr & \therefore \,\,\frac{{{q_2}}}{{{q_1}}} = \frac{{K\,A\left( {100} \right)}}{{2\,\ell }} \times \frac{\ell }{{K\,2\,A\left( {100} \right)}} \cr & = \frac{1}{4} \cr} $$

Let $${K_1}$$ and $${K_2}$$ be the thermal conductivities of larger and smaller sphere.
For bigger sphere,
$$\frac{{{K_1}4\pi {{\left( {3r} \right)}^2} \times 100}}{{\frac{t}{4}}} = \frac{{\frac{4}{3}\pi {{\left( {3r} \right)}^3}\rho L}}{{30 \times 60}}$$
For smaller sphere,
$$\eqalign{ & \frac{{{K_2} \times 4\pi {r^2} \times 100}}{t} = \frac{{\frac{4}{3}\pi {r^3}\rho L}}{{20 \times 60}} \cr & \therefore \frac{{{K_1}}}{{{K_2}}} = \frac{1}{2} \cr} $$

Heat lost by I^st body = heat gained by II^nd body. Body at $${100^ \circ }C$$  temperature has greater heat capacity than body at $${0^ \circ }C$$  so final temperature will be closer to $${100^ \circ }C.$$  So, $${T_c} > {50^ \circ }C.$$

Given $${H_{\text{I}}} = {H_{{\text{II}}}}$$
$$\eqalign{ & {k_1}\frac{{A\,\Delta T}}{{\ell + \ell }} \times {t_1} = {k_2}\frac{{\left( {A + A} \right)\Delta T}}{\ell } \times {t_2} \cr & \therefore \,\,{t_2} = \frac{{{k_1}}}{{{k_2}}} \times \frac{{{t_1}}}{4}\,\,\,\,\,.....\left( {\text{i}} \right) \cr} $$
Where $${{k_1}}$$ and $${{k_2}}$$ are the equivalent conductivities in configuration I and II respectively.
For configuration I :
$$\eqalign{ & \frac{{\ell + \ell }}{{{k_1}}} = \frac{\ell }{k} + \frac{\ell }{{2\,k}} \cr & \therefore \,\,\frac{2}{{{k_1}}} = \frac{3}{{2\,k}} \cr & \therefore \,\,{k_1} = \frac{{4\,k}}{3}\,\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
For configuration II :
$$\eqalign{ & {k_2}\left( {A + A} \right) = kA + 2\,kA \cr & \therefore \,\,{k_2} = \frac{{3\,k}}{2}\,\,\,\,\,\,\,.....\left( {{\text{iii}}} \right) \cr} $$
From (i), (ii) and (iii)
$$\eqalign{ & {t_2} = \frac{{\frac{{4\,k}}{3}}}{{\frac{{3\,k}}{2}}} \times \frac{9}{4} \cr & = 2\,\sec \cr} $$
option (A) is correct

Thermal resistance $$R = \frac{l}{{KA}}$$
Where $$l = 20\,cm.\& A\left( {{\text{cylindrical rod}}} \right)$$
$$ = \pi {r^2} = 40\pi c{m^2}$$
So $$R = \frac{{20}}{{0.5 \times 40\pi }} = 0.318\frac{{cal}}{{\sec { \times ^ \circ }C}}$$

$$\eqalign{ & \frac{\ell }{{{K_1}A}} + \frac{\ell }{{2{K_2}A}} + \frac{\ell }{{{K_3}A}} = \frac{\ell }{{KA}} + \frac{\ell }{{2KA}} + \frac{\ell }{{KA}} \cr & \frac{5}{{2K}} = \left( {\frac{1}{{{K_1}}} + \frac{1}{{2{K_2}}} + \frac{1}{{{K_3}}}} \right) \cr & {K_{{\text{eq}}}} = \frac{5}{{2\left( {\frac{1}{{{K_1}}} + \frac{1}{{2{K_2}}} + \frac{1}{{{K_3}}}} \right)}} \cr} $$

$$\eqalign{ & \frac{{{K_1}A\left( {{T_1} - T} \right)}}{{{l _1}}} = \frac{{{K_2}A\left( {T - {T_2}} \right)}}{{{l _2}}}, \cr & \therefore \,\,T = \frac{{{K_1}{T_1}{l _2} + {K_2}{T_2}{l _1}}}{{{K_2}{l _1} + {K_1}{l _2}}} \cr & = \frac{{{K_1}{l _2}{T_1} + {K_2}{l _1}{T_2}}}{{{K_1}{l _2} + {K_2}{l _1}}}. \cr} $$

Let $${\theta ^ \circ }C$$  be the temperature at $$B.$$ Let $$Q$$ is the heat flowing per second from $$A$$ to $$B$$ on account of temperature difference.
$$\therefore \,\,Q = \frac{{KA\left( {90 - \theta } \right)}}{\ell }\,\,\,.....\left( {\text{i}} \right)$$

By symmetry, the same will be the case for heat flow from $$C$$ to $$B.$$
∴ The heat flowing per second from $$B$$ to $$D$$ will be
$$2\,Q = \frac{{KA\left( {\theta - 0} \right)}}{\ell }\,\,\,\,.....\left( {{\text{ii}}} \right)$$
Dividing eq. (ii) by eq. (i)
$$\eqalign{ & 2 = \frac{\theta }{{90 - \theta }} \cr & \Rightarrow \,\,\theta = {60^ \circ } \cr} $$

At, $$P$$
$$\eqalign{ & H = {H_1} + {H_{2s}} \cr & \frac{{kA\left( {{T_A} - {T_P}} \right)}}{{\frac{L}{2}}} \cr & = \frac{{kA\left( {{T_P} - {T_Q}} \right)}}{{\frac{{3L}}{2}}} + \frac{{kA\left( {{T_P} - {T_Q}} \right)}}{L} \cr & \therefore \,\,2\left( {{T_A} - {T_P}} \right) = \frac{2}{3}\left( {{T_P} - {T_Q}} \right) + \left( {{T_P} - {T_Q}} \right) \cr & \therefore \,\,2\left( {{T_A} - {T_P}} \right) = \frac{5}{3}\left( {{T_P} - {T_Q}} \right)\,\,\,.....\left( {\text{i}} \right) \cr & {\text{At, }}Q \cr & {H_1} + {H_2} = H \cr & \therefore \,\,\frac{{kA\left( {{T_P} - {T_Q}} \right)}}{{\frac{{3L}}{2}}} + \frac{{kA\left( {{T_P} - {T_Q}} \right)}}{L} \cr & = \frac{{kA\left( {{T_Q} - {T_B}} \right)}}{{\frac{L}{2}}}\,\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr & \therefore \,\,2\left( {{T_Q} - {T_P}} \right) = \frac{5}{3}\left( {{T_P} - {T_Q}} \right) \cr} $$
From (i) & (ii)
$$\eqalign{ & 2\left( {{T_A} - {T_P}} \right) + 2\left( {{T_Q} - {T_B}} \right) = \frac{{10}}{3}\left( {{T_P} - {T_Q}} \right) \cr & {T_A} - {T_B} = \frac{8}{3}\left( {{T_P} - {T_Q}} \right) \cr & \therefore \,\,{T_P} - {T_Q} = \frac{3}{8} \times 120 \cr & = 45{\,^ \circ }C \cr} $$

Here, $$\Delta {T_1} = 110 - 100 = {10^ \circ }C$$
$$\frac{{d{Q_1}}}{{dt}} = 4\,J/s \Rightarrow \Delta {T_2} = 210 - 200 = {10^ \circ }C$$
$$\frac{{d{Q_2}}}{{dt}} = ?$$
As the rate of heat flow is directly proportional to the temperature difference and the temperature difference in both the cases is same i.e. $${10^ \circ }C.$$  So, the same rate of heat will flow in the second case.
Hence, $$\frac{{d{Q_2}}}{{dt}} = 4\,J/s$$