QBManager

Heat Transfer

149352 An aluminium rod of length $1 \mathrm{~m}$ and a steel rod of length $2 \mathbf{m}$ both having same cross-sectional area, are soldered together end-to-end. The thermal conductivity of aluminium rod and steel rod is $200 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ and $50 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ respectively. The temperatures of the free ends are maintained at $300 \mathrm{~K}$ and $500 \mathrm{~K}$. What is the temperature of the junction?

1 $322 \mathrm{~K}$

2 $350 \mathrm{~K}$

3 $367 \mathrm{~K}$

4 $400 \mathrm{~K}$

TS- EAMCET-05.05.2018

Heat Transfer

149353 Three conducting rods of same material and cross-section are connected as shown in figure. Temperatures of $A, D$ and $C$ are maintained at $20^{\circ} \mathrm{C}, 90^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$. If there is no flow of heat in $A B$, then ratio of the lengths of $B C$ and $B D$ is

1 $2 / 9$

2 $9 / 2$

3 $2 / 7$

4 $7 / 2$

CG PET-22.05.2022

Heat Transfer

149354 The temperature gradient in a rod of $0.5 \mathrm{~m}$ long is $80^{\circ} \mathrm{C} / \mathrm{m}$. If the temperature of hotter end of the rod is $30^{\circ} \mathrm{C}$, then the temperature of the cooler end is

1 $0^{\circ} \mathrm{C}$

2 $-10^{\circ} \mathrm{C}$

3 $10^{\circ} \mathrm{C}$

4 $40^{\circ} \mathrm{C}$

JCECE-2012

Heat Transfer

149355 A doubled layered wall has layer $A, 10 \mathrm{~cm}$ thick and $B, 20 \mathrm{~cm}$ thick, The thermal conductivity of $A$ is thrice that of $B$. In the steady state, the temperature difference across the wall is $35^{\circ} \mathrm{C}$. The temperature difference across the layer $A$ is

1 $28^{\circ} \mathrm{C}$

2 $14^{\circ} \mathrm{C}$

3 $7^{\circ} \mathrm{C}$

4 $5^{\circ} \mathrm{C}$

JCECE-2011

Heat Transfer

149356 Two solid spheres $A$ and $B$ made of the same material have radii $r_{A}$ and $r_{B}$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature $A$ and $B$ is

1 $\frac{r_{A}}{r_{B}}$

2 $\frac{r_{B}}{r_{A}}$

3 $\frac{\mathrm{r}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{B}}^{2}}$

4 $\frac{\mathrm{r}_{\mathrm{B}}^{2}}{\mathrm{r}_{\mathrm{A}}^{2}}$

Explanation:

B Let the mass of sphere are $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}$
$\therefore \quad \mathrm{m}_{\mathrm{A}}=\rho \mathrm{V}_{\mathrm{A}}=\frac{4}{3} \pi \mathrm{r}_{\mathrm{A}}^{3} \rho$
$\mathrm{m}_{\mathrm{B}}=\rho \mathrm{V}_{\mathrm{B}}=\frac{4}{3} \pi \mathrm{r}_{\mathrm{B}}^{3} \rho$
Since we know that,
$\mathrm{Q}=\mathrm{ms} \Delta \mathrm{T}$
$\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}=\mathrm{m}_{\mathrm{A}} \mathrm{s}\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}$
$\text { and } \quad \left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{B}}=\mathrm{m}_{\mathrm{B}} \mathrm{s}\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}$
On dividing equation (i) by equation (ii) we get-
$\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}} \frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}$
According to Stefan-Boltzmann law,
$\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}=\sigma \varepsilon \mathrm{A}_{\mathrm{A}}\left(\mathrm{T}^{4}-\mathrm{T}_{\mathrm{o}}^{4}\right)$
and $\quad\left(\frac{d Q}{d t}\right)_{B}=\sigma \varepsilon A_{B}\left(T^{4}-T_{o}^{4}\right)$
On dividing equation (iv) by equation (v) we get-
$\frac{\left(\frac{d Q}{d t}\right)_{A}}{\left(\frac{d Q}{d t}\right)_{B}}=\frac{A_{A}}{A_{B}}=\frac{4 \pi r_{A}^{2}}{4 \pi r_{B}^{2}}$
From equation (iii) and (vi) -
$\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}} \frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{4 \pi \mathrm{r}_{\mathrm{A}}^{2}}{4 \pi \mathrm{r}_{\mathrm{B}}^{2}}$
$\frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{\mathrm{r}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{B}}^{2}} \times \frac{\frac{4}{3} \pi \mathrm{r}_{\mathrm{B}}^{3} \times \rho}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{A}}^{3} \times \rho}$
$\frac{\left(\frac{d T}{d t}\right)_{A}}{\left(\frac{d T}{d t}\right)_{B}}=\frac{r_{B}}{r_{A}}$

JCECE-2009

Heat Transfer

149352 An aluminium rod of length $1 \mathrm{~m}$ and a steel rod of length $2 \mathbf{m}$ both having same cross-sectional area, are soldered together end-to-end. The thermal conductivity of aluminium rod and steel rod is $200 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ and $50 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ respectively. The temperatures of the free ends are maintained at $300 \mathrm{~K}$ and $500 \mathrm{~K}$. What is the temperature of the junction?

1 $322 \mathrm{~K}$

2 $350 \mathrm{~K}$

3 $367 \mathrm{~K}$

4 $400 \mathrm{~K}$

TS- EAMCET-05.05.2018

Heat Transfer

149353 Three conducting rods of same material and cross-section are connected as shown in figure. Temperatures of $A, D$ and $C$ are maintained at $20^{\circ} \mathrm{C}, 90^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$. If there is no flow of heat in $A B$, then ratio of the lengths of $B C$ and $B D$ is

1 $2 / 9$

2 $9 / 2$

3 $2 / 7$

4 $7 / 2$

CG PET-22.05.2022

Heat Transfer

149354 The temperature gradient in a rod of $0.5 \mathrm{~m}$ long is $80^{\circ} \mathrm{C} / \mathrm{m}$. If the temperature of hotter end of the rod is $30^{\circ} \mathrm{C}$, then the temperature of the cooler end is

1 $0^{\circ} \mathrm{C}$

2 $-10^{\circ} \mathrm{C}$

3 $10^{\circ} \mathrm{C}$

4 $40^{\circ} \mathrm{C}$

JCECE-2012

Heat Transfer

149355 A doubled layered wall has layer $A, 10 \mathrm{~cm}$ thick and $B, 20 \mathrm{~cm}$ thick, The thermal conductivity of $A$ is thrice that of $B$. In the steady state, the temperature difference across the wall is $35^{\circ} \mathrm{C}$. The temperature difference across the layer $A$ is

1 $28^{\circ} \mathrm{C}$

2 $14^{\circ} \mathrm{C}$

3 $7^{\circ} \mathrm{C}$

4 $5^{\circ} \mathrm{C}$

JCECE-2011

Heat Transfer

149356 Two solid spheres $A$ and $B$ made of the same material have radii $r_{A}$ and $r_{B}$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature $A$ and $B$ is

1 $\frac{r_{A}}{r_{B}}$

2 $\frac{r_{B}}{r_{A}}$

3 $\frac{\mathrm{r}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{B}}^{2}}$

4 $\frac{\mathrm{r}_{\mathrm{B}}^{2}}{\mathrm{r}_{\mathrm{A}}^{2}}$

Explanation:

B Let the mass of sphere are $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}$
$\therefore \quad \mathrm{m}_{\mathrm{A}}=\rho \mathrm{V}_{\mathrm{A}}=\frac{4}{3} \pi \mathrm{r}_{\mathrm{A}}^{3} \rho$
$\mathrm{m}_{\mathrm{B}}=\rho \mathrm{V}_{\mathrm{B}}=\frac{4}{3} \pi \mathrm{r}_{\mathrm{B}}^{3} \rho$
Since we know that,
$\mathrm{Q}=\mathrm{ms} \Delta \mathrm{T}$
$\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}=\mathrm{m}_{\mathrm{A}} \mathrm{s}\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}$
$\text { and } \quad \left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{B}}=\mathrm{m}_{\mathrm{B}} \mathrm{s}\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}$
On dividing equation (i) by equation (ii) we get-
$\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}} \frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}$
According to Stefan-Boltzmann law,
$\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}=\sigma \varepsilon \mathrm{A}_{\mathrm{A}}\left(\mathrm{T}^{4}-\mathrm{T}_{\mathrm{o}}^{4}\right)$
and $\quad\left(\frac{d Q}{d t}\right)_{B}=\sigma \varepsilon A_{B}\left(T^{4}-T_{o}^{4}\right)$
On dividing equation (iv) by equation (v) we get-
$\frac{\left(\frac{d Q}{d t}\right)_{A}}{\left(\frac{d Q}{d t}\right)_{B}}=\frac{A_{A}}{A_{B}}=\frac{4 \pi r_{A}^{2}}{4 \pi r_{B}^{2}}$
From equation (iii) and (vi) -
$\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}} \frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{4 \pi \mathrm{r}_{\mathrm{A}}^{2}}{4 \pi \mathrm{r}_{\mathrm{B}}^{2}}$
$\frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{\mathrm{r}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{B}}^{2}} \times \frac{\frac{4}{3} \pi \mathrm{r}_{\mathrm{B}}^{3} \times \rho}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{A}}^{3} \times \rho}$
$\frac{\left(\frac{d T}{d t}\right)_{A}}{\left(\frac{d T}{d t}\right)_{B}}=\frac{r_{B}}{r_{A}}$

JCECE-2009

Heat Transfer

149352 An aluminium rod of length $1 \mathrm{~m}$ and a steel rod of length $2 \mathbf{m}$ both having same cross-sectional area, are soldered together end-to-end. The thermal conductivity of aluminium rod and steel rod is $200 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ and $50 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ respectively. The temperatures of the free ends are maintained at $300 \mathrm{~K}$ and $500 \mathrm{~K}$. What is the temperature of the junction?

1 $322 \mathrm{~K}$

2 $350 \mathrm{~K}$

3 $367 \mathrm{~K}$

4 $400 \mathrm{~K}$

TS- EAMCET-05.05.2018

Heat Transfer

149353 Three conducting rods of same material and cross-section are connected as shown in figure. Temperatures of $A, D$ and $C$ are maintained at $20^{\circ} \mathrm{C}, 90^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$. If there is no flow of heat in $A B$, then ratio of the lengths of $B C$ and $B D$ is

1 $2 / 9$

2 $9 / 2$

3 $2 / 7$

4 $7 / 2$

CG PET-22.05.2022

Heat Transfer

149354 The temperature gradient in a rod of $0.5 \mathrm{~m}$ long is $80^{\circ} \mathrm{C} / \mathrm{m}$. If the temperature of hotter end of the rod is $30^{\circ} \mathrm{C}$, then the temperature of the cooler end is

1 $0^{\circ} \mathrm{C}$

2 $-10^{\circ} \mathrm{C}$

3 $10^{\circ} \mathrm{C}$

4 $40^{\circ} \mathrm{C}$

JCECE-2012

Heat Transfer

149355 A doubled layered wall has layer $A, 10 \mathrm{~cm}$ thick and $B, 20 \mathrm{~cm}$ thick, The thermal conductivity of $A$ is thrice that of $B$. In the steady state, the temperature difference across the wall is $35^{\circ} \mathrm{C}$. The temperature difference across the layer $A$ is

1 $28^{\circ} \mathrm{C}$

2 $14^{\circ} \mathrm{C}$

3 $7^{\circ} \mathrm{C}$

4 $5^{\circ} \mathrm{C}$

JCECE-2011

Heat Transfer

149356 Two solid spheres $A$ and $B$ made of the same material have radii $r_{A}$ and $r_{B}$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature $A$ and $B$ is

1 $\frac{r_{A}}{r_{B}}$

2 $\frac{r_{B}}{r_{A}}$

3 $\frac{\mathrm{r}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{B}}^{2}}$

4 $\frac{\mathrm{r}_{\mathrm{B}}^{2}}{\mathrm{r}_{\mathrm{A}}^{2}}$

Explanation:

B Let the mass of sphere are $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}$
$\therefore \quad \mathrm{m}_{\mathrm{A}}=\rho \mathrm{V}_{\mathrm{A}}=\frac{4}{3} \pi \mathrm{r}_{\mathrm{A}}^{3} \rho$
$\mathrm{m}_{\mathrm{B}}=\rho \mathrm{V}_{\mathrm{B}}=\frac{4}{3} \pi \mathrm{r}_{\mathrm{B}}^{3} \rho$
Since we know that,
$\mathrm{Q}=\mathrm{ms} \Delta \mathrm{T}$
$\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}=\mathrm{m}_{\mathrm{A}} \mathrm{s}\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}$
$\text { and } \quad \left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{B}}=\mathrm{m}_{\mathrm{B}} \mathrm{s}\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}$
On dividing equation (i) by equation (ii) we get-
$\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}} \frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}$
According to Stefan-Boltzmann law,
$\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}=\sigma \varepsilon \mathrm{A}_{\mathrm{A}}\left(\mathrm{T}^{4}-\mathrm{T}_{\mathrm{o}}^{4}\right)$
and $\quad\left(\frac{d Q}{d t}\right)_{B}=\sigma \varepsilon A_{B}\left(T^{4}-T_{o}^{4}\right)$
On dividing equation (iv) by equation (v) we get-
$\frac{\left(\frac{d Q}{d t}\right)_{A}}{\left(\frac{d Q}{d t}\right)_{B}}=\frac{A_{A}}{A_{B}}=\frac{4 \pi r_{A}^{2}}{4 \pi r_{B}^{2}}$
From equation (iii) and (vi) -
$\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}} \frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{4 \pi \mathrm{r}_{\mathrm{A}}^{2}}{4 \pi \mathrm{r}_{\mathrm{B}}^{2}}$
$\frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{\mathrm{r}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{B}}^{2}} \times \frac{\frac{4}{3} \pi \mathrm{r}_{\mathrm{B}}^{3} \times \rho}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{A}}^{3} \times \rho}$
$\frac{\left(\frac{d T}{d t}\right)_{A}}{\left(\frac{d T}{d t}\right)_{B}}=\frac{r_{B}}{r_{A}}$

JCECE-2009

Heat Transfer

149352 An aluminium rod of length $1 \mathrm{~m}$ and a steel rod of length $2 \mathbf{m}$ both having same cross-sectional area, are soldered together end-to-end. The thermal conductivity of aluminium rod and steel rod is $200 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ and $50 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ respectively. The temperatures of the free ends are maintained at $300 \mathrm{~K}$ and $500 \mathrm{~K}$. What is the temperature of the junction?

1 $322 \mathrm{~K}$

2 $350 \mathrm{~K}$

3 $367 \mathrm{~K}$

4 $400 \mathrm{~K}$

TS- EAMCET-05.05.2018

Heat Transfer

149353 Three conducting rods of same material and cross-section are connected as shown in figure. Temperatures of $A, D$ and $C$ are maintained at $20^{\circ} \mathrm{C}, 90^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$. If there is no flow of heat in $A B$, then ratio of the lengths of $B C$ and $B D$ is

1 $2 / 9$

2 $9 / 2$

3 $2 / 7$

4 $7 / 2$

CG PET-22.05.2022

Heat Transfer

149354 The temperature gradient in a rod of $0.5 \mathrm{~m}$ long is $80^{\circ} \mathrm{C} / \mathrm{m}$. If the temperature of hotter end of the rod is $30^{\circ} \mathrm{C}$, then the temperature of the cooler end is

1 $0^{\circ} \mathrm{C}$

2 $-10^{\circ} \mathrm{C}$

3 $10^{\circ} \mathrm{C}$

4 $40^{\circ} \mathrm{C}$

JCECE-2012

Heat Transfer

149355 A doubled layered wall has layer $A, 10 \mathrm{~cm}$ thick and $B, 20 \mathrm{~cm}$ thick, The thermal conductivity of $A$ is thrice that of $B$. In the steady state, the temperature difference across the wall is $35^{\circ} \mathrm{C}$. The temperature difference across the layer $A$ is

1 $28^{\circ} \mathrm{C}$

2 $14^{\circ} \mathrm{C}$

3 $7^{\circ} \mathrm{C}$

4 $5^{\circ} \mathrm{C}$

JCECE-2011

Heat Transfer

149356 Two solid spheres $A$ and $B$ made of the same material have radii $r_{A}$ and $r_{B}$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature $A$ and $B$ is

1 $\frac{r_{A}}{r_{B}}$

2 $\frac{r_{B}}{r_{A}}$

3 $\frac{\mathrm{r}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{B}}^{2}}$

4 $\frac{\mathrm{r}_{\mathrm{B}}^{2}}{\mathrm{r}_{\mathrm{A}}^{2}}$

Explanation:

B Let the mass of sphere are $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}$
$\therefore \quad \mathrm{m}_{\mathrm{A}}=\rho \mathrm{V}_{\mathrm{A}}=\frac{4}{3} \pi \mathrm{r}_{\mathrm{A}}^{3} \rho$
$\mathrm{m}_{\mathrm{B}}=\rho \mathrm{V}_{\mathrm{B}}=\frac{4}{3} \pi \mathrm{r}_{\mathrm{B}}^{3} \rho$
Since we know that,
$\mathrm{Q}=\mathrm{ms} \Delta \mathrm{T}$
$\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}=\mathrm{m}_{\mathrm{A}} \mathrm{s}\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}$
$\text { and } \quad \left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{B}}=\mathrm{m}_{\mathrm{B}} \mathrm{s}\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}$
On dividing equation (i) by equation (ii) we get-
$\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}} \frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}$
According to Stefan-Boltzmann law,
$\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}=\sigma \varepsilon \mathrm{A}_{\mathrm{A}}\left(\mathrm{T}^{4}-\mathrm{T}_{\mathrm{o}}^{4}\right)$
and $\quad\left(\frac{d Q}{d t}\right)_{B}=\sigma \varepsilon A_{B}\left(T^{4}-T_{o}^{4}\right)$
On dividing equation (iv) by equation (v) we get-
$\frac{\left(\frac{d Q}{d t}\right)_{A}}{\left(\frac{d Q}{d t}\right)_{B}}=\frac{A_{A}}{A_{B}}=\frac{4 \pi r_{A}^{2}}{4 \pi r_{B}^{2}}$
From equation (iii) and (vi) -
$\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}} \frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{4 \pi \mathrm{r}_{\mathrm{A}}^{2}}{4 \pi \mathrm{r}_{\mathrm{B}}^{2}}$
$\frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{\mathrm{r}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{B}}^{2}} \times \frac{\frac{4}{3} \pi \mathrm{r}_{\mathrm{B}}^{3} \times \rho}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{A}}^{3} \times \rho}$
$\frac{\left(\frac{d T}{d t}\right)_{A}}{\left(\frac{d T}{d t}\right)_{B}}=\frac{r_{B}}{r_{A}}$

JCECE-2009

Heat Transfer

149352 An aluminium rod of length $1 \mathrm{~m}$ and a steel rod of length $2 \mathbf{m}$ both having same cross-sectional area, are soldered together end-to-end. The thermal conductivity of aluminium rod and steel rod is $200 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ and $50 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ respectively. The temperatures of the free ends are maintained at $300 \mathrm{~K}$ and $500 \mathrm{~K}$. What is the temperature of the junction?

1 $322 \mathrm{~K}$

2 $350 \mathrm{~K}$

3 $367 \mathrm{~K}$

4 $400 \mathrm{~K}$

TS- EAMCET-05.05.2018

Heat Transfer

149353 Three conducting rods of same material and cross-section are connected as shown in figure. Temperatures of $A, D$ and $C$ are maintained at $20^{\circ} \mathrm{C}, 90^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$. If there is no flow of heat in $A B$, then ratio of the lengths of $B C$ and $B D$ is

1 $2 / 9$

2 $9 / 2$

3 $2 / 7$

4 $7 / 2$

CG PET-22.05.2022

Heat Transfer

149354 The temperature gradient in a rod of $0.5 \mathrm{~m}$ long is $80^{\circ} \mathrm{C} / \mathrm{m}$. If the temperature of hotter end of the rod is $30^{\circ} \mathrm{C}$, then the temperature of the cooler end is

1 $0^{\circ} \mathrm{C}$

2 $-10^{\circ} \mathrm{C}$

3 $10^{\circ} \mathrm{C}$

4 $40^{\circ} \mathrm{C}$

JCECE-2012

Heat Transfer

149355 A doubled layered wall has layer $A, 10 \mathrm{~cm}$ thick and $B, 20 \mathrm{~cm}$ thick, The thermal conductivity of $A$ is thrice that of $B$. In the steady state, the temperature difference across the wall is $35^{\circ} \mathrm{C}$. The temperature difference across the layer $A$ is

1 $28^{\circ} \mathrm{C}$

2 $14^{\circ} \mathrm{C}$

3 $7^{\circ} \mathrm{C}$

4 $5^{\circ} \mathrm{C}$

JCECE-2011

Heat Transfer

149356 Two solid spheres $A$ and $B$ made of the same material have radii $r_{A}$ and $r_{B}$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature $A$ and $B$ is

1 $\frac{r_{A}}{r_{B}}$

2 $\frac{r_{B}}{r_{A}}$

3 $\frac{\mathrm{r}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{B}}^{2}}$

4 $\frac{\mathrm{r}_{\mathrm{B}}^{2}}{\mathrm{r}_{\mathrm{A}}^{2}}$

Explanation:

B Let the mass of sphere are $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}$
$\therefore \quad \mathrm{m}_{\mathrm{A}}=\rho \mathrm{V}_{\mathrm{A}}=\frac{4}{3} \pi \mathrm{r}_{\mathrm{A}}^{3} \rho$
$\mathrm{m}_{\mathrm{B}}=\rho \mathrm{V}_{\mathrm{B}}=\frac{4}{3} \pi \mathrm{r}_{\mathrm{B}}^{3} \rho$
Since we know that,
$\mathrm{Q}=\mathrm{ms} \Delta \mathrm{T}$
$\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}=\mathrm{m}_{\mathrm{A}} \mathrm{s}\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}$
$\text { and } \quad \left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{B}}=\mathrm{m}_{\mathrm{B}} \mathrm{s}\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}$
On dividing equation (i) by equation (ii) we get-
$\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}} \frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}$
According to Stefan-Boltzmann law,
$\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{A}}=\sigma \varepsilon \mathrm{A}_{\mathrm{A}}\left(\mathrm{T}^{4}-\mathrm{T}_{\mathrm{o}}^{4}\right)$
and $\quad\left(\frac{d Q}{d t}\right)_{B}=\sigma \varepsilon A_{B}\left(T^{4}-T_{o}^{4}\right)$
On dividing equation (iv) by equation (v) we get-
$\frac{\left(\frac{d Q}{d t}\right)_{A}}{\left(\frac{d Q}{d t}\right)_{B}}=\frac{A_{A}}{A_{B}}=\frac{4 \pi r_{A}^{2}}{4 \pi r_{B}^{2}}$
From equation (iii) and (vi) -
$\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}} \frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{4 \pi \mathrm{r}_{\mathrm{A}}^{2}}{4 \pi \mathrm{r}_{\mathrm{B}}^{2}}$
$\frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{A}}}{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{\mathrm{B}}}=\frac{\mathrm{r}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{B}}^{2}} \times \frac{\frac{4}{3} \pi \mathrm{r}_{\mathrm{B}}^{3} \times \rho}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{A}}^{3} \times \rho}$
$\frac{\left(\frac{d T}{d t}\right)_{A}}{\left(\frac{d T}{d t}\right)_{B}}=\frac{r_{B}}{r_{A}}$

JCECE-2009