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Heat Transfer

149337 Two spheres of different materials one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the larger sphere is $\mathbf{2 5}$ minute and for smaller one is $\mathbf{1 6}$ minute, the ratio of thermal conductivities of the materials of larger spheres to that of smaller sphere is

1 $4: 5$

2 $5: 4$

3 $25: 8$

4 $8: 25$

CG PET-2010

Heat Transfer

149339 If the temperature of a hot body is increased by $\mathbf{5 0} \%$ then the amount of radiation emitted by it increases approximately by

1 $400 \%$

2 $225 \%$

3 $250 \%$

4 $500 \%$

CG PET -2018

Heat Transfer

149340 A composite bar consists of a cylinder of radius $R$ and thermal conductivity $K_{1}$ fitted inside a cylindrical shell of internal radius $R$ and external radius $2 R$. If the thermal conductivity of shell is $K_{2}$, then the equivalent thermal conductivity of the composite bar is

1 $\mathrm{K}_{1}+\mathrm{K}_{2}$

2 $\frac{\mathrm{K}_{1}+3 \mathrm{~K}_{2}}{4}$

3 $\mathrm{K}_{1}+3 \mathrm{~K}_{2}$

4 $\frac{\mathrm{K}_{2}+3 \mathrm{~K}_{2}}{4}$

CG PET -2016

Heat Transfer

149341 Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or II as shown in figure. One of the blocks has thermal conductivity $K$ and the other $2 \mathrm{~K}$. The temperature difference between the ends along the $X$ - axis is the same in both the configurations. It takes $9 \mathrm{~s}$ to transport a certain amount of heat from the hot end to the cold end in configuration $I$. The time to transport the same amount of heat in the configuration II is

1 $2.0 \mathrm{~s}$

2 $3.0 \mathrm{~s}$

3 $4.5 \mathrm{~s}$

4 $6.0 \mathrm{~s}$

Explanation:

A original image
Thermal resistance circuit looks like electricalresistance circuit.
Then for - I
Equivalent thermal resistance,
$\mathrm{R}_{\mathrm{eq}} =\mathrm{R}_{1}+\mathrm{R}_{2} \quad \text { [Series combination] }$
$=\frac{l}{\mathrm{~K}_{1} \mathrm{~A}}+\frac{l}{\mathrm{~K}_{2} \mathrm{~A}}$
$=\frac{l}{\mathrm{~A}}\left(\frac{\mathrm{K}_{1}+\mathrm{K}_{2}}{\mathrm{~K}_{1} \mathrm{~K}_{2}}\right)=\frac{l}{\mathrm{~A}}\left(\frac{\mathrm{K}+2 \mathrm{~K}}{\mathrm{~K} \times 2 \mathrm{~K}}\right)=\frac{3}{2}\left(\frac{l}{\mathrm{~A}}\right) \frac{1}{\mathrm{~K}}$
For case - II
Equivalent thermal Resistance
$\frac{1}{\mathrm{R}_{\mathrm{eq}}^{\prime}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}} \quad \text { [Parallel combination] }$
$\mathrm{R}_{\mathrm{eq}}^{\prime}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$
$= \frac{\frac{l}{\mathrm{~K}_{1} \mathrm{~A}} \times \frac{l}{\mathrm{~K}_{2} \mathrm{~A}}}{\mathrm{~K}_{1} \mathrm{~A}}+\frac{l}{\mathrm{~K}_{2} \mathrm{~A}}$
$\mathrm{R}_{\mathrm{eq}}^{\prime}=\frac{\left(\frac{l}{\mathrm{~A}}\right)^{2}\left(\frac{1}{\mathrm{~K}_{1}} \frac{1}{\mathrm{~K}_{2}}\right)}{\left(\frac{l}{\mathrm{~A}}\right)\left(\frac{\mathrm{K}_{1}+\mathrm{K}_{2}}{\mathrm{~K}_{1} \mathrm{~K}_{2}}\right)}=\frac{l}{\mathrm{~A}}\left(\frac{1}{\mathrm{~K}_{1}+\mathrm{K}_{2}}\right)=\frac{l}{\mathrm{~A}}\left(\frac{1}{3 \mathrm{~K}}\right)$
Heat flow rate
$\frac{\mathrm{Q}}{\mathrm{t}_{1}}=\frac{\Delta \mathrm{Q}}{\mathrm{R}_{\text {eq. }}}$
For case-I
$\frac{\mathrm{Q}}{\mathrm{t}_{1}}=\frac{\Delta \mathrm{Q}}{\frac{3}{2}(\mathrm{l} / \mathrm{A}) \frac{1}{\mathrm{~K}}}$
For case - II
$\frac{\mathrm{Q}}{\mathrm{t}_{2}}=\frac{\Delta \mathrm{Q}}{\frac{l}{\mathrm{~A}}\left(\frac{1}{3 \mathrm{~K}}\right)}$
Dividing equation (i) to (ii)
$\frac{\mathrm{Q} / \mathrm{t}_{1}}{\mathrm{Q} / \mathrm{t}_{2}}=\frac{\frac{\Delta \mathrm{Q}}{\frac{3}{2}(\mathrm{l} / \mathrm{A}) \frac{1}{\mathrm{~K}}}}{\frac{\Delta \mathrm{Q}}{\frac{1}{3} \frac{l}{\mathrm{~A}}\left(\frac{1}{\mathrm{~K}}\right)}}$
$\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}=\frac{2}{9}$
$\mathrm{t}_{2}=\frac{2}{9} \times 9=2 \mathrm{sec} \quad\left(\because \mathrm{t}_{1}=9 \mathrm{sec}\right)$

CG PET -2016

Heat Transfer

149337 Two spheres of different materials one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the larger sphere is $\mathbf{2 5}$ minute and for smaller one is $\mathbf{1 6}$ minute, the ratio of thermal conductivities of the materials of larger spheres to that of smaller sphere is

1 $4: 5$

2 $5: 4$

3 $25: 8$

4 $8: 25$

CG PET-2010

Heat Transfer

149339 If the temperature of a hot body is increased by $\mathbf{5 0} \%$ then the amount of radiation emitted by it increases approximately by

1 $400 \%$

2 $225 \%$

3 $250 \%$

4 $500 \%$

CG PET -2018

Heat Transfer

149340 A composite bar consists of a cylinder of radius $R$ and thermal conductivity $K_{1}$ fitted inside a cylindrical shell of internal radius $R$ and external radius $2 R$. If the thermal conductivity of shell is $K_{2}$, then the equivalent thermal conductivity of the composite bar is

1 $\mathrm{K}_{1}+\mathrm{K}_{2}$

2 $\frac{\mathrm{K}_{1}+3 \mathrm{~K}_{2}}{4}$

3 $\mathrm{K}_{1}+3 \mathrm{~K}_{2}$

4 $\frac{\mathrm{K}_{2}+3 \mathrm{~K}_{2}}{4}$

CG PET -2016

Heat Transfer

149341 Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or II as shown in figure. One of the blocks has thermal conductivity $K$ and the other $2 \mathrm{~K}$. The temperature difference between the ends along the $X$ - axis is the same in both the configurations. It takes $9 \mathrm{~s}$ to transport a certain amount of heat from the hot end to the cold end in configuration $I$. The time to transport the same amount of heat in the configuration II is

1 $2.0 \mathrm{~s}$

2 $3.0 \mathrm{~s}$

3 $4.5 \mathrm{~s}$

4 $6.0 \mathrm{~s}$

Explanation:

A original image
Thermal resistance circuit looks like electricalresistance circuit.
Then for - I
Equivalent thermal resistance,
$\mathrm{R}_{\mathrm{eq}} =\mathrm{R}_{1}+\mathrm{R}_{2} \quad \text { [Series combination] }$
$=\frac{l}{\mathrm{~K}_{1} \mathrm{~A}}+\frac{l}{\mathrm{~K}_{2} \mathrm{~A}}$
$=\frac{l}{\mathrm{~A}}\left(\frac{\mathrm{K}_{1}+\mathrm{K}_{2}}{\mathrm{~K}_{1} \mathrm{~K}_{2}}\right)=\frac{l}{\mathrm{~A}}\left(\frac{\mathrm{K}+2 \mathrm{~K}}{\mathrm{~K} \times 2 \mathrm{~K}}\right)=\frac{3}{2}\left(\frac{l}{\mathrm{~A}}\right) \frac{1}{\mathrm{~K}}$
For case - II
Equivalent thermal Resistance
$\frac{1}{\mathrm{R}_{\mathrm{eq}}^{\prime}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}} \quad \text { [Parallel combination] }$
$\mathrm{R}_{\mathrm{eq}}^{\prime}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$
$= \frac{\frac{l}{\mathrm{~K}_{1} \mathrm{~A}} \times \frac{l}{\mathrm{~K}_{2} \mathrm{~A}}}{\mathrm{~K}_{1} \mathrm{~A}}+\frac{l}{\mathrm{~K}_{2} \mathrm{~A}}$
$\mathrm{R}_{\mathrm{eq}}^{\prime}=\frac{\left(\frac{l}{\mathrm{~A}}\right)^{2}\left(\frac{1}{\mathrm{~K}_{1}} \frac{1}{\mathrm{~K}_{2}}\right)}{\left(\frac{l}{\mathrm{~A}}\right)\left(\frac{\mathrm{K}_{1}+\mathrm{K}_{2}}{\mathrm{~K}_{1} \mathrm{~K}_{2}}\right)}=\frac{l}{\mathrm{~A}}\left(\frac{1}{\mathrm{~K}_{1}+\mathrm{K}_{2}}\right)=\frac{l}{\mathrm{~A}}\left(\frac{1}{3 \mathrm{~K}}\right)$
Heat flow rate
$\frac{\mathrm{Q}}{\mathrm{t}_{1}}=\frac{\Delta \mathrm{Q}}{\mathrm{R}_{\text {eq. }}}$
For case-I
$\frac{\mathrm{Q}}{\mathrm{t}_{1}}=\frac{\Delta \mathrm{Q}}{\frac{3}{2}(\mathrm{l} / \mathrm{A}) \frac{1}{\mathrm{~K}}}$
For case - II
$\frac{\mathrm{Q}}{\mathrm{t}_{2}}=\frac{\Delta \mathrm{Q}}{\frac{l}{\mathrm{~A}}\left(\frac{1}{3 \mathrm{~K}}\right)}$
Dividing equation (i) to (ii)
$\frac{\mathrm{Q} / \mathrm{t}_{1}}{\mathrm{Q} / \mathrm{t}_{2}}=\frac{\frac{\Delta \mathrm{Q}}{\frac{3}{2}(\mathrm{l} / \mathrm{A}) \frac{1}{\mathrm{~K}}}}{\frac{\Delta \mathrm{Q}}{\frac{1}{3} \frac{l}{\mathrm{~A}}\left(\frac{1}{\mathrm{~K}}\right)}}$
$\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}=\frac{2}{9}$
$\mathrm{t}_{2}=\frac{2}{9} \times 9=2 \mathrm{sec} \quad\left(\because \mathrm{t}_{1}=9 \mathrm{sec}\right)$

CG PET -2016

Heat Transfer

149337 Two spheres of different materials one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the larger sphere is $\mathbf{2 5}$ minute and for smaller one is $\mathbf{1 6}$ minute, the ratio of thermal conductivities of the materials of larger spheres to that of smaller sphere is

1 $4: 5$

2 $5: 4$

3 $25: 8$

4 $8: 25$

CG PET-2010

Heat Transfer

149339 If the temperature of a hot body is increased by $\mathbf{5 0} \%$ then the amount of radiation emitted by it increases approximately by

1 $400 \%$

2 $225 \%$

3 $250 \%$

4 $500 \%$

CG PET -2018

Heat Transfer

149340 A composite bar consists of a cylinder of radius $R$ and thermal conductivity $K_{1}$ fitted inside a cylindrical shell of internal radius $R$ and external radius $2 R$. If the thermal conductivity of shell is $K_{2}$, then the equivalent thermal conductivity of the composite bar is

1 $\mathrm{K}_{1}+\mathrm{K}_{2}$

2 $\frac{\mathrm{K}_{1}+3 \mathrm{~K}_{2}}{4}$

3 $\mathrm{K}_{1}+3 \mathrm{~K}_{2}$

4 $\frac{\mathrm{K}_{2}+3 \mathrm{~K}_{2}}{4}$

CG PET -2016

Heat Transfer

149341 Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or II as shown in figure. One of the blocks has thermal conductivity $K$ and the other $2 \mathrm{~K}$. The temperature difference between the ends along the $X$ - axis is the same in both the configurations. It takes $9 \mathrm{~s}$ to transport a certain amount of heat from the hot end to the cold end in configuration $I$. The time to transport the same amount of heat in the configuration II is

1 $2.0 \mathrm{~s}$

2 $3.0 \mathrm{~s}$

3 $4.5 \mathrm{~s}$

4 $6.0 \mathrm{~s}$

Explanation:

A original image
Thermal resistance circuit looks like electricalresistance circuit.
Then for - I
Equivalent thermal resistance,
$\mathrm{R}_{\mathrm{eq}} =\mathrm{R}_{1}+\mathrm{R}_{2} \quad \text { [Series combination] }$
$=\frac{l}{\mathrm{~K}_{1} \mathrm{~A}}+\frac{l}{\mathrm{~K}_{2} \mathrm{~A}}$
$=\frac{l}{\mathrm{~A}}\left(\frac{\mathrm{K}_{1}+\mathrm{K}_{2}}{\mathrm{~K}_{1} \mathrm{~K}_{2}}\right)=\frac{l}{\mathrm{~A}}\left(\frac{\mathrm{K}+2 \mathrm{~K}}{\mathrm{~K} \times 2 \mathrm{~K}}\right)=\frac{3}{2}\left(\frac{l}{\mathrm{~A}}\right) \frac{1}{\mathrm{~K}}$
For case - II
Equivalent thermal Resistance
$\frac{1}{\mathrm{R}_{\mathrm{eq}}^{\prime}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}} \quad \text { [Parallel combination] }$
$\mathrm{R}_{\mathrm{eq}}^{\prime}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$
$= \frac{\frac{l}{\mathrm{~K}_{1} \mathrm{~A}} \times \frac{l}{\mathrm{~K}_{2} \mathrm{~A}}}{\mathrm{~K}_{1} \mathrm{~A}}+\frac{l}{\mathrm{~K}_{2} \mathrm{~A}}$
$\mathrm{R}_{\mathrm{eq}}^{\prime}=\frac{\left(\frac{l}{\mathrm{~A}}\right)^{2}\left(\frac{1}{\mathrm{~K}_{1}} \frac{1}{\mathrm{~K}_{2}}\right)}{\left(\frac{l}{\mathrm{~A}}\right)\left(\frac{\mathrm{K}_{1}+\mathrm{K}_{2}}{\mathrm{~K}_{1} \mathrm{~K}_{2}}\right)}=\frac{l}{\mathrm{~A}}\left(\frac{1}{\mathrm{~K}_{1}+\mathrm{K}_{2}}\right)=\frac{l}{\mathrm{~A}}\left(\frac{1}{3 \mathrm{~K}}\right)$
Heat flow rate
$\frac{\mathrm{Q}}{\mathrm{t}_{1}}=\frac{\Delta \mathrm{Q}}{\mathrm{R}_{\text {eq. }}}$
For case-I
$\frac{\mathrm{Q}}{\mathrm{t}_{1}}=\frac{\Delta \mathrm{Q}}{\frac{3}{2}(\mathrm{l} / \mathrm{A}) \frac{1}{\mathrm{~K}}}$
For case - II
$\frac{\mathrm{Q}}{\mathrm{t}_{2}}=\frac{\Delta \mathrm{Q}}{\frac{l}{\mathrm{~A}}\left(\frac{1}{3 \mathrm{~K}}\right)}$
Dividing equation (i) to (ii)
$\frac{\mathrm{Q} / \mathrm{t}_{1}}{\mathrm{Q} / \mathrm{t}_{2}}=\frac{\frac{\Delta \mathrm{Q}}{\frac{3}{2}(\mathrm{l} / \mathrm{A}) \frac{1}{\mathrm{~K}}}}{\frac{\Delta \mathrm{Q}}{\frac{1}{3} \frac{l}{\mathrm{~A}}\left(\frac{1}{\mathrm{~K}}\right)}}$
$\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}=\frac{2}{9}$
$\mathrm{t}_{2}=\frac{2}{9} \times 9=2 \mathrm{sec} \quad\left(\because \mathrm{t}_{1}=9 \mathrm{sec}\right)$

CG PET -2016

Heat Transfer

149337 Two spheres of different materials one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the larger sphere is $\mathbf{2 5}$ minute and for smaller one is $\mathbf{1 6}$ minute, the ratio of thermal conductivities of the materials of larger spheres to that of smaller sphere is

1 $4: 5$

2 $5: 4$

3 $25: 8$

4 $8: 25$

CG PET-2010

Heat Transfer

149339 If the temperature of a hot body is increased by $\mathbf{5 0} \%$ then the amount of radiation emitted by it increases approximately by

1 $400 \%$

2 $225 \%$

3 $250 \%$

4 $500 \%$

CG PET -2018

Heat Transfer

149340 A composite bar consists of a cylinder of radius $R$ and thermal conductivity $K_{1}$ fitted inside a cylindrical shell of internal radius $R$ and external radius $2 R$. If the thermal conductivity of shell is $K_{2}$, then the equivalent thermal conductivity of the composite bar is

1 $\mathrm{K}_{1}+\mathrm{K}_{2}$

2 $\frac{\mathrm{K}_{1}+3 \mathrm{~K}_{2}}{4}$

3 $\mathrm{K}_{1}+3 \mathrm{~K}_{2}$

4 $\frac{\mathrm{K}_{2}+3 \mathrm{~K}_{2}}{4}$

CG PET -2016

Heat Transfer

149341 Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or II as shown in figure. One of the blocks has thermal conductivity $K$ and the other $2 \mathrm{~K}$. The temperature difference between the ends along the $X$ - axis is the same in both the configurations. It takes $9 \mathrm{~s}$ to transport a certain amount of heat from the hot end to the cold end in configuration $I$. The time to transport the same amount of heat in the configuration II is

1 $2.0 \mathrm{~s}$

2 $3.0 \mathrm{~s}$

3 $4.5 \mathrm{~s}$

4 $6.0 \mathrm{~s}$

Explanation:

A original image
Thermal resistance circuit looks like electricalresistance circuit.
Then for - I
Equivalent thermal resistance,
$\mathrm{R}_{\mathrm{eq}} =\mathrm{R}_{1}+\mathrm{R}_{2} \quad \text { [Series combination] }$
$=\frac{l}{\mathrm{~K}_{1} \mathrm{~A}}+\frac{l}{\mathrm{~K}_{2} \mathrm{~A}}$
$=\frac{l}{\mathrm{~A}}\left(\frac{\mathrm{K}_{1}+\mathrm{K}_{2}}{\mathrm{~K}_{1} \mathrm{~K}_{2}}\right)=\frac{l}{\mathrm{~A}}\left(\frac{\mathrm{K}+2 \mathrm{~K}}{\mathrm{~K} \times 2 \mathrm{~K}}\right)=\frac{3}{2}\left(\frac{l}{\mathrm{~A}}\right) \frac{1}{\mathrm{~K}}$
For case - II
Equivalent thermal Resistance
$\frac{1}{\mathrm{R}_{\mathrm{eq}}^{\prime}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}} \quad \text { [Parallel combination] }$
$\mathrm{R}_{\mathrm{eq}}^{\prime}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$
$= \frac{\frac{l}{\mathrm{~K}_{1} \mathrm{~A}} \times \frac{l}{\mathrm{~K}_{2} \mathrm{~A}}}{\mathrm{~K}_{1} \mathrm{~A}}+\frac{l}{\mathrm{~K}_{2} \mathrm{~A}}$
$\mathrm{R}_{\mathrm{eq}}^{\prime}=\frac{\left(\frac{l}{\mathrm{~A}}\right)^{2}\left(\frac{1}{\mathrm{~K}_{1}} \frac{1}{\mathrm{~K}_{2}}\right)}{\left(\frac{l}{\mathrm{~A}}\right)\left(\frac{\mathrm{K}_{1}+\mathrm{K}_{2}}{\mathrm{~K}_{1} \mathrm{~K}_{2}}\right)}=\frac{l}{\mathrm{~A}}\left(\frac{1}{\mathrm{~K}_{1}+\mathrm{K}_{2}}\right)=\frac{l}{\mathrm{~A}}\left(\frac{1}{3 \mathrm{~K}}\right)$
Heat flow rate
$\frac{\mathrm{Q}}{\mathrm{t}_{1}}=\frac{\Delta \mathrm{Q}}{\mathrm{R}_{\text {eq. }}}$
For case-I
$\frac{\mathrm{Q}}{\mathrm{t}_{1}}=\frac{\Delta \mathrm{Q}}{\frac{3}{2}(\mathrm{l} / \mathrm{A}) \frac{1}{\mathrm{~K}}}$
For case - II
$\frac{\mathrm{Q}}{\mathrm{t}_{2}}=\frac{\Delta \mathrm{Q}}{\frac{l}{\mathrm{~A}}\left(\frac{1}{3 \mathrm{~K}}\right)}$
Dividing equation (i) to (ii)
$\frac{\mathrm{Q} / \mathrm{t}_{1}}{\mathrm{Q} / \mathrm{t}_{2}}=\frac{\frac{\Delta \mathrm{Q}}{\frac{3}{2}(\mathrm{l} / \mathrm{A}) \frac{1}{\mathrm{~K}}}}{\frac{\Delta \mathrm{Q}}{\frac{1}{3} \frac{l}{\mathrm{~A}}\left(\frac{1}{\mathrm{~K}}\right)}}$
$\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}=\frac{2}{9}$
$\mathrm{t}_{2}=\frac{2}{9} \times 9=2 \mathrm{sec} \quad\left(\because \mathrm{t}_{1}=9 \mathrm{sec}\right)$

CG PET -2016

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