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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 $25$ minute and for smaller one is $16$ 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

Explanation:

D We assume that,
Radius of small sphere $= r$
Thickness of small sphere $= t$
Radius of bigger sphere $= 2 r$
Thickness of bigger sphere $= t / 4$
Let $K_{1}$ and $K_{2}$ be the thermal conductivities of larger and smaller sphere.
Now, mass of melted ice $=$ Volume of sphere $\times$ Density of ice
$m = V \cdot ρ$
For bigger sphere, if
Heat lost $=$ Heat gained by ice
$\frac{K_{2} A_{2} (θ_{1} - θ_{2}) \times time}{\frac{t}{4}} = \frac{4}{3} π (2 r)^{3} \cdot ρ$
$\frac{4 [K_{2} 4 π (2 r)^{2} \times 100] \times 25 \times 60}{t} = \frac{4}{3} π 8 \times r^{3} ρ$
Similarly,
For smaller sphere,
$\frac{[K_{1} \times 4 π (r)^{2} \times 100] \times 16 \times 60}{t} = \frac{4}{3} π r^{3} ρ$
Dividing equation (i) from equation (ii), we get
$\frac{4 \times K_{2} \times 4 \times 25}{K_{1} \times 16} = \frac{8}{1}$
$\frac{K_{2}}{K_{1}} = \frac{8}{100} \times \frac{16}{4}$
$K_{2} : K_{1} = 8 : 25$

CG PET-2010

Heat Transfer

149339 If the temperature of a hot body is increased by $50 %$ then the amount of radiation emitted by it increases approximately by

1

400 %

2

225 %

3

250 %

4

500 %

Explanation:

A According to Stefan's law,
$E = σ T^{4}$
If temperature of hot body is increased by $50 %$ , then percentage increase in amount of radiation
$= \frac{E_{2} - E_{1}}{E_{1}} \times 100$
$= \frac{σ {(\frac{150}{100} T_{1})}^{4} - E σ {(T_{1})}^{4}}{σ T_{1}^{4}} \times 100$
$= 406.25 % = 400 %$

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

K_{1} + K_{2}

2

\frac{K_{1} + 3 K_{2}}{4}

3

K_{1} + 3 K_{2}

4

\frac{K_{2} + 3 K_{2}}{4}

Explanation:

B
Equivalent thermal resistance of cylinder and shell is,
$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{K 4 π R^{2}}{l} = \frac{K_{1} π R^{2}}{l} + \frac{K_{2} 3 π R^{2}}{l}$
$K = \frac{K_{1} + 3 K_{2}}{4}$

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 $2 K$ . The temperature difference between the ends along the $X$ - axis is the same in both the configurations. It takes $9 s$ 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 s

2

3.0 s

3

4.5 s

4

6.0 s

Explanation:

A original image
Thermal resistance circuit looks like electricalresistance circuit.
Then for - I
Equivalent thermal resistance,
$R_{eq} = R_{1} + R_{2} [Series combination]$
$= \frac{l}{K_{1} A} + \frac{l}{K_{2} A}$
$= \frac{l}{A} (\frac{K_{1} + K_{2}}{K_{1} K_{2}}) = \frac{l}{A} (\frac{K + 2 K}{K \times 2 K}) = \frac{3}{2} (\frac{l}{A}) \frac{1}{K}$
For case - II
Equivalent thermal Resistance
$\frac{1}{R_{eq}^{'}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} [Parallel combination]$
$R_{eq}^{'} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
$= \frac{\frac{l}{K_{1} A} \times \frac{l}{K_{2} A}}{K_{1} A} + \frac{l}{K_{2} A}$
$R_{eq}^{'} = \frac{{(\frac{l}{A})}^{2} (\frac{1}{K_{1}} \frac{1}{K_{2}})}{(\frac{l}{A}) (\frac{K_{1} + K_{2}}{K_{1} K_{2}})} = \frac{l}{A} (\frac{1}{K_{1} + K_{2}}) = \frac{l}{A} (\frac{1}{3 K})$
Heat flow rate
$\frac{Q}{t_{1}} = \frac{Δ Q}{R_{eq.}}$
For case-I
$\frac{Q}{t_{1}} = \frac{Δ Q}{\frac{3}{2} (l / A) \frac{1}{K}}$
For case - II
$\frac{Q}{t_{2}} = \frac{Δ Q}{\frac{l}{A} (\frac{1}{3 K})}$
Dividing equation (i) to (ii)
$\frac{Q / t_{1}}{Q / t_{2}} = \frac{\frac{Δ Q}{\frac{3}{2} (l / A) \frac{1}{K}}}{\frac{Δ Q}{\frac{1}{3} \frac{l}{A} (\frac{1}{K})}}$
$\frac{t_{2}}{t_{1}} = \frac{2}{9}$
$t_{2} = \frac{2}{9} \times 9 = 2 \sec (∵ t_{1} = 9 \sec)$

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 $25$ minute and for smaller one is $16$ 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

Explanation:

D We assume that,
Radius of small sphere $= r$
Thickness of small sphere $= t$
Radius of bigger sphere $= 2 r$
Thickness of bigger sphere $= t / 4$
Let $K_{1}$ and $K_{2}$ be the thermal conductivities of larger and smaller sphere.
Now, mass of melted ice $=$ Volume of sphere $\times$ Density of ice
$m = V \cdot ρ$
For bigger sphere, if
Heat lost $=$ Heat gained by ice
$\frac{K_{2} A_{2} (θ_{1} - θ_{2}) \times time}{\frac{t}{4}} = \frac{4}{3} π (2 r)^{3} \cdot ρ$
$\frac{4 [K_{2} 4 π (2 r)^{2} \times 100] \times 25 \times 60}{t} = \frac{4}{3} π 8 \times r^{3} ρ$
Similarly,
For smaller sphere,
$\frac{[K_{1} \times 4 π (r)^{2} \times 100] \times 16 \times 60}{t} = \frac{4}{3} π r^{3} ρ$
Dividing equation (i) from equation (ii), we get
$\frac{4 \times K_{2} \times 4 \times 25}{K_{1} \times 16} = \frac{8}{1}$
$\frac{K_{2}}{K_{1}} = \frac{8}{100} \times \frac{16}{4}$
$K_{2} : K_{1} = 8 : 25$

CG PET-2010

Heat Transfer

149339 If the temperature of a hot body is increased by $50 %$ then the amount of radiation emitted by it increases approximately by

1

400 %

2

225 %

3

250 %

4

500 %

Explanation:

A According to Stefan's law,
$E = σ T^{4}$
If temperature of hot body is increased by $50 %$ , then percentage increase in amount of radiation
$= \frac{E_{2} - E_{1}}{E_{1}} \times 100$
$= \frac{σ {(\frac{150}{100} T_{1})}^{4} - E σ {(T_{1})}^{4}}{σ T_{1}^{4}} \times 100$
$= 406.25 % = 400 %$

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

K_{1} + K_{2}

2

\frac{K_{1} + 3 K_{2}}{4}

3

K_{1} + 3 K_{2}

4

\frac{K_{2} + 3 K_{2}}{4}

Explanation:

B
Equivalent thermal resistance of cylinder and shell is,
$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{K 4 π R^{2}}{l} = \frac{K_{1} π R^{2}}{l} + \frac{K_{2} 3 π R^{2}}{l}$
$K = \frac{K_{1} + 3 K_{2}}{4}$

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 $2 K$ . The temperature difference between the ends along the $X$ - axis is the same in both the configurations. It takes $9 s$ 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 s

2

3.0 s

3

4.5 s

4

6.0 s

Explanation:

A original image
Thermal resistance circuit looks like electricalresistance circuit.
Then for - I
Equivalent thermal resistance,
$R_{eq} = R_{1} + R_{2} [Series combination]$
$= \frac{l}{K_{1} A} + \frac{l}{K_{2} A}$
$= \frac{l}{A} (\frac{K_{1} + K_{2}}{K_{1} K_{2}}) = \frac{l}{A} (\frac{K + 2 K}{K \times 2 K}) = \frac{3}{2} (\frac{l}{A}) \frac{1}{K}$
For case - II
Equivalent thermal Resistance
$\frac{1}{R_{eq}^{'}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} [Parallel combination]$
$R_{eq}^{'} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
$= \frac{\frac{l}{K_{1} A} \times \frac{l}{K_{2} A}}{K_{1} A} + \frac{l}{K_{2} A}$
$R_{eq}^{'} = \frac{{(\frac{l}{A})}^{2} (\frac{1}{K_{1}} \frac{1}{K_{2}})}{(\frac{l}{A}) (\frac{K_{1} + K_{2}}{K_{1} K_{2}})} = \frac{l}{A} (\frac{1}{K_{1} + K_{2}}) = \frac{l}{A} (\frac{1}{3 K})$
Heat flow rate
$\frac{Q}{t_{1}} = \frac{Δ Q}{R_{eq.}}$
For case-I
$\frac{Q}{t_{1}} = \frac{Δ Q}{\frac{3}{2} (l / A) \frac{1}{K}}$
For case - II
$\frac{Q}{t_{2}} = \frac{Δ Q}{\frac{l}{A} (\frac{1}{3 K})}$
Dividing equation (i) to (ii)
$\frac{Q / t_{1}}{Q / t_{2}} = \frac{\frac{Δ Q}{\frac{3}{2} (l / A) \frac{1}{K}}}{\frac{Δ Q}{\frac{1}{3} \frac{l}{A} (\frac{1}{K})}}$
$\frac{t_{2}}{t_{1}} = \frac{2}{9}$
$t_{2} = \frac{2}{9} \times 9 = 2 \sec (∵ t_{1} = 9 \sec)$

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 $25$ minute and for smaller one is $16$ 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

Explanation:

D We assume that,
Radius of small sphere $= r$
Thickness of small sphere $= t$
Radius of bigger sphere $= 2 r$
Thickness of bigger sphere $= t / 4$
Let $K_{1}$ and $K_{2}$ be the thermal conductivities of larger and smaller sphere.
Now, mass of melted ice $=$ Volume of sphere $\times$ Density of ice
$m = V \cdot ρ$
For bigger sphere, if
Heat lost $=$ Heat gained by ice
$\frac{K_{2} A_{2} (θ_{1} - θ_{2}) \times time}{\frac{t}{4}} = \frac{4}{3} π (2 r)^{3} \cdot ρ$
$\frac{4 [K_{2} 4 π (2 r)^{2} \times 100] \times 25 \times 60}{t} = \frac{4}{3} π 8 \times r^{3} ρ$
Similarly,
For smaller sphere,
$\frac{[K_{1} \times 4 π (r)^{2} \times 100] \times 16 \times 60}{t} = \frac{4}{3} π r^{3} ρ$
Dividing equation (i) from equation (ii), we get
$\frac{4 \times K_{2} \times 4 \times 25}{K_{1} \times 16} = \frac{8}{1}$
$\frac{K_{2}}{K_{1}} = \frac{8}{100} \times \frac{16}{4}$
$K_{2} : K_{1} = 8 : 25$

CG PET-2010

Heat Transfer

149339 If the temperature of a hot body is increased by $50 %$ then the amount of radiation emitted by it increases approximately by

1

400 %

2

225 %

3

250 %

4

500 %

Explanation:

A According to Stefan's law,
$E = σ T^{4}$
If temperature of hot body is increased by $50 %$ , then percentage increase in amount of radiation
$= \frac{E_{2} - E_{1}}{E_{1}} \times 100$
$= \frac{σ {(\frac{150}{100} T_{1})}^{4} - E σ {(T_{1})}^{4}}{σ T_{1}^{4}} \times 100$
$= 406.25 % = 400 %$

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

K_{1} + K_{2}

2

\frac{K_{1} + 3 K_{2}}{4}

3

K_{1} + 3 K_{2}

4

\frac{K_{2} + 3 K_{2}}{4}

Explanation:

B
Equivalent thermal resistance of cylinder and shell is,
$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{K 4 π R^{2}}{l} = \frac{K_{1} π R^{2}}{l} + \frac{K_{2} 3 π R^{2}}{l}$
$K = \frac{K_{1} + 3 K_{2}}{4}$

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 $2 K$ . The temperature difference between the ends along the $X$ - axis is the same in both the configurations. It takes $9 s$ 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 s

2

3.0 s

3

4.5 s

4

6.0 s

Explanation:

A original image
Thermal resistance circuit looks like electricalresistance circuit.
Then for - I
Equivalent thermal resistance,
$R_{eq} = R_{1} + R_{2} [Series combination]$
$= \frac{l}{K_{1} A} + \frac{l}{K_{2} A}$
$= \frac{l}{A} (\frac{K_{1} + K_{2}}{K_{1} K_{2}}) = \frac{l}{A} (\frac{K + 2 K}{K \times 2 K}) = \frac{3}{2} (\frac{l}{A}) \frac{1}{K}$
For case - II
Equivalent thermal Resistance
$\frac{1}{R_{eq}^{'}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} [Parallel combination]$
$R_{eq}^{'} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
$= \frac{\frac{l}{K_{1} A} \times \frac{l}{K_{2} A}}{K_{1} A} + \frac{l}{K_{2} A}$
$R_{eq}^{'} = \frac{{(\frac{l}{A})}^{2} (\frac{1}{K_{1}} \frac{1}{K_{2}})}{(\frac{l}{A}) (\frac{K_{1} + K_{2}}{K_{1} K_{2}})} = \frac{l}{A} (\frac{1}{K_{1} + K_{2}}) = \frac{l}{A} (\frac{1}{3 K})$
Heat flow rate
$\frac{Q}{t_{1}} = \frac{Δ Q}{R_{eq.}}$
For case-I
$\frac{Q}{t_{1}} = \frac{Δ Q}{\frac{3}{2} (l / A) \frac{1}{K}}$
For case - II
$\frac{Q}{t_{2}} = \frac{Δ Q}{\frac{l}{A} (\frac{1}{3 K})}$
Dividing equation (i) to (ii)
$\frac{Q / t_{1}}{Q / t_{2}} = \frac{\frac{Δ Q}{\frac{3}{2} (l / A) \frac{1}{K}}}{\frac{Δ Q}{\frac{1}{3} \frac{l}{A} (\frac{1}{K})}}$
$\frac{t_{2}}{t_{1}} = \frac{2}{9}$
$t_{2} = \frac{2}{9} \times 9 = 2 \sec (∵ t_{1} = 9 \sec)$

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 $25$ minute and for smaller one is $16$ 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

Explanation:

D We assume that,
Radius of small sphere $= r$
Thickness of small sphere $= t$
Radius of bigger sphere $= 2 r$
Thickness of bigger sphere $= t / 4$
Let $K_{1}$ and $K_{2}$ be the thermal conductivities of larger and smaller sphere.
Now, mass of melted ice $=$ Volume of sphere $\times$ Density of ice
$m = V \cdot ρ$
For bigger sphere, if
Heat lost $=$ Heat gained by ice
$\frac{K_{2} A_{2} (θ_{1} - θ_{2}) \times time}{\frac{t}{4}} = \frac{4}{3} π (2 r)^{3} \cdot ρ$
$\frac{4 [K_{2} 4 π (2 r)^{2} \times 100] \times 25 \times 60}{t} = \frac{4}{3} π 8 \times r^{3} ρ$
Similarly,
For smaller sphere,
$\frac{[K_{1} \times 4 π (r)^{2} \times 100] \times 16 \times 60}{t} = \frac{4}{3} π r^{3} ρ$
Dividing equation (i) from equation (ii), we get
$\frac{4 \times K_{2} \times 4 \times 25}{K_{1} \times 16} = \frac{8}{1}$
$\frac{K_{2}}{K_{1}} = \frac{8}{100} \times \frac{16}{4}$
$K_{2} : K_{1} = 8 : 25$

CG PET-2010

Heat Transfer

149339 If the temperature of a hot body is increased by $50 %$ then the amount of radiation emitted by it increases approximately by

1

400 %

2

225 %

3

250 %

4

500 %

Explanation:

A According to Stefan's law,
$E = σ T^{4}$
If temperature of hot body is increased by $50 %$ , then percentage increase in amount of radiation
$= \frac{E_{2} - E_{1}}{E_{1}} \times 100$
$= \frac{σ {(\frac{150}{100} T_{1})}^{4} - E σ {(T_{1})}^{4}}{σ T_{1}^{4}} \times 100$
$= 406.25 % = 400 %$

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

K_{1} + K_{2}

2

\frac{K_{1} + 3 K_{2}}{4}

3

K_{1} + 3 K_{2}

4

\frac{K_{2} + 3 K_{2}}{4}

Explanation:

B
Equivalent thermal resistance of cylinder and shell is,
$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{K 4 π R^{2}}{l} = \frac{K_{1} π R^{2}}{l} + \frac{K_{2} 3 π R^{2}}{l}$
$K = \frac{K_{1} + 3 K_{2}}{4}$

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 $2 K$ . The temperature difference between the ends along the $X$ - axis is the same in both the configurations. It takes $9 s$ 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 s

2

3.0 s

3

4.5 s

4

6.0 s

Explanation:

A original image
Thermal resistance circuit looks like electricalresistance circuit.
Then for - I
Equivalent thermal resistance,
$R_{eq} = R_{1} + R_{2} [Series combination]$
$= \frac{l}{K_{1} A} + \frac{l}{K_{2} A}$
$= \frac{l}{A} (\frac{K_{1} + K_{2}}{K_{1} K_{2}}) = \frac{l}{A} (\frac{K + 2 K}{K \times 2 K}) = \frac{3}{2} (\frac{l}{A}) \frac{1}{K}$
For case - II
Equivalent thermal Resistance
$\frac{1}{R_{eq}^{'}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} [Parallel combination]$
$R_{eq}^{'} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
$= \frac{\frac{l}{K_{1} A} \times \frac{l}{K_{2} A}}{K_{1} A} + \frac{l}{K_{2} A}$
$R_{eq}^{'} = \frac{{(\frac{l}{A})}^{2} (\frac{1}{K_{1}} \frac{1}{K_{2}})}{(\frac{l}{A}) (\frac{K_{1} + K_{2}}{K_{1} K_{2}})} = \frac{l}{A} (\frac{1}{K_{1} + K_{2}}) = \frac{l}{A} (\frac{1}{3 K})$
Heat flow rate
$\frac{Q}{t_{1}} = \frac{Δ Q}{R_{eq.}}$
For case-I
$\frac{Q}{t_{1}} = \frac{Δ Q}{\frac{3}{2} (l / A) \frac{1}{K}}$
For case - II
$\frac{Q}{t_{2}} = \frac{Δ Q}{\frac{l}{A} (\frac{1}{3 K})}$
Dividing equation (i) to (ii)
$\frac{Q / t_{1}}{Q / t_{2}} = \frac{\frac{Δ Q}{\frac{3}{2} (l / A) \frac{1}{K}}}{\frac{Δ Q}{\frac{1}{3} \frac{l}{A} (\frac{1}{K})}}$
$\frac{t_{2}}{t_{1}} = \frac{2}{9}$
$t_{2} = \frac{2}{9} \times 9 = 2 \sec (∵ t_{1} = 9 \sec)$

CG PET -2016