QBManager

Heat Transfer

149343 The end $A$ of a rod $A B$ of length $1 m$ is maintained at $100^{\circ}$ and the end $B$ at $10^{\circ} C$ . The temperature at a distance of $60 cm$ from the end $B$ is

1

64^{\circ} C

2

36^{\circ} C

3

46^{\circ} C

4

72^{\circ} C

Explanation:

A Given, length of $Rod A B = 1 m$
$T_{A} = 100^{\circ} C$
$T_{B} = 10^{\circ} C$

Let $θ$ is the temperature at the point $C$ . Heat rate is same through the length of the rod because of steady state heat transfer.
$\frac{Q}{t} = \frac{KA (θ_{1} - θ_{2})}{L}$
As point ' $C$ '
$\frac{(100^{\circ} C - θ) KA}{40} = \frac{(θ - 10^{\circ} C) KA}{60}$
$300^{\circ} C - 3 θ = 2 θ - 20^{\circ} C$
$5 θ = 320^{\circ} C$
$θ = 64^{\circ} C$

CG PET- 2008

Heat Transfer

149344 Two identical vessels made up of same material are filled with same amount of ice. If in the vessels the ice melts in time $t_{1}$ and $t_{2}$ respectively, then the ratio of their thermal conductivities will be

1

t_{2} : t_{1}

2

t_{1} : t_{2}

3

t_{2}^{2} : t_{1}^{2}

4

t_{1}^{2} : t_{2}^{2}

Explanation:

A We know
$Q = \frac{KA Δ θ t}{l}$
$\frac{Q}{t} = \frac{KA Δ θ}{l}$
Here, $A, Δ θ, l$ and $Q$ are same.
$\frac{(\frac{Q}{t_{1}})}{(\frac{Q}{t_{2}})} = \frac{(\frac{K_{1} A Δ θ}{l})}{(\frac{K_{2} A Δ θ}{l})}$
$\frac{t_{2}}{t_{1}} = \frac{K_{1}}{K_{2}}$
$\frac{K_{1}}{K_{2}} = \frac{t_{2}}{t_{1}}$

CG PET- 2007

Heat Transfer

149345 A cylinder of radius $R$ and made of a material of thermal conductivity $K_{1}$ is surrounded by a cylindrical shell of inner radius $R$ and outer radius $2 R$ made of a material of thermal conductivity $K_{2}$ . The two ends of the combined system are maintained at two different temperature. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is

1

K_{1} + K_{2}

2

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

3

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

4

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

Explanation:

C Rate of heat flow in inner cylinder
$H_{1} = \frac{K_{1} A_{1} Δ θ}{l} = \frac{K_{1} π R^{2} Δ θ}{l}$
Rate of heat flow in outer cylinder
$H_{2} = \frac{K_{2} A_{2} Δ θ}{l} = \frac{K_{2} [π (2 R)^{2} - π R^{2}] Δ θ}{l}$
$H_{2} = \frac{K_{2} 3 π R^{2} Δ θ}{l}$
Total rate of heat flow
$H = H_{1} + H_{2}$
$= \frac{K_{1} π R^{2} Δ θ}{l} + \frac{K_{2} 3 π R^{2} Δ θ}{l}$
$= \frac{(K_{1} + 3 K_{2}) π R^{2} Δ θ}{l}$
If $K$ is the resultant thermal conductivity, then
$H = \frac{K π (2 R)^{2} Δ θ}{l} = \frac{K 4 R^{2} Δ θ}{l}$
From equation (iii) and (iv)
$K = \frac{K_{1} + 3 K_{2}}{4}$

AMU-2016

Heat Transfer

149346 Two rods (one semi-circular and other straight) of same material and of same cross-sectional area are joined as shown in the figure. The points $A$ and $B$ are maintained at different temperature. The ratio of the heat transferred through a cross-section of the straight rod in a given time is

1

2 : π

2

1 : 2

3

π : 2

4

3 : 2

Explanation:

A
Let, $R$ , radius of semicircle and $2 R$ is length of straight rod.
We know,
Heat transferred rate.
$\frac{dQ}{dt} = \frac{KA Δ θ}{l}$
For the semi-circle and straight rod K,A and $Δ θ$ is same then
$\frac{Δ Q}{dt} \propto \frac{1}{l}$
$\frac{{(\frac{dQ}{dt})}_{semi-circular}}{{(\frac{dQ}{dt})}_{straight rod}} = \frac{l_{straight rod}}{l_{semi-circular rod}} = \frac{2 R}{π R} = \frac{2}{π}$

Manipal UGET-2014

Heat Transfer

149343 The end $A$ of a rod $A B$ of length $1 m$ is maintained at $100^{\circ}$ and the end $B$ at $10^{\circ} C$ . The temperature at a distance of $60 cm$ from the end $B$ is

1

64^{\circ} C

2

36^{\circ} C

3

46^{\circ} C

4

72^{\circ} C

Explanation:

A Given, length of $Rod A B = 1 m$
$T_{A} = 100^{\circ} C$
$T_{B} = 10^{\circ} C$

Let $θ$ is the temperature at the point $C$ . Heat rate is same through the length of the rod because of steady state heat transfer.
$\frac{Q}{t} = \frac{KA (θ_{1} - θ_{2})}{L}$
As point ' $C$ '
$\frac{(100^{\circ} C - θ) KA}{40} = \frac{(θ - 10^{\circ} C) KA}{60}$
$300^{\circ} C - 3 θ = 2 θ - 20^{\circ} C$
$5 θ = 320^{\circ} C$
$θ = 64^{\circ} C$

CG PET- 2008

Heat Transfer

149344 Two identical vessels made up of same material are filled with same amount of ice. If in the vessels the ice melts in time $t_{1}$ and $t_{2}$ respectively, then the ratio of their thermal conductivities will be

1

t_{2} : t_{1}

2

t_{1} : t_{2}

3

t_{2}^{2} : t_{1}^{2}

4

t_{1}^{2} : t_{2}^{2}

Explanation:

A We know
$Q = \frac{KA Δ θ t}{l}$
$\frac{Q}{t} = \frac{KA Δ θ}{l}$
Here, $A, Δ θ, l$ and $Q$ are same.
$\frac{(\frac{Q}{t_{1}})}{(\frac{Q}{t_{2}})} = \frac{(\frac{K_{1} A Δ θ}{l})}{(\frac{K_{2} A Δ θ}{l})}$
$\frac{t_{2}}{t_{1}} = \frac{K_{1}}{K_{2}}$
$\frac{K_{1}}{K_{2}} = \frac{t_{2}}{t_{1}}$

CG PET- 2007

Heat Transfer

149345 A cylinder of radius $R$ and made of a material of thermal conductivity $K_{1}$ is surrounded by a cylindrical shell of inner radius $R$ and outer radius $2 R$ made of a material of thermal conductivity $K_{2}$ . The two ends of the combined system are maintained at two different temperature. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is

1

K_{1} + K_{2}

2

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

3

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

4

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

Explanation:

C Rate of heat flow in inner cylinder
$H_{1} = \frac{K_{1} A_{1} Δ θ}{l} = \frac{K_{1} π R^{2} Δ θ}{l}$
Rate of heat flow in outer cylinder
$H_{2} = \frac{K_{2} A_{2} Δ θ}{l} = \frac{K_{2} [π (2 R)^{2} - π R^{2}] Δ θ}{l}$
$H_{2} = \frac{K_{2} 3 π R^{2} Δ θ}{l}$
Total rate of heat flow
$H = H_{1} + H_{2}$
$= \frac{K_{1} π R^{2} Δ θ}{l} + \frac{K_{2} 3 π R^{2} Δ θ}{l}$
$= \frac{(K_{1} + 3 K_{2}) π R^{2} Δ θ}{l}$
If $K$ is the resultant thermal conductivity, then
$H = \frac{K π (2 R)^{2} Δ θ}{l} = \frac{K 4 R^{2} Δ θ}{l}$
From equation (iii) and (iv)
$K = \frac{K_{1} + 3 K_{2}}{4}$

AMU-2016

Heat Transfer

149346 Two rods (one semi-circular and other straight) of same material and of same cross-sectional area are joined as shown in the figure. The points $A$ and $B$ are maintained at different temperature. The ratio of the heat transferred through a cross-section of the straight rod in a given time is

1

2 : π

2

1 : 2

3

π : 2

4

3 : 2

Explanation:

A
Let, $R$ , radius of semicircle and $2 R$ is length of straight rod.
We know,
Heat transferred rate.
$\frac{dQ}{dt} = \frac{KA Δ θ}{l}$
For the semi-circle and straight rod K,A and $Δ θ$ is same then
$\frac{Δ Q}{dt} \propto \frac{1}{l}$
$\frac{{(\frac{dQ}{dt})}_{semi-circular}}{{(\frac{dQ}{dt})}_{straight rod}} = \frac{l_{straight rod}}{l_{semi-circular rod}} = \frac{2 R}{π R} = \frac{2}{π}$

Manipal UGET-2014

Heat Transfer

149343 The end $A$ of a rod $A B$ of length $1 m$ is maintained at $100^{\circ}$ and the end $B$ at $10^{\circ} C$ . The temperature at a distance of $60 cm$ from the end $B$ is

1

64^{\circ} C

2

36^{\circ} C

3

46^{\circ} C

4

72^{\circ} C

Explanation:

A Given, length of $Rod A B = 1 m$
$T_{A} = 100^{\circ} C$
$T_{B} = 10^{\circ} C$

Let $θ$ is the temperature at the point $C$ . Heat rate is same through the length of the rod because of steady state heat transfer.
$\frac{Q}{t} = \frac{KA (θ_{1} - θ_{2})}{L}$
As point ' $C$ '
$\frac{(100^{\circ} C - θ) KA}{40} = \frac{(θ - 10^{\circ} C) KA}{60}$
$300^{\circ} C - 3 θ = 2 θ - 20^{\circ} C$
$5 θ = 320^{\circ} C$
$θ = 64^{\circ} C$

CG PET- 2008

Heat Transfer

149344 Two identical vessels made up of same material are filled with same amount of ice. If in the vessels the ice melts in time $t_{1}$ and $t_{2}$ respectively, then the ratio of their thermal conductivities will be

1

t_{2} : t_{1}

2

t_{1} : t_{2}

3

t_{2}^{2} : t_{1}^{2}

4

t_{1}^{2} : t_{2}^{2}

Explanation:

A We know
$Q = \frac{KA Δ θ t}{l}$
$\frac{Q}{t} = \frac{KA Δ θ}{l}$
Here, $A, Δ θ, l$ and $Q$ are same.
$\frac{(\frac{Q}{t_{1}})}{(\frac{Q}{t_{2}})} = \frac{(\frac{K_{1} A Δ θ}{l})}{(\frac{K_{2} A Δ θ}{l})}$
$\frac{t_{2}}{t_{1}} = \frac{K_{1}}{K_{2}}$
$\frac{K_{1}}{K_{2}} = \frac{t_{2}}{t_{1}}$

CG PET- 2007

Heat Transfer

149345 A cylinder of radius $R$ and made of a material of thermal conductivity $K_{1}$ is surrounded by a cylindrical shell of inner radius $R$ and outer radius $2 R$ made of a material of thermal conductivity $K_{2}$ . The two ends of the combined system are maintained at two different temperature. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is

1

K_{1} + K_{2}

2

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

3

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

4

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

Explanation:

C Rate of heat flow in inner cylinder
$H_{1} = \frac{K_{1} A_{1} Δ θ}{l} = \frac{K_{1} π R^{2} Δ θ}{l}$
Rate of heat flow in outer cylinder
$H_{2} = \frac{K_{2} A_{2} Δ θ}{l} = \frac{K_{2} [π (2 R)^{2} - π R^{2}] Δ θ}{l}$
$H_{2} = \frac{K_{2} 3 π R^{2} Δ θ}{l}$
Total rate of heat flow
$H = H_{1} + H_{2}$
$= \frac{K_{1} π R^{2} Δ θ}{l} + \frac{K_{2} 3 π R^{2} Δ θ}{l}$
$= \frac{(K_{1} + 3 K_{2}) π R^{2} Δ θ}{l}$
If $K$ is the resultant thermal conductivity, then
$H = \frac{K π (2 R)^{2} Δ θ}{l} = \frac{K 4 R^{2} Δ θ}{l}$
From equation (iii) and (iv)
$K = \frac{K_{1} + 3 K_{2}}{4}$

AMU-2016

Heat Transfer

149346 Two rods (one semi-circular and other straight) of same material and of same cross-sectional area are joined as shown in the figure. The points $A$ and $B$ are maintained at different temperature. The ratio of the heat transferred through a cross-section of the straight rod in a given time is

1

2 : π

2

1 : 2

3

π : 2

4

3 : 2

Explanation:

A
Let, $R$ , radius of semicircle and $2 R$ is length of straight rod.
We know,
Heat transferred rate.
$\frac{dQ}{dt} = \frac{KA Δ θ}{l}$
For the semi-circle and straight rod K,A and $Δ θ$ is same then
$\frac{Δ Q}{dt} \propto \frac{1}{l}$
$\frac{{(\frac{dQ}{dt})}_{semi-circular}}{{(\frac{dQ}{dt})}_{straight rod}} = \frac{l_{straight rod}}{l_{semi-circular rod}} = \frac{2 R}{π R} = \frac{2}{π}$

Manipal UGET-2014

Heat Transfer

149343 The end $A$ of a rod $A B$ of length $1 m$ is maintained at $100^{\circ}$ and the end $B$ at $10^{\circ} C$ . The temperature at a distance of $60 cm$ from the end $B$ is

1

64^{\circ} C

2

36^{\circ} C

3

46^{\circ} C

4

72^{\circ} C

Explanation:

A Given, length of $Rod A B = 1 m$
$T_{A} = 100^{\circ} C$
$T_{B} = 10^{\circ} C$

Let $θ$ is the temperature at the point $C$ . Heat rate is same through the length of the rod because of steady state heat transfer.
$\frac{Q}{t} = \frac{KA (θ_{1} - θ_{2})}{L}$
As point ' $C$ '
$\frac{(100^{\circ} C - θ) KA}{40} = \frac{(θ - 10^{\circ} C) KA}{60}$
$300^{\circ} C - 3 θ = 2 θ - 20^{\circ} C$
$5 θ = 320^{\circ} C$
$θ = 64^{\circ} C$

CG PET- 2008

Heat Transfer

149344 Two identical vessels made up of same material are filled with same amount of ice. If in the vessels the ice melts in time $t_{1}$ and $t_{2}$ respectively, then the ratio of their thermal conductivities will be

1

t_{2} : t_{1}

2

t_{1} : t_{2}

3

t_{2}^{2} : t_{1}^{2}

4

t_{1}^{2} : t_{2}^{2}

Explanation:

A We know
$Q = \frac{KA Δ θ t}{l}$
$\frac{Q}{t} = \frac{KA Δ θ}{l}$
Here, $A, Δ θ, l$ and $Q$ are same.
$\frac{(\frac{Q}{t_{1}})}{(\frac{Q}{t_{2}})} = \frac{(\frac{K_{1} A Δ θ}{l})}{(\frac{K_{2} A Δ θ}{l})}$
$\frac{t_{2}}{t_{1}} = \frac{K_{1}}{K_{2}}$
$\frac{K_{1}}{K_{2}} = \frac{t_{2}}{t_{1}}$

CG PET- 2007

Heat Transfer

149345 A cylinder of radius $R$ and made of a material of thermal conductivity $K_{1}$ is surrounded by a cylindrical shell of inner radius $R$ and outer radius $2 R$ made of a material of thermal conductivity $K_{2}$ . The two ends of the combined system are maintained at two different temperature. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is

1

K_{1} + K_{2}

2

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

3

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

4

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

Explanation:

C Rate of heat flow in inner cylinder
$H_{1} = \frac{K_{1} A_{1} Δ θ}{l} = \frac{K_{1} π R^{2} Δ θ}{l}$
Rate of heat flow in outer cylinder
$H_{2} = \frac{K_{2} A_{2} Δ θ}{l} = \frac{K_{2} [π (2 R)^{2} - π R^{2}] Δ θ}{l}$
$H_{2} = \frac{K_{2} 3 π R^{2} Δ θ}{l}$
Total rate of heat flow
$H = H_{1} + H_{2}$
$= \frac{K_{1} π R^{2} Δ θ}{l} + \frac{K_{2} 3 π R^{2} Δ θ}{l}$
$= \frac{(K_{1} + 3 K_{2}) π R^{2} Δ θ}{l}$
If $K$ is the resultant thermal conductivity, then
$H = \frac{K π (2 R)^{2} Δ θ}{l} = \frac{K 4 R^{2} Δ θ}{l}$
From equation (iii) and (iv)
$K = \frac{K_{1} + 3 K_{2}}{4}$

AMU-2016

Heat Transfer

149346 Two rods (one semi-circular and other straight) of same material and of same cross-sectional area are joined as shown in the figure. The points $A$ and $B$ are maintained at different temperature. The ratio of the heat transferred through a cross-section of the straight rod in a given time is

1

2 : π

2

1 : 2

3

π : 2

4

3 : 2

Explanation:

A
Let, $R$ , radius of semicircle and $2 R$ is length of straight rod.
We know,
Heat transferred rate.
$\frac{dQ}{dt} = \frac{KA Δ θ}{l}$
For the semi-circle and straight rod K,A and $Δ θ$ is same then
$\frac{Δ Q}{dt} \propto \frac{1}{l}$
$\frac{{(\frac{dQ}{dt})}_{semi-circular}}{{(\frac{dQ}{dt})}_{straight rod}} = \frac{l_{straight rod}}{l_{semi-circular rod}} = \frac{2 R}{π R} = \frac{2}{π}$

Manipal UGET-2014

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