MyQuestionIcon

14

You visited us 14 times! Enjoying our articles? Unlock Full Access!

Conduction Law

Three rods ...

Question

Three rods $A, B$ and $C$ have the same dimensions. Their thermal conductivities $k_{A}, k_{B}$ and $k_{C}$ respectively. $A$ and $B$ are placed end to end, with the free ends kept at a certain temperature difference. $C$ is placed separately with its ends kept at same temperature difference. The two arrangements conduct heat at the same rate $k_{C}$ equal to

A

k_{A} + k_{B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

\frac{k_{A} k_{B}}{k_{A} + k_{B}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

\frac{1}{2} (k_{A} + k_{B})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

2 \frac{k_{A} k_{B}}{k_{A} + k_{B}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{k_{A} k_{B}}{k_{A} + k_{B}}$
Given that three rods A, B and C have same dimensions. Their thermal conductivities $k_{A}$ , $k_{B}$ and $k_{C}$ respectively. A and B are placed end to end,
with the free ends kept at a certain temperature difference. C is placed separately with its ends kept at same temperature difference.
Heat flow through A and B rods at thermal equilibrium is equal
$H_{A} = H_{B} = \frac{k_{A} a (T - T_{1})}{L} = \frac{k_{B} a (T_{2} - T)}{L} (k_{A} + k_{B}) T = k_{A} T_{1} + k_{B} T_{2} T = \frac{k_{A} T_{1} + k_{B} T_{2}}{k_{A} + k_{B}}$
where T is the temperature of the common end Of A and B, $T_{1}$ and $T_{2}$ are temperatures at the other ends of A and B, $a$ is area and $L$ is the length of the each rod
Heat flow through C
$H_{C} = \frac{k_{C} a (T_{2} - T_{1})}{L}$
If the rate of heat flow for the above two arrangements are equal
$H_{A} = H_{B} = H_{C}$
$\frac{k_{A} a (\frac{k_{A} T_{1} + k_{B} T_{2}}{k_{A} + k_{B}} - T_{1})}{L} = \frac{k_{C} a (T_{2} - T_{1})}{L} \frac{k_{A} (k_{A} T_{1} + k_{B} T_{2} - k_{A} T_{1} - k_{B} T_{1})}{k_{A} + k_{B}} = k_{C} (T_{2} - T_{1}) \frac{k_{A} k_{B}}{k_{A} + k_{B}} = k_{C}$
option (B) is the correct answer.

Suggest Corrections

thumbs-up

0

Similar questions

Q. Three rods A,B and C have the same dimensions.Their conductivities are

K_{A}, K_{B}

K_{C}

respectively. A and B are placed end to end, with their free ends kept at certain temperature difference. C is placed separately with its ends kept at same temperature difference. The two arrangements conduct heat at the same rate

K_{c}

must be equal to

A, B

C

have the same dimension. Their thermal conductivities are

k_{A}, k_{B}

k_{C}

. They are placed individually, with their ends kept at the same temperature difference. The rate of heat flow through

C

is equal to the rate of combined heat flow through

A

B

k_{C}

must be equal to:

Q. The kinetic energies of a planet in an elliptical orbit about the sun at positions

A

B

C

K_{A}

K_{B}

K_{C}

A C

is the major axis and

S B

is perpendicular to

A C

at the position of the sun

S

as shown in the figure with

S A < S B

Q. Two reactions,

A \to p r o d u c t s

B \to p r o d u c t s

, have rate constant

K_{a}

K_{b}

at temperature T and activation energies

E_{a}

E_{b}

respectively. If

K_{a} > K_{b}

E_{a} > E_{b}

and assuming that A for both the reactions is same then:

Q. Two reactions,

A \to

B \to

Products, have rate constants

k_{A}

k_{B}

T

and activation energies

E_{A}

E_{B}

respectively. If

k_{A} > k_{B}

E_{A} < E_{B}

and assuming that collision frequency for both the reactions is same, then:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Conduction

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon