(B) The rate of heat flow through a rod is given by: $\dot{Q} = \frac{KA \Delta T}{l}$.Since the rods are connected in series,the rate of heat flow $\

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

$\frac{T_A}{T_B} = \frac{1}{3}$

(B) The rate of heat flow through a rod is given by: $\dot{Q} = \frac{KA \Delta T}{l}$.Since the rods are connected in series,the rate of heat flow $\dot{Q}$ through both rods must be the same.For rod $A$ (thermal conductivity $3K$): $\dot{Q}_A = \frac{(3K)A T_A}{l}$.For rod $B$ (thermal conductivity $K$): $\dot{Q}_B = \frac{KA T_B}{l}$.Since $\dot{Q}_A = \dot{Q}_B$,we have:$\frac{3KA T_A}{l} = \frac{KA T_B}{l}$$3 T_A = T_B$Therefore,$\frac{T_A}{T_B} = \frac{1}{3}$.

Class 11 Physics 10-2.Heat Transfer Heat Conduction and Thermal Conductivity

Difficult

Two rods $A$ and $B$ of the same cross-sectional area $A$ and length $l$ are connected in series between a source $(T_1 = 100^{\circ}C)$ and a sink $(T_2 = 0^{\circ}C)$ as shown in the figure. The rods are laterally insulated. If the thermal conductivities of rods $A$ and $B$ are $3K$ and $K$ respectively,and $T_A$ and $T_B$ are the temperature drops across rods $A$ and $B$,then:

A
$\frac{T_A}{T_B} = \frac{3}{1}$
B
$\frac{T_A}{T_B} = \frac{1}{3}$
C
$\frac{T_A}{T_B} = \frac{3}{4}$
D
$\frac{T_A}{T_B} = \frac{4}{3}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

10-2.Heat Transfer

Subtopic

Heat Conduction and Thermal Conductivity

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_{1}$ and $T_{2}$ $(T_{1} > T_{2})$. The rate of heat transfer,$\frac{dQ}{dt}$ through the rod in a steady state is given by:

Medium

View Solution

If $K_{1}$ and $K_{2}$ are the thermal conductivities,$L_{1}$ and $L_{2}$ are the lengths,and $A_{1}$ and $A_{2}$ are the cross-sectional areas of steel and copper rods respectively,such that $\frac{K_{2}}{K_{1}}=9$,$\frac{A_{1}}{A_{2}}=2$,and $\frac{L_{1}}{L_{2}}=2$. For the arrangement shown in the figure,the value of the temperature $T$ of the steel-copper junction in the steady state will be ........... $^{\circ}C$.

JEE MAIN 2022

View Solution

Two concentric spheres of radii $r_1$ and $r_2$ are maintained at temperatures $T_1$ and $T_2$ respectively. The rate of radial heat flow between the two concentric spheres is proportional to:

Difficult

View Solution

On a cold morning,a metal surface will feel colder to touch than a wooden surface because

EasyAIIMS 1998

View Solution

Two spheres made of different materials have radii in the ratio $2:1$ and wall thicknesses in the ratio $1:4$. They are completely filled with ice. If the time taken to melt the ice completely in the larger sphere is $25 \ min$ and in the smaller sphere is $16 \ min$,then the ratio of the thermal conductivities of the materials of the larger and smaller spheres is ..........

Difficult

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_{1}$ and $T_{2}$ $(T_{1} > T_{2})$. The rate of heat transfer,$\frac{dQ}{dt}$ through the rod in a steady state is given by:

Two concentric spheres of radii $r_1$ and $r_2$ are maintained at temperatures $T_1$ and $T_2$ respectively. The rate of radial heat flow between the two concentric spheres is proportional to:

On a cold morning,a metal surface will feel colder to touch than a wooden surface because

Vedclass Test Series

Exam Paper Generator

Online Exam Module