(D) The thermal resistance $R$ of a rod is given by the formula:$R = \frac{l}{KA}$where $l$ is the length of the rod,$K$ is the thermal conductivity,a

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

(D) The thermal resistance $R$ of a rod is given by the formula:$R = \frac{l}{KA}$where $l$ is the length of the rod,$K$ is the thermal conductivity,and $A$ is the cross-sectional area.Given:Length $l = 2 \ m$Radius $r = 1 \ cm = 10^{-2} \ m$Thermal conductivity $K = 401 \ W/m-K$Cross-sectional area $A = \pi r^2 = 3.14 \times (10^{-2})^2 = 3.14 \times 10^{-4} \ m^2$Substituting the values into the formula:$R = \frac{2}{401 \times 3.14 \times 10^{-4}}$$R = \frac{2}{0.125914} \approx 15.88 \ K/W$Rounding to one decimal place,we get $R \approx 15.9 \ K/W$.

Class 11 Physics 10-2.Heat Transfer Thermal Resistance and it's Combination

Medium

$A$ copper rod $2 \ m$ long has a circular cross-section of radius $1 \ cm$. One end is kept at $100^{\circ}C$ and the other at $0^{\circ}C$,and the surface is covered by non-conducting material to prevent heat losses through the surface. The thermal resistance of the bar in Kelvin per Watt is (Take thermal conductivity $K = 401 \ W/m-K$ of copper):- (in $.9$)

A
$12$
B
$13$
C
$14$
D
$15$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

10-2.Heat Transfer

Subtopic

Thermal Resistance and it's Combination

Six identical conducting rods are joined as shown in the figure. Points $A$ and $D$ are maintained at temperatures of $200^oC$ and $20^oC$ respectively. The temperature of junction $B$ will be ....... $^oC$.

Medium

View Solution

$A$ wall has two layers $A$ and $B,$ each made of different material. Both the layers have the same thickness. The thermal conductivity for $A$ is twice that of $B$ and under steady condition,the temperature difference across the wall is $36\,^{\circ}C.$ The temperature difference across the layer $A$ is....... $^{\circ}C$

Difficult

View Solution

Two rods of cross-sectional area $A$ and $2A$ and equal length having thermal conductivities $2K$ and $3K$ are joined in parallel. The equivalent thermal conductivity of their combination will be:

Difficult

View Solution

$A$ slab consists of two parallel layers of copper and brass of equal thickness. The ratio of their thermal conductivities is $1:4$. If the temperature of the free side of the brass is $100^{\circ}C$ and that of the copper is $0^{\circ}C$,find the temperature of the interface in $^{\circ}C$.

Medium

View Solution

$A$ wall consists of two layers $A$ and $B$ of equal thickness. Their thermal resistances are $R_1$ and $R_2$. Find the temperature of the interface.

Medium

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

Six identical conducting rods are joined as shown in the figure. Points $A$ and $D$ are maintained at temperatures of $200^oC$ and $20^oC$ respectively. The temperature of junction $B$ will be ....... $^oC$.

Two rods of cross-sectional area $A$ and $2A$ and equal length having thermal conductivities $2K$ and $3K$ are joined in parallel. The equivalent thermal conductivity of their combination will be:

$A$ slab consists of two parallel layers of copper and brass of equal thickness. The ratio of their thermal conductivities is $1:4$. If the temperature of the free side of the brass is $100^{\circ}C$ and that of the copper is $0^{\circ}C$,find the temperature of the interface in $^{\circ}C$.

$A$ wall consists of two layers $A$ and $B$ of equal thickness. Their thermal resistances are $R_1$ and $R_2$. Find the temperature of the interface.

Vedclass Test Series

Exam Paper Generator

Online Exam Module