(C) The change in length of a rod due to thermal expansion is given by $\Delta l = l \alpha \Delta T$.Given that the change in length for both rods is

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

$\frac{\alpha_s}{\alpha_a + \alpha_s}$

(C) The change in length of a rod due to thermal expansion is given by $\Delta l = l \alpha \Delta T$.Given that the change in length for both rods is the same,we have $\Delta l_1 = \Delta l_2$.Substituting the formula,we get $l_1 \alpha_a t = l_2 \alpha_s t$.Canceling $t$ from both sides,we get $l_1 \alpha_a = l_2 \alpha_s$,which implies $\frac{l_1}{l_2} = \frac{\alpha_s}{\alpha_a}$.To find the ratio $\frac{l_1}{l_1 + l_2}$,we use the property of ratios: if $\frac{a}{b} = \frac{c}{d}$,then $\frac{a}{a+b} = \frac{c}{c+d}$.Applying this to our equation,we get $\frac{l_1}{l_1 + l_2} = \frac{\alpha_s}{\alpha_a + \alpha_s}$.

Class 11 Physics 10-1.Thermometry, Thermal Expansion and Calorimetry Thermal Expansion for Solid

Medium

Two rods,one of aluminum and the other made of steel,having initial lengths $l_1$ and $l_2$ are connected together to form a single rod of length $l_1 + l_2$. The coefficients of linear expansion for aluminum and steel are $\alpha_a$ and $\alpha_s$ respectively. If the length of each rod increases by the same amount when their temperature is raised by $t ^\circ C$,then find the ratio $\frac{l_1}{l_1 + l_2}$.

IIT 2003 All IIT

A
$\frac{\alpha_s}{\alpha_a}$
B
$\frac{\alpha_a}{\alpha_s}$
C
$\frac{\alpha_s}{\alpha_a + \alpha_s}$
D
$\frac{\alpha_a}{\alpha_a + \alpha_s}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

10-1.Thermometry, Thermal Expansion and Calorimetry

Subtopic

Thermal Expansion for Solid

Previous Year

IIT 2003 Paper All IIT years →

$A$ hole is drilled in a copper sheet. The diameter of the hole is $4.24 \; cm$ at $27.0 \; ^{\circ}C$. What is the change in the diameter of the hole when the sheet is heated to $227 \; ^{\circ}C$? (Coefficient of linear expansion of copper $\alpha = 1.70 \times 10^{-5} \; K^{-1}$)

Medium

View Solution

The density of a substance at $0^{\circ}C$ is $10 \, g/cm^3$ and at $100^{\circ}C$,its density is $9.7 \, g/cm^3$. The coefficient of linear expansion of the substance is:

Medium

View Solution

$A$ clock which keeps correct time at $20\,^{\circ}C$ has a pendulum rod made of brass. How many seconds will it gain or lose per day when the temperature falls to $0\,^{\circ}C$? $(\alpha = 18 \times 10^{-6}/^{\circ}C)$

Difficult

View Solution

The area of a circular copper coin increases by $0.4 \%$ when its temperature is raised by $100^{\circ} C$. The coefficient of linear expansion of the coin is:

EasyTS EAMCET 2019

View Solution

Three rods of length $\ell$ each are joined to form an equilateral triangle $PQR$. $O$ is the midpoint of $PQ$. For a small change in temperature,the distance $OR$ remains constant. If the coefficient of linear expansion for $PR$ and $RQ$ is $\alpha_2$ and for $PQ$ is $\alpha_1$,then:

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

$A$ hole is drilled in a copper sheet. The diameter of the hole is $4.24 \; cm$ at $27.0 \; ^{\circ}C$. What is the change in the diameter of the hole when the sheet is heated to $227 \; ^{\circ}C$? (Coefficient of linear expansion of copper $\alpha = 1.70 \times 10^{-5} \; K^{-1}$)

The density of a substance at $0^{\circ}C$ is $10 \, g/cm^3$ and at $100^{\circ}C$,its density is $9.7 \, g/cm^3$. The coefficient of linear expansion of the substance is:

$A$ clock which keeps correct time at $20\,^{\circ}C$ has a pendulum rod made of brass. How many seconds will it gain or lose per day when the temperature falls to $0\,^{\circ}C$? $(\alpha = 18 \times 10^{-6}/^{\circ}C)$

The area of a circular copper coin increases by $0.4 \%$ when its temperature is raised by $100^{\circ} C$. The coefficient of linear expansion of the coin is:

Three rods of length $\ell$ each are joined to form an equilateral triangle $PQR$. $O$ is the midpoint of $PQ$. For a small change in temperature,the distance $OR$ remains constant. If the coefficient of linear expansion for $PR$ and $RQ$ is $\alpha_2$ and for $PQ$ is $\alpha_1$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module