(D) The change in length (diameter) required is $\Delta L = L_2 - L_1 = 6.241 \,cm - 6.230 \,cm = 0.011 \,cm$.The formula for linear thermal expansion

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

(D) The change in length (diameter) required is $\Delta L = L_2 - L_1 = 6.241 \,cm - 6.230 \,cm = 0.011 \,cm$.The formula for linear thermal expansion is $\Delta L = L_1 \alpha_L \Delta T$,where $\Delta T = T_f - T_i$.Substituting the given values: $0.011 = 6.230 \times (1.4 \times 10^{-5}) \times (T_f - 27)$.Solving for $\Delta T$: $\Delta T = \frac{0.011}{6.230 \times 1.4 \times 10^{-5}} = \frac{0.011 \times 10^5}{8.722} \approx 126.11^{\circ} C$.Therefore,the final temperature $T_f = 27 + 126.11 = 153.11^{\circ} C$.Comparing with the given options,the closest value is $152.7^{\circ} C$.

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

Medium

At what temperature (in $ ^{\circ} C$) should a gold ring of diameter $6.230 \,cm$ be heated so that it can be fitted on a wooden bangle of diameter $6.241 \,cm$ (in $.7$)? Both diameters were measured at room temperature $(27^{\circ} C)$. (Given: coefficient of linear thermal expansion of gold $\alpha_{L}=1.4 \times 10^{-5} \,K ^{-1}$)

JEE MAIN 2022 All JEE MAIN

A
$125$
B
$91$
C
$425$
D
$152$

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

JEE MAIN 2022 Paper All JEE MAIN years →

The area of cross-section of a railway track is $0.01\, m^2$. The temperature variation is $10^{\circ}C$. The coefficient of linear expansion of the material of the track is $10^{-5} /^{\circ}C$. The energy stored per meter in the track is ...... $J/m$. (Young's modulus of the material of the track is $10^{11}\, Nm^{-2}$)

MediumJEE MAIN 2021

View Solution

The coefficient of apparent expansion of mercury in a glass vessel is $153 \times 10^{-6} /{ }^{\circ} C$ and in a steel vessel is $144 \times 10^{-6} /{ }^{\circ} C$. If $\alpha$ for steel is $12 \times 10^{-6} /{ }^{\circ} C$, then the coefficient of linear expansion $\alpha$ for glass is:

MediumAIIMS 2019

View Solution

$A$ brass disc fits snugly in a hole of a steel plate. The disc can be loosened from the hole if the system:

Easy

View Solution

The coefficients of linear expansion of brass and steel are ${\alpha _1}$ and ${\alpha _2}$ respectively. If we take a brass rod of length ${l_1}$ and a steel rod of length ${l_2}$ at $0^{\circ}C$,their difference in length $({l_2} - {l_1})$ will remain the same at any temperature if:

Difficult

View Solution

An iron sphere having diameter $D$ and mass $M$ is immersed in hot water so that the temperature of the sphere increases by $\delta T$. If $\alpha$ is the coefficient of linear expansion of the iron,then the change in the surface area of the sphere is:

EasyAP EAMCET 2022

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 coefficient of apparent expansion of mercury in a glass vessel is $153 \times 10^{-6} /{ }^{\circ} C$ and in a steel vessel is $144 \times 10^{-6} /{ }^{\circ} C$. If $\alpha$ for steel is $12 \times 10^{-6} /{ }^{\circ} C$, then the coefficient of linear expansion $\alpha$ for glass is:

$A$ brass disc fits snugly in a hole of a steel plate. The disc can be loosened from the hole if the system:

The coefficients of linear expansion of brass and steel are ${\alpha _1}$ and ${\alpha _2}$ respectively. If we take a brass rod of length ${l_1}$ and a steel rod of length ${l_2}$ at $0^{\circ}C$,their difference in length $({l_2} - {l_1})$ will remain the same at any temperature if:

An iron sphere having diameter $D$ and mass $M$ is immersed in hot water so that the temperature of the sphere increases by $\delta T$. If $\alpha$ is the coefficient of linear expansion of the iron,then the change in the surface area of the sphere is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module