A steel ring of radius r and cross-section ar | vedclass

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

Question

(B) The initial length (circumference) of the ring is $L = 2\pi r$.The final length (circumference) of the ring after fitting it onto the disc is $L'

Vedclass · Accepted Answer

$AE\left( \frac{R - r}{r} \right)$

Vedclass · Answer

(B) The initial length (circumference) of the ring is $L = 2\pi r$.The final length (circumference) of the ring after fitting it onto the disc is $L' = 2\pi R$.The change in length is $\Delta L = 2\pi R - 2\pi r = 2\pi(R - r)$.The longitudinal strain is defined as $	ext{strain} = \frac{\Delta L}{L} = \frac{2\pi(R - r)}{2\pi r} = \frac{R - r}{r}$.According to the definition of Young's modulus $E = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta L/L}$.Substituting the values,we get $E = \frac{F/A}{(R - r)/r}$.Rearranging for force $F$,we get $F = AE\left( \frac{R - r}{r} ight)$.

$A$ steel ring of radius $r$ and cross-section area $A$ is fitted onto a wooden disc of radius $R$ $(R > r)$. If Young's modulus is $E$,then the force with which the steel ring is expanded is:

Similar Questions

The oxide of a metal contains $40 \%$ of oxygen. The valency of the metal is $2$. What is the atomic weight of the metal?

The probability that a number selected at random from the set of numbers ${1, 2, 3, \ldots, 100}$ is a cube, is

Let $A$ and $B$ be two symmetric matrices of the same order. Then,the matrix $AB - BA$ is

$A$ conducting circular loop is placed in a uniform magnetic field,$B = 0.025 \, T$,with its plane perpendicular to the magnetic field. The radius of the loop is made to shrink at a constant rate of $1 \, mm/s$. The induced $e.m.f.$ when the radius is $2 \, cm$ is

$H_3BO_3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module