A metallic ring of radius r and cross-section | vedclass

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

Question

(B) The Young's modulus $Y$ is defined as the ratio of stress to strain: $Y = \frac{F/A}{\Delta L/L}$.Here,the original length of the ring is $L = 2 \

Vedclass · Accepted Answer

$\frac{A Y(R-r)}{r}$

Vedclass · Answer

(B) The Young's modulus $Y$ is defined as the ratio of stress to strain: $Y = \frac{F/A}{\Delta L/L}$.Here,the original length of the ring is $L = 2 \pi r$.The change in length required to fit the ring onto the disc of radius $R$ is $\Delta L = 2 \pi R - 2 \pi r = 2 \pi (R - r)$.Rearranging the formula for force $F$,we get $F = \frac{Y A \Delta L}{L}$.Substituting the values,$F = \frac{Y A [2 \pi (R - r)]}{2 \pi r}$.Simplifying this,we obtain $F = \frac{A Y (R - r)}{r}$.

$A$ metallic ring of radius $r$ and cross-sectional area $A$ is fitted into a wooden circular disc of radius $R$ $(R > r)$. If the Young's modulus of the material of the ring is $Y$,the force with which the metal ring expands is

Similar Questions

Crystalline sodium chloride is an insulator of electricity because $......$

$A$ monoatomic gas is suddenly compressed to $(1/8)^{\text{th}}$ of its initial volume adiabatically. The ratio of the final pressure to initial pressure of the gas is (given $\gamma = 5/3$).

$\int_0^{\pi / 2} \frac{200 \sin x+100 \cos x}{\sin x+\cos x} d x$ is equal to (in $pi$)

$A$ particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_c$ is varying with time $t$ as $a_c = k^2rt^2$. The power delivered to the particle by the forces acting on it is

Which of the following does not exist?

Vedclass Test Series

Exam Paper Generator

Online Exam Module