Question

A thin steel ring of inner radius r and cross-sectional area A is to be mounted on a wheel of radius R (R > r). If the Young modulus of the metal ring is Y, then the tension in the steel ring is:

(a) YAR/R²−r²
(b) YAR/R+r​
(c) YAR/R−r​
(d) YAR/2R​

Answer 1

The tension in the steel ring when mounted on the wheel is given by option (c) which is the force calculated as YA(R - r)/r, where Y is the Young's modulus, A is the cross-sectional area, R is the radius of the wheel, and r is the radius of the ring.

Step-by-step explanation:

The question is asking to find the tension in a steel ring when it is mounted on a wheel, given the radius of the ring r, the radius of the wheel R, the cross-sectional area A, and the Young's modulus of the ring Y.

Using the formula for tensile strain and understanding that the ring must be stretched by an amount ΔL to fit onto the wheel, we can relate the change in length of the ring to its circumference. The initial circumference of the ring is 2πr, and the final circumference must be 2πR. Therefore, the change in length ΔL is 2πR - 2πr.

According to the relation between force F, change in length ΔL, Young's modulus Y, cross-sectional area A, and original length Lo (which is 2πr here), we can write the relation as:

F = ΔLYA
---
Lo

Substituting the values, we get:

F = (2πR - 2πr)YA
-------------
2πr
F = YA(R - r)
--------
r

So, the correct option for the tension in the steel ring, represented by the force F, is (c) YA(R - r)/r.