Question

5.
A steel ring of radius 10 cm and cross-section area
1 cm>2 is fitted to a wooden disc of radius 10.5 cm.
If Young's modulus of steel is 2 x 10>11 N/m2. then
the force with which the steel ring is expanded is
(in N)
​

Answer 1

Answer:

The force which steel is expanded is 10⁶ N

Explanation:

The Young's modulus Y is given by

$\boxed{Y=\frac{FL}{Ae}}$

Where F is the force

A is the area of the ring

L is the circumference of the ring = 2πr

e = expansion in the circumference

  = 2πR - 2πr

  = 2π(R - r)

Here, given that

R = 10.5 cm

r = 10 cm

A = 1 cm² = 10⁻⁴ m²

L = 2πr = 2π × 10 = 20π cm

e = 2π(R - r) = 2π(10.5 - 10) = π cm

Thus Force is given by

$F=\frac{YAe}{L}$

$\implies F=\frac{2\times 10^{11}\times 10^{-4}\times\pi}{20\pi}$

$\implies F=\frac{2\times 10^{7}}{20}$

$\implies F=10^6$ N

Answer 2

Explanation:

hope this will help you............................