# An elastic wire is found to be elongated by 1cm when a mass is gently hung from it. It extends by 3cm more, …..

Q: An elastic wire is found to be elongated by 1 cm when a mass is gently hung from it. It extends by 3 cm more, when the attached mass is made to move uniformly in a horizontal circle performing 2 rev s-1. The length of the unstretched wire (in cm) is (Given, g = 9.81 ms-2)

(a) $\displaystyle \frac{24-\sqrt{15}}{\sqrt{15}}\times 4$

(b) $\displaystyle \frac{24}{\sqrt{15}}\times 4$

(c) $\displaystyle \frac{4}{\sqrt{5}}$

(d) $\displaystyle \frac{4-\sqrt{15}}{2}$

Ans: (a)

Sol:

Let the length of unstretched wire is L0

here , ω = 2πν = 4 π

T cosθ = m g …(i)

T sinθ = m ω2 r

T sinθ = m (4 π)2 r …(ii)

On dividing ‘

tanθ = 16 π2 r/g

As wire is elastic. So, it obeys Hooke’s law.

Initially , mg = k(1)

when mass rotates ,

T = k(4)

T = 4 mg

From (i)

4 mg cosθ = mg

cosθ = 1/4

hence , tanθ = √(15)

√(15) = 16 π2 r/g

$\displaystyle r = \frac{g \sqrt{15}}{16 \pi^2}$

$\displaystyle r = \frac{981 \sqrt{15}}{16 (3.14)^2}$

r = 24 cm

From figure ,

$\displaystyle sin\theta = \frac{24}{L_0 + 4}$

$\displaystyle \frac{\sqrt{15}}{4} = \frac{24}{L_0 + 4}$

$\displaystyle L_0 = \frac{24-\sqrt{15}}{\sqrt{15}}\times 4$