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 L_{0}

here , ω = 2πν = 4 π

T cosθ = m g …(i)

T sinθ = m ω^{2} r

T sinθ = m (4 π)^{2} r …(ii)

On dividing ‘

tanθ = 16 π^{2} 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 π^{2} 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 $