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 $