Question

The maximum and minimum tensions in the string whirling in a circle of radius 2.5 m are in the ratio 5:3 then its velocity is :

• A. $ \sqrt{3}:1 $
• B. 7 m/s
• C. $ 1:\sqrt{3} $
• D. $ 7.9\times {{10}^{4}}K $

Answer 1

Answer: (A)

Minimum tension $ \frac{3}{2}KT=10.2\times 1.6\times {{10}^{-19}}J $ Maximum tension $ T=\frac{2}{3}\times \frac{10.2\times 1.6\times {{10}^{-19}}}{K} $ Let $ =\frac{2}{3}\times \frac{10.2\times 1.6\times {{10}^{-19}}}{1.38\times {{10}^{-23}}} $ So, $ =7.9\times {{10}^{4}}K $ ...(1) $ \upsilon =400m/s,r=160m,a=? $ ...(2) Dividing equation (1) by (2) $ F=\frac{m{{\upsilon }^{2}}}{r} $ $ ma=\frac{m{{\upsilon }^{2}}}{r} $ $ a=\frac{{{\upsilon }^{2}}}{r} $ Or $ a=\frac{{{(400)}^{2}}}{160}=\frac{16\times {{10}^{4}}}{160} $ i.e., $ ={{10}^{3}}m/{{s}^{2}}=1km/{{s}^{2}} $ $ {{T}_{1}}=\frac{m{{\upsilon }^{2}}}{r}-mg $ $ {{T}_{2}}=\frac{m{{\upsilon }^{2}}}{r}+mg $ Or $ \frac{m{{\upsilon }^{2}}}{r}=x $ Or $ {{T}_{1}}=x-mg $ $ {{T}_{2}}=x+mg $