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)

$\sqrt{3}:1$

Answer 2

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 ${{T}_{1}}=x-mg$ ${{T}_{2}}=x+mg$