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 $T_1=x-mg$ $T_2=x+mg$