Question

Two point size dense particles of same mass are knotted to a single massless string at different points, are whirled along concentric circles in horizontal plane. The ratio of distances of particles from centre of circles is $1:2$. If the tension in the string between two particles is $4N$, the tension in the remaining string is:

• A. 2 N
• B. 3 N
• C. 5 N
• D. 6 N

Answer 1

The correct option is D 6 N

$T_2=4N$

as $\omega$ is same than velocities are

$V_1=\omega l,V_2=\omega \left(2l\right)$

For particle on the right

$T=\frac{m{V}_2^2}{2l}$

$4=\frac{m(2\omega l\times 2\omega l)}{2l}\Rightarrow {\omega }^2=\frac{2}{ml}$

$\leftarrow {\omega }^{2}l$

$T_1\leftarrow \cdot \to T_2$

for 1st particle

$T_1-T_2=m{\omega }^{2}l$

$T_1=4+ml\frac{2}{ml}$

$T_1=4+2=6N.$