Question

The blocks $1$ and $2$ in the arrangement have mass $m$ each. The strings $AB$ and $BC$ are $T_1$ and $T_2$ respectively. The system is in equilibrium with a constant horizontal force $mg$. Then prove $\tan {\theta }_1=\frac{1}{2}$, $\tan {\theta }_2=1$, $T_1=\sqrt{5}mg$, $T_2=\sqrt{2}mg$.

Answer 1

$\sum F_x=0mg=T_2\sin {\theta }_2=0mg=T_2\sin {\theta }_2(\theta =45°)T_2\sqrt{2}mg$

Ring of man $=m$

Radius$=R$

$x-y$ plane $I_0$ Uniform external magnetic field of strength $\overline{B}=B_0(2i-2j+5k)$

$t=0$

$B_0=constant$

$\int Edl=-\frac{dQ}{dt}=\pi r^2\frac{dB}{dt}$

Let $\lambda =9/2\pi R$ be the charge per unit length of the ring

The torque is given by

$N=\int \lambda Edl$

Angular momentum$=I\omega =\int Ndt$

On solving the aboveset of sequential equation

$R\lambda {\pi }^{2}B=i\omega R\pi R^{2}Bq/2\pi R=m{R}^2\omega \omega =\frac{9B}{2m}$