Question

Two wires $AC$ and $BC$ are tied at $C$ of small sphere of mass $5kg$, which revolves at a constant speed $v$ in the horizontal circle of radius $1.6m$. The minimum value of $v$ is?

• A. $3.01m/s$
• B. $4.01m/s$
• C. $8.2m/s$
• D. $3.96m/s$

Answer 1

The correct option is D $3.96m/s$

From force diagram shown in figure,
$T_1\cos {30}^o+T_2\cos {45}^o=mg$ .........(i)
$T_1\sin {30}^o+T_2\sin {45}^o=\frac{m{v}^2}{r}$ .......(ii)
After solving eqs. (i) and (ii), we get
$T_1=\frac{mg-\frac{m{v}^2}{r}}{\frac{(\sqrt{3}-1)}{2}}$
But $T_1>0$
$\therefore \frac{mg-\frac{m{v}^2}{r}}{\frac{\sqrt{3}-1}{2}}>0$
or $mg>\frac{m{v}^2}{r}$ or $v<\sqrt{rg}$
$\therefore v_{min}=\sqrt{rg}=\sqrt{1.6\times 9.8}=3.96m/s$.