Question

Two wires $\text{AC}$ and $\text{BC}$ are tied at $\text{C}$ to a small sphere of mass $5\text{kg}$, which revolves a constant speed $v$ in the horizontal circle of radius $1.6\text{m}$. If both wires are to remain taut and if the tension in either of the wires is not to exceed $60\text{N}$, then minimum value of $v$ is
[Take, $g=10{\text{m/s}}^2$]

• A. $3.0\text{m/s}$
• B. $4.0\text{m/s}$
• C. $8.2\text{m/s}$
• D. $3.96\text{m/s}$

Answer 1

The correct option is A $3.0\text{m/s}$

Tension in both wired will be different, let it be $T_1$ and $T_2$ in $\text{AC}$ and $\text{BC}$ respectively.


Let the vertical direction be represented by $y-$axis. The body is moving in horizontal plane ($xz-$plane), thus applying equilibrium condition along vertical direction:

$T_1\sin {60}^{\circ }+T_2\sin {45}^{\circ }=mg$

$\frac{\sqrt{3}}{2}{T}_1+\frac{T_2}{\sqrt{2}}=mg.......\left(1\right)$

Applying equation of dynamics towards centre,

$T_1\cos {60}^{\circ }+T_2\cos {45}^{\circ }=\frac{m{v}^2}{r}$

$\Rightarrow \frac{T_1}{2}+\frac{T_2}{\sqrt{2}}=\frac{m{v}^2}{r}....\left(2\right)$

Subtracting eqs. (2) to (1)

$\frac{\sqrt{3}{T}_1}{2}-\frac{T_1}{2}=mg-\frac{m{v}^2}{r}$

$\Rightarrow T_1=\frac{mg-\frac{m{v}^2}{r}}{\left[\frac{\sqrt{3}-1}{2}\right]}$

Here, $T_1=60\text{N};m=5\text{kg};r=1.6\text{m};v=v_1.$

$\Rightarrow 60=\frac{(5\times 10)-\frac{5\times v_1^2}{1.6}}{\left[\frac{\sqrt{3}-1}{2}\right]}$

$\therefore v_1=3\text{m/s}$

From question, $T_1\le 60\text{N}$

$\therefore v\ge v_1\Rightarrow v\ge 3\text{m/s}$

Again from eqs. $\left(1\right)\&\left(2\right)$

$T_2=\frac{\sqrt{2}}{(\sqrt{3}-1)}[\frac{\sqrt{3}m{v}^2}{r}-mg]$

Here, $T_2=60\text{N};m=5\text{kg};r=1.6\text{m};v=v_2$

$\Rightarrow 60=\frac{\sqrt{2}}{(\sqrt{3}-1)}[\frac{\sqrt{3}\times 5v_2^2}{1.6}-(5\times 10)]$

$\therefore v_2=3.86\text{m/s}$

Again from question, $T_2\le 60\text{N}$

$\therefore v\le v_2\Rightarrow v\le 3.86\text{m/s}$

So, from both condition, we conclude that,

$v\ge 3\text{m/s};v\le 3.86\text{m/s}$

$\therefore v_{min}=3\text{m/s}$

Hence, option $\left(a\right)$ is correct.