Physics - W.E.P. Question with Solution | TestHub

For a particle whirling in a vertical circle with a constant speed $v$, the tension in the string is maximum at the lowest point and minimum at the highest point.

The maximum tension is given by:
$T_{\text{max}}=\frac{m{v}^2}{r}+mg$

The minimum tension is given by:
$T_{\text{min}}=\frac{m{v}^2}{r}-mg$

Given the ratio of maximum to minimum tension is $5:3$, we have:
$\frac{T_{\text{max}}}{T_{\text{min}}}=\frac{5}{3}$

$\frac{\frac{m{v}^2}{r}+mg}{\frac{m{v}^2}{r}-mg}=\frac{5}{3}$

Canceling the mass $m$ from the numerator and denominator:
$\frac{\frac{v^2}{r}+g}{\frac{v^2}{r}-g}=\frac{5}{3}$

Cross-multiplying yields:
$3\left(\frac{v^2}{r}+g\right)=5\left(\frac{v^2}{r}-g\right)$

$3\frac{v^2}{r}+3g=5\frac{v^2}{r}-5g$

$2\frac{v^2}{r}=8g$

$\frac{v^2}{r}=4g$

$v=\sqrt{4gr}$

Substituting the given values $r=2.5\text{m}$ and $g=10{\text{m/s}}^2$:
$v=\sqrt{4\times 10\times 2.5}$

$v=\sqrt{100}=10\text{m/s}$

Answer: $10\text{m/s}$