JEE Main 2024 — Laws of Motion Question with Solution

Given that 

Given the mass $m=900 \mathrm{g}=\frac{900}{1000} \mathrm{kg}=\frac{9}{10} \mathrm{kg}$,

The radius of the circular path $r=1 \mathrm{m}$

And, the angular velocity

$\omega =\frac{2\mathrm{\pi N}}{60}=\frac{2\mathrm{\pi }\left(10\right)}{60}=\frac{\mathrm{\pi }}{3} \mathrm{rad} {\mathrm{s}}^{-1}$

With reference to the above diagram, the formula to calculate the tension at the lowermost point can be written as

$$\begin{gathered} T-mg=mr{\omega }^2 \\ \Rightarrow T=mg+mr{\omega }^2 ...\left(1\right) \end{gathered}$$

From equation (1), it follows that$$

$\begin{array}{rcl}T& =& \frac{9}{10}\times 9.8+\frac{9}{10}\times 1{\left(\frac{\pi }{3}\right)}^2\\ & =& 8.82+\frac{9}{10}\times \frac{{\pi }^2}{9}\\ & =& 8.82+0.98\\ & =& 9.8 \mathrm{N}\end{array}$