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QJEEM24Phbb69: JEE Main 2024 Physics - Class 11 JEEM24Phbb69 Solution: A stone of mass is tied to a string and moved in a vertical…

A stone of mass 900 g900 \mathrm{~g} is tied to a string and moved in a vertical circle of radius 1 m1 \mathrm{~m} making 10 rpm10 \mathrm{~rpm} . The tension in the string, when the stone is at the lowest point is (if π2=9.8\pi^2=9.8 and g=9.8 m/s2g=9.8 \mathrm{~m} / \mathrm{s}^2 ) :

Multiple Choice Options

  • A 17.8 N
  • B 97 N
  • C 9.8 N
  • D 8.82 N

Step-by-Step Verified Solution

Given that

JEE Main 2024 (Online) 29th January Evening Shift Physics - Circular Motion Question 12 English Explanation

amp;m=900 gm=9001000 kg=910 kgamp;r=1 mamp;ω=2πN60=2π(10)60=π3rad/secamp;T−mg=mrω2amp; T=mg+mrω2amp;=910×9.8+910×1(π3)2amp;=8.82+910×π29amp;=8.82+0.98amp;=9.80 N\begin{aligned}& \mathrm{m}=900 \mathrm{~gm}=\frac{900}{1000} \mathrm{~kg}=\frac{9}{10} \mathrm{~kg} \\& \mathrm{r}=1 \mathrm{~m} \\& \omega=\frac{2 \pi \mathrm{N}}{60}=\frac{2 \pi(10)}{60}=\frac{\pi}{3} \mathrm{rad} / \mathrm{sec} \\& \mathrm{T}-\mathrm{mg}=\mathrm{mr} \omega^2 \\& \mathrm{~T}=\mathrm{mg}+\mathrm{mr} \omega^2 \\& =\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 \\& =9.80 \mathrm{~N}\end{aligned}

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