A 1-kg stone at the end of 1 m long string is whirled in a vertical circle at a constant speed of 4 ms -1. The tension in the string is 6 N when the stone is

Question

A 1-kg stone at the end of 1 m long string is whirled in a vertical circle at a constant speed of 4 ms -1. The tension in the string is 6 N when the stone is (A) At the top of the circle (B) At the bottom of the circle (C) Half way down (D) None of above

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e9fc00606f08c2d1e46fb460731ba041-.png" /> (m v2/R) =(1 × 42/1)=16 N m g=1 × 10=10 N ⇒ T =m g cos θ+(m v2/R) 6 =10 cos θ+16 operatornamecosec θ=-1 θ =180°

Q. A $1 - k g$ stone at the end of $1 m$ long string is whirled in a vertical circle at a constant speed of $4 m s^{- 1}$ . The tension in the string is $6 N$ when the stone is

At the top of the circle

At the bottom of the circle

Half way down

None of above

$\frac{m v ^{2}}{R} = \frac{1 \times 4 ^{2}}{1} = 16 N$
$m g = 1 \times 10 = 10 N$
$\Rightarrow T = m g cos θ + \frac{m v ^{2}}{R}$
$6 = 10 cos θ + 16$
$cosec θ = - 1$
$θ = 18 0^{\circ}$

Q. A 1−kg stone at the end of 1m long string is whirled in a vertical circle at a constant speed of 4ms−1. The tension in the string is 6N when the stone is