Question

A body of mass $100\,g$ is tied to one end of a $2\,m$ long string. The other end of the string is at the centre of the horizontal circle. The maximum revolution in one minute is $200$. The maximum tensile strength of the string is approx.
A. $8.76\,N$
B. $8.9\,N$
C. $87\,dyne$
D. $87.64\,N$

Answer 1

Hint:In order to solve this question, we will use the concept that centripetal force on a body moving in circular path acts in inward direction towards the centre and this force is balanced by the tension acting on the string attached to the body so here we will solve for centripetal force which will exactly equal to the tension on the string.

Formula used:
Centripetal force is given by,
$F = mr{\omega ^2}$
where r is the radius of the circular path, m is the mass of the body having angular frequency $\omega $.

Complete step by step solution:
We have given that, length of the string which will be radius of the circular path of the body is $r = 2m$ and mass of the body is $m = 100g = 0.1kg$ and frequency is given to us,
$f = 200r.p.m \\
\Rightarrow f = \dfrac{{20}}{6}\,r.p.s \\ $
And so angular frequency is calculated using,
$\omega = 2\pi f \\
\Rightarrow \omega = 2\pi \dfrac{{20}}{6}\,rad{\sec ^{ - 1}} \\
\Rightarrow \omega = 20.9\,rad{\sec ^{ - 1}} \\ $
Now, if T is the tension on the string then we know that,
$T = F = mr{\omega ^2}$
On putting the values and solving for T we get,
$T = 0.1(2){(20.9)^2} \\
\therefore T \approx 87.64\,N $

Hence, the correct answer is option D.

Note: It should be remembered that, basic unit of conversions are used as $1\,kg = 1000\,g$ and frequency were given to us in the form of revolutions per minute so we changed it into revolutions per minute by converting $1\min = 60\sec $ so, always remembered all the basic conversions in order to solve numerical easily.