Question:

A $1\, kg$ stone at the end of $1\, m$ long string is whirled in a vertical circle at constant speed of $4\, m / s$. The tension in the string is $6\, N$, when the stone at $\left(g=10\, m / s ^{2}\right)$

Updated On: Jul 2, 2022
  • top of the circle
  • bottom of the circle
  • half way down
  • None of the above
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

The force on string $F =m g=1 \times 10$ $=10\, N$ Centripetal force $=\frac{m v^{2}}{r}=\frac{1 \times(4)^{2}}{1}=16$ Tension at the top of circle $=\frac{m v^{2}}{r}-m g=16-10=6\, N$ Tension at the bottom of circle $=\frac{m v^{2}}{r}+m g$ $=10+16=2.6\, N$
Was this answer helpful?
0
0

Questions Asked in AIIMS exam

View More Questions

Concepts Used:

Tension

A force working along the length of a medium, especially if this force is carried by a flexible medium like cable or rope is called tension.  The flexible cords which bear muscle forces to other parts of the body are called tendons.

Net force = 𝐹𝑛𝑒𝑡 = 𝑇−𝑊=0,

where,

T and W are the magnitudes of the tension and weight and their signs indicate a direction, be up-front positive here.

Read More: Tension Formula