A body of mass 1 kg is connected to a string and is undergoing a complete vertical circular motion of radius 1 m . If it is barely able to complete the vertical circle, then the difference in its kinetic energy at the top and bottom of the circle is (Take g=10 ms- 2 )

Question

A body of mass 1 kg is connected to a string and is undergoing a complete vertical circular motion of radius 1 m . If it is barely able to complete the vertical circle, then the difference in its kinetic energy at the top and bottom of the circle is (Take g=10 ms- 2 ) (A) 10J (B) 20J (C) 30J (D) 50J

Tardigrade Admin · Accepted Answer

Difference in K E = (1/2) m [. (. √5 g r .)2 - (. √g r .)2 ]. =2mgr=20J

Q. A body of mass $1 k g$ is connected to a string and is undergoing a complete vertical circular motion of radius $1 m$ . If it is barely able to complete the vertical circle, then the difference in its kinetic energy at the top and bottom of the circle is (Take $g = 10 m s^{- 2}$ )

$10 J$

$20 J$

$30 J$

$50 J$

Difference in $K E = \frac{1}{2} m [(5 g r)^{2} - (g r)^{2}]$
$= 2 m g r = 20 J$

Q. A body of mass 1kg is connected to a string and is undergoing a complete vertical circular motion of radius 1m . If it is barely able to complete the vertical circle, then the difference in its kinetic energy at the top and bottom of the circle is (Take g=10ms−2 )

10J

20J

30J

50J