Q. A small object of mass $0.5\, kg$ is attached to an end of a mass less $2\, m$ long rope. It is rotated under gravity in a vertical circle with the other end of the rope being at the centre of the circle. The motion is started from the lowest point. Match columns I and II.
Column I Column II
i If the speed of the object at lowest point is $3.5\, ms ^{-1}$ a there will be some point on the circle at which speed of the object is zero but tension in the rope is not zero.
ii If the speed of the object at lowest point is $8\, ms ^{-1}$ b there will be some point on the circle at which tension in the rope is zero but speed of the object is
iii If the maximum tension in the rope is $15 \,N$ c the object will not be able to reach the highest point
iv If the maximum tension in the rope is $30\, N$ d the object will be able to reach the highest point

Work, Energy and Power Report Error

Solution:

$\sqrt{5 g r}=\sqrt{5 \times 10 \times 2}=10\, ms ^{-1}$
$\sqrt{2 g r}=\sqrt{2 \times 10 \times 2}=\sqrt{40}\, ms ^{-1}$
(i) $3.5\, ms ^{-1}$ is less than $\sqrt{2 g r}$.
So the particle will stop before the string becomes horizontal.
(ii) $8\, ms ^{-1}$ lies between $\sqrt{2 g r}$ and $\sqrt{5 g r}$.
The particle will cross the horizontal position but will not be able to reach the highest point.
(iii) Tension is maximum at the lowest point. Hence,
$15=m g+\frac{m v_{L}^{2}}{r}$
$\Rightarrow v_{2}=\sqrt{40}\, ms ^{-1}$
In this case particle will just be able to reach the horizontal level. Here both speed and tension become zero.
(iv) $30=m g+\frac{m v_{2}^{2}}{r}$
$\Rightarrow v_{2}=10\, m s ^{-1}$
Here particle will be able to reach the highest point.
Tension will become zero at the highest point.
Hence (b) and (d) are matching.