### JEE Main 12 Jan 2019 Evening Question 30

##### Question: A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be (Take $ g=10m/s^{2} $ ) [JEE Main Online Paper Held On 12-Jan-2019 Evening]

#### Options:

A) $ 6kg-m^{2}/s $

B) $ 8kg-m^{2}s $

C) $ 3kg-m^{2}s $

D) $ 2kg-m^{2}/s $

## Show Answer

#### Answer:

Correct Answer: A

#### Solution:

Applying law of conservation of energy $ \frac{1}{2}mv_A^{2}+mgh=\frac{1}{2}mv_B^{2} $

$ \Rightarrow $ $ \frac{1}{2}(20\times {10^{-3}}){{(5)}^{2}}+(20\times {10^{-3}})(10)(10) $ $ =\frac{1}{2}(20\times {10^{-3}})(V_B^{2}) $

$ \Rightarrow $ $ v _{B}=15m/s $

So, the angular momentum of the particle about point O is $ =6kg-m^{2}/s $