wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A particle is projected along the inner surface of a smooth vertical circle of radius R, its velocity at the lowest point being 1595Rg. It will leave the circle at angle θ with the verticle. The angle θ is

A
37
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
53
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
60
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
30
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 53
Let B be the point where the particle leaves the circle and its velocity at point B be v

Using conservation of mechanical energy between points A and B
12mu2=mgR(1+cosθ)+12mv2

Using second law at point B,

mgcosθ=mv2R ( N=0)

v2=Rgcosθ

12m(1595Rg)2=mgR(1+cosθ)+12mgRcosθ

9525=2+2cosθ+cosθ

3cosθ=4525

cosθ=1525=35

or, θ=53

Hence, option (b) is the correct answer.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon