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

A frictionless wire is fixed between A and B inside a ring of radius R. A bead slips along the wire. The time taken by the bead to slip from A to B will be

A
2R/g
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

gR/gcosθ

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

2gRgcosθ

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2gRcosθg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2R/g
Here, AB=2Rcosθ
Component of acceleration of bead along AB is,
a=gcosθ
since wire is frictionless, a = constant = gcosθ
initial speed of bead, u=0
(released from rest)

Applying kinematic equation,
s=ut+12at2

AB=0+12gcosθt2

or,2Rcosθ=12gcosθt2

t=4Rg=2Rg
It gives required time for bead to slip from A to B.

Why this question?

Tip: Bead is constrained to move along AB, hence acceleration along AB is provided by the component of weight along AB, i.e. mgcosθ.

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