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

A particle of mass 1 kg is fastened to one end of a string of length 2 m and a mass 2 kg is attached to the middle point, the other end of the string is fastened to a fixed point on a smooth horizontal table. The particles are then projected so that the two portions of the string are always in the same straight line and describes horizontal circles. Then the ratio of tensions (T1T2) in the two parts of the string as shown in diagram is

A
2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
23
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
32
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2
Given mA=1 kg;mB=2 kg
rA= length of mA from O=2 m
rB= length of mB from O=1 m

By the FBD of the mass A: let T2 be the tension force acting on the string connected with mass A

By the force equation, we know that (Fc=mω2r)

T2=mArAω2, as (ω=2 rad/s) given

T2=1×2×(2)2=8 N

By the FBD of the mass B: let T1 be the tension force acting on the left part of the string connected with mass B

T1T2=mBrBω2
T1T2=2×1×(2)2
=8 N
T1=T2+8=16 N
(T1T2)=2

Hence option A is the correct answer

flag
Suggest Corrections
thumbs-up
1
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Relative
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon