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

Two blocks of masses 1 kg and 2 kg are connected by a metal wire going over a smooth pulley as shown in figure. The breaking stress of the metal is 2×109 N m2. What should be the minimum radius of the wire used if it is not to break? Take g = 10 m s2.


A

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

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

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

The correct option is C


The stress in the wire = TensionArea of cross section
To avoid breaking, this stress should not exceed the breaking stress.
Let the tension in the wire be T. The equations of motion of the two blocks are,
T - 10 N = (1 kg)a
And 20 N - T = (2kg) a.
Eliminating a from these equations,
T = (40/3) N.
The stress =FA=Fπr2
If the minimum radius needed to avoid breaking is r =403π×2×109

Solving this,

r = 4.6 × 105 m.


flag
Suggest Corrections
thumbs-up
7
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Hooke's Law
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon