JEE Physics/ Lakshmikanta Satapathy/ Laws of Motion QA part 7/ Question on Breaking stress of wire connected in a pulley block system solved with the related concepts

1. Physics Helpline
L K Satapathy Laws of Motion 7
Pulley Block System
2M M

2. Physics Helpline
L K Satapathy Laws of Motion 7
Question : Two blocks of masses M and 2M are connected by a
metal wire of cross sectional area passing over a light
smooth pulley as shown in the figure . The breaking stress of the wire
is . The system is released from rest . The maximum
value of M for which the wire will not break is [Take ]
( ) 2 ( ) 4 ( ) 6 ( ) 8a kg b kg c kg d kg
Answer :
9 2
2 10 .N m

8 2
4 10 m

2M M
Let the common acceleration of the system = a
 The block of mass 2M will move down with acceleration a
& The block of mass M will move up with acceleration a
Tension in the wire should not exceed the breaking force of the wire
2
10g m s

3. Physics Helpline
L K Satapathy Laws of Motion 7
Let the tension in the wire = T
Mg
T
a M
(i) For mass 2M :
2Mg
T
a 2M
2 2 . . . (1)Mg T Ma 
(ii) For mass M : . . . (2)T Mg Ma 
To find T : We multiply equation (2) by 2
2 2 2 . . . (3)T Mg Ma  
(3) (1) (2 2 ) (2 ) 0T Mg Mg T     
3 4 0T Mg  
4
3
Mg
T 
We can also use
1 2
1 2
1 2
2
2 ,
m m g
T
m m
m M m M


 
[Free body diagram]

4. Physics Helpline
L K Satapathy Laws of Motion 7
[6 ]M kg Ans 
Correct option = (c)
Breaking stress of the wire = 9 2
2 10 .N m

Cross sectional area of the wire = 8 2
4 10 m

Breaking force of the wire = 9 2 8 2
(2 10 . )(4 10 ) 80N m m N 
  
 Tension in the wire should be  80N
In the limiting case , T = 80N
4
80
3
Mg
 
40
80
3
M
 
2
10g m s  

5. Physics Helpline
L K Satapathy
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