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6 years ago

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```										Dear Souvik ,

PLease see the above free body diagram.
writing equations--
for mass 4m : Fr  =  4ma2   ----------- (1)
for mass m placed above 4m : T - Fr = ma1 ---------------(2)
For hanging mass m : mg -T = ma1 ( acceleration is same as string has to be tight )-------------(3)
Fr = µkmg ---------------------(4)
solving we get , a2 = µkg /4
a1 =g -µkg /2
When only half of the block is still on the 4m mass block , distance covered  = 3l/2
time taken  = (2s/a)1/2  =(3l/a1)1/2
in that time distance travelled by m mass = 1/2 (a2)t2  = 1/2 *(µkg /4)*(6l/(1-µk) g) =  3µkl/4(1 - µk)   m
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation. Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

```
6 years ago
```										Respected Sir,
I have also proceeded like this way and arrived to the same answer provided by you, but the ans. given in the material is something like this: x= 7μl/[8(2-3μ)]
I hope you will clear this confusion.
```
6 years ago
```										how didi u come to the conclusion that distance travelled by the m block is 3L/2?
```
6 years ago

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