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Question

Two containers of sand are arranged like the block as shown in figure. The containers alone have negligible mass; the sand in these containers has a total mass Mtot; the sand in the hanging container H has mass m.
To measure the magnitude of the acceleration of the system, a large number of experiments have been carried out where m varies from experiment to experiment but Mtot does not; that is, sand is shifted between the containers before each trial.


Which of the curves gives the tension in the connecting cord (the vertical axis is for tension)?

A
1
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B
2
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C
3
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D
4
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Solution

The correct option is D 4
For calculating the equation of graph, we need the relation between tension in spring (T) and mMtot.
Direction and magnitude of acceleration are shown below.


F.B.D of blocks are shown below,
For M mass block,
Using Newton's second law,
T=Ma ----------(1)
Similarly for m mass block we can write,
mg−T=ma
Replace the expression of T from equation (1),
mg−Ma=ma
⟹a=g−Mam
a[1+Mm]=g
⟹a=g1+Mm=mgM+m
From question,
Mtot=m+M
Hence,
a=mgMtot--------(2)
From equation (1),
T=Ma
Using expression of a from equation (2) we get,
T=MmgMtot
Replace M in terms of m and Mtot,
M=Mtot−m
⟹T=(Mtot−m)mgMtot
Now we can write,
T=MtotgmMtot−m2M2totMtotg
Replace T from x and mMtotfrom y.
⟹y=Mtotgx−Mtotgx2
It is the equation of parabola having concave downward (for more detail see video solution). Hence graph number 4 is the correct answer.

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