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 $M_{tot}$; 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 $M_{tot}$ 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. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

For calculating the equation of graph, we need the relation between tension in spring ($T$) and $\frac{m}{M_{tot}}$.
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),
$mg-Ma=ma$
$⟹a=g-\frac{Ma}{m}$
$a[1+\frac{M}{m}]=g$
$⟹a=\frac{g}{1+\frac{M}{m}}=\frac{mg}{M+m}$
From question,
$M_{tot}=m+M$
Hence,
$a=\frac{mg}{M_{tot}}$--------(2)
From equation (1),
$T=Ma$
Using expression of $a$ from equation (2) we get,
$T=\frac{Mmg}{M_{tot}}$
Replace $M$ in terms of $m$ and $M_{tot}$,
$M=M_{tot}-m$
$⟹T=\frac{(M_{tot}-m)mg}{M_{tot}}$
Now we can write,
$T=M_{tot}g\frac{m}{M_{tot}}-\frac{m^2}{M_{tot}^2}{M}_{tot}g$
Replace $T$ from $x$ and $\frac{m}{M_{tot}}$from $y$.
$⟹y=M_{tot}gx-M_{tot}g{x}^2$
It is the equation of parabola having concave downward (for more detail see video solution). Hence graph number 4 is the correct answer.