Physics - NLM Question with Solution | TestHub

Now, we are given that block 5m remains stationary, so from F.B.D of 5m block.

$$\mathrm{T}=5\mathrm{mg}$$

Let acceleration of the rope be 'a' in counter clockwise direction, as shown. Yatin is just holding the rope, so its acceleration be 'a ${}_{\mathrm{y}}$ ' From string constraint :

$$a=\frac{a_y+0}{2}=a_y=2a$$

Now from F.B.D. of Yatin

$$\begin{array}{rlrlr}& \mathrm{T}-\mathrm{mg}={\mathrm{ma}}_{\mathrm{y}}& 5\mathrm{mg}-\mathrm{mg}=\mathrm{m}(2\mathrm{a})& 4\mathrm{mg}=\mathrm{m}2\mathrm{a}& \mathrm{a}=2\mathrm{g}\end{array}$$

Now, let the acceleration of Ronit w.r.t. rope be ' ${\mathrm{a}}_{\mathrm{R}}$ '. So acceleration of Ronit in ground frame $={\mathrm{a}}_{\mathrm{R}}-\mathrm{a}$ So from F.B.D of Ronit :

$$\begin{array}{rlr}& 2\mathrm{T}-\mathrm{mg}=\mathrm{m}\left[{\mathrm{a}}_{\mathrm{R}}-\mathrm{a}\right]& 2[5\mathrm{mg}]-\mathrm{mg}=\mathrm{m}\left[{\mathrm{a}}_{\mathrm{R}}-2\mathrm{g}\right]\end{array}$$

$9\mathrm{mg}=\mathrm{m}\left[{\mathrm{a}}_{\mathrm{R}}-2\mathrm{g}\right]$

${\mathrm{a}}_{\mathrm{R}}=9\mathrm{g}+2\mathrm{g}={\mathrm{a}}_{\mathrm{R}}=11\mathrm{g}$

Thus, acceleration of Ronit w.r.t. rope is 11 g that is $11\times 10=110\mathrm{m}/{\mathrm{s}}^2$