A rope with two mass-less platforms at its two ends passes over a fixed pulley as shown in the figure. Discs with narrow slots and having equal weight of 20N each can be placed on the platforms. The number of discs placed on the left side platform is 𝑛 and that on the right side platform is 𝑚.

It is found that for 𝑛 = 5 and 𝑚 = 0, a force 𝐹= 200 N (refer to part (i) of the figure) is just sufficient to initiate upward motion of the left side platform. If the force 𝐹 is removed then the minimum value of 𝑚 (refer to part (ii) of the figure) required to prevent downward motion of the left side platform is______ (in integer).

The correct answer is 3.

1. Analyzing Condition (i):
In the first setup, there are $n=5$ discs on the left-side platform. Since each disc weighs $20\text{N}$ and the platform is massless, the total weight on the left side is:
$$W_L=5\times 20\text{N}=100\text{N}$$
A downward force of $F=200\text{N}$ is applied to the right-side platform to initiate the upward motion of the left-side platform. In this state, the rope is on the verge of slipping clockwise (towards the right). Therefore, the tension on the right side ($T_{\text{right}}$) is the tight-side tension, and the tension on the left side ($T_{\text{left}}$) is the slack-side tension:
$$T_{\text{right}}=200\text{N}$$
$$T_{\text{left}}=100\text{N}$$
The contact angle of the rope around the fixed pulley is $\theta =\pi \text{rad}$. Using the belt friction equation:
$$\frac{T_{\text{right}}}{T_{\text{left}}}=e^{\mu \theta }$$
Substituting the tension values gives:
$$\frac{200}{100}$$

2. Analyzing Condition (ii):
When the external force $F$ is removed, we place $m$ discs on the right-side platform to prevent the left-side platform from moving downward.
If the left platform starts moving downward, the rope will slip counter-clockwise (towards the left). Thus, the left side becomes the tight side and the right side becomes the slack side:
$$T_{\text{left}}=100\text{N}$$
Let $T_{\text{right}}$ be the tension supporting the right platform with $m$ discs:
$$T_{\text{right}}=m\times 20\text{N}$$
To prevent downward motion, the tension on the right side must satisfy the limiting friction condition:
$$\frac{T_{\text{left}}}{T_{\text{right}}}\le e^{\mu \pi }$$
Substituting $T_{\text{left}}=100\text{N}$ and $e^{\mu \pi }=2$:
$$\frac{100}{T_{\text{right}}}\le 2\Rightarrow T_{\text{right}}\ge 50\text{N}$$

3. Finding the Minimum Integer Value of $m$:
Using the relation for the weight of the discs on the right side:
$$20m\ge 50\Rightarrow m\ge 2.5$$
Since the number of discs $m$ must be a whole number, the minimum integer value of $m$ required to prevent downward motion of the left-side platform is:
$$m=3$$