Question

List one gives certain physical quantity and List - II gives there magnitude.



$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(I) Acceleration of 2 kg}& \text{(P) 0}\\ \text{(II) Acceleration of 4 kg}& \text{(Q) 3}\\ \text{(III) Acceleration of 6 kg}& \text{(R) 6}\\ \text{(IV) Friction between 2 kg \& 4 kg}& \text{(S)}\frac{8}{3}\\ & \text{(T)}\frac{16}{3}\\ & \text{(U)}\frac{18}{5}\end{array}$
If external force of 40 N is applied horizontly on 4 kg block, Correct match for acceleration and friction is

• A. $I\to Q;II\to S;III\to P;IV\to R$
• B. $I\to S;II\to S;III\to P;IV\to T$
• C. $I\to Q;II\to Q;III\to P;IV\to R$
• D. $I\to R;II\to Q;III\to S;IV\to P$

Answer 1

The correct option is B $I\to S;II\to S;III\to P;IV\to T$

Limiting friction between 2 kg & 4 kg is $f_{24}=6N$
Limiting friction between 4 kg & 6 kg is $f_{46}=24N$
Which means, the maximum force the 4 kg block can apply on the 6 kg block is 24 N.
However, the limiting friction between 6 kg & ground is $f_6=24N$. This means, the 6 Kg block will not move.
The two possible cases are,
The 2 kg and 4 kg blocks move together or they slip between each other.
Say, they move together, then
$a=\frac{40-24}{2+4}=\frac{8}{3}$
The net force on 2 kg block is $2\times \frac{8}{3}=\frac{16}{3}=5.33$
This is less than the limiting friction. So, they two blocks move together, and the acceleration was found to be $\frac{8}{3}$.