Question

In the system shown in figure, masses of the blocks are such that when system is released, acceleration of pulley $P_1$ is $a$ upwards and acceleration of block 1 is $a_1$ upwards, such that $a<a_1<2a$.
It is found that acceleration of block 3 is same as that of 1 both in magnitude and direction.

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{a) Acceleration of}2& p)2a+a_1\\ \text{b) Acceleration of}4& q)2a-a_1\\ \text{c) Acceleration of}2\text{w.r.t.}3& \text{r) upwards}\\ \text{d) Acceleration of}2\text{w.r.t.}4& \text{s) downwards}\end{array}$

• A. $a-(q,r)b-(p,s)c-\left(s\right)d-\left(r\right)$
• B. $a-\left(r\right)b-\left(r\right)c-\left(s\right)d-\left(r\right)$
• C. $a-\left(q\right)b-(p,r)c-\left(s\right)d-\left(r\right)$
• D. $a-(q,r)b-(p,s)c-(s,r)d-(r,s)$

Answer 1

The correct option is A $a-(q,r)b-(p,s)c-\left(s\right)d-\left(r\right)$

Let accelerations of various blocks are as shown in figure.


$a_2$ is acceleration of block $2,$ acceleration of pulley is $a$
Now $a=\frac{a_1+a_2}{2}\Rightarrow a_2=2a-a_1>0$
So acceleration of 2 is upwards.
Hence $\left(a\right)\to (q,r)$

Pullet $P_2$ will have downwards acceleration $a$.
and $a=\frac{-a_1+a_4}{2}\Rightarrow a_4=2a+a_1>0$
So acceleration of 4 is downwards
Hence $\left(b\right)\to (p,s)$

Acceleration of 2 w.r.t. 3:
$a_{23}=a_2-a_3=a_2-a_1=2(a-a_1)<0$
This is downwards.
Hence $\left(c\right)\to \left(s\right)$
Acceleration of 2 w.r.t. 4
$a_{24}=a_2-(-a_4)=4a>0$
This is upwards Hence $\left(d\right)\to \left(r\right)$