Question

In the figure, all pulleys are massless and strings are light. Take $g=10m{s}^{-2}$
$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{(a) 1 kg block}& \text{(p) will remain stationary}\\ \text{(b) 2 kg block}& \text{(q) will move down}\\ \text{(c) 3 kg block}& \text{(r) will move up}\\ \text{(d) 4 kg block}& \text{(s)}5m/s^2\\ & \text{(t)}10m/s^2\end{array}$

• A. $a-p;b-q;c-r;d-p$
• B. $a-p,t;b-q;c-r;d-p,s$
• C. $a-r,t;b-p;c-q;d-q,s$
• D. $a-p;b-q;c-r,s;d-p$

Answer 1

The correct option is C $a-r,t;b-p;c-q;d-q,s$

$\left(a\right)\to \left(r\right),\left(t\right);\left(b\right)\to \left(p\right)$
$\left(c\right)\to \left(q\right);\left(d\right)\to \left(q\right),\left(s\right)$
As pulleys are massless,
$80=F=2T_1\Rightarrow T_1=40\text{N}$
$T=T_2=\frac{T_1}{2}=40\text{N}$
For FBD of $1\text{kg}$
$20-g=a\Rightarrow a=10{\text{m/s}}^2\uparrow$
For FBD of $2\text{kg}$
$20-2g=a\Rightarrow a=0{\text{m/s}}^2$
For FBD of $3\text{kg}$
$3g-20=3a\Rightarrow a=\frac{10}{3}{\text{m/s}}^2\downarrow$
For FBD of $4\text{kg}$
$4g-20=4a\Rightarrow a=5{\text{m/s}}^2\downarrow$