Physics - NLM Question with Solution | TestHub

From FBD of 4kg

$$T-40=4a_1$$

Now FBD of 2kg block

$$\mathrm{T}-2\mathrm{g}=2{\mathrm{a}}_2$$

Now using virtual work method.

$$\begin{array}{rlrl}& \Sigma \mathrm{T}.\mathrm{a}=0& -2{\mathrm{Ta}}_{\mathrm{p}}+{\mathrm{Ta}}_2+{\mathrm{Ta}}_1=0& \Rightarrow {\mathrm{a}}_{\mathrm{p}}=\frac{{\mathrm{a}}_1+{\mathrm{a}}_2}{2}\end{array}$$

FBD of pulley

$$100=2\mathrm{T}\Rightarrow \mathrm{T}=50\mathrm{N}$$

Putting the value of T in above equation

$$\begin{array}{rlrlr}& \mathrm{T}-40=4{\mathrm{a}}_1\Rightarrow 50-40=4{\mathrm{a}}_1& {\mathrm{a}}_1=\frac{10}{4}=2.5\mathrm{m}/{\mathrm{s}}^2& \mathrm{T}-20=2{\mathrm{a}}_2\Rightarrow 50-20=2{\mathrm{a}}_2& {\mathrm{a}}_2=15\mathrm{m}/{\mathrm{s}}^2\end{array}$$

Now ${\mathrm{a}}_{\mathrm{p}}=\frac{{\mathrm{a}}_1+{\mathrm{a}}_2}{2}=\frac{15+2.5}{2}=\frac{35}{4}\mathrm{m}/{\mathrm{s}}^2$ Since the values of accelerations of both the blocks are positive, hence the direction, we assumed is correct, hence acceleration of both the blocks is upwards. Hence, all the four options (A,B,C \& D) are Correct.