Question 28 of 210

Three masses $M=100kg,{m}_{1}=10kg$ and ${m}_{2}=20kg$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force $F$ is applied on the system so that the mass ${m}_{2}$ moves upward with an acceleration of $2{ms}^{-2}$. The value of $F$ is
(Take $g=10{ms}^{-2}$)

A$3360N$B$3380N$C$3120N$D$3240N$

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Answer: (A) $3360N$

Let acceleration of $100 \mathrm{~kg}$ block $=a_1$
FBD of $100 \mathrm{~kg}$ block w.r.t ground

$\mathrm{F}-\mathrm{T}-\mathrm{N}_1=100 \mathrm{a}_1$ ... (i)
FBD of 20 block wrt $100 \mathrm{~kg}$

$\mathrm{T}-20 \mathrm{~g}=20(2)$
$\mathrm{T}=240$ ... (ii)
$\mathrm{~N}_1=20 \mathrm{a}_1$ ... (iii)
FBD of $10 \mathrm{~kg}$ block wrt $100 \mathrm{~kg}$

$10 \mathrm{a}_1-240=10(2)$
$\mathrm{a}_1=26 \mathrm{~m} / \mathrm{s}^2$
$\mathrm{~F}-240-20(26)=100 \times 26$
$\Rightarrow \mathrm{F}=3360 \mathrm{~N}$