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Three masses M= 100kg, m₁= 10kg and m₂=20kg are arranged in a system as shown in figure. AII the surface 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 m₂ moves upward with an acceleration of 2 ms^-2. The value of F is
(Take g= 10 ms^-2)

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3120 N

3360 N

3380 N

3240 N

answer is A.

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Detailed Solution

$Let acceleration of 100 kg block = a_{1}$

$F - T - N_{1} = 100 a_{1}$ ..................(1)

$FBD of 20 kg block wrt 100 kg$

In vertical direction,

$T - 20 g = 20 (2)$

$\Rightarrow T = 240 N$ ............(2)

In horizontal direction,

$N_{1} = 20 a_{1}$ .....................(3)

$FBD of 10 kg block wrt 100 kg$

$10 a_{1} - T = 10 (2)$

$\Rightarrow 10 a_{1} - 240 = 10 (2)$

$\Rightarrow a_{1} = 26 m / s^{2}$

Now using (1), (2) and (3),

$F - 240 - 20 (26) = 100 \times 26$

$\Rightarrow F = 3360 N$

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