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Three masses $\mathrm{\mathrm{M}=100 \mathrm{~kg}, \mathrm{~m}_{1}=10 \mathrm{~kg}}$ and $\mathrm{\mathrm{m}_{2}=20 \mathrm{~kg}}$ 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 $\mathrm{F}$ is applied on the system so that the mass $\mathrm{\mathrm{m}_{2}}$ moves upward with an acceleration of $2 \mathrm{~ms}^{-2}.$ The value of $\mathrm{F}$ is : $\mathrm{\left(\text { Take } \mathrm{g}=10 \mathrm{~ms}^{-2}\right)}$

Option: 1
$\mathrm{3360 \mathrm{~N}}$

Option: 2
$\mathrm{3380 \mathrm{~N}}$

Option: 3
$\mathrm{3120 \mathrm{~N}}$

Option: 4
$\mathrm{3240 \mathrm{~N}}$

Answers (1)

best_answer

$\mathrm{f\rightarrow}$ Pseudo force on $\mathrm{m_{1}}$ in the frame of block of mass $\mathrm{M}$

$\mathrm{f=m_{1}a}\rightarrow (1)$

Let the acceleration of system and block of mass $\mathrm{m_{1}}$ be $\mathrm{a}$ and $\mathrm{a_{1}}$ respectively

$\mathrm{f-T=m_{1}a_{1}}\rightarrow (2)$

$\mathrm{T-m_{2}g=m_{2}a_{2}}\rightarrow (3)$

Common acceleration $\mathrm{=a=\frac{F}{m_{1}+m_{2}+m}}$

$\mathrm{a=\frac{F}{130}}\rightarrow (4)$

from eqⁿ 1,2 & 4

$\mathrm{\frac{F}{13}-T=10a_{1},=10(2)=20}$

$\mathrm{(a_{1}=2m/s^{2})}----\mathrm{(Given)}$

$\mathrm{T-200=20(2)}$

$\mathrm{T=240}$

$\mathrm{\frac{F}{13}-240=20}$

$\mathrm{F=260\times 13}$

$\mathrm{F=3380N}$

Hence (2) is correct option

Posted by

Ritika Kankaria

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