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  • Need clarity, kindly explain! - A mass of 10kg is suspended vertically by a rope from the roof. When a horizontal force is applied o - Laws of motion - JEE Main

A mass of 10kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45^{\circ}at the roof point. If suspended mass is at equilibrium, the magnitude of force applied is(g = 10ms-2)

  • Option 1)

    200 N

  • Option 2)

    70N

  • Option 3)

    140 N

  • Option 4)

    100 N

Answers (1)

best_answer

 

Newton's 2nd Law -

F\propto \frac{dp}{dt}

F=\tfrac{kdp}{dt} 

F=\tfrac{d\left (mv \right )}{dt} 

F=\tfrac{m\left (dv \right )}{dt}

\frac{dv}{dt}=a

Therefore  F=ma

- wherein

K=1 in C.G.S & S.I

Force can be defined as rate of change of momentum.

 

\tan 45^{\circ}=\frac{100}{F}

F=100N


Option 1)

200 N

Option 2)

70N

Option 3)

140 N

Option 4)

100 N
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