JEE PYQ: Laws Of Motion Question 30
Question 30 - 2019 | 09 Jan Shift 2] (MCQ)
A mass of 10 kg 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 $45°$ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is ($g = 10,\text{ms}^{-2}$)
(1) 200 N
(2) 140 N
(3) 70 N
(4) 100 N [Correct]
Show Answer
Answer: (4)
Solution
$F = mg\tan 45° = 100$ N.
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