JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 15)
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 45o at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms$$-$$2
200 N
140 N
70 N
100 N
Explanation
_9th_January_Evening_Slot_en_15_1.png)
tan 45o = $${F \over {mg}}$$
$$ \therefore $$ F = mg
= 10 $$ \times $$ 10
= 100 N
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