JEE PYQ: Laws Of Motion Question 20
Question 20 - 2020 | 07 Jan Shift 2] (MCQ)
A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of $45°$ with the vertical. Then $F$ equals:
(1) 100 N [Correct]
(2) 90 N
(3) 70 N
(4) 75 N
Show Answer
Answer: (1)
Solution
$T\cos 45° = 100$ N, $T\sin 45° = F$. $\tan 45° = \frac{F}{100}$. $F = 100$ N.
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