JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A mass of \(10 \,kg\) is suspended vertically by a rope of length \(5 \,m\) from the roof. A force of \(30 \,N\) is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is \(\theta=\tan ^{-1}\left(x \times 10^{-1}\right)\). The value of \(x\) is ................ \(\text { (Given } g =10 \,m / s ^{2} \text { ) }\)
- A \(2\)
- B \(5\)
- C \(4\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\( T \sin \theta=30 \) \( T \cos \theta=100 \) \( \Rightarrow \quad \tan \theta=0.3 \)
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