JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A \(1 \mathrm{~kg}\) mass is suspended from the ceiling by a rope of length \(4 \mathrm{~m}\). A horizontal force ' \(F\) ' is applied at the mid point of the rope so that the rope makes an angle of \(45^{\circ}\) with respect to the vertical axis as shown in figure. The magnitude of \(F\) is _______.
- A \(\frac{10}{\sqrt{2}} \mathrm{~N}\)
- B \(1 \mathrm{~N}\)
- C \(\frac{1}{10 \times \sqrt{2}} \mathrm{~N}\)
- D \(10 \mathrm{~N}\)
Answer & Solution
Correct Answer
(D) \(10 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{T}_1 \sin 45^{\circ}=\mathrm{F}\) \(\mathrm{T}_1 \cos 45^{\circ}=\mathrm{T}_2=1 \times \mathrm{g}\) \(\therefore \tan 45^{\circ}=\frac{\mathrm{F}}{\mathrm{g}}\) \(\therefore \mathrm{F}=10 \mathrm{~N}\)
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