JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2024
- Hard
- MCQ
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 _______.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(10 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{T}_1 \sin 45^{\circ}=\mathrm{F}\)
\(\mathrm{T}_1 \cos 45^{\circ}=\mathrm{T}_2=1 \times \mathrm{g}\)
\(\therefore \tan 45^{\circ}=\frac{\mathrm{F}}{\mathrm{g}}\)
\(\therefore \mathrm{F}=10 \mathrm{~N}\)
\(\mathrm{T}_1 \cos 45^{\circ}=\mathrm{T}_2=1 \times \mathrm{g}\)
\(\therefore \tan 45^{\circ}=\frac{\mathrm{F}}{\mathrm{g}}\)
\(\therefore \mathrm{F}=10 \mathrm{~N}\)
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