A 1 mathrm{~kg} mass is suspended from ceiling by a rope of length 4 mathrm{~m} . A horizontal force

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4-1.Newton's Laws of Motion

hard

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

$10 \mathrm{~N}$

(JEE MAIN-2024)

Solution

$\mathrm{T}_1 \sin 45^{\circ}=\mathrm{F}$

$\mathrm{T}_1 \cos 45^{\circ}=\mathrm{T}_2=1 \times \mathrm{g}$

$\therefore \tan 45^{\circ}=\frac{\mathrm{F}}{\mathrm{g}}$

$\therefore \mathrm{F}=10 \mathrm{~N}$

Standard 11

Physics

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