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JEE Main 20199 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Evening Shift
Subject: Physics
Chapter: Laws of Motion

JEE Main 2019 · 9 January · Evening Shift · Physics · Laws of Motion

A mass of 10kg is suspended vertically by a rope from the roof....

A mass of

10kg

is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of

45 °

at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is

\left(g = 10m{s}^{- 2}\right)

A mass of 10kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some

Options

A $100N$
B $200N$
C $140N$
D $70N$

Correct answer

A. $100N$

Step-by-step solution

A mass of 10kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some

Tcos 45 ° = mg

<br /><br />Tsin 45 ° = F

\tan 45 ° = \dfrac{F}{mg}

F = mg = 100N

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