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Standard XII

Physics

Archimedes' Principle

A mass of 6...

Question

A mass of $6 k g$ is suspended by a rope of length $2 m$ from a ceiling. A force of $50 N$ is applied in horizontal direction at the mid point of the rope. What is the angle between the rope and the vertical in equilibrium.

{tan}^{- 1} (\frac{4}{5})

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{tan}^{- 1} (\frac{5}{4})

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{tan}^{- 1} (\frac{5}{6})

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Solution

The correct option is C ${tan}^{- 1} (\frac{5}{6})$

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Similar questions

A mass of 6 kg is suspended by a rope of length 2 m from a ceiling. A force of 50 N in the horizontal direction is applied at the mid-point of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium?

Q. A block of mass

5 k g

is suspended by a massless rope of length

2 m

from the ceiling. A force of

50 N

is applied in the horizontal direction at the midpoint

P

of the rope, as shown in the figure.
The angle made by the rope with the vertical in equilibrium is (Take

g = 10 m s^{- 2}

)

Q. A mass of

6

kg is suspended by a rope of length

2

m from ceiling. A force of

60

N is applied in the horizontal direction at the mid-point of the rope. The angle made by the rope, with the vertical, in equilibrium position will be :

(Take

g = 10 m s^{- 2}

, neglect the mass of the rope)

Q. A mass of

6

k g

is suspended by a rope of length

2

m

from the ceiling. A force of

50

N

horizontally is applied at the mid point

P

of the rope. Calculate the angle of rope makes with the vertical. Neglect the mass of rope.

(g = 9.8

m s^{- 2})

Q. A mass of

10 kg

is suspended by a rope of length

4 m

, from the ceiling. A force

F

is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of

45^{\circ}

with the vertical. Then

F

equals:
(Take

g = 10 {ms}^{- 2}

and the rope to be massless)

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A mass of 6 kg is suspended by a rope of length 2 m from a ceiling. A force of 50 N is applied in horizontal direction at the mid point of the rope. What is the angle between the rope and the vertical in equilibrium.

The correct option is C tan−1(56)

A mass of $6 k g$ is suspended by a rope of length $2 m$ from a ceiling. A force of $50 N$ is applied in horizontal direction at the mid point of the rope. What is the angle between the rope and the vertical in equilibrium.

The correct option is C ${tan}^{- 1} (\frac{5}{6})$