A mass of $M k g$ is suspended by a weightless string. The horizontal force that is required to keep the string and the mass at an angle of $45^{0}$ with the initial vertical direction is:

\frac{M g}{\sqrt{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

M g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

M g (\sqrt{2} + 1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

M g \sqrt{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $M g$
In equilibrium , $F cos θ = M g sin θ$
Horizontal force $= F = M g cot θ = M g cot 45^{o} = M g$

Suggest Corrections

Similar questions

A mass of M kg is suspended by a weightless string. The horizontal force required to displace it until the string makes an angel of $45^{\circ}$ with the initial vertical direction is

Q. A mass of

M k g

is suspended by a weightless string. The minimum horizontal force that is required to displace it slowly until the string makes an angle of

45^{0}

with the initial vertical direction is

Q. A mass of M kg is suspended by a weightless string . the horizontal force that is required to displace it unitl the string makes an angle of

45^{0}

with the initial verticle direction is

A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of $45^{o}$ with the initial vertical direction:

Q. A mass

M k g

is suspended by a weightless string. The horizontal force required to hold the mass at

60^{o}

with the vertical is

Join BYJU'S Learning Program

A mass of Mkg is suspended by a weightless string. The horizontal force that is required to keep the string and the mass at an angle of 450 with the initial vertical direction is:

The correct option is C MgIn equilibrium , Fcosθ=Mgsinθ Horizontal force =F=Mgcotθ=Mgcot45o=Mg

A mass of $M k g$ is suspended by a weightless string. The horizontal force that is required to keep the string and the mass at an angle of $45^{0}$ with the initial vertical direction is:

The correct option is C $M g$
In equilibrium , $F cos θ = M g sin θ$
Horizontal force $= F = M g cot θ = M g cot 45^{o} = M g$