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:

M g (\sqrt{2} - 1)

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M g \sqrt{2}

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

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None of these

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Solution

The correct option is C $\frac{M g}{\sqrt{2}}$
$tan 4 5^{\circ} = \frac{f}{m g} \Rightarrow f = \frac{m g}{2}$
Hence, the options $C$ is the correct answer.

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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 45o with the initial vertical direction:

The correct option is C Mg√2 tan⁡45∘ =fmg ⇒f=mg2 tan45∘=fmg⇒f=mg√2tan45∘ =mgf​ ⇒f=2​ mg​ ​Hence, the options CCC is the correct answer.

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:

The correct option is C $\frac{M g}{\sqrt{2}}$
$tan 4 5^{\circ} = \frac{f}{m g} \Rightarrow f = \frac{m g}{2}$
Hence, the options $C$ is the correct answer.