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

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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M g (\sqrt{2} - 1)

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Solution

The correct option is B $M g \sqrt{2}$
$sin 45^{o} = \frac{O A}{O C}$
$\Rightarrow O C = \frac{O A}{sin 45^{o}}$
$\Rightarrow O C = \frac{M g}{1 / \sqrt{2}}$
$\Rightarrow O C = M g \sqrt{2}$
$\Rightarrow O C = F = M g \sqrt{2}$

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

The correct option is B Mg√2sin45o=OAOC⇒OC=OAsin45o⇒OC=Mg1/√2⇒OC=Mg√2⇒OC=F=Mg√2

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

The correct option is B $M g \sqrt{2}$
$sin 45^{o} = \frac{O A}{O C}$
$\Rightarrow O C = \frac{O A}{sin 45^{o}}$
$\Rightarrow O C = \frac{M g}{1 / \sqrt{2}}$
$\Rightarrow O C = M g \sqrt{2}$
$\Rightarrow O C = F = M g \sqrt{2}$