Question

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. $Mg(\sqrt{2}+1)$
• B. $Mg\sqrt{2}$
• C. $\frac{Mg}{\sqrt{2}}$
• D. $Mg(\sqrt{2}-1)$

Answer 1

The correct option is D $Mg(\sqrt{2}-1)$

$\begin{array}{c}\text{1et the length of}\\ \text{the String}=l\\ \text{from work energy theorem,}\end{array}$

$\begin{array}{c}F\times 1\cos {45}^{\circ }=\mathrm{mg}(1-1\cos {95}^{\circ })\\ \text{on}\frac{F}{\sqrt{2}}=\mathrm{mg}(1-\frac{1}{\sqrt{2}})\\ \text{on}F=\mathrm{mg}(\sqrt{2}-1)\end{array}$

Ans Option D