One end of an elastic string is attached to a block of mass m and the other end is held keeping the string relaxed in vertical. Relaxed length of the string is $l_{0}$ and there is a mark P at height $0.8 l_{0}$ from the lower end. Now the upper end is slowly raised. When the block leaves the ground, the mark P reaches the point where upper end of the relaxed string was. Expression for the work done by the force applied by the hand is $n m g l_{0}$ . Then value of n is_______

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$∵ K = \frac{A Y}{l} \Rightarrow A Y = K l = c o n s \tan t K_{1} \times 0.8 l_{0} = K l_{0} \Rightarrow K_{1} = \frac{5 K}{4} a L s o m g = K_{1} x = \frac{5 K}{4} \times 0.2 l_{0} \Rightarrow K = \frac{4 m g}{l_{0}} ∵ K x_{n e t} = K_{1} 0.2 l_{0}$

$\Rightarrow x_{n e t} = \frac{5 K}{4} \times \frac{0.2 l_{0}}{K} = \frac{l_{0}}{4} \Rightarrow W_{F} = \frac{1}{2} (\frac{4 m g}{l_{0}}) {(\frac{l_{0}}{4})}^{2} = \frac{m g l_{0}}{8} = n m g l_{0}$