$\begin{array}{1 1}160\;N.cm\\170\;N.cm\\156\;N.cm\\150\;N.cm\end{array} $

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Answer : 160 N.cm

We have $F(35-30)=50N=k(35-30)cm\large\frac{N}{cm}$$=k (5cm)$

So $k= \large\frac{50}{5}$$=10$

The work required to extend the spring from 32 to 36 cm is

$W=\int\limits_{(32-30)cm}^{(36-30)cm} 10x \large\frac{N}{cm}$$dx$

$\;\;\;\;=10\large\frac{N}{cm}$$\int\limits_{2cm}^{6cm} xdx$

$\;\;\;\;=10\large\frac{N}{cm}\big(\large\frac{x^2}{2}\big)\bigg|_{2cm}^{6cm}$

$\;\;\;\;=5\large\frac{N}{cm}$$\big((6cm)^2-(2cm)^2\big)$

$\;\;\;\;=5\large\frac{N}{cm}$$(32cm^2)$

$\;\;\;\;=160 \;N.cm$

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