- A$kx_1^2$
- ✓$\frac{1}{2}kx_1^2$
- C$2kx_1^2$
- D$2k{x_1}$
$d w=F \cdot d x$
$\int d x=\int_{0}^{x_1}-k x d x$
$w=-k \int_{0}^{x_1}xdx$
$=-k\left[\frac{x^{2}}{2}\right]_{0}^{x_{1}}$
$=-K\left[\frac{x_{1}^{2}}{2}\right]$
$w=-\frac{-k x_{1}^{2}}{2}$
$w=\frac{-1}{2} k x_{1}^{2}$
work done $=-(w)$
$=-\left[\frac{-1}{2} k x_{1}^{2}\right]$
$=\frac{1}{2} k x_{1}^{2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Statement $I$ : When speed of liquid is zero everywhere, pressure difference at any two points depends on equation $\mathrm{P}_1-\mathrm{P}_2=\rho \mathrm{g}\left(\mathrm{h}_2-\mathrm{h}_1\right)$
Statement $II$ : In ventury tube shown $2 \mathrm{gh}=v_1^2-v_2^2$
In the light of the above statements, choose the most appropriate answer from the options given below.

