A 2m long scale with least count of 0.2cm is used to measure the locations of objects on an optical bench. While measuring the foal length of a convex lens, the object pin and the convex lens are placed at 80cm mark and 1m mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at 180cm mark. The % error in the estimation of focal length is:

The formula for a lens is given by $\frac{1}{f}=\frac{1}{\upsilon }-\frac{1}{u}\mathrm{.......}\left(i\right)$ 
$u=(100\pm 0.2)cm-(80\pm 0.2)cm=(20\pm 0.4)cm$ 
$\upsilon =(180\pm 0.2)cm-(100\pm 0.2)cm=(80\pm 0.4)cm$ 
Substituting the values in the equation of focal length,
$\frac{1}{f}=\frac{1}{\upsilon }-\frac{1}{u}=\frac{1}{80}-(-\frac{1}{20})=\frac{5}{80}\Rightarrow f=16cm$ 
Differentiating the equation (i)
$\frac{df}{f^2}=\frac{d\upsilon }{{\upsilon }^2}+\frac{du}{u^2}$ (Only positive sign is considered for error calculations)
$=\frac{0.4}{6400}+\frac{0.4}{400}=0.4\times \frac{17}{6400}$ 
$\Rightarrow df=0.4\times \frac{17}{6400}\times {16}^2=0.272$ 
The percentage error is $\frac{df}{f}\times 100=\frac{0.272}{16}\times 100=1.70\%$