The refractive index ($\mu$) of glass is found to have the values $1.49,1.50,1.52,1.54and1.48$. Calculate percentage error.

The correct option is B $1.32\%$
Given,

$n_1=1.49$

$n_2=1.50$

$n_3=1.52$

$n_4=1.54$

$n_5=1.48$

Mean value, $n_m=\frac{n_1+n_2+n_3+n_4+n_5}{5}$

$n_m=\frac{1.49+1.50+1.52+1.54+1.48}{5}=1.505$

$n_m=1.51$

Absolute error,

$|\Delta n_1|=n_m-n_1=1.51-1.49=0.02$

$|\Delta n_2|=n_m-n_2=1.51-1.50=0.01$

$|\Delta n_3|=n_3-n_m=1.51-1.52=0.01$

$|\Delta n_4|=n_m-n_4=1.54-1.51=0.03$

$|\Delta n_5|=n_m-n_5=1.51-1.48=0.03$

The mean of absolute error,

$\Delta n_m=\frac{0.02+0.01+0.01+0.03+0.03}{5}$

$\Delta n_m=0.02$

Percentage error, $\frac{\Delta n_m}{n_m}\times 100=\frac{0.02}{1.51}\times 100=1.32$%

The correct option is B.