Numerical Problems Based on Errors in Measurements for Class 11 Physics

(i) Mean value of diameter ($d_m$):

(ii) Mean absolute error ($\Delta d_m$):

Calculate the absolute errors ($|\Delta d_i| = |d_i – d_m|$):
$|\Delta d_1| = |0.026 – 0.027| = 0.001$
$|\Delta d_2| = |0.028 – 0.027| = 0.001$
$|\Delta d_3| = |0.029 – 0.027| = 0.002$
$|\Delta d_4| = |0.027 – 0.027| = 0.000$
$|\Delta d_5| = |0.024 – 0.027| = 0.003$
$|\Delta d_6| = |0.027 – 0.027| = 0.000$

(iii) Relative error:

(iv) Percentage error:

Final Results:
In terms of absolute error: $(d_m \pm \Delta d_m) = (0.027 \pm 0.001) \text{ cm}$
In terms of percentage error: $(0.027 \text{ cm} \pm 3.7\%)$

(i) Mean value of $\mu$ ($\mu_m$):

(ii) Mean absolute error ($\Delta\mu_m$):

Calculate the absolute errors ($|\Delta\mu_i| = |\mu_i – \mu_m|$):
$|1.29 – 1.33| = 0.04$
$|1.33 – 1.33| = 0.00$
$|1.34 – 1.33| = 0.01$
$|1.35 – 1.33| = 0.02$
$|1.32 – 1.33| = 0.01$
$|1.36 – 1.33| = 0.03$
$|1.30 – 1.33| = 0.03$
$|1.33 – 1.33| = 0.00$

(iii) Relative error:

(iv) Percentage error:

Mean value of focal length ($f_m$):

Mean absolute error ($\Delta f_m$):

Calculate the absolute errors ($|\Delta f_i| = |f_i – f_m|$):
$|17.3 – 17.9| = 0.6$
$|17.8 – 17.9| = 0.1$
$|18.3 – 17.9| = 0.4$
$|18.2 – 17.9| = 0.3$
$|17.9 – 17.9| = 0.0$
$|18.0 – 17.9| = 0.1$

Percentage error:

Result expression: