Question

A diameter of wire measured by screw gauge is given by 1.012cm,1.015cm, 1.018cm,1.017cm,1.013cm. Find the diameter of wire and mean error and mean absolute error

Answer 1

Answer:

(i) 0.027cm 9ii) 0.001 cm(iii) 0.037 (iv) 3.7 % (0.027 +- 0.001) cm

0.027 cm +- 3.7 %`

Explanation:

Solution :

(i) Mean value of diameter,

D=0.026+0.028+0.029+0.027+0.024+0.0276

D=0.1616=0.268=0.027cm

Now, ΔD1=0.027−0.026=0.001

ΔD2=0.027−0.028=0.002

ΔD3=0.027−0.029=0.002

ΔD4=0.027−0.027=0.000

ΔD5=0.027−0.024=0.003

ΔD6=0.027−0.027=0.000

(ii) Mean absolute arror ΔD1=0.027−0.026=0.001

=|ΔD1|+|ΔD2|+|ΔD3+|ΔD4|+|ΔD5|+|ΔD6∣∣6 ΔD1=0.027−0.026=0.001

=0.001+0.001+0.002+0.000+0.003+0.0006 ΔD1=0.027−0.026=0.001

=0.0076=0.001cm

(iii) Relative error =0.0010.027=0.037

(iv) Percentage error =0.037×10%=3.7%

(v) Diameter of wire =(0.27±0.001)cm

=0.027cm±3.7%