Question

Radius of wire is measured with a screw gauge, the values are 3.04 mm, 3.06 mm, 3.08 mm, 3.07 mm and 3.05 mm. Find (i) Average Radius of wire (ii) absolute error in each case (iii) Mean absolute Error (iv) Relative Error (v) Percentage Error.

Answer 1

Answer:

The values of the Radius of wire measured in screw gauge are,3.04mm,3.06mm,3.08mm,3.07mm,3.05mm

1) Average radius of wire,

Average=3.04+3.06+3.08+3.07+3.05/5

=15.3mm/5

=12.83mm

2) Now, Absolute error in each case,

∆x1=12.83-3.04=9.79

∆x2=12.83-3.06=9.77

∆x3=12.83-3.08=9.75

∆x4=12.83-3.07=9.76

∆x5=12.83-3.05=9.78

3) Mean absolute error,

∆X=∆x1+∆x2+∆x3+∆x4+∆x5/5

∆X=9.79+9.77+9.75+9.76+9.78/5

∆X=48.85/5

∆X=41.01mm

4) Relative error of the wire,=±∆X/X

±∆X/X=±41.01/12.83=±3.196mm

5) percentage error of the wire=±3.196×100

=±31.9%

Explanation:

Hope it helps you frnd.....

Answer 2

Given Observations:

a1 = 3.04 mm

a2 = 3.06 mm

a3 = 3.08 mm

a4 = 3.07 mm

a5 = 3.05 mm

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(i) Average radius of wire = $\frac {Sum\:of\:all\:Observations}{Total\:no.\:of\:Observations}$

= $\frac {3.04+3.06+3.08+3.07+3.05}{5}$

= $\frac {15.3}{5}$

= 3.06 mm

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(ii) Errors in each case :

Δ a1 = 3.06 - 3.04 = 0.02

Δ a2 = 3.06 - 3.06 = 0

Δ a3 = 3.06 - 3.08 = 0.02

Δ a4 = 3.06 - 3.07 = 0.01

Δ a5 = 3.06 - 3.05 = 0.01

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(iii) Absolute error = $\frac {0.02+0+0.02+0.01+0.01}{5}$

= $\frac {0.06}{5}$

= 0.01 mm

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(iv) Relative error = $\frac {Absolute\:Error}{True\:Value}$

= $\frac {0.01}{3.06}$

= 0.003

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(v) Percentage Error = Relative error × 100

= 0.003 × 100

= 0.3 %

PLEASE MARK IT BRAINILIEST