Question

vi) In a workshop a worker measures the length
of a steel plate with a Vernier callipers
having a least count 0.01 cm. Four such
measurements of the length yielded the
following values: 3.11 cm, 3.13 cm, 3.14
cm, 3.14 cm. Find the mean length, the
mean absolute error and the percentage
error in the measured value of the
length
please help me to solve this problems ​

Answer 1

four measurements of the length yielded the following values : 3.11cm, 3.13cm, 3.14cm , 3.14 cm

so, mean length = sum of observations/number of observations.

(3.11 + 3.13 + 3.14 + 3.14)/4

= 3.13 cm

hence, mean length, m = 3.13cm

let, a1 = 3.11, a2 = 3.13 , a3 = 3.14 and a4 = 3.14

then, ∆a1 = m - a1 = 3.13 - 3.11 = 0.02

∆a2 = m - a2 = 3.13 - 3.13 = 0.00

∆a3 = m - a3 = 3.13 - 3.14 = 0.01

∆a4 = m - a4 = 3.13 - 3.14 = 0.01

now, mean absolute error , ∆m = (∆a1 + ∆a2 + ∆a3 + ∆a4)/4

= (0.02 + 0.00 + 0.01 + 0.01)/4

= 0.04/4 = 0.01

hence, mean absolute error , ∆m = 0.01

percentage error in the measured value of the length = ∆m/m × 100

= 0.01/3.13 × 100

= 1/3.13

≈ 0.32 %

Answer 2

Explanation:

id Kaya h..... name.........wo Sara delete hogaya h phir say Instagram Start kiye h n tho....ya phir massage Kara old wala pay....jipay phaly thay.....