Question 126: Units , Dimensions and Error Analysis

Question

A student determined Young's Modulus of elasticity using the formula $$Y = {{Mg{L^3}} \over {4b{d^3}\delta }}$$. The value of g is taken to be 9.8 m/s
2
, without any significant error, his observation are as following.
Physical
Quantity
Least count of the
Equipment used
for measurement
Observed value
Mass (M)
1 g
2 kg
Length of bar (L)
1 mm
1 m
Breadth of bar (b)
0.1 mm
4 cm
Thickness of bar (d)
0.01 mm
0.4 cm
Depression ($$\delta $$)
0.01 mm
5 mm
Then the fractional error in the measurement of Y is :

Accepted Answer

$$y = {{Mg{L^3}} \over {4b{d^3}\delta }}$$
$${{\Delta y} \over y} = {{\Delta M} \over M} + {{3\Delta L} \over L} + {{\Delta b} \over b} + {{3\Delta d} \over d} + {{\Delta \delta } \over \delta }$$
$${{\Delta y} \over y} = {{{{10}^{ - 3}}} \over 2} + {{3 	imes {{10}^{ - 3}}} \over 1} + {{{{10}^{ - 2}}} \over 4} + {{3 	imes {{10}^{ - 2}}} \over 4} + {{{{10}^{ - 2}}} \over 5}$$
$$ = {10^{ - 3}}[0.5 + 3 + 2.5 + 7.5 + 2] = 0.0155$$
Option (b)

Answer

0.0083

Answer

0.0155

Answer

0.155

Answer

0.083

Units , Dimensions and Error Analysis

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