Question 307: Units , Dimensions and Error Analysis

Question

During Searle's experiment, zero of the Vernier scale lies between 3.20 $$ 	imes $$ 10
-2
m and 3.25 $$ 	imes $$ 10
-2
m of the
main scale. The 20
th
division of the Vernier scale exactly coincides with one of the main scale divisions.
When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 $$ 	imes $$ 10
-2
m and 3.25 $$ 	imes $$ 10
-2
m of the main scale but now the 45
th
division of Vernier scale coincides with one of
the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 $$ 	imes $$ 10
-7
m
2
. The least count of the Vernier scale is 1.0 $$ 	imes $$ 10
-5
m. The maximum percentage error in the Young's
modulus of the wire is

Accepted Answer

The difference between the two measurements by Vernier scale gives elongation of the wire caused by the additional load of 2 kg. In the first measurement, main scale reading is MSR = 3.20 $$	imes$$ 10
$$-$$2
m and Vernier scale reading is VSR = 20. The least count of Vernier scale is LC = 1 $$	imes$$ 10
$$-$$5
m. Thus, the first measurement by Vernier scale is
L
1
= MSR + VSR $$	imes$$ LC
= 3.20 $$	imes$$ 10
$$-$$2
+ 20(1 $$	imes$$ 10
$$-$$5
)
= 3.220 $$	imes$$ 10
$$-$$2
m.
In the second measurement, MSR = 3.20 $$	imes$$ 10
$$-$$2
m and VSR = 45. Thus, the second measurement by Vernier scale is
L
2
= 3.20 $$	imes$$ 10
$$-$$2
+ 45(1 $$	imes$$ 10
$$-$$5
)
= 3.245 $$	imes$$ 10
$$-$$2
m.
The elongation of the wire due to force F = 2g is
l = L
2
$$-$$ L
1
= 0.025 $$	imes$$ 10
$$-$$2
m.
The maximum error in the measurement of l is $$\Delta$$l = LC = 1 $$	imes$$ 10
$$-$$5
m. Young's modulus is given by Y = $${{FL} \over {lA}}$$. The maximum percentage error in the measurement of Y is
$${{\Delta Y} \over Y} 	imes 100 = {{\Delta l} \over l} 	imes 100 = {{1 	imes {{10}^{ - 5}}} \over {0.025 	imes {{10}^{ - 2}}}} 	imes 100 = 4\% $$.

Units , Dimensions and Error Analysis

Integer / Numeric Answer Type