if least count of an instrument is taken as absolute
error then the relative error in the measurement of
length 32.4 cm by a metre scale is
1.
0.01/32.4
2.
0.1/32.4
3.
1.0/32.4
4.
0.001/32.4
Answers
Answer: Relative error = ∆a/a
Here ∆a equals to least count which will be 0.1 in case of a metre scale
And a is 32.4
Therefore relative error will be 0.1/32.4
Hope this helps
Answer:
Least Count
The least count of an instrument is the smallest measurement that can be taken accurately with it.
DEFINITION
Least Count Error
The smallest value that can be measured by the measuring instrument is called its least count. Measured values are good only up to this value. The least count error is the error associated with the resolution of the instrument.
Example: During Searle's experiment, zero of the Vernier scale lies between 3.20×10
−2
m and 3.25×10
−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 3.20×10
−2
m and 3.25×10
−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 8×10
−7
m
2
. The least count of the Vernier scale is 1.0×10
−5
m. What is the maximum percentage error in the Young's modulus of the wire?
Solution:
Y=
lA
FL
since the experiment measures only change in the length of wire
∴
Y
ΔY
×100=
l
Δl
×100
From the obervation l
1
=MSR+20(LC) (MSR-Main Scale Reading)
l
2
=MSR+45(LC)
⇒ change in lengths = 25 (LC)
and the maximum permissible percentage error in elongation is one LC
∴
Y
ΔY
×100=
25(LC)
(LC)
×100=