
Which of the following instruments has minimum least count?
A. A vernier callipers with 20 divisions on the vernier scale coinciding with 19 main scale divisions.
B. A screw gauge of pitch 1mm and 100 divisions on the circular scale.
C. A spherometer of pitch 0.1m and 100 divisions on the circular scale.
D. An optical instrument that can measure length to within a wavelength of light
Answer
522.6k+ views
Hint: Least Count of an instrument is the smallest change or value of a physical parameter an instrument can measure or detect.
Complete step by step solution:
For vernier calliper, the least count is calculated by the
1(Main Scale Division) - 1 (Vernier Scale Division) =
\[1MSD - \dfrac{{19}}{{20}}MSD = \dfrac{1}{{20}}MSD\]
$ = \dfrac{1}{{20}}mm = \dfrac{1}{{200}}cm = 0.005cm$
For screw gauge, the least count is calculated by the
$Least\;count\;of\;screw\;gauge = \dfrac{{Pitch}}{{Number\;of\,division\,on\;circular{\text{ }}scale}}$
$\dfrac{1}{{100}}mm = \dfrac{1}{{1000}}cm = 0.001m$
For spherometer, the least count is calculated by the
$Least\;count\;of\;screw\;gauge = \dfrac{{Pitch}}{{Number\;of\,division\,on\;circular{\text{ }}scale}}$
$\dfrac{{0.1mm}}{{100}} = \dfrac{1}{{1000}}mm = 0.0001m$
For optical instrument, the least count is calculated by the
Let’s take an example of Red Light, The wavelength of the red light is 6000Angstrom
So, the screw gauge has minimum least count.
So, the correct answer is “Option B”.
Note: With the help of the least count we measure the smallest value that can be measured using the instrument.
Least count gives us the resolution of the instrument.
Ammeter or Voltmeter is said to have zero error if their pointer doesn't read zero when it is not connected in the circuit.
The least count error is the error associated with the resolution of the instrument.
A meter ruler may have graduations at 1 mm division scale spacing or interval.
Complete step by step solution:
For vernier calliper, the least count is calculated by the
1(Main Scale Division) - 1 (Vernier Scale Division) =
\[1MSD - \dfrac{{19}}{{20}}MSD = \dfrac{1}{{20}}MSD\]
$ = \dfrac{1}{{20}}mm = \dfrac{1}{{200}}cm = 0.005cm$
For screw gauge, the least count is calculated by the
$Least\;count\;of\;screw\;gauge = \dfrac{{Pitch}}{{Number\;of\,division\,on\;circular{\text{ }}scale}}$
$\dfrac{1}{{100}}mm = \dfrac{1}{{1000}}cm = 0.001m$
For spherometer, the least count is calculated by the
$Least\;count\;of\;screw\;gauge = \dfrac{{Pitch}}{{Number\;of\,division\,on\;circular{\text{ }}scale}}$
$\dfrac{{0.1mm}}{{100}} = \dfrac{1}{{1000}}mm = 0.0001m$
For optical instrument, the least count is calculated by the
Let’s take an example of Red Light, The wavelength of the red light is 6000Angstrom
So, the screw gauge has minimum least count.
So, the correct answer is “Option B”.
Note: With the help of the least count we measure the smallest value that can be measured using the instrument.
Least count gives us the resolution of the instrument.
Ammeter or Voltmeter is said to have zero error if their pointer doesn't read zero when it is not connected in the circuit.
The least count error is the error associated with the resolution of the instrument.
A meter ruler may have graduations at 1 mm division scale spacing or interval.
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