Which of the following is the most precise device for measuring length:
$(a)$ a vernier callipers with $20$ divisions on the sliding scale
$(b)$ a screw gauge of pitch $1\; mm$ and $100$ divisions on the circular scale
$(c)$ an optical instrument that can measure length to within a wavelength of light?
Which of the following is the most precise device for measuring length:
$(a)$ a vernier callipers with $20$ divisions on the sliding scale
$(b)$ a screw gauge of pitch $1\; mm$ and $100$ divisions on the circular scale
$(c)$ an optical instrument that can measure length to within a wavelength of light?
A device with minimum count is the most suitable to measure length.
Least count of vernier callipers
$=1$ standard division $(SD) -1$ vernier division $(VD)$ $=1-\frac{9}{10}=\frac{1}{10}=0.01 cm$
Least count of screw gauge = $\frac{\text { Pitch }}{\text { Number of divisions }}=\frac{1}{1000}=0.001 cm$
Least count of an optical device $=$ Wavelength of light $\sim 10^{-5} cm$
$=0.00001 cm$
Hence, it can be inferred that an optical instrument is the most suitable device to measure length.
Similar Questions
Asseretion $A:$ If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5\, {mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001\, {cm}$.
Reason $R:$ Least Count $=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}$
In the light of the above statements, choose the most appropriate answer from the options given below:
Asseretion $A:$ If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5\, {mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001\, {cm}$.
Reason $R:$ Least Count $=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}$
In the light of the above statements, choose the most appropriate answer from the options given below:
- [JEE MAIN 2021]
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A screw gauge with a pitch of $0.5 \ mm$ and a circular scale with $50$ divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the $45^{th} $division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet (in $mm$) if the main scale reading is $0.5\ mm$ and the $25^{th}$ division coincides with the main scale line
- [JEE MAIN 2016]
In an experiment the angles are required to be measured using an instrument, $29$ divisions of the main scale exactly coincide with the $30$ divisions of the vernier scale. If the smallest division of the main scale is half- a degree $(= 0.5^o )$, then the least count of the instrument is
In an experiment the angles are required to be measured using an instrument, $29$ divisions of the main scale exactly coincide with the $30$ divisions of the vernier scale. If the smallest division of the main scale is half- a degree $(= 0.5^o )$, then the least count of the instrument is
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While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is $1 \mathrm{~mm}$ and circular scale reading is equal to $42$ divisions. Pitch of screw gauge is $1 \mathrm{~mm}$ and it has $100$ divisions on circular scale. The diameter of the wire is $\frac{x}{50} \mathrm{~mm}$. The value of $x$ is :
- [JEE MAIN 2024]
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A vernier callipers has $20$ divisions on the vernier scale, which coincides with $19^{\text {th }}$ division on the main scale. The least count of the instrument is $0.1 \mathrm{~mm}$. One main scale division is equal to $. . . . . ..$ $\mathrm{mm}$
- [JEE MAIN 2024]