
A screw gauge has a pitch of 0.1 mm, how many divisions must be there on its head to measure accurately up to 0.0005 mm?
A. 200
B. 100
C. 500
D.1000
Answer
495.9k+ views
Hint:We know that screw gauge is a measuring device used to make accurate measurements of small readings (less than 1 mm or nearly in that range usually). Least count is the minimum measurable value in an instrument. The Pitch of a screw gauge is the forward movement in the linear scale for one complete rotation of the circular scale. By careful consideration of each option, we can find the answer.
Complete step by step answer:
Given:
The Pitch of the screw gauge is P=0.1 mm.
The least count of screw gauge is LC=0.0005 mm.
We will now write the expression for Number of divisions N.
$N = \dfrac{P}{{LC}}$
We will then substitute the given values, and we will get,
$\begin{array}{l}
\Rightarrow N = \dfrac{{0.1\;{\rm{mm}}}}{{0.0005\;{\rm{mm}}}}\\
\Rightarrow N = 200
\end{array}$
Therefore, the correct option is (A).
Additional Information:We have studied that least count is defined as the minimum value that can be measured by an instrument. The Pitch of a screw gauge is the forward distance travelled by the spindle per revolution. The circular scale performs this one complete rotation. The only major difference between the two measuring devices is that the screw gauge is used only for measuring external measurements. In contrast, the Vernier calliper is used for both internal as well as external measurements.
Note:The key point that we need to keep in mind is that the zero of the main scale should be in the same position as of the zero in the circular scale for a zero error. Zero of the circular scale below the zero in the main scale is considered to be a positive error and vice versa. We should also remember the principal on which the screw gauge works. The principal is that if we rotate the screw head then, in that case, we can obtain the linear movement of the main scale. This linear movement that is obtained is used to calculate the diameter of any wire or even the thickness of a metal plate.
Complete step by step answer:
Given:
The Pitch of the screw gauge is P=0.1 mm.
The least count of screw gauge is LC=0.0005 mm.
We will now write the expression for Number of divisions N.
$N = \dfrac{P}{{LC}}$
We will then substitute the given values, and we will get,
$\begin{array}{l}
\Rightarrow N = \dfrac{{0.1\;{\rm{mm}}}}{{0.0005\;{\rm{mm}}}}\\
\Rightarrow N = 200
\end{array}$
Therefore, the correct option is (A).
Additional Information:We have studied that least count is defined as the minimum value that can be measured by an instrument. The Pitch of a screw gauge is the forward distance travelled by the spindle per revolution. The circular scale performs this one complete rotation. The only major difference between the two measuring devices is that the screw gauge is used only for measuring external measurements. In contrast, the Vernier calliper is used for both internal as well as external measurements.
Note:The key point that we need to keep in mind is that the zero of the main scale should be in the same position as of the zero in the circular scale for a zero error. Zero of the circular scale below the zero in the main scale is considered to be a positive error and vice versa. We should also remember the principal on which the screw gauge works. The principal is that if we rotate the screw head then, in that case, we can obtain the linear movement of the main scale. This linear movement that is obtained is used to calculate the diameter of any wire or even the thickness of a metal plate.
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