Answer the following:
$(a)$ You are given a thread and a metre scale. How will you estimate the diameter of the thread?
$(b)$ $A$ screw gauge has a pitch of $1.0\; mm$ and $200$ divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
$(c)$ The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of $100$ measurements of the diameter expected to yield a more reliable estimate than a set of $5$ measurements only?

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(N/A) Part $(a)$: Wrap the thread closely around a uniform smooth rod such that the coils are touching each other without overlapping. Measure the total length $(L)$ of the coiled portion using a metre scale. If $(n)$ is the number of turns,the diameter $(d)$ of the thread is given by $d = \frac{L}{n}$.
Part $(b)$: No,it is not possible to increase the accuracy arbitrarily. While increasing the number of divisions decreases the least count,the accuracy is limited by other factors such as the mechanical errors of the instrument,the flexibility of the screw,and the precision of the human observer.
Part $(c)$: $A$ set of $100$ measurements is more reliable because the random errors in measurement follow a statistical distribution. As the number of observations $(N)$ increases,the random error in the mean value decreases by a factor of $\frac{1}{\sqrt{N}}$. Thus,$100$ measurements provide a much smaller uncertainty compared to $5$ measurements.

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Similar Questions

$A$ tiny metallic rectangular sheet has a length and breadth of $5 \ mm$ and $2.5 \ mm$,respectively. Using a specially designed screw gauge which has a pitch of $0.75 \ mm$ and $15$ divisions on the circular scale,you are asked to find the area of the sheet. In this measurement,the maximum fractional error will be $\frac{x}{100}$ where $x$ is . . . . . .

In a vernier callipers,$20$ $VSD$ coincide with $16$ $MSD$ (each division of length $1 \text{ mm}$). The least count of the vernier callipers is: (in $\text{ cm}$)

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: In a Vernier calliper,if a positive zero error exists,then while taking measurements,the reading taken will be more than the actual reading.
Reason $(R)$: The zero error in a Vernier calliper might have happened due to a manufacturing defect or due to rough handling.
In the light of the above statements,choose the correct answer from the options given below:

$A$ Vernier calipers has $1 \,mm$ marks on the main scale. It has $20$ equal divisions on the Vernier scale which match with $16$ main scale divisions. For this Vernier calipers, the least count is (in $\,mm$)

The least count of a screw gauge is $0.01 \ mm$. If the pitch is increased by $75\%$ and the number of divisions on the circular scale is reduced by $50\%$,the new least count will be . . . . . . $\times 10^{-3} \ mm$.

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