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?
$(a)$ 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?
(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.
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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$(D)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.005 \ mm$.
$(A)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.01 \ mm$.
$(B)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.005 \ mm$.
$(C)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.01 \ mm$.
$(D)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.005 \ mm$.
In a Vernier calipers, when both jaws touch each other, the zero of the Vernier scale is shifted to the right of the zero of the main scale and the $7^{th}$ Vernier division coincides with a main scale division. If the value of $1$ main scale division is $1 \text{ mm}$ and there are $10$ Vernier scale divisions, then the Vernier caliper has
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