One main scale division of a Vernier calliper is equal to $1 \text{ mm}$ and the number of divisions on the Vernier scale is $10$. When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that $4^{th}$ Vernier division coincides with a division of the main scale. If the Vernier calliper measures the length of a wire to be $1 \text{ cm}$, the actual length of the wire is : (in $\text{ cm}$)

Identify the physical quantity that cannot be measured using a spherometer:

Thickness of a pencil measured by a screw gauge (least count $0.001 \ cm$) comes out to be $0.802 \ cm$. The percentage error in the measurement is (in $\%$)

The smallest division on the main scale of a Vernier calipers is $0.1 \text{ cm}$. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is (in $\text{ cm}$)

In a screw gauge, the zero of the main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are $100$ divisions on the circular scale and the pitch of the screw gauge is $0.1 \text{ mm}$. When the diameter of a sphere is measured, the reading of the main scale is $5 \text{ mm}$ and the $50^{th}$ division of the circular scale coincides with the reference line of the main scale. The diameter of the sphere is . . . . . . $\text{mm}$.

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?