$A$ student measured the length of a rod and wrote it as $3.50\;cm$. Which instrument did he use to measure it?

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 . . . . . . $\text{mm}$.

$A$ plano-convex lens of unknown material and unknown focal length is given. With the help of a Spherometer,we can measure the,

In a vernier callipers,$10$ divisions of vernier scale coincide with $9$ divisions of main scale,the least count of which is $0.1\,cm$. If in the measurement of inner diameter of a cylinder,the zero of the vernier scale lies between $1.3\,cm$ and $1.4\,cm$ of the main scale and the $2^{nd}$ division of the vernier scale coincides with a main scale division,then the diameter will be .......... $cm$.

Diameter of a steel ball is measured using a Vernier callipers which has divisions of $0.1\,cm$ on its main scale $(MS)$ and $10$ divisions of its vernier scale $(VS)$ match $9$ divisions on the main scale. Three such measurements for a ball are given as:

$S$.No.	$MS\;(cm)$	$VS$ divisions
$(1)$	$0.5$	$8$
$(2)$	$0.5$	$4$
$(3)$	$0.5$	$6$

If the zero error is $-0.03\,cm,$ then the mean corrected diameter is ........... $cm$.

When the gap is closed without placing any object in a screw gauge whose least count is $0.005 \ mm$,the $5^{th}$ division on its circular scale coincides with the reference line on the main scale. When a small sphere is placed,the reading on the main scale advances by $4$ divisions,whereas the circular scale reading advances by five times the corresponding reading when no object was placed. There are $200$ divisions on the circular scale. The radius of the sphere is .......... $mm$.