$A$ screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : $0 \ mm$
Circular scale reading : $52 \ divisions$
Given that $1 \ mm$ on the main scale corresponds to $100$ divisions of the circular scale. The diameter of the wire from the above data is: (in $cm$)

For the determination of the refractive index of a glass slab,a travelling microscope is used whose main scale contains $300$ equal divisions equal to $15 \ cm$. The vernier scale attached to the microscope has $25$ divisions equal to $24$ divisions of the main scale. The least count $(LC)$ of the travelling microscope is (in $cm$):

The diameter of a spherical bob is measured using a vernier callipers. $9$ divisions of the main scale,in the vernier callipers,are equal to $10$ divisions of vernier scale. One main scale division is $1\, {mm}$. The main scale reading is $10\, {mm}$ and $8^{\text{th}}$ division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of $0.04\, {cm}$,then the radius of the bob is $...... \,\times 10^{-2} \,{cm}$.

Assertion $A$: If in five complete rotations of the circular scale,the distance travelled on the main scale of the screw gauge is $5 \, mm$ and there are $50$ total divisions on the circular scale,then the least count is $0.001 \, cm$.
Reason $R$: $\text{Least Count} = \frac{\text{Pitch}}{\text{Total divisions on circular scale}}$
In the light of the above statements,choose the most appropriate answer from the options given below:

The vernier calliper used for measurement has a positive zero error of $0.3 \ mm$. While taking measurement for the internal diameter of a vessel,it was observed that the zero of the vernier scale lies between $9.5 \ cm$ and $9.6 \ cm$ of the main scale and the $6^{th}$ division of the vernier scale coincides with a main scale division. If the least count of the vernier calliper is $0.01 \ cm$,the correct value of the diameter will be: (in $cm$)

$A$ student measured the diameter of a small steel ball using a screw gauge of least count $0.001 \, cm$. The main scale reading is $5 \, mm$ and the circular scale division coinciding with the reference level is $25$. If the screw gauge has a zero error of $-0.004 \, cm$,the correct diameter of the ball is: (in $, cm$)