The vernier constant of Vernier callipers is $0.1 \,mm$ and it has zero error of $(-0.05) \,cm$. While measuring the diameter of a sphere,the main scale reading is $1.7 \,cm$ and the coinciding vernier division is $5$. The corrected diameter will be ........... $\times 10^{-2} \,cm$.

The smallest division on the main scale of a vernier callipers is $1 \ mm$,and $10$ vernier divisions coincide with $9$ main scale divisions. While measuring the diameter of a sphere,the zero mark of the vernier scale lies between $2.0 \ cm$ and $2.1 \ cm$ of the main scale,and the fifth division of the vernier scale coincides with a main scale division. The diameter of the sphere is:

$A$ student measured the diameter of a wire using a screw gauge with the least count $0.001\, cm$ and listed the measurements. The measured value should be recorded as (in $, cm$)

$A$ vernier callipers has $20$ divisions on the vernier scale, which coincide with $19$ divisions on the main scale. The least count of the instrument is $0.1 \,mm$. One main scale division is equal to $...$ $mm$.

There are two Vernier calipers,both of which have $1 \ cm$ divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers $(C_1)$ has $10$ equal divisions that correspond to $9$ main scale divisions. The Vernier scale of the other caliper $(C_2)$ has $10$ equal divisions that correspond to $11$ main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in $cm$) by calipers $C_1$ and $C_2$,respectively,are

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