$A$ screw gauge of pitch $0.5\,mm$ is used to measure the diameter of a uniform wire of length $6.8\,cm$. The main scale reading is $1.5\,mm$ and the circular scale reading is $7$. The calculated curved surface area of the wire to appropriate significant figures is $......\,cm^2$. [Screw gauge has $50$ divisions on the circular scale]

While measuring the diameter of a wire using a screw gauge, the following readings were noted. The main scale reading is $1 \,mm$ and the circular scale reading is equal to $42$ divisions. The pitch of the screw gauge is $1 \,mm$ and it has $100$ divisions on the circular scale. The diameter of the wire is $\frac{x}{50} \,mm$. The value of $x$ is:

$A$ screw gauge has $50$ divisions on its circular scale. The circular scale is $4$ units ahead of the pitch scale marking,prior to use. Upon one complete rotation of the circular scale,a displacement of $0.5\, mm$ is noticed on the pitch scale. The nature of zero error involved,and the least count of the screw gauge,are respectively

The least count of a screw gauge is $0.01 \ mm$. If the pitch is increased by $75\%$ and the number of divisions on the circular scale is reduced by $50\%$,the new least count will be . . . . . . $\times 10^{-3} \ mm$.

The vernier scale of a travelling microscope has $50$ divisions which coincide with $49$ main scale divisions. If each main scale division is $0.5 \ mm$,calculate the minimum inaccuracy in the measurement of distance. (in $mm$)

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: