NEET Practice Questions
A screw gauge has the least count of $0.01~\text{mm}$ and there are $50$ divisions in its circular scale. The pitch of the screw gauge is:
A screw gauge gives the following readings 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 on the circular scale, the diameter of the wire that can be inferred from the given data is:
If force $[F]$ , acceleration $[A]$ and time $[T]$ are chosen as the fundamental physical quantities, then find the dimensions of energy:
(Given, $g=\frac{4\pi^2L}{T^2},L=(10\pm0.1)~\text{cm, }T=(100\pm1)~\text s$ )
The determination of the value of acceleration due to gravity $(g)$ by simple pendulum method employs the formula,
$g=4\pi^2\dfrac{L}{T^2}$
The expression for the relative error in the value of $g$ is:
When the circular scale of a screw gauge completes $2$ rotations, it covers $1$ mm over the pitch scale. The total number of circular scale divisions is $50.$ The least count of the screw gauge in metres is:
