A screw gauge with a pitch of $$0.5 \,mm$$ and a circular scale with $$50$$ divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the $${45^{th}}$$ division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is $$0.5 \,mm$$ and the $${25^{th}}$$ division coincides with the main scale line?
A.
$$0.70 \,mm$$
B.
$$0.50 \,mm$$
C.
$$0.75 \,mm$$
D.
$$0.80 \,mm$$
Answer :
$$0.80 \,mm$$
Solution :
$$L.C.$$ = 50 = 0.01 $$mm$$
Zero error = 5 x 0.01 = 0.05 mm (Negative)
Reading = (0.5 + 25 x 0.01) + 0.05 = 0.80 $$mm$$
Releted MCQ Question on Basic Physics >> Unit and Measurement
Releted Question 1
The dimension of $$\left( {\frac{1}{2}} \right){\varepsilon _0}{E^2}$$ ($${\varepsilon _0}$$ : permittivity of free space, $$E$$ electric field)
A quantity $$X$$ is given by $${\varepsilon _0}L\frac{{\Delta V}}{{\Delta t}}$$ where $${ \in _0}$$ is the permittivity of the free space, $$L$$ is a length, $$\Delta V$$ is a potential difference and $$\Delta t$$ is a time interval. The dimensional formula for $$X$$ is the same as that of-
Pressure depends on distance as, $$P = \frac{\alpha }{\beta }exp\left( { - \frac{{\alpha z}}{{k\theta }}} \right),$$ where $$\alpha ,$$ $$\beta $$ are constants, $$z$$ is distance, $$k$$ is Boltzman’s constant and $$\theta $$ is temperature. The dimension of $$\beta $$ are-