Question

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 main scale corresponds to $$100$$ divisions of the circular scale. The diameter of wire from the above data is:

A. $$0.052 \,cm$$  
B. $$0.026 \,cm$$
C. $$0.005 \,cm$$
D. $$0.52 \,cm$$
Answer :   $$0.052 \,cm$$
Solution :
L.C. = $$\frac{1}{{100}}$$ mm
Diameter of wire = MSR + CSR × L.C. = 0 + $$\frac{1}{{100}}$$ × 52 = 0.52 $$mm$$ = 0.052 $$cm$$

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. $$ML{T^{ - 1}}$$
B. $$M{L^2}{T^{ - 2}}$$
C. $$M{L^{ - 1}}{T^{ - 2}}$$
D. $$M{L^2}{T^{ - 1}}$$
Releted Question 2

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-

A. resistance
B. charge
C. voltage
D. current
Releted Question 3

A cube has a side of length $$1.2 \times {10^{ - 2}}m$$  . Calculate its volume.

A. $$1.7 \times {10^{ - 6}}{m^3}$$
B. $$1.73 \times {10^{ - 6}}{m^3}$$
C. $$1.70 \times {10^{ - 6}}{m^3}$$
D. $$1.732 \times {10^{ - 6}}{m^3}$$
Releted Question 4

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-

A. $${M^0}{L^0}{T^0}$$
B. $${M^{ - 1}}{L^{ - 1}}{T^{ - 1}}$$
C. $${M^0}{L^2}{T^0}$$
D. $${M^{ - 1}}{L^1}{T^2}$$

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