Unit and Measurement MCQ Questions & Answers in Basic Physics | Physics

$$\eqalign{ & E = hv \cr & \Rightarrow h = {\text{Planck's constant}} = \frac{{{\text{Energy}}\left( E \right)}}{{{\text{frequency}}\left( v \right)}} \cr & \therefore \left[ h \right] = \frac{E}{v} = \frac{{\left[ {M{L^2}{T^{ - 1}}} \right]}}{{\left[ {{T^{ - 1}}} \right]}} = \left[ {M{L^2}{T^{ - 1}}} \right] \cr & \left( {\text{A}} \right){\text{ Linear momentum}} = {\text{Mass}} \times {\text{velocity}} \cr & {\text{or}}\,p = m \times v = \left[ M \right]\left[ {L{T^{ - 1}}} \right] = \left[ {ML{T^{ - 1}}} \right] \cr & \left( {\text{B}} \right){\text{Angular momentum}} = {\text{Moment of inertia}} \times {\text{angular velocity}} \cr & {\text{or,}}\,L = I \times \omega = m{r^2}\omega \,\left[ {\because I = m{r^2}} \right] \cr & \therefore \left[ L \right] = \left[ M \right]\left[ {{L^2}} \right]\left[ {{T^{ - 1}}} \right] = \left[ {M{L^2}{T^{ - 1}}} \right] \cr & \left( {\text{C}} \right){\text{Energy}}\,\left[ E \right] = \left[ {M{L^2}{T^{ - 2}}} \right] \cr & \left( {\text{D}} \right)\,{\text{Power}} = {\text{Force}} \times {\text{velocity}} \cr & {\text{or,}}\,P = F \times v \cr & \therefore \left[ P \right] = \left[ {ML{T^{ - 2}}} \right]\left[ {L{T^{ - 1}}} \right] = \left[ {M{L^2}{T^{ - 3}}} \right] \cr} $$
Hence, option (B) is correct.
NOTE
According to homogeneity of dimensions, the dimensions of all the terms in a physical expression should be same. e.g. in the physical expression $$s = ut{\text{ + }}\frac{1}{2}a{t^2},$$    the dimensions of $$s, ut$$  and $$\frac{1}{2}a{t^2}$$  all are same.

In, $$1 - {v^2},{v^2}$$   should be dimensionsless, so it should be $${1 - \frac{{{v^2}}}{{{c^2}}}}.$$

$$\eqalign{ & \rho = \frac{m}{{\ell \pi {r^2}}} \cr & \frac{{\Delta \rho }}{\rho } = \frac{{\Delta m}}{m} + \frac{{2\Delta r}}{r} + \frac{{\Delta \ell }}{\ell } \cr} $$
Putting the values :
$$\eqalign{ & \Delta \ell = 0.06{\mkern 1mu} \,cm,\,\ell = 6\,{\mkern 1mu} cm;\,\, \cr & \Delta r = 0.005\,{\mkern 1mu} cm,\,r = 0.5\,{\mkern 1mu} cm; \cr & \Delta m = 0.003{\mkern 1mu} gm;\,m = 0.3{\mkern 1mu} gm \cr & \therefore \frac{{\Delta \rho }}{\rho } = \frac{4}{{100}} \cr & \therefore \frac{{\Delta \rho }}{\rho } \times 100 = 4\% \cr} $$

Only same dimensional quantity can be added or subtracted.

$${\left( {{\mu _0}{\varepsilon _0}} \right)^{ - \frac{1}{2}}}$$   is the expression for velocity of light.
As $$c = \frac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$$
So, dimension of $$c = \left[ {L{T^{ - 1}}} \right]$$

$$\eqalign{ & R = \frac{V}{I}\,{\text{so,}}\,\frac{{\Delta R}}{R} \times 100 = \frac{{\Delta V}}{V} \times 100 + \frac{{\Delta I}}{I} \times 100 \cr & \Rightarrow \frac{{\Delta R}}{R} \times 100 = \frac{5}{{100}} \times 100 + \frac{{0.2}}{{10}} \times 100 = 7\% \cr} $$

The value of 1 division of main scale $$ = \frac{1}{{10}} = 0.1\,cm$$
The value of 1 division of vernier scale $$ = \frac{{8 \times 0.1}}{{10}} = 0.08\,cm$$
Thus, $$L.C. = 0.1 - 0.08 = 0.02\,cm$$

$$\eqalign{ & M = {\text{Pole}}\,\,{\text{strength}} \times {\text{length}} \cr & = amp - metre \times metre \cr & = amp - metr{e^2} \cr} $$

$$\eqalign{ & \left( {\text{i}} \right)\,{\text{Dimensions of velocity}} = \left[ {{M^0}{L^1}{T^{ - 1}}} \right] \cr & \,\,\,\,\,\,{\text{Here,}}\,a = 0,b = 1,c = - 1 \cr & \left( {{\text{ii}}} \right)\,{\text{Dimensions of acceleration }} = \left[ {{M^0}{L^1}{T^{ - 2}}} \right] \cr & \,\,\,\,\,{\text{Here,}}\,a = 0,b = 1,c = - 2 \cr & \left( {{\text{iii}}} \right)\,{\text{Dimensions of force}} = \left[ {{M^1}{L^1}{T^{ - 2}}} \right] \cr & \,\,\,\,{\text{Here,}}\,a = 1,b = 1,T = - 2 \cr & \left( {{\text{iv}}} \right)\,{\text{Dimensions of pressure}} = \left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right] \cr & \,\,\,\,\therefore {\text{Here,}}\,a = 1,b = - 1,c = - 2 \cr & \therefore {\text{The physical quantity is pressure}}{\text{.}} \cr} $$

For $${{\bf{C}}_{\bf{1}}}$$
L.C. = 1 MSD $$-$$ 1 VSD
$$= 1\,mm - 0.9\,mm$$
$$=0.1\,mm$$
$$= 0.01\, cm$$   [10 VSD $$= 9\,mm$$  ]
Reading = MSR + L.C. × Vernier scale division coinciding the Main scale division = 2.8 + (0.01) × 7 = 2.87 $$cm$$
For $${{\bf{C}}_{\bf{2}}}$$
L.C. = 1 $$mm-$$  1.1 $$mm$$ [10 VSD $$= 11\, mm$$  ]
L.C. = $$-$$ 0.1 $$mm$$ = $$-$$ 0.01 $$cm$$
Reading = 2.8 + (10 $$-$$ 7) × 0.01= 2.83 $$cm$$