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

$$\eqalign{ & {\text{Dimensions of energy per unit volume}} \cr & = \frac{{{\text{Dimensions of energy}}}}{{{\text{Dimensions of volume}}}} = \frac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {{L^3}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right] \cr & {\text{Dimensions of force per unit area}} \cr & = \frac{{{\text{Dimensions of force}}}}{{{\text{Dimensions of area}}}} = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right] \cr & {\text{Voltage}} \times \frac{{{\text{Charge}}}}{{{\text{Volume}}}} = \frac{{\left( {\frac{W}{q}} \right) \times \left( {it} \right)}}{{{l^3}}} \cr & = \frac{{\left( W \right)}}{{\left( {{l^3}} \right)}} = \frac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {{L^3}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right] \cr & {\text{Angular momentum}} \cr & = \left( r \right)\left( p \right) = \left( r \right)\left( {mv} \right) = \left[ L \right]\left[ M \right]\left[ {L{T^{ - 1}}} \right] = \left[ {M{L^2}{T^{ - 1}}} \right] \cr} $$
So, dimensions of angular momentum is different from other three.

$$\eqalign{ & {\text{Energy density = }}\frac{{{\text{Energy}}}}{{{\text{Volume}}}} \Rightarrow u = \frac{E}{V} \cr & {\text{Dimensions of }}u = \frac{{{\text{Dimensions of E}}}}{{{\text{Dimensions of }}V}} \cr & = \frac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {{L^3}} \right]}} \cr & = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right] \cr} $$
Refractive index is a dimensionless quantity. Dielectric constant is a dimensionless quantity.
$$\eqalign{ & {\text{Young's modulus}}\,{\text{ = }}\frac{{{\text{Longitudinal stress}}}}{{{\text{Longitudinal strain}}}} = \frac{{\frac{F}{A}}}{{\frac{{\Delta l}}{l}}} \cr & = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right] \cr & {\text{Magnetic field = }}\frac{{{\text{Force}}}}{{{\text{Charge}} \times {\text{Velocity}}}} = \frac{F}{{qv}} \cr & = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {AT} \right]\left[ {L{T^{ - 1}}} \right]}} \cr & = \left[ {M{T^{ - 2}}{A^{ - 1}}} \right] \cr} $$

Pressure $$ = \frac{{ML{T^{ - 2}}}}{{{L^2}}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$$
$$ \Rightarrow a = 1,b = - 1,c = - 2.$$

$$\eqalign{ & T = \sqrt {\frac{{2L}}{g}} + \frac{L}{v} \cr & {\text{with error limits}} \cr & T + \delta T = \sqrt {\frac{{2\left( {L + \delta L} \right)}}{g}} + \frac{{L + \delta L}}{v} \cr & \therefore T + \delta T = \sqrt {\frac{{2L}}{g}\left( {1 + \frac{{\delta L}}{L}} \right)} + \frac{L}{v}\left( {1 + \frac{{\delta L}}{L}} \right) \cr & \therefore T + \delta T = \sqrt {\frac{{2L}}{g}} \times \left( {1 + \frac{{\delta L}}{{2L}}} \right) + \frac{L}{v}\left( {1 + \frac{{\delta L}}{L}} \right) \cr & \therefore T + \delta T = \sqrt {\frac{{2L}}{g}} + \sqrt {\frac{{2L}}{g}} \frac{{\delta L}}{{2L}} + \frac{L}{v} + \frac{L}{v}\frac{{\delta L}}{L} \cr & T + \delta T = T + \sqrt {\frac{{2L}}{g}} \frac{{\delta L}}{{2L}} + \frac{L}{v}\frac{{\delta L}}{L} \cr & \delta T = \frac{{\delta L}}{L}\left[ {\frac{1}{2}\sqrt {\frac{{2L}}{g}} + \frac{L}{v}} \right] \cr & {\text{Substituting }}\delta T = 0.015,\,L = 20\,m,\, \cr & g = 10\,m{s^{ - 2}},v = 300\,m{s^{ - 1}} \cr & {\text{We get}} \cr & \frac{{\delta L}}{L} = \frac{{15}}{{1600}} \cr & \therefore \frac{{\delta L}}{L} \times 100 = \frac{{15}}{{1600}} \times 100 = \frac{{15}}{{16}}\% \approx 1\% \cr} $$

We know that $$\frac{{{Q^2}}}{{2C}}$$  is energy of capacitor, so it represent the dimension of energy $$ = \left[ {M{L^2}{T^{ - 2}}} \right].$$

From stokes law
$$\eqalign{ & F = 6\pi \eta rv \Rightarrow \eta = \frac{F}{{6\pi rv}} \cr & \therefore \eta = \frac{{ML{T^{ - 2}}}}{{\left[ L \right]\left[ {L{T^{ - 1}}} \right]}} \Rightarrow \eta = \left[ {M{L^{ - 1}}{T^{ - 1}}} \right] \cr} $$

$$F = 6\pi \eta vr \Rightarrow \eta = \frac{F}{{6\pi vr}}$$
∴ Unit of coefficient of viscosity in S.I. system $$kg/m - s.$$

$$\eqalign{ & {\text{We know that}} \cr & {\text{Surface tension }}\left( S \right) = \frac{{{\text{Force}}\left[ F \right]}}{{{\text{Length}}\left[ L \right]}} \cr & {\text{So,}}\,\left[ S \right] = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ L \right]}} = \left[ {M{L^0}\;{T^{ - 2}}} \right] \cr & {\text{Energy}}\,\left( E \right) = {\text{Force}} \times {\text{displacement}} \cr & \Rightarrow \left[ E \right] = \left[ {M{L^2}\;{T^{ - 2}}} \right] \cr & {\text{Velocity}}\,\left( v \right) = \frac{{{\text{ displacement }}}}{{{\text{ time }}}} \cr & \Rightarrow \left[ v \right] = \left[ {L{T^{ - 1}}} \right] \cr & {\text{As,}}\,S \propto {E^a}{v^b}{T^c} \cr & {\text{where,}}\,a,b,c\,{\text{are}}\,{\text{constants}}{\text{.}} \cr & {\text{From the principle of homogeneity,}} \cr & \left[ {LHS} \right] = \left[ {RHS} \right] \cr & \Rightarrow \left[ {M{L^0}{T^{ - 2}}} \right] = {\left[ {M{L^2}\;{T^{ - 2}}} \right]^a}{\left[ {L{T^{ - 1}}} \right]^b}{\left[ T \right]^c} \cr & \Rightarrow \left[ {M{L^0}\;{T^{ - 2}}} \right] = \left[ {{M^a}{L^{2a + b}}\;{T^{ - 2a - b + c}}} \right] \cr & {\text{Equating the power on both sides, we get}} \cr & a = 1,2a + b = 0,b = - 2 \cr & \Rightarrow - 2a - b + c = - 2 \cr & \Rightarrow c = \left( {2a + b} \right) - 2 = 0 - 2 = - 2 \cr & {\text{So}}\,\left[ S \right] = \left[ {E{v^{ - 2}}\;{T^{ - 2}}} \right] \cr} $$

$$\left[ A \right] \ne \left[ {{B^2}} \right]$$   so, cannot be added or subtracted

Diameter
$$\eqalign{ & D = M.S.R. + \left( {C.S.R} \right) \times L.C. \cr & D = 2.5 + 20 \times \frac{{0.5}}{{50}} \cr & D = 2.70\,\,mm \cr} $$
The uncertainty in the measurement of diameter
$$\Delta D = 0.01\, mm$$
we know that
$$\eqalign{ & \rho = \frac{{{\text{Mass}}}}{{{\text{Volume}}}} = \frac{M}{V} = \frac{M}{{\frac{4}{3}\pi {{\left( {\frac{D}{2}} \right)}^3}}} \cr & \therefore \frac{{\Delta \rho }}{\rho } \times 100 = \frac{{\Delta M}}{M} \times 100 + 3\frac{{\Delta D}}{\Delta } \times 100 \cr & = 2 + 3 \times \frac{{0.01}}{{2.70}} \times 100 \cr & = 3.1\% \cr} $$