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

$$\eqalign{ & m \propto {v^x}{\rho ^y}{g^z} \cr & M \equiv {\left[ {L{T^{ - 1}}} \right]^x}{\left[ {M{L^{ - 3}}} \right]^y}{\left[ {L{T^{ - 2}}} \right]^z} \cr & {M^1}{L^0}{T^0} = {M^y}\,{L^{x - 3y + z}}\,{T^{ - x - 2z}} \cr & y = 1,x - 3y + z = 0, - x - 2z = 0 \cr} $$
Solving above equations
$$x - 3 - \frac{x}{2} = 0 \Rightarrow \frac{x}{2} = 3 \Rightarrow x = 6$$

$$\eqalign{ & {\text{As }}F = - \eta A\frac{{dv}}{{dz}} \Rightarrow \eta = - \frac{F}{{A\left( {\frac{{dv}}{{dz}}} \right)}} \cr & {\text{As}}\,F = \left[ {ML{T^{ - 2}}} \right],A = \left[ {{L^2}} \right] \cr & dv = \left[ {L{T^{ - 1}}} \right],dz = \left[ L \right] \cr & \therefore \eta = \frac{{\left[ {ML{T^{ - 2}}} \right]\left[ L \right]}}{{\left[ {{L^2}} \right]\left[ {L{T^{ - 1}}} \right]}} = \left[ {M{L^{ - 1}}\;{T^{ - 1}}} \right] \cr} $$

Let $$m$$ and $$d$$ be the mass and diameter of the sphere, then the density $$\rho $$ of the sphere is given by
$$\eqalign{ & \rho = \frac{{{\text{mass}}}}{{{\text{volume}}}} = \frac{m}{{\frac{4}{3}\pi {{\left( {\frac{d}{2}} \right)}^3}}} = \frac{{6m}}{{\pi {d^3}}} \cr & {\text{or}}\,\,{\left( {\frac{{d\rho }}{\rho } \times 100} \right)_{\max }} = \left| {\frac{{dm}}{m} \times 100} \right| + \left| {\frac{{3d\left( d \right)}}{d} \times 100} \right| \cr & = 2 + 3\left( 3 \right) \cr & = 11\% \cr} $$

Note: Here $$\left( {\frac{1}{2}} \right){e_0}{E^2}$$   represents energy per unit volume.
$$\left[ {{e_0}} \right]\left[ {{E^2}} \right] = \frac{{\left[ {{\text{Energy}}} \right]}}{{\left[ {{\text{volume}}} \right]}} = \frac{{M{L^2}{T^{ - 2}}}}{{{L^3}}} = M{L^{ - 1}}{T^{ - 2}}$$

$$\eqalign{ & P = \frac{{{a^3}{b^2}}}{{cd}},\frac{{\Delta P}}{P} \times 100\% = 3\frac{{\Delta a}}{a} \times 100\% + 2\frac{{\Delta b}}{b} \times 100\% + \frac{{\Delta c}}{c} \times 100\% + \frac{{\Delta d}}{d} \times 100\% \cr & = 3 \times 1\% + 2 \times 2\% + 3\% + 4\% \cr & = 14\% \cr} $$

$$\eqalign{ & {\text{We know that}} \cr & F = ma \cr & \Rightarrow F = \frac{{mv}}{t} \cr & \Rightarrow m = \frac{{Ft}}{v} \cr & \left[ M \right] = \frac{{\left[ F \right]\left[ T \right]}}{{\left[ v \right]}} = \left[ {F{v^{ - 1}}T} \right] \cr} $$

$$\eqalign{ & \because C = \frac{q}{V} = \frac{q}{{\frac{W}{q}}} = \frac{{{q^2}}}{W} = \frac{{{{\left( {it} \right)}^2}}}{{F \cdot x}} = \frac{{{{\left[ {AT} \right]}^2}}}{{\left[ {M{L^2}{T^{ - 2}}} \right]}} \cr & = \left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}} \right]\,\,{\text{and }}R = \frac{V}{i} = \frac{W}{{qi}} = \frac{{F \cdot x}}{{{i^2}t}} \cr & = \frac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {AT} \right]\left[ A \right]}} = \left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right] \cr & \therefore {\text{Dimensional formula of }}CR = \left[ {{M^{ - 1}}\;{L^{ - 2}}\;{T^4}\;{A^2}} \right]\left[ {M{L^2}\;{T^{ - 3}}\;{A^{ - 2}}} \right] \cr & = \left[ {{M^0}\;{L^0}\;T} \right] \cr} $$

If $$z = \frac{A}{B}\,{\text{then}}$$
$$\frac{{\Delta z}}{z} = \frac{{\Delta A}}{A} + \frac{{\Delta B}}{B}$$

$$\eqalign{ & F = \frac{{GMm}}{{{R^2}}} \cr & \therefore G = \frac{{F{R^2}}}{{Mm}} \Rightarrow G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right] \cr} $$

The maximum possible error in $$Y$$ due to $$l$$ and $$d$$ are $$\frac{{\Delta Y}}{Y} = \frac{{\Delta l}}{l} + \frac{{2\Delta d}}{d}$$
$$\eqalign{ & {\text{Least count}} = \frac{{{\text{Pitch}}}}{{{\text{Number of division on circular scale}}}} \cr & = \frac{{0.5}}{{100}}\,mm = 0.005\,mm \cr & {\text{Error contribution of }}l \cr & = \frac{{\Delta l}}{l} = \frac{{0.005\,mm}}{{0.25\,mm}} = \frac{1}{{50}} \cr & {\text{Error contribution of }}d \cr & = \frac{{2\Delta d}}{d} = \frac{{2 \times 0.005\,mm}}{{0.5\,mm}} = \frac{1}{{50}} \cr} $$