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

$$\eqalign{ & \frac{{\Delta Q}}{Q} \times 100 = \frac{{2\Delta I}}{I} \times 100 + \frac{{\Delta R}}{R} \times 100 + \frac{{\Delta t}}{t} \times 100 \cr & = 2 \times 2\% + 1\% + 1\% \cr & = 6\% \cr} $$

$$\eqalign{ & {M_E} = K + U = 300 \pm 3J = E \pm \Delta E \cr & \therefore \frac{{\Delta E}}{E} \times 100 = 1\% \cr} $$

$$\left[ x \right] = \left[ A \right] \times \left[ y \right] = \left[ B \right] \Rightarrow \left[ x \right] \ne \left[ A \right]$$

$$\eqalign{ & {\text{Kinetic energy }}K = \frac{1}{2}m{v^2} \cr & \therefore \frac{{\Delta K}}{K} \times 100 = \frac{{\Delta m}}{m} \times 100 + 2 \times \frac{{\Delta v}}{v} \times 100 \cr & {\text{Here,}}\,\frac{{\Delta m}}{m} \times 100 = 2\% \Rightarrow \frac{{\Delta v}}{v} \times 100 = 3\% \cr & \therefore \frac{{\Delta K}}{K} \times 100 = 2\% + 2 \times 3\% = 8\% \cr} $$

Since percentage increase in length $$ = 2\% $$
Hence, percentage increase in area of square sheet $$ = 2 \times 2\% = 4\% $$

$$\eqalign{ & {\text{Area}} = {\left( {{\text{Length}}} \right)^2} \cr & {\text{or}}\,{\text{length}} = {\left( {{\text{Area}}} \right)^{\frac{1}{2}}} = {\left( {100 \pm 2} \right)^{\frac{1}{2}}} \cr & = {\left( {100} \right)^{\frac{1}{2}}} \pm \frac{1}{2} \times 2 \cr & = \left( {10 \pm 1} \right)m \cr} $$

$$\eqalign{ & W = \vec F \cdot \vec s = F\,s\cos \theta \cr & = \left[ {ML{T^{ - 2}}} \right]\left[ L \right] = \left[ {M{L^2}{T^{ - 2}}} \right]; \cr & \vec \tau = \vec r \times \vec F \Rightarrow \tau = r\,F\sin \theta \cr & = \left[ L \right]\left[ {ML{T^{ - 2}}} \right] = \left[ {M{L^2}{T^{ - 2}}} \right] \cr} $$

Radiation pressure, $$p = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$$
Velocity of light, $$c = \left[ {L{T^{ - 1}}} \right]$$
Energy striking unit area per second
$$S = \frac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {{L^2}T} \right]}} = \left[ {M{T^{ - 3}}} \right]$$
Now, $${p^x}{S^y}{c^z}$$  is dimensionless.
$$\eqalign{ & \therefore \left[ {{M^0}\;{L^0}\;{T^0}} \right] = {p^x}{S^y}{c^z} \cr & {\text{or}}\,\left[ {{M^0}\;{L^0}\;{T^0}} \right] = {\left[ {{M^1}\;{L^{ - 1}}\;{T^{ - 2}}} \right]^x}{\left[ {{M^1}\;{T^{ - 3}}} \right]^y}{\left[ {\;{L^1}\;{T^{ - 1}}} \right]^z} \cr & {\text{or}}\,\left[ {{M^0}\;{L^0}\;{T^0}} \right] = {\left[ M \right]^{x + y}}\;{\left[ L \right]^{ - x + z}}{\left[ T \right]^{ - 2x - 3y - z}} \cr} $$
From principle of homogeneity of dimensions
$$\eqalign{ & x + y = 0\,......\left( {\text{i}} \right) \cr & - x + z = 0\,......\left( {{\text{ii}}} \right) \cr & - 2x - 3y - z = 0\,......\left( {{\text{iii}}} \right) \cr} $$
Solving Eqs. (i), (ii) and (iii), we get
$$x = 1,y = - 1,z = 1$$

Energy incident per unit area per second $$ = \frac{{{\text{Energy}}}}{{{\text{area}} \times {\text{second}}}} = \frac{{M{L^2}{T^{ - 2}}}}{{{L^2}T}} = M{T^{ - 3}}$$

$$\eqalign{ & {\text{Density}}\left( {{d}} \right){\text{ = }}\frac{{{\text{Mass}}\left( {{M}} \right)}}{{{\text{Volume}}\left( {{V}} \right)}}{\text{ = }}\frac{{{M}}}{{{{{L}}^{\text{3}}}}} \cr & \therefore {\text{Error in density}}, \cr & \frac{{\Delta d}}{d} = \frac{{\Delta M}}{M} + \frac{{3\Delta L}}{L} \cr & = 1.5\% + 3\left( {1\% } \right) \cr & = 4.5\% \cr} $$