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

Volume of a sphere, $$V = \frac{4}{3}\pi {r^3}$$
$$\eqalign{ & \therefore \frac{{\Delta V}}{V} \times 100 = \frac{{3 \times \Delta r}}{r} \times 100 \cr & {\text{Here}}\,\frac{{\Delta r}}{r} \times 100 = 2\% \cr & \therefore \frac{{\Delta V}}{V} \times 100 = 3 \times 2\% = 6\% \cr} $$

The time period of a simple pendulum is given by
$$\eqalign{ & T = 2\pi \sqrt {\frac{\ell }{g}} \cr & \therefore {T^2} = 4{\pi ^2}\frac{\ell }{g} \cr & \Rightarrow g = 4{\pi ^2}\frac{\ell }{{{T^2}}} \cr & \Rightarrow \frac{{\Delta g}}{g} \times 100 = \frac{{\Delta \ell }}{\ell } \times 100 + 2\frac{{\Delta T}}{T} \times 100 \cr & {\bf{Case}}\,{\bf{(i)}} \cr & \Delta \ell = 0.1\,cm,\,\ell = 64\,cm,\,\Delta T = 0.1s,\,T = 128s \cr} $$

Unit of energy will be $$kg - {m^2}/se{c^2}$$
whereas kg.m/s is the unit of momentum.

$$F = ma = \rho \,{\text{volume}}\,'a'$$
To write volume in terms of $$'a'$$ and $$'f'$$
Volume $$ = {L^3} = {\left( {\frac{L}{{{T^2}}}} \right)^3}{T^6} = {a^3}{f^{ - 6}}$$
$$\therefore F = \rho {a^4}{f^{ - 6}}$$

$$\eqalign{ & {\text{As force }}F = \frac{{{e^2}}}{{4\pi {\varepsilon _0}{r^2}}} \Rightarrow \frac{{{e^2}}}{{4\pi {\varepsilon _0}}} = {r^2} \cdot F \cr & {\text{Putting dimensions of }}r{\text{ and }}F,{\text{ we get,}} \cr & \Rightarrow \left[ {\frac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right] = \left[ {M{L^3}\;{T^{ - 2}}} \right]\,......\left( {\text{i}} \right) \cr & {\text{Also, force,}}\,F = \frac{{G{m^2}}}{{{r^2}}} \cr & \Rightarrow \left[ G \right] = \frac{{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}{{\left[ {{M^2}} \right]}} \cr & \Rightarrow \left[ G \right] = \left[ {{{\text{M}}^{ - 1}}\;{{\text{L}}^3}\;{{\text{T}}^{ - 2}}} \right]\,......\left( {{\text{ii}}} \right) \cr & {\text{and}}\,\left[ {\frac{1}{{{c^2}}}} \right] = \frac{1}{{\left[ {\;{L^2}\;{T^{ - 2}}} \right]}} = \left[ {{L^{ - 2}}\;{T^2}} \right]\,......\left( {{\text{iii}}} \right) \cr & {\text{Now, checking optionwise,}} \cr & = \frac{1}{{{c^2}}}{\left( {\frac{{G{e^2}}}{{4\pi {\varepsilon _0}}}} \right)^{\frac{1}{2}}} \cr & = \left[ {{L^{ - 2}}\;{T^2}} \right]{\left[ {{L^6}\;{T^{ - 4}}} \right]^{\frac{1}{2}}} = \left[ L \right] \cr} $$

In multiplication or division the final result should return as many significant figures as there are in the original number with the least significant figures.

If number is < 1, then zero(s) on right of decimal but left of first non-zero digit not significant.

It is given that Young's modulus $$\left( Y \right)$$  is,
$$\eqalign{ & Y = 1.9 \times {10^{11}}\,N/{m^2} \cr & 1\,N = {10^5}dyne \cr & {\text{So,}}\,\,Y = 1.9 \times {10^{11}} \times {10^5}\,dyne/{m^2} \cr} $$
Convert meter to centimeter $$\because 1\,m = 100\,cm$$
$$\eqalign{ & Y = 1.9 \times {10^{11}} \times {10^5}\,dyne/\left( {{{100}^2}\,c{m^2}} \right) \cr & \,\,\,\,\,\, = 1.9 \times {10^{12}}\,dyne/c{m^2} \cr} $$

$$\eqalign{ & V = {\ell ^3} = {\left( {1.2 \times {{10}^{ - 2}}m} \right)^3} = 1.728 \times {10^{ - 6}}{m^3} \cr & V = 1.7 \times {10^{ - 6}}{m^3} \cr} $$

Dimensionally $${\varepsilon _0}L = C$$
where $$C =$$  capacitance
Now the given expression reduces to $$\frac{{C\Delta V}}{{\Delta t}}.$$
Dimensionally $$C\Delta V = q$$   where $$q$$ is charge.
Again the given expression reduces to $$\frac{q}{{\Delta t}}.$$
Dimensionally $$\frac{q}{{\Delta t}} = I$$   where $$I$$ is current