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

$$\frac{L}{R}$$ is time constant of $$R-L$$  circuit so, dimensions of $$\frac{L}{R}$$ is same as that of time.
Alternative
$$\frac{{{\text{Dimensions of}}\,L}}{{{\text{Dimensions of}}\,R}} = \frac{{\left[ {M{L^2}{T^{ - 2}}{A^{ - 2}}} \right]}}{{\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]}} = \left[ T \right]$$

Union of $$k$$ is joules per kelvin or dimensional formula of $$k$$ is $$\left[ {M{L^2}{T^{ - 2}}{\theta ^{ - 1}}} \right]$$
Note: The power of an exponent is a number.
Therefore, dimensionally $$\frac{{\alpha z}}{{k\theta }} = {M^ \circ }{L^ \circ }{T^ \circ }$$
$$\eqalign{ & \therefore \alpha = \frac{{k\theta }}{z} \cr & \therefore \alpha = \frac{{\left[ {M{L^2}{T^{ - 2}}{\theta ^{ - 1}}} \right]\left[ \theta \right]}}{{\left[ L \right]}} = \left[ {ML{T^{ - 2}}} \right] \cr} $$
and dimensionally $$P = \frac{\alpha }{\beta } \Rightarrow \beta = \frac{\alpha }{P}$$
$$\therefore \left[ \beta \right] = \frac{{ML{T^{ - 2}}}}{{M{L^{ - 1}}{T^{ - 2}}}} = {M^0}{L^2}{T^0}$$

$$\left[ x \right] = \left[ {b{t^2}} \right].{\text{Hence}}\,\left[ b \right] = \left[ {\frac{x}{{{t^2}}}} \right] = km/{s^2}$$

According to Newton's law of gravitation, the force of attraction between two masses $${m_1}$$ and $${m_2}$$  separated by a distance $$r$$ is,
$$F = \frac{{G{m_1}{m_2}}}{{{r^2}}} \Rightarrow G = \frac{{F{r^2}}}{{{m_1}{m_2}}}$$
Substituting the dimensions for the quantities on the right hand side, we obtain
Dimensions of $$G = \frac{{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}{{{{\left[ M \right]}^2}}} = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$$

Dimensionally $${e_0}L = $$   Capacitance (c)
$$\therefore {e_0}L\frac{{\Delta V}}{{\Delta t}} = \frac{{C\Delta V}}{{\Delta t}} = \frac{q}{{\Delta t}} = I$$

As given that,
$$\eqalign{ & A = 1.0\,m \pm 0.2\,m,B = 2.0\,m \pm 0.2\,m \cr & {\text{So,}}\,\,X = \sqrt {AB} = \sqrt {\left( {1.0} \right)\left( {2.0} \right)} = 1.414\,m \cr} $$
Rounding off upto two significant digit
$$X = 1.4\,m = \left( r \right)\,\,\left( {{\text{Let}}} \right)$$
$$\eqalign{ & \frac{{\Delta x}}{x} = \frac{1}{2}\left[ {\frac{{\Delta A}}{A} + \frac{{\Delta B}}{B}} \right] \cr & = \frac{1}{2}\left[ {\frac{{0.2}}{{1.0}} + \frac{{0.2}}{{2.0}}} \right] \cr & = 0.212 \cr} $$
Rounding off upto one significant digit
$$\Delta x = 0.2\,m = \Delta r\,\,\left( {{\text{Let}}} \right)$$
So, correct value of $$\sqrt {AB} = r + \Delta r = \left( {1.4 \pm 0.2} \right)m$$

In CGS system, $$d = 4\frac{g}{{c{m^3}}}$$
The unit of mass is $$100\,g$$  and unit of length is $$10\,cm,$$  so
$$\eqalign{ & {\text{density}} = \frac{{4\left( {\frac{{100g}}{{100}}} \right)}}{{{{\left( {\frac{{10}}{{10}}cm} \right)}^3}}} = \frac{{\left( {\frac{4}{{100}}} \right)}}{{{{\left( {\frac{1}{{10}}} \right)}^3}}}\frac{{\left( {100g} \right)}}{{{{\left( {10\,cm} \right)}^3}}} \cr & = \frac{4}{{100}} \times {\left( {10} \right)^3} \cdot \frac{{100g}}{{{{\left( {10\,cm} \right)}^3}}} = 40\,unit \cr} $$

$$s = \frac{Q}{{m\theta }} = \frac{{M{L^2}{T^{ - 2}}}}{{MK}} = \left[ {{L^2}{T^{ - 2}}{K^{ - 1}}} \right]$$

$$\frac{N}{{{m^2}}} = \frac{{{{10}^5}dynes}}{{{{10}^4}c{m^2}}} = 10\,dynes/c{m^2} = 10\beta $$

Mathematically, magnetic flux
$$\phi = BA\,......\left( {\text{i}} \right)$$
but magnetic force
$$F = Bil\,{\text{or}}\,B = \frac{F}{{il}}$$
Putting the value of $$B$$ in Eq. (i), we have
$$\phi = \frac{F}{{il}}A$$
Thus, dimensions of $$\phi = \frac{{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}{{\left[ {AL} \right]}}$$
$$ = \left[ {M{L^2}{T^{ - 2}}{A^{ - 1}}} \right]$$