Heat produced in a current carrying conducting wire depends on current $I$, resistance $R$ of the wire and time $t$ for which current is passed. Using these facts, obtain the formula for heat energy.
Suppose, heat energy $\mathrm{H} \propto \mathrm{I}^{a} \mathrm{R}^{b} t^{c}$
$\therefore \mathrm{H}=k \mathrm{I}^{a} \mathrm{R}^{b} \mathrm{t}^{c} \ldots$ $(i)$ (where $k$ is a dimensionless constant.)
Now, writing the dimensional formula on both sides,
$\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}=(\mathrm{A})^{a}\left(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right)^{b}(\mathrm{~T})^{c}$
$\therefore \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}=(\mathrm{A})^{a-2 b} \mathrm{M}^{b} \mathrm{~L}^{2 b} \mathrm{~T}^{c-3 b}$
${[\mathrm{H}]=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}}$
${[\mathrm{I}]=\mathrm{A}[t]=\mathrm{T}}$
${[\mathrm{R}]=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}}$
Equating the indices on both sides,
$a-2 b=0, b=1 \therefore a-2(1)=0 \therefore a=2$
$-3 b+c=-2$
$\therefore-3+c=-2$
$\therefore c=-2+3=1$
Thus, $a=2, b=1$ and $c=1$.
Substituting these values in equation $(i)$, we get
$\mathrm{H}=k \mathrm{I}^{2} \mathrm{R} t$
$\mathrm{H}=k \mathrm{I}^{2} \mathrm{R} t \text { But, } k=1 \quad \therefore \mathrm{H}=\mathrm{I}^{2} \mathrm{R} t$
Given below are two statements: One is labelled as Assertion $(A)$ and other is labelled as Reason $(R)$.
Assertion $(A)$ : Time period of oscillation of a liquid drop depends on surface tension $(S)$, if density of the liquid is $p$ and radius of the drop is $r$, then $T = k \sqrt{ pr ^{3} / s ^{3 / 2}}$ is dimensionally correct, where $K$ is dimensionless.
Reason $(R)$: Using dimensional analysis we get $R.H.S.$ having different dimension than that of time period.
In the light of above statements, choose the correct answer from the options given below.
Using dimensional analysis, the resistivity in terms of fundamental constants $h, m_{e}, c, e, \varepsilon_{0}$ can be expressed as
In the relation $y = a\cos (\omega t - kx)$, the dimensional formula for $k$ is
if Energy is given by $U = \frac{{A\sqrt x }}{{{x^2} + B}},\,$, then dimensions of $AB$ is