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1.Units, Dimensions and Measurement
medium
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.
ABoth $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$
BBoth $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A)$
C$(A)$ is true but $(R)$ is false
D$(A)$ is false but $(R)$ is true
(JEE MAIN-2022)
Solution
$T=k \sqrt{\frac{\rho r^{3}}{s^{3 / 2}}}$
Dimensions of $RHS =\frac{\left[ M ^{1 / 2} L ^{-3 / 2}\right]\left[ L ^{3 / 2}\right]}{\left[ MT ^{-2}\right]^{3 / 4}}= M ^{1 / 3} L ^{0} T ^{3 / 2}$
Dimensions of L.H.S $\neq$ Dimensions of R.H.S
$\therefore$ option (D)
Dimensions of $RHS =\frac{\left[ M ^{1 / 2} L ^{-3 / 2}\right]\left[ L ^{3 / 2}\right]}{\left[ MT ^{-2}\right]^{3 / 4}}= M ^{1 / 3} L ^{0} T ^{3 / 2}$
Dimensions of L.H.S $\neq$ Dimensions of R.H.S
$\therefore$ option (D)
Standard 11
Physics
