Question 119: Units , Dimensions and Error Analysis

Question

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 $$\rho$$ and radius of the drop is r, then $$\mathrm{T}=\mathrm{K} \sqrt{\rho \mathrm{r}^{3} / \mathrm{S}^{3 / 2}}$$ 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.

Accepted Answer

We know,
$$
\begin{gathered}
{[S]=\left[\mathrm{MT}^{-2}ight]} \\
{[ho]=\left[\mathrm{ML}^{-3}ight]} \\
{[r]=\left[\mathrm{L}]ight.}
\end{gathered}
$$
$$
\begin{aligned}
	herefore 	ext { Dimension of } \mathrm{RHS} &=\frac{\left[\mathrm{M}^{\frac{1}{2}} \mathrm{~L}^{-\frac{3}{2}}ight]\left[\mathrm{L}^{\frac{3}{2}}ight]}{\left[\mathrm{MT}^{-2}ight]^{\frac{3}{4}}} \\
&=\left[\mathrm{M}^{\left(\frac{1}{2}-\frac{3}{4}ight)} \mathrm{L}^{\left(-\frac{3}{2}+\frac{3}{2}ight)} \mathrm{T}^{\frac{6}{4}}ight]=\mathrm{M}^{-\frac{1}{4}} \mathrm{~L}^{0} \mathrm{~T}^{\frac{3}{2}}
\end{aligned}
$$
dimension of L.H.S. = $[\mathrm{T}]$
$	herefore$ Dimension of LHS $
eq$ Dimension of RHS.

Answer

Both (A) and (R) are true and (R) is the correct explanation of (A)

Answer

Both (A) and (R) are true but (R) is not the correct explanation of (A)

Answer

(A) is true but (R) is false

Answer

(A) is false but (R) is true

Units , Dimensions and Error Analysis

Choose the correct answer: