(C) The dimensions of the physical quantities are: Time period $t = [T]$,Density $d = [ML^{-3}]$,Radius $r = [L]$,Surface tension $s = [MT^{-2}]$.Give

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(C) The dimensions of the physical quantities are: Time period $t = [T]$,Density $d = [ML^{-3}]$,Radius $r = [L]$,Surface tension $s = [MT^{-2}]$.Given formula: $t = r^b s^{c/2} d^{a/4}$.Substituting dimensions: $[T] = [L]^b [MT^{-2}]^{c/2} [ML^{-3}]^{a/4}$.Equating powers of $M$: $0 = c/2 + a/4 \implies a = -2c$.Equating powers of $T$: $1 = -c \implies c = -1$.Substituting $c = -1$ into $a = -2c$,we get $a = 2$.Equating powers of $L$: $0 = b - 3(a/4) = b - 3(2/4) = b - 3/2$.Therefore,$b = 3/2$.

Class 11 Physics Units and Measurements Dimensional Analysis, Uses and Limitations

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If the time period $t$ of the oscillation of a drop of liquid of density $d$,radius $r$,vibrating under surface tension $s$ is given by the formula $t = \sqrt{r^{2b} s^c d^{a/2}}$. It is observed that the time period is directly proportional to $\sqrt{\frac{d}{s}}$. The value of $b$ should therefore be

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A
$3/4$
B
$\sqrt{3}$
C
$3/2$
D
$2/3$

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