If the time period ‘t’ of the oscillation of a drop of liquid of density ‘d’ radius ‘r’ vibrating under surface tension ‘s’ as given by the formula t=r2bscda/2. It is observed that the time period is directly proportional to ds. The value of ‘b’ should be

[t]=[n]b[S]c/2[d]a/4 [T]=[2]b[MT−2]c/2=[m2−3]a/4Comparing m dimensions c2+a4=0Comparing time dimensions 1=−C⇒C=−1Comparing lengthb=3a4=0 b=3a4 ∴ b=32

If the time period ‘t’ of the oscillation of a drop of liquid of density ‘d’ radius ‘r’ vibrating under surface tension ‘s’ as given by the formula t=r2bscda/2. It is observed that the time period is directly proportional to ds. The value of ‘b’ should be

[t]=[n]b[S]c/2[d]a/4 [T]=[2]b[MT−2]c/2=[m2−3]a/4Comparing m dimensions c2+a4=0Comparing time dimensions 1=−C⇒C=−1Comparing lengthb=3a4=0 b=3a4 ∴ b=32

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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’ as given by the formula $t = \sqrt{r^{2 b} s^{c} d^{a / 2}}$ . It is observed that the time period is directly proportional to $\sqrt{\frac{d}{s}}$ . The value of ‘b’ should be

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$\frac{3}{4}$

$\sqrt{3}$

$\frac{3}{2}$

$\frac{2}{3}$

answer is C.

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Detailed Solution

$[t] = {[n]}^{b} {[S]}^{c / 2} {[d]}^{a / 4} [T] = {[2]}^{b} {[M T^{- 2}]}^{c / 2} = {[m 2^{- 3}]}^{a / 4}$

Comparing m dimensions
$\frac{c}{2} + \frac{a}{4} = 0$
Comparing time dimensions
$1 = - C \Rightarrow C = - 1$
Comparing length
$b = \frac{3 a}{4} = 0 b = \frac{3 a}{4} ∴ b = \frac{3}{2}$

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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’ as given by the formula t=r2bscda/2. It is observed that the time period is directly proportional to ds. The value of ‘b’ should be

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