ExamBro
ExamBro
JEE Mains · Physics · STD 11 - Units,dimensions and measurement
  • JEE Mains 2013
  • Hard
  • MCQ

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

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{3}{2}\)

Step-by-step Solution

Detailed explanation

\(T=\sqrt{\left(M L^{-3}\right)^{a} r^{b}[M T^ {-2}] c}\)

\(= \sqrt{M L^{-3} \quad r^{b} \quad M^{-1} T^{+2}}\)

\(b-3=0\)

\(b=3\)

Same subject
Explore more questions on app