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JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
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
- A \(\frac{3}{4}\)
- B \(\sqrt 3 \)
- C \(\frac{3}{2}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
\(T=\sqrt{\left(M L^{-3}\right)^{a} r^{b}[M T^ {-2}] c}\) \(= \sqrt{M L^{-3} \quad r^{b} \quad M^{-1} T^{+2}}\) \(b-3=0\) \(b=3\)
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