- Home
- Standard 11
- Physics
1.Units, Dimensions and Measurement
hard
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$\frac{2}{3}$
(JEE MAIN-2013)
Solution
$T=\sqrt{\left(M L^{-3}\right)^{a} r^{b}[M T^ {-2}] c}$
$= \sqrt{M L^{-3} \quad L^{b} \quad M^{-1} T^{+2}}$
$b-3=0$
$b=3$
$= \sqrt{M L^{-3} \quad L^{b} \quad M^{-1} T^{+2}}$
$b-3=0$
$b=3$
Standard 11
Physics
