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Two spherical bobs of masses $M_A$ and $M_B$ are hung vertically from two strings of length $l_A$ and $l_B$ respectively. They are executing SHM with frequency relation $f_A = 2f_B$, then :
A
$l_A = \dfrac{l_B}{4}$
B
$l_A = 4l_B$
C
$l_A = 2l_B$ and $M_A = 2M_B$
D
$l_A = \dfrac{l_B}{2}$ and $M_A = \dfrac{M_B}{2}$
Detailed Solution
Frequency of a simple pendulum: $f = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$, so $f \propto \dfrac{1}{\sqrt{l}}$; it does not depend on the mass of the bob.
$\dfrac{f_A}{f_B} = \sqrt{\dfrac{l_B}{l_A}}$
$2 = \sqrt{\dfrac{l_B}{l_A}} \Rightarrow 4 = \dfrac{l_B}{l_A}$
$l_A = \dfrac{l_B}{4}$
$\dfrac{f_A}{f_B} = \sqrt{\dfrac{l_B}{l_A}}$
$2 = \sqrt{\dfrac{l_B}{l_A}} \Rightarrow 4 = \dfrac{l_B}{l_A}$
$l_A = \dfrac{l_B}{4}$