The bob of a simple pendulum executes simple | vedclass

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Question

(c) $\because \,\,{t_o} = 2\,\pi \sqrt {\frac{l}{g}} $

Effective weight of bob inside water,
$W' = mg - {\rm{thrust}} = V\rho g - V\rho 'g

Vedclass · Accepted Answer

$t = 2{t_0}$

Vedclass · Answer

(c) $\because \,\,{t_o} = 2\,\pi \sqrt {\frac{l}{g}} $

Effective weight of bob inside water,
$W' = mg - {m{thrust}} = Vho g - Vho 'g$
$ \Rightarrow V\,\,ho {g_{eff}} = V(ho - ho ')g,$ where, $ho $ = Density of bob
$ \Rightarrow {g_{eff}} = \left( {1 - \frac{{ho '}}{ho }} ight)\,g$ and $ho '$ = Density of water
$	herefore t = 2\,\pi \sqrt {\frac{l}{{{g_{eff}}}}} = 2\,\pi \sqrt {\frac{l}{{(1 - ho '/ho )g}}} $    $  (\because ho ' = {10^3}kg/{m^3}  \,\,ho  = \frac{4}{3} 	imes {10^3}kg/{m^3}) $

$	herefore \frac{t}{{{t_0}}} = \sqrt {\frac{1}{{1 - ho '/ho }}} = \sqrt {\frac{1}{{1 - \frac{3}{4}}}} $

$ \Rightarrow t = 2\,{t_0}$.

column$-I$	column $-II$
$(a)$ ${T^2} \to l$	$(i)$ Linear
$(b)$ ${T^2} \to g$	$(ii)$ Parabolic
$(c)$ ${T} \to l$	$(iii)$ Hyperbolic

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