JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 28)
A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field
E, as shown in figure. Its bob has mass m and charge q. The time period of the pendulum is given by :
_10th_April_Evening_Slot_en_28_1.png)
$$2\pi \sqrt {{L \over {\sqrt {{g^2} - {{{q^2}{E^2}} \over {{m^2}}}} }}} $$
$$2\pi \sqrt {{L \over {\left( {g + {{qE} \over m}} \right)}}} $$
$$2\pi \sqrt {{L \over {\sqrt {{g^2} + {{{q^2}{E^2}} \over {{m^2}}}} }}} $$
$$2\pi \sqrt {{L \over {\left( {g - {{qE} \over m}} \right)}}} $$
Explanation
_10th_April_Evening_Slot_en_28_2.png)
$$t = 2\pi \sqrt {{L \over {{g_{eff}}}}} $$$$ \Rightarrow {g_{eff}} = \sqrt {{g^2} + {{\left( {{{gE} \over m}} \right)}^2}} $$
$$ \Rightarrow t = 2\pi \sqrt {{L \over {\sqrt {{g^2} + {{\left( {{{qE} \over m}} \right)}^2}} }}} $$
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