Question
Assertion : If three capacitors of capacitances $C_1 < C_2 < C_3$ are connected in parallel then their equivalent capacitance $C_P > C_S$.
Reason : $\frac{1}{{{C_p}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}$
Reason : $\frac{1}{{{C_p}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}$
$C_p > C_s$
$\frac{1}{{{C_p}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}$ is incorrect.
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$E=200\left[\sin \left(6 \times 10^{15}\right) t+\sin \left(9 \times 10^{15}\right) t\right] \,Vm ^{-1}$
Given : $h=4.14 \times 10^{-15} \,eVs$
If this light falls on a metal surface having a work function of $2.50 \,eV$, the maximum kinetic energy of the photoelectrons will be ........... $eV$
$Reason$ : Acceleration due to gravity acting on a body having free fall is zero.
