An ac source is connected to a capacitor $C$ . Due to decrease in its operating frequency:

Option 1 - Capacitive reactance remains constant

Option 2 - Capacitive reactance decreases.

Option 3 - Displacement current increases

Option 4 - Displacement current decreases

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Alok Kumar Singh | Contributor-Level 10

5 months ago

Correct Option - 3

Detailed Solution:

Speed of light in air is

$C = \frac{x}{t_{1}}$

Speed of light in another denser medium.

$C_{2} = \frac{10 x}{t_{2}}$

$\Rightarrow μ = \frac{C}{C_{2}} = \frac{x}{t_{1}} * \frac{t_{2}}{10 x}$

$\Rightarrow μ = \frac{t_{2}}{10 t_{1}}$

(d) $Δ θ = \frac{λ}{d}$

$Δ θ$ is proportional to $λ$ but independent of $D$ .

for minimum wavelength, maximum loss of energy

$e V = (\frac{h c}{λ})$

$λ \propto (\frac{1}{V})$

(b) Incident energy $= 2.20 e V$

If $? < 2.20 e V$ electron will emit.

$? > 2.20 e V$ No electron emission

Only caesium will emit electron

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An ac source is connected to a capacitor C . Due to decrease in its operating frequency:

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