An ac source is connected to a capacitor <math> <mi>C</mi> </math> . Due to decrease in its operating frequency:

Speed of light in air is C = x t 1 Speed of light in another denser medium. C 2 = 10 x t 2  ⇒ μ = C C 2 = x t 1 × t 2 10 x ⇒ μ = t 2 10 t 1 (d) Δ θ = λ d  Δ θ is proportional to λ but independent of D .(c) e V = Energy of electronfor minimum wavelength, maximum loss of energy e V = h c λ  λ ∝ 1 V (b) Incident energy = 2.20 e V If ? < 2.20 e V electron will emit. ? > 2.20 e V No electron emissionOnly caesium will emit electron

An ac source is connected to a capacitor <math> <mi>C</mi> </math> . Due to decrease in its operating frequency:

Displacement current increases

Capacitive reactance remains constant

Capacitive reactance decreases.

Displacement current decreases

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Physics NCERT Exemplar Solutions Class 11th Chapter Eight Alternating Current Physics Ncert Solutions Class 12th Physics Class 12th

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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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

Speed of light in air is

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

Speed of light in another denser medium.

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

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

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

(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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