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This is a multiple choice answer as classified in NCERT Exemplar

(a, d) inductive reactance XL= 2 $\pi \vartheta$ L

 

For an LCR circuit Z= $\sqrt[]{R^2+(X_L-X_C)}$ 2

As frequency (ν) increases, Z decreases and at certain value of frequency know as resonant frequency (ν₀), impedance Z is minimum that is Z_min= R current varies inversely with impedance and at Z_min current is maximum.

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This is a Long Answer Type Questions as classified in NCERT Exemplar

The group 1 elements are called alkali metals which possess outermost electronic configuration as ns1 .All the elements that have same outermost electron configuration possess similar properties and are placed in the same group.

This is a multiple choice answer as classified in NCERT Exemplar

Secondary voltage V_s= 24V

Power associated with secondary P_s = 12W

I_s= $\frac{P_S}{V_S}$ = 12/24= 0.5A

I₀= Is $\sqrt[]{2}$ = 1/ $\sqrt[]{2}$ A

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This is a multiple choice answer as classified in NCERT Exemplar

Average power dissipated is E_rmsI_rmscos $\varnothing$

Z= $\sqrt[]{R^2+X_L^2}$

= $\sqrt[]{1+4}$ = $\sqrt[]{5}$

I_rms= 6/ $\sqrt[]{5}$ A

cos $\varnothing$ =R/Z=2/ $\sqrt[]{5}$

p_av= 6 $\times \frac{6}{\sqrt[]{5}}\times \frac{2}{\sqrt[]{5}}$ = 14.4W

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This is a multiple choice answer as classified in NCERT Exemplar

(c) As we know that Q= $\frac{1}{R}\surd \frac{L}{C}$ to make Q high R should be low, L should be high and C should be low.

This is a multiple choice answer as classified in NCERT Exemplar

As we know resonant frequency is v₀= $\frac{1}{2\pi \sqrt[]{Lc}}$

To increase capacitance we must connect it to another capacitor parallel to the first.

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This is a multiple choice answer as classified in NCERT Exemplar

(c) The voltmeter connected to AC Mains reads mean value (2>) and is calibrated in such a way that it gives value of, which is multiplied by form factor to give rms value.

This is a multiple choice answer as classified in NCERT Exemplar

(c) For delivering maximum power from the generator to the load, total internal reactance must be equal to conjugate of total external reactance.

X_L=-X_g

This is a multiple choice answer as classified in NCERT Exemplar

(b) I₀= $\sqrt[]{2}$ Irms= $\sqrt[]{2}$  (5)A

I=I₀sinwt = 5 $\sqrt[]{2}$ sin $\pi /3$ = 5 $\frac{\sqrt[]{3}}{2}$ A

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, b)

Explanation- Due to the point source light propagates in all directions symmetrically and hence, wavefront will be spherical as shown in the diagram.

If power of the source is P, then intensity of the source will be I= p/4 $\pi r$ ²

 where, r is radius of the wavefront at anytime.

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