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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

An inductor of reactance 1 ? and a resistor of 2 ? are connected in series to the terminals of a 6 V (rms) AC source. The power dissipated in the circuit is

(a) 8 W

(b) 12 W

(c) 14.4 W

(d) 18 W

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Average power dissipated is E_rmsI_rmscos $\emptyset$

Z= $\sqrt[]{R^{2} + X_{L}^{2}}$

= $\sqrt[]{1 + 4}$ = $\sqrt[]{5}$

I_rms= 6/ $\sqrt[]{5}$ A

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

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

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New answer posted

a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?

(a) R = 20 ?, L = 1.5 H, C = 35μF

(b) R = 25 ?, L = 2.5 H, C = 45 μF

(c) R=15?, L = 3.5H, C = 30 μF

(d) R = 25 ?, L = 1.5 H, C = 45 μF

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

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

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a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

To reduce the resonant frequency in an L-C-R series circuit with a generator

(a) The generator frequency should be reduced

(b) Another capacitor should be added in parallel to the first

(c) The iron core of the inductor should be removed

(d) Dielectric in the capacitor should be removed

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

As we know resonant frequency is v₀= $\frac{1}{2 π \sqrt[]{L c}}$

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

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a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220 V. This means

(a) Input voltage cannot be AC voltage, but a DC voltage

(b) Maximum input voltage is 220 V

(c) The meter reads not v but < v²> and is calibrated to read v²

(d) The pointer of the meter is stuck by some mechanical defect

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Payal Gupta

Contributor-Level 10

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.

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a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

An alternating current generator has an internal resistance R_gand an internal reactance X_gIt is used to supply power to a passive load consisting of a resistance R_g and a reactance X_L. For maximum power to be delivered from the generator to the load, the value of X_L is equal to

(a) Zero

(b) X_g

(c) -X_g

(d) R_g

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Payal Gupta

Contributor-Level 10

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

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a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

If the rms current in a 50 Hz AC circuit is 5 A, the value of the current 1/300 s after its value becomes zero is

(a) 5 $\sqrt[]{2} A$

(b) 5 $\frac{\sqrt[]{3}}{2}$ A

(c) 5/6A

(d) 5/ $\sqrt[]{2}$ A

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Payal Gupta

Contributor-Level 10

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 $π / 3$ = 5 $\frac{\sqrt[]{3}}{2}$ A

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a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Ten

For light diverging from a point source,
(a) The wavefront is spherical
(b) The intensity decreases in proportion to the distance squared
(c) The wavefront is parabolic
(d) The intensity at the wavefront does not depend on the distance

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Vishal Baghel

Contributor-Level 10

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 $π r$ ²

where, r is radius of the wavefront at anytime.

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New answer posted

a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Ten

Consider the diffraction pattern for a small pinhole. As the size of the hole is increased
(a) The size decreases

(b) The intensity increases
(c) The size increases

(d) The intensity decreases

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, b)

Explanation- (a) When a decreases w increases. So, size decreases.

(b) Now, light energy is distributed over a small area and intensity?1/Area is decreasing so intensity increases

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New answer posted

a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Ten

Consider sunlight incident on a pinhole of width 10³ Å. The image of the pinhole seen on a screen shall be
(a) A sharp white ring
(b) Different from a geometrical image
(c) A diffused central spot, white in colour
(d) Diffused coloured region around a sharp central white spot

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (b, d)

Explanation- We know that wavelength of sunlight ranges from 4000 Å to 8000 Å.

Clearly, wavelength λ < width of the slit.

Hence, light is diffracted from the hole. Due to diffraction from the slight the image formed On the screen will be different from the geometrical image.

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New answer posted

a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

An inductor opposes flow of current through it by developing a back emf according to Lenz's law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice -versa.

Since, the induced emf is proportional to the rate of change of current, it will provide greater reactance to the flow of current if the rate of change is faster, i.e., if the frequency is higher. The reactance of an inductor, therefore, is proportional to the frequency. Mat

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