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New answer posted
7 months agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
Secondary voltage Vs= 24V
Power associated with secondary Ps = 12W
Is= = 12/24= 0.5A
I0= Is = 1/ A
New answer posted
7 months agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
Average power dissipated is ErmsIrmscos
Z=
= =
Irms= 6/ A
cos =R/Z=2/
pav= 6 = 14.4W
New answer posted
7 months agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
(c) As we know that Q= to make Q high R should be low, L should be high and C should be low.
New answer posted
7 months agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
As we know resonant frequency is v0=
To increase capacitance we must connect it to another capacitor parallel to the first.
New answer posted
7 months agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
(c) The voltmeter connected to AC Mains reads mean value (
New answer posted
7 months agoContributor-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.
XL=-Xg
New answer posted
7 months agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
(b) I0= Irms= (5)A
I=I0sinwt = 5 sin = 5 A
New answer posted
7 months agoContributor-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 2
where, r is radius of the wavefront at anytime.
New answer posted
7 months agoContributor-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
New answer posted
7 months agoContributor-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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