Physics

Get insights from 5.6k questions on Physics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics

Follow Ask Question
5.6k

Questions

0

Discussions

30

Active Users

0

Followers

New answer posted

7 months ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Secondary voltage Vs= 24V

Power associated with secondary Ps = 12W

IsPSVS = 12/24= 0.5A

I0= Is 2 = 1/ 2 A

New answer posted

7 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Average power dissipated is ErmsIrmscos

Z= R2+XL2

1+4 = 5

Irms= 6/ 5 A

cos  =R/Z=2/ 5

pav= 6 *65*25 = 14.4W

New answer posted

7 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) As we know that Q= 1RLC to make Q high R should be low, L should be high and C should be low.

New answer posted

7 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

As we know resonant frequency is v0= 12πLc

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

New answer posted

7 months ago

0 Follower 2 Views

P
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.

New answer posted

7 months ago

0 Follower 1 View

P
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.

XL=-Xg

New answer posted

7 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) I02 Irms= 2  (5)A

I=I0sinwt = 5 2 sin π/3 = 5 32 A

New answer posted

7 months ago

0 Follower 3 Views

V
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.

New answer posted

7 months ago

0 Follower 3 Views

V
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

New answer posted

7 months ago

0 Follower 3 Views

V
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.

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 684k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.