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

6 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

E 2 > E 1

For statement 1

              hf = E1  - E2

Since, E2 > E1

So, it should be

hf = E2 – E1

Therefore statement 1 is wrong

For statement 2

For the jumping of electron from higher energy orbit

(E2) to lower energy orbit (E1)

hf = E2 – E1

  f = E 2 E 1 h              

Statement (2) is correct

New answer posted

6 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

  μ 2 > μ 1

μ 2 μ 1 > 1               

μ 2 = c v 2 , μ 1 = c v 1               

μ 2 μ 1 = v 1 v 2 > 1 v 1 > v 2               

Frequency remains constant while refraction since energy is constant

υ = v λ               

λ α v               

λ 1 > λ 2               

λ decreases

wavelength and speed decreases but frequency remains constant

New question posted

6 months ago

0 Follower

New answer posted

6 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

R 1 = R  

R 2 = R       

P = 1 f = ( μ 1 ) ( 1 R 1 1 R 2 )               

P = 1 f = ( μ 1 ) ( 1 R ( 1 R ) )

P = ( μ 1 ) 2 R              

L2

R1 = R

R 2 =               

R 1 =               

R2 = ­­-R

Power of  L 1 = 1 f ' ' = ( μ 1 ) ( 1 R 1 1 R 2 )

Power of L 1 = ( μ 1 ) * 2 R = P  

New answer posted

6 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

Statement 1 is true

Speed of EM ware in vaccum is given by,

C = 1 μ 0 0                

Speed of EM wave in a material medium

v = 1 u r μ 0 r 0                

So, statement (2) is false.

New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Sol. m (L)=m1S1 (ΔT)

m3.4*105= (200) (4200) (25)

m=61.7

New answer posted

6 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Sol.

BA=μ0Iθ4πR

BABB=IAθARBIBθBRA

23π2 (4)35π3 [2]

65 .

New answer posted

6 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

 

Sol. V?1=2Ucm?-U1?2m/2Vom2+m3-Vo

6 5 V o - V o V 0 5

λ o = h c m 2 V o λ f = h c M 2 V o 5

Δ λ = 8 h c m V o

New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Sol. K=QrΔxΔAT

ML2T-2 (L)L2 (θ) (T)M1L1-T-3θ-1

New answer posted

6 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

V 2 = U 2 + 2 g S V 2 = 0 + 2 g ( h - y ) V 2 = 2 g h - 2 g y V = 2 g h - 2 g y

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.