Class 12th

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New answer posted

8 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

For first fringe

d y 1 D = 1 * λ 1 [ n = 1 ] λ 1 Red wavelength

d y 2 D = 1 * λ 2 [ n = 1 ] λ 2 violet wavelength

d D ( y 1 y 2 ) = λ 1 λ 2           

y 1 y 2 = D d ( λ 1 λ 2 )        

= 0 . 3 * 1 0 3 1 . 5 ( 3 . 5 2 ) * 1 0 3

= 300 * 10-9

= 300 nm

New answer posted

8 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

2nqV = 1 2 m v 2

n = m v 2 4 q V

= 1 . 6 7 * 1 0 2 7 * ( 0 . 5 * 1 0 8 ) 2 4 * 1 . 6 * 1 0 1 9 * 1 2 * 1 0 3 = 5 4 3  

             

New answer posted

8 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

  τ 2 = B 1 M 2 S m 9 0 °

= B1 M2

= ( μ 0 4 π * m 1 r 3 ) * m 2        

= 1 0 7 1 1 3 * 1

= 10-7 N – m

New answer posted

8 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

l 2 c o s 2 ? = 3 l 8

? c o s ? = 3 2

? ? = 3 0 °

New answer posted

8 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

f ' = 3 4 0 2 0 3 4 0 + 2 0 * f 0

1 8 0 0 = 3 2 0 3 6 0 * f 0

f0 = 2025 Hz

New answer posted

8 months ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

C e q = C + C 1 . C 2 C 1 + C 2              

= ε 0 A 2 d + k 1 ε 0 A 2 d 2 . k 2 ε 0 A 2 d 2 k 1 ε 0 A 2 d 2 + k 2 ε 0 A 2 d 2              

= ε 0 A d ( 1 2 + k 1 k 2 k 1 + k 2 )  

 

New answer posted

8 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

R 1 R 2 R 1 + R 2 = 3

( ρ 1 l A ) ( ρ 2 l A ) ( ρ 1 l A ) + ( ρ 2 l A ) = 3

= 3 * [ π 4 * ( 2 * 1 0 3 ) 2 ] [ 1 2 + 5 1 ] * 1 0 6 * 1 0 2 1 2 * 1 0 6 * 1 0 2 * 5 1 * 1 0 6 * 1 0 2

= 9 7

New answer posted

8 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Range of communication

= d1 + d2

= 2 R h T + 2 R h R  

= 2 * 6 4 0 0 ( 0 . 0 5 + 0 . 0 8 )  

= 57.28 km

 

 

 

New answer posted

8 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Pentavalent materials have more electrons and so electron density increase. But overall semiconductor is neutral

New answer posted

8 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

A            B            X            Y            Z

1            0            0           

1            0            1     

...more

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