Class 12th

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New answer posted

a year ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a year ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Au + NaCN + O2 ® Na [Au (CN)2]

Z n + N a [ A u ( C N ) 2 ] → N a 2 [ Z n ( C N ) 4 ] + A u

A     i s     [ A u ( C N ) 2 ] −     a n d     B     i s     [ Z n ( C N ) 4 ] − 2

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

v = 1 ε μ = 1 2 ε 0 . 2 μ 0 = 1 2 ε 0 μ 0 = c 2 = 1 5 * 1 0 7 m / s .

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  N p N s = V p V s ⇒ N p = V p V s * N s = 2 2 0 1 2 * 2 4 = 4 4 0

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

l = l 0 c o s 2 θ = 1 0 0 * c o s 2 ( 3 0 ° ) = 7 5     L u m e n s  

Note : As the incident light is unpolarized, intensity of emerging light does not depend on the polarization axis of polarizer.

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Quality factor = ω L R = 2 * 3 . 1 4 * 1 0 * 1 0 6 * 2 * 1 0 − 4 6 . 2 8 = 2 0 0 0

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Percentage Modulation =   A m A c * 1 0 0 = 2 0 8 0 * 1 0 0 = 2 5 %

New answer posted

a year ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

Potential difference across 2k Ω  is 5V, thus current through it,

i = 5 2 * 1 0 3 = 2 5 * 1 0 − 4 A .           

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Amplitude is proportional to the slit – width, thus,

A 1 A 2 = 3

l m a x l m i n = ( A 1 + A 2 ) 2 ( A 1 − A 2 ) 2 = ( A 1 A 2 + 1 ) 2 ( A 1 A 2 − 1 ) 2 = ( 3 + 1 ) 2 ( 3 − 1 ) 2 = 4 .

New answer posted

a year ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

Given, L = 2H

l = 2 sin (t2)

u = ∫ 0 2 L i d i = L 2 [ i 2 ] 0 2 = L 2 [ 4 − 0 ]

u = 2 2 * 4 = 4 J

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