Class 12th

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New answer posted

a year ago

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R
Raj Pandey

Contributor-Level 9

  1. Density of solid decreases in Schottky defect and vacancy defect.
  2. Density of solid increases in interstitial defect.
  3. Density of solid remain unchanged in Frenkel defect.

New answer posted

a year ago

0 Follower 4 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

C = 10-6 f

  f = 0.5 * 103 = 500

z = x R 2 + ( x L − x C ) 2  

for minimum impedance :

xL = xC

⇒ ω 2 = 1 L C  

⇒ L = 1 π 2 = 1 1 0 * 1 0 0 0 = 1 0 0 m H

 

New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

 

Aniline show acid-base reaction with AlCl3

aniline is a Lewis base while AlCl3 acts as lewis acid.

New question posted

a year ago

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New answer posted

a year ago

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R
Raj Pandey

Contributor-Level 9

r = m v q B = 2 k m q B

2 * 5 * 1 0 3 * 1 . 6 * 1 0 − 1 9 * 2 4 * 1 . 6 6 * 1 0 − 2 7 1 . 6 * 1 0 − 1 9 * 1 2

= 9.975 * 10-2 cm

= 9.975 cm

r ≈ 1 0 cm

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

With AlCl3, alkyl halide will form cabocation which will show rearrangement.

 

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a year ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

x R = 1 2 0 Ω

x L = ω L = 1 0 0 * 1 0 0 * 1 0 − 3 1 0 Ω

x C = 1 ω c = 1 1 0 0 * 1 0 0 * 1 0 − 6 = 1 0 0 Ω

z = 1 2 0 2 + ( 1 0 0 − 1 0 ) 2 = 1 2 0 2 + 9 0 2

= 150  Ω

⇒ 3 2 = 1 6 0 7 5 * t ⇒ t = 3 2 * 7 5 1 6 0 = 1 5

New answer posted

a year ago

0 Follower 7 Views

R
Raj Pandey

Contributor-Level 9

Using whetstone

R 3 = 4 6 ⇒ R = 2 Ω

R e q u = 5 4 1 5 Ω

i = 3 6 5 4 1 5 = 1 0 A

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