Class 12th

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New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

[h] = ML2T-1

[E] = ML2T-2

[V] = ML2T-2C-1

[P] = MLT-1

New answer posted

a year ago

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P
Payal Gupta

Contributor-Level 10

I > III > IV > II

[FeF6]3−→Fe3+−3d5− all are unpaired (n = 5)

[Co (NH3)6]3+→Co3+−3d6− all are paired (n = 0)

[NiCl4]2−→Ni2+−3d8→ (n=2)

[Cu (NH3)4]2+→Cu++→3d9→ (n=1)

New answer posted

a year ago

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Payal Gupta

Contributor-Level 10

Indium and other semi conducting materials behaves different characteristic of impurities and pure particle at high temperature zone. Hence in metallurgical process it is refined by the zone refining.

New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

Buna-S required nascent oxygen

Butadiene + styrene → [ 0 ] (Buna-S)

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a year ago

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V
Vishal Baghel

Contributor-Level 10

Magnetic field on the axis of a circular coil at distance x from centre, B = μ 0 N i r 2 2 ( r 2 + x 2 ) 3 2

( r 2 + ( 0 . 2 ) 2 ) 3 2 ( r 2 + ( 0 . 0 5 ) 2 ) 3 2 = 8 r 2 + ( 0 . 2 ) 2 r 2 + ( 0 . 0 5 ) 2 = 4 ⇒ r 2 = 0 . 0 1 ⇒ r = 0 . 1 m .

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

Number of half lives of Y = 3

Number of half lives of X = 6 [As half life of X is half of that of Y].

N 1 2 6 = N 2 2 3 ⇒ N 1 N 2 = 8

New answer posted

a year ago

0 Follower 32 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a year ago

0 Follower 1 View

P
Payal Gupta

Contributor-Level 10

Conversion form

New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

D1 is in forward bias and D2 is in reverse bias.

Current, I = 5 − 0 . 7 1 0 = 0 . 4 3 A

New answer posted

a year ago

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A
Aadit Singh Uppal

Contributor-Level 10

Yes. Determinants can be calculated for any square matrix of n-order, and it is done by expansion of rows and columns. Even in higher dimensions, their job is to define hyper volumes and transformations.

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