Class 12th

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New answer posted

11 months ago

0 Follower 111 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

11 months ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

11 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

11 months ago

0 Follower 26 Views

V
Vishal Baghel

Contributor-Level 10

Kindly comsider the following figure

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11 months ago

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V
Vishal Baghel

Contributor-Level 10

Sodium fusion extract is boiled with conc.   HNO3 to remove NaCN and Na2S.

New answer posted

11 months ago

0 Follower 15 Views

A
alok kumar singh

Contributor-Level 10

ω = 2 π f = 2 π * 5 0 = 1 0 0 π

X L = w L = 1 0 0 π * 1 0 0 * 1 0 3

X C = 1 w C = 1 1 0 0 π * 1 0 0 * 1 0 6 = 1 0 0 π

z = 1 0 . 0 0 8 6 Ω

l = v z = 2 2 0 1 0 . 0 0 8 = 2 2 A

New answer posted

11 months ago

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V
Vishal Baghel

Contributor-Level 10

Benzoic acid & naphthalene can be separated by crystallization. Benzoic acid is soluble is hot water but naphthalene is insoluble.

Assertion is incorrect & Reason is correct.

 

New answer posted

11 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

volume of wire will remains constant

π r 1 2 l 1 = π r 2 2 l 2

l 2 = 2 l 1 ( G i v e n )

r 2 = r 1 2

R 2 = ρ l 2 π r 2 2 = ρ 2 l 1 π ( r 1 2 ) 2

= R 2 R 1 R 1 * 1 0 0 = 4 R 1 R 1 R 1 * 1 0 0 = 3 R 1 R 1 * 1 0 0 = 3 0 0 %

New answer posted

11 months ago

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V
Vishal Baghel

Contributor-Level 10

Assertion : Clean water has BOD less than 5ppm. Highly polluted water has BOD greater than or equal to 17 ppm.

Reason : BOD is measure of oxygen required to oxidize only bio-degradable organic matter.

Thus Assertion (A) is correct & Reason (R) is wrong.

 

New answer posted

11 months ago

0 Follower 15 Views

A
alok kumar singh

Contributor-Level 10

Volume of 27 identical drops = volume of a bigger drop

2 7 * 4 3 π r 3 = 4 3 π R 3              

R3 = 27r3

R = 3r

Given potential of a small drop = 22v

V b i g g e r = k ( 2 7 q ) R = 2 7 k q 3 r = 9 k q r = 9 * 2 2 = 1 9 8 v o l t

               

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