Class 12th

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New answer posted

a year ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Nucleophilic addition on carbonyl compounds decreases by increasing steric hindrance.

New answer posted

a year ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Due to Intramolecular hydrogen bonding, O-nitrophenol fails to spread its surface, so that its b.p is less than para-nitrophenol.

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

k = [ a →     b →     c → ] + 2 [ a →     b →     c → ] + [ a →     b →     c → ] − [ a →     b →     c → ] [ a →     b →     c → ]            

k = 3

New answer posted

a year ago

0 Follower 37 Views

A
alok kumar singh

Contributor-Level 10

a r e a = 2 * ( 1 2 * 1 * 1 ) = 1 = k

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  l i m x → 1 − f ( x ) = l i m x → 1 + f ( x ) ⇒ a + b = 2   

  ∴ l i m x → 3 − f ( x ) = l i m x → 3 + f ( x ) ⇒ 3 a + b = 6 t a n 3 π 1 2          

∴ a = 2, b = 0

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

  | a ¯ * ( a ¯ * c ¯ ) | = | 3 b ¯ | = 3 | b ¯ |

⇒ 3 = 3 . 2 s i n θ ⇒ s i n θ = 1 2 ⇒ c o s 2 θ = 3 4  

            

New answer posted

a year ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

l i m x → 0 [ s i n 2 ( π 2 − 3 x ) ] s e c 2 ( π 2 − 5 x )

e l i m x → 0 [ s i n 2 ( π 2 − 3 x ) − 1 ] s e c 2 ( π 2 − 5 x )

= e l i m x → 0 − s i n 2 ( − 3 π x 2 ( 2 − 3 x ) ) s i n 2 ( − 5 π x 2 ( 2 − 5 x ) ) = e − 9 2 5

New answer posted

a year ago

0 Follower 10 Views

R
Raj Pandey

Contributor-Level 9

v = f u u - f = ( - 10 ) ( - 15 ) - 15 + 10 c m = - 30 c m

v image   = d v d t = - v u 2 d u d t = - - 30 - 15 2 ( 10 ) c m s - 1 = - 40 c m s - 1

V object   = d u d t = + 10 c m s - 1

Relative velocity = v object   - v image  

= 10 - ( - 40 ) = 50 c m s - 1

New answer posted

a year ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

Key point: Fehling solution do not oxidise aromatic aldehydes, except when EWG present 
at O/P position So (I), (III), (IV) not show this test

New answer posted

a year ago

0 Follower 1 View

V
Vikash Kumar Vishwakarma

Contributor-Level 10

You can use the position vector to find the real displacement of a point from a reference point.

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