Class 12th

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New answer posted

8 months ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

d x ( x 1 x + 2 ) 3 / 4 ( x + 2 ) 3 / 4 ( x + 2 ) 5 / 4 = d x ( x 1 x + 2 ) 3 / 4 ( x + 2 ) 2 p u t x 1 x + 2 = t 3 ( x + 2 ) 2 d x = d t

1 3 d t t 3 / 4 = 1 3 . t 1 / 4 1 4 + C = 4 3 ( x 1 x + 2 ) 1 / 4 + C

New answer posted

8 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Comparing E = 20 cos (2 * 1010 t – 200x) V/m to

E = E 0 c o s ( ω t k x ) v / m              

ω = 2 * 1 0 1 0 , K = 2 0 0               

Speed = 2 * 1 0 1 0 2 0 0 = 1 0 8 m / s  

R.I. = C s p e e d = 3 * 1 0 8 1 0 8 = 3  

N o w R . I . = ε r μ r

3 = ε r * 1

ε r = 9              

            

 

New answer posted

8 months ago

0 Follower 3 Views

J
Jaya Sinha

Beginner-Level 5

Students should prepare complete syllabus when they have time to prepare. However, you can use the list of highweightage chapters in last minute revision for scoring well.

  • The p-Block Elements: This chapter holds a high weightage of 8–10 marks. (in the latest syllabus this is deleted)
  • Aldehydes, Ketones, and Carboxylic Acids: This chapter contributes around 8–10 marks.

  • Biomolecules: This chapter accounts for around 8 marks.

  • Chemical Kinetics: This chapter holds a high weightage of 5-6 marks. 

  • The d- and f-Block Elements: This chapter contributes around 5-6 marks.

  • Amines: This chapter contributes around 5-6 marks.

New answer posted

8 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Capacitor makes potential difference constant.

New answer posted

8 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Using faraday's law magnetic field should be outward and decreasing with time

New answer posted

8 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

L H S = l i m x 0 f ( x ) = l i m h 1 h l n ( 1 h a 1 + h b ) = l i m h 0 l n ( 1 h a ) a ( h a ) + l i m h 0 l n ( 1 + h b ) b ( h b ) = 1 a + 1 b . . . . . . . . . . . . . ( i )

R H S = l i m x 0 + f ( x ) = l i m x 0 c o s 2 x 1 x 2 + 1 1 ( x 2 + 1 + 1 ) = l i m x 0 2 s i n x x x 2 * 2 = 4 . . . . . . . . . . . . . . . ( i i )

a n d l i m x 0 f ( x ) = k . . . . . . . . . . . . . ( i i i )

f ( 0 ) = f ( 0 + ) = f ( 0 )

1 a + 1 b = 4 = k

From (i), (ii) and (iii) we get 1 a + 1 b + 4 k = k + 4 k = 4 1 = 5

New answer posted

8 months ago

Directions for questions: Five people joined different engineering colleges. Their first names were Komal (Ms.), Mansi (Ms.), Kabeer, Ram and Rupali (Ms.). The surnames were Pandey, Kumar, Verma, Sharma and Gupta. Except for one college which was rated 3 star, all other colleges were rated either 4 star or 5 star.

The “Techno Institute” had a higher rating than college where Rupali studied. The three-star college was not “Decan College”. Ram's last name was Kumar but he didn't study at “Barla College.” Komal, whose last name wasn't Verma, joined “Techno Institute.” Ms. Sharma and Kabeer both studied at four-star college

...more
0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

Name

Surname

College

Star

Ram

Kumar

Chemical

4

Komal

Sharma

Techno

4

Rupali

Pandey

Anipal

3

Kabeer

Gupta

Deccan

4

Mansi

Verma

Barla

5

New answer posted

8 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

Energy required to melt

Q = M S Δ T + M L  

1 0 1 * 2 * 1 0 3 * 1 0 + 1 0 1 * 3 . 3 3 * 1 0 5      

->3.53 * 104 J

Heat produce in wire

H = l2RT

  Q = 3 . 5 3 * 1 0 4 = ( 1 2 ) 2 * ( 4 * 1 0 3 ) * t

t = 3 . 5 3 * 1 0 4 * 4 4 * 1 0 3 = 3 5 . 3 s e c             

             

New answer posted

8 months ago

0 Follower 1 View

N
Nishtha Datta

Beginner-Level 5

Shiksha's NCERT notes are extremely useful for efficient preparation. We offer structured chapter-wise notes for the latest CBSE  syllabus and provide concise summaries and key formulas. These notes are designed for quick and last-minute revision. These short revision notes offer step-by-step explanations for conceptual clarity and include important questions from previous years' papers and NCERT Textbooks. Our notes are equally beneficial for competitive exams like JEE Mains, NEET and other exams as well.

New answer posted

8 months ago

0 Follower 1 View

P
Payal Gupta

Contributor-Level 10

Room

Person

101

Krish, Ishan

201

Moksh/Nitin

301

Pari, Niti

401

Nitin/Moksh

501

Arjun

601

Jyoti

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