Class 12th

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New answer posted

a year ago

0 Follower 48 Views

V
Vishal Baghel

Contributor-Level 10

Diode, in forward biased condition only, will allow current to flow through it.

Pot. different across resistor is

Δ V = ( 1 0 s i n ω t − 3 ) v o l t  

But in reverse biased condition of diode,

Δ V = 0 (across diode)

New answer posted

a year ago

0 Follower 18 Views

A
alok kumar singh

Contributor-Level 10

( a − 1 ) x + 0 y + z = α        ……(1)

x + ( b − 1 ) y + 0 z = β                       ……(2)

0 x + y + ( c − 1 ) z = γ                       ……(3)

| ( a − 1 ) 0 1 1 ( b − 1 ) 0 0 1 ( c − 1 ) | = 0           For no unique solution D = 0

( a − 1 ) ( b − 1 ) ( c − 1 ) + 1 = 0            

∴ a = 2 ;       b = 2 ;         c = 0            

Hence, |a + b + c| = 4

 

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

V = 10 ( 1 ? e ? t / R C )

2 = 20 ( 1 ? e ? t / R C )

1 1 0 = 1 ? e ? t / R C

e t / R C = 1 0 9

t R C = l n ( 1 0 9 ) = 0 . 1 0 5

C = t R * 0 . 1 0 5 = 1 0 ? 6 1 0 * 0 . 1 0 5 = 0 . 9 5 ? F

 

New answer posted

a year ago

0 Follower 75 Views

A
alok kumar singh

Contributor-Level 10

  A ( a ) = 2 ∫ 0 1 − a ( ( 1 − x 2 ) − a ) d x = 4 3 ( 1 − a ) 3 / 2  

  ∴ A ( 0 ) = 4 3            

and    A ( 1 2 ) = 4 3 ( 1 2 ) 3 2 ⇒ A ( 0 ) A ( 1 2 ) = 2 2

New answer posted

a year ago

0 Follower 15 Views

V
Vishal Baghel

Contributor-Level 10

As observer is at O So height of water observed by observer

⇒ H μ ω = H ( 4 / 3 ) = 3 H 4

given diagram (17.5 H) is height of observer

S o 3 H 4 = 1 7 . 5 − H

7 H 4 = 1 7 . 5

7H = 70

H = 10

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

| a → + b → + c → + d → | = ∑ | a → | 2 + 2 ∑ a → . b →  

= 4 + 2 d → . ( a → + b → + c → )  (a, b, c are mutually ^ r)

Let   d → = λ a → + μ b → + v c →

Also   λ 2 + μ 2 + v 2 = 1 = 3 c o s 2 θ       o r     c o s θ = 1 3

∴ | a → + b → + c → + d → | 2 = 4 ± 2 . 3 3        

= 4 ± 2 3 3       

New answer posted

a year ago

0 Follower 80 Views

R
Raj Pandey

Contributor-Level 9

y ( y d x + x d y ) = x 5 x d y - y d x x 2

⇒ d ( x y ) = x 5 y - 1 x d y - y d x x 2 ⇒ d ( x y ) = ( x y ) α y x β d y x

α - β = 5 α + β = - 1

2 α = 4 α = 2 β = - 3

⇒ ( x y ) - 2 d ( x y ) = y x - 3 d y x ⇒ - ( x y ) - 1 = - 1 2 y x - 2 + c

⇒ - ( x y ) - 1 = - 1 2 y x - 2 + c Passing ( 1,1 )

⇒ - 1 = - 1 2 + c ⇒ c = - 1 2 ; - ( x y ) - 1 = - 1 2 y x - 2 - 1 2

New answer posted

a year ago

0 Follower 17 Views

A
alok kumar singh

Contributor-Level 10

I n     x ∈ ( − 2 , 2 )  

 |x|- 1| is not differentiable at x = -1, 0, 1

|cospx| is not differentiable at x =  3 2 , − 1 2 , 1 2 , 3 2 -  

New answer posted

a year ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

∫ c o t - 1 ? 1 1 + x d x

= ∫ t a n - 1 ? ( 1 + x ) d x

= x t a n - 1 ? ( 1 + x ) - ∫ x 1 + ( 1 + x ) 2 d x ; = x t a n - 1 ? ( 1 + x ) - ∫ x x 2 + 2 x + 2 d x

= x t a n - 1 ? ( 1 + x ) - ∫ 1 2 ( 2 x + 2 ) - 1 x 2 + 2 x + 2 d x = x t a n - 1 ? ( 1 + x ) - 1 2 l o g ? x 2 + 2 x + 2 + ∫ d x ( x + 1 ) 2 + 1

= x t a n - 1 ? ( 1 + x ) - 1 2 l o g ? x 2 + 2 x + 2 + t a n - 1 ? ( x + 1 ) + c

New answer posted

a year ago

0 Follower 53 Views

A
alok kumar singh

Contributor-Level 10

A 2 | A | 2 − | B | 3 . 2 B

= 2 5 A 2 − | B | . B

= 2 5 A 2 + | B | A 2 = 0

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