Class 12th

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New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

[ a →     b →       c → ] = 0

| 1 6 3 3 2 1 α + 1 β − 1 1 | = 0

8 β − 2 4 = 0 ⇒ β = 3

| c → | = 6

( α + 1 ) 2 + ( β − 1 ) 2 + 1 = 6 ⇒ ( α + 1 ) 2 = 1

a = 1 = 1, 1 ⇒ α = − 2 , 0

+ = 1, 3

New answer posted

a year ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

f ( x ) = [ − 1 x x ∈ ( − ∞ , − 1 ) a x 2 + b x ∈ ( − 1 , 1 ) 1 x x ∈ [ 1 , ∞ )

cont at x = 1, a + b = 1

diff at x = 1, 2a = 1 ⇒ a = − 1 2 ⇒ b = 3 2

New answer posted

a year ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-Level 10

∫ 0 π / 2 ( 1 ( 1 + t a n 2 x ) ( 1 + 2 x ) + 1 ( 1 + t a n 2 x ) ( 1 + 2 − x ) ) d x

∫ 0 π / 2 d x ( 1 + t a n 2 x ) = ∫ 0 π / 2 c o s 2 x     d x = π 4

New answer posted

a year ago

0 Follower 19 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

a year ago

0 Follower 32 Views

V
Vishal Baghel

Contributor-Level 10

Eq. = no. of faraday

1 0 * 0 . 2 * 1 0 − 2 * 1 9 . 7 1 9 7 * 3 = 0 . 1 9 3 * t 9 6 5 0 0

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

2 a = 2 5 8 . 3 6 2

α = 2 5 8 . 3 6 = 2 r

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Kindly go through the answers 

(8.00)

New answer posted

a year ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

t = t 1 / 2 l o g 2 . l o g V ∞ V ∞ − V t 2 0 = 1 0 l o g 2 . l o g V ∞ V ∞ − 2 5

∴ V ∞ = 3 3 . 3 3     m l

New answer posted

a year ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

T↓ (300Kto70K)

T↓Rmeta1↓R↑               ( A l )                   ( Si ) semi? conductor

New answer posted

a year ago

0 Follower 7 Views

R
Raj Pandey

Contributor-Level 9

mg In equilibrium, Fe=T sin θ

mg=T cos θ

 tan θ=Femg=q24π∈0x2*mg

also  tan θ≈s?=x/2l

Hence, x 2 l = q 2 4π∈0x2*mg

⇒ x 3 = 2 q 2 p 4πε0mg

x=(q2l2π∈0mg)1/3

Therefore x∝l1/3

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