Class 12th

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New answer posted

6 months ago

0 Follower 6 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

6 months ago

0 Follower 6 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

6 months ago

0 Follower 18 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

6 months ago

0 Follower 13 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

6 months ago

0 Follower 2 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

6 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Kindly go through the answers 

(8.00)

New answer posted

6 months ago

0 Follower 8 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

6 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

T (300Kto70K)

TRmeta1R ( A l ) ( Si ) semi? conductor

New answer posted

6 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

mg In equilibrium, Fe=T sin θ

mg=T cos θ

 tan θ=Femg=q24π0x2*mg

also  tan θs?=x/2l

Hence, x 2 l = q 2 4π0x2*mg

x 3 = 2 q 2 p 4πε0mg

x=(q2l2π0mg)1/3

Therefore xl1/3

New answer posted

6 months ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

Ammonical AgNO3   

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