Class 12th

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New answer posted

6 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

 hυ1=hυ0+K.E1

3.8eV=0.6eV+K.E1

K. E1 = 3.2 eV (i)

hυ2=hυ01+K.E2

1.4 = 0.6 + K.E2

KE2 = 0.8 eV (ii)

K.E1K.E2=3.20.8=4

v1v2=2:1

New answer posted

6 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

6 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

  f ( x ) + f ( x + k ) = n x R , & k > 0 . . . . . . . . . . . . ( i )

Replace x by x + k.          

f ( x + k ) + f ( x + 2 k ) = n . . . . . . . . . . . . . . . ( i i )

From (i) & (ii), f(x + 2k) = f(x).

f ( x )  is periodic with period = 2k.

I 1 = 0 4 n k f ( x ) d x = 2 x 0 2 k f ( x ) d x . . . . . . . . . . . . . . . . . . . ( i i )

I 2 = k 3 k f ( x ) d x put x = t + k

= 2 k 2 k t ( t + k ) d t = 2 0 2 k f ( t + k ) d t

= 2 n 0 2 k n d x = 4 n 2 k .

               

               

New answer posted

6 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

IR = IB (due to magnetic field)

(due to Radiation)

PA*Efficiency=B022μ0c

2004π*42*3.5100=B022*4π*107*3*108

B0=1.71*108T

New answer posted

6 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

FC=Fq

mv2R=ρR20q

mv2=ρR2q2ε012mv2=k.E.=14ρR2q0

New answer posted

6 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Ui=Q22C

Uf = 1.44 Ui

Qf = Q + 2

1.44Ui= (Q+2)22C

1.44Q22C= (Q+2)22C

1.2Q=Q+2

0.2 Q = 2

Q = 10 C

New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

  f ( r 4 ) = 2 , f ( r 2 ) = 0 & f ' ( r 2 ) = 1

g(x)   ( f ' ( t ) s e c t + t a n t s e c t f ( t ) ) d t

= [ f ( t ) s e c t ] x π / 4

=   2 * 2 f ( x ) s e c x

= 2 c o s x f ( x ) c o s x

=   2 s i n x + 1 s i n x = 3

New answer posted

6 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Sphalarite           ZnS

Calamine            ZnCO3

Galena                PbS

Siderite               FeCO3

New answer posted

6 months ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

f ( x ) = { [ x ] , x < 0 | 1 x | , x 0 g ( x ) = { c x x , x < 0 ( x 1 ) 2 1 , x 0

fog

f o g ( [ e x x ] , ( e x x < 0 ) ( x < 0 ) [ ( x 1 ) 2 1 ] ( x 1 ) 2 1 < 0 x 0 | 1 e x + x | , e x x 0 x < 0 | 1 ( x + 1 ) 2 + 1 | , ( x 1 ) 2 1 0 x 0 , x ( 2 , ) (

 Not possible as of inequalities give ? .  

ex – x < 0, x < 0                 (x 1)2 – 1 < 0

Not possible                     (x – 1 + 1)(x – 1 – 1) < 0

(x)(x – 2) < 0

x   ( 0 , 2 )

continuous  x < 0

f o g { | 1 e x + x | , x < 0 | 1 ( x 1 ) 2 + 1 | , x = 0 | 1 ( x 1 ) 2 + 1 | , x 2 Discontinuous at 0

continuous   x

   fog is discontinuous at 0

New answer posted

6 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Emulsions gets separated into two layers on standing.

For stabilization of emulsion. Emulsifying agents added into it but not electrolyte.

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