Class 12th

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New answer posted

5 months ago

0 Follower 20 Views

A
alok kumar singh

Contributor-Level 10

In given following phenol and its derivaties only para-methyl phenol does not produce any colour with phthalic anhydride in the presence of conc. H2SO4.

New answer posted

5 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

The interaction energy of London forces ; E 1 r 6 o r E r 6 . Hence x = -6.

New answer posted

5 months ago

0 Follower 23 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

5 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Current through zener diode,

l z = 2 4 1 0 1 1 0 5 m A = 1 2 m A              

Power across zener diode,

P z = V z . l z = 1 0 * 1 2 = 1 2 0 m W          

New answer posted

5 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

τ = M B s i n θ

= 0 . 2 * 3 4 * ( 0 . 1 ) 2 * 2 0 * 1 0 3 * s i n 9 0 °

= 3 * 1 0 5 N m              

 

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

ε m a x = n B A ω

= 20 * 3.0 * 10-2 * 3.14 * (0.08)2 * 50

= 60.3 * 10-2 V

New answer posted

5 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

ω m o d u l a t i o n = 2 π f m o d u l a t i o n        

f m o d u l a t i o n = ω m o d u l a t i o n 2 π

= 1 2 5 6 0 2 π         

= 2000 Hz

= 2kHz

New answer posted

5 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

P = 2 l C P radiation pressure

= 2 C ( 1 2 ε 0 E 0 2 C )

= ε 0 E 0 2 = ( 8 . 8 5 * 1 0 1 2 ) * ( 2 0 0 ) 2 = 3 5 4 * 1 0 9 N / m 2

New answer posted

5 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

As image coincides with the object, image formed by convex lens should be at the centre of curvature of the convex mirror. If the convex mirror is removed, then, also, image formed the convex lens is at the same location.

Distance of image from object = 12 + 8 + 2 * 15 = 50 cm

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

For first fringe

d y 1 D = 1 * λ 1 [ n = 1 ] λ 1 Red wavelength

d y 2 D = 1 * λ 2 [ n = 1 ] λ 2 violet wavelength

d D ( y 1 y 2 ) = λ 1 λ 2           

y 1 y 2 = D d ( λ 1 λ 2 )        

= 0 . 3 * 1 0 3 1 . 5 ( 3 . 5 2 ) * 1 0 3

= 300 * 10-9

= 300 nm

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