Class 12th

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New answer posted

5 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

Key is open.

i r m s = 1 5 6 0 = 1 4 A

2 0 = 1 4 * 1 0 0 * L L = 0 . 8 H

1 0 = 1 4 . 1 1 0 0 C C = 2 . 5 * 1 0 4 F = 2 5 0 μ F

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

K = qV

p = 2 m k = 2 m q V

λ = h p = h 2 m q V

λ p λ α = m α m p . q α q p = 4 * 2 = 2 * 1 . 4 = 2 . 8

New answer posted

5 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

l = 1 2 c ε 0 E 0 2 8 4 π ( 1 0 ) 2 * 1 2 = 1 4 * c * 1 c 2 μ 0 * E 0 2 E 0 = 2 1 0 μ 0 c π x = 2

New answer posted

5 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

After contact, charges an each sphere will be

q1+q22=1ncF=kq1q2r2=36*109N

New answer posted

5 months ago

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P
Payal Gupta

Contributor-Level 10

Energy corresponding to a particle is mc2

hcλ=mc2hcλ= (x3h)c2x=3λc=310*1010*3*108 = 1101=10

New answer posted

5 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

i2 = 2i1

i1 + i2 = 6

i1 = 2A.

New answer posted

5 months ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

Tsinθ=kq2r2

Tcosθ=mg

tanθ=kq2mgr2

Using the above equation find the value of 'q'

New question posted

5 months ago

0 Follower 3 Views

New answer posted

5 months ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

Since force will act in horizontal direction so vertical component of speed will be same

v1 cos α = v2 cos β

K 1 K 2 = ( v 1 v 2 ) 2 = c o s 2 β c o s 2 α

New answer posted

5 months ago

0 Follower 1 View

P
Payal Gupta

Contributor-Level 10

In ferromagnetic material, below Curie's temperature, a domain is defined as macroscopic region with zero magentisation.

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