Class 12th

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New answer posted

a year ago

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P
Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Potential energy of system of charges be algebraic sum

U = 1 4 π ? ? ( q q r - q q r - q q r )

9 * 109/10-15 (1.6 * 10 -19)2 (1/3)2- (2/3) (1/3)- (2/3) (1/3)

-7.68 * 10-14J

-7.68 * 10-14 * 1.6 * 10 -19= 0.48Mev

New answer posted

a year ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- B (z) points in the same direction on z-axis and hence, J (L) isa monotonically function of L so B.dl = B.dl as cos 0 = 1

(b) B.dl=μoI but when L  

So B  1/r3

(c) the magnetic field due to a circular current carrying loop of radius in the xy plane with centre at origin at any point lying at a distance of from origin.

 

B= μ0IR22 (Z2+R2)3/2

Z=Rtan θ

dz=Rsec2 θdθ

-Bzdz=μ0I2-π/2π/2cosθdθ = μ0I

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- as we know magnetic moment = nIA

For equilateral triangle M= nIA= 4I ( 34a2 )

M= Ia23

For square, n=3 so total length of wire is 12a

M= nIA= 3I (a2) = 3Ia2

For regular hexagon of side a, n=2 so total length = 12a

M= nIA=2 ( 634a2 )= 3 3 a2I

New answer posted

a year ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- Since, B is along the x-axis, for a circular orbit the momenta of the two particles are in the y-z plane. Let P1 and P2 be the momentum of the electron and positron, respectively. Both traverse a circle of radius R of opposite sense. Let P1 make an angle? with the y-axis P2 must make the same angle.

The centres of the respective circles must be perpendicular to the momenta and at a distance R. Let the centre of the electron be at Ce and of the positron at Cp . The coordinates of Ce is

Ce= (0, -Rsin θ, Rcosθ )

Cp= (0, -Rsin θ, -32R-Rcosθ )

The circles of the two

...more

New answer posted

a year ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- the thicker wire has a resistance R, then the other wire has a resistance @R as the wires are of the same material but different area

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – for equilibrium balance net torque should be zero

Mgl= Wcoill

500gl = Wcoill

Wcoil= 500 * 9.8N

Taking moment of force about mid point then magnetic field

Mgl+mgl=Wcoill+IBlsin90

Mgl=BILl

M= 0.2 * 4.9 * 1 * 10 - 2 9.8 = 10 -3kg= 1g

New question posted

a year ago

0 Follower 7 Views

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- |A+B|=|A-B|

A | 2 + B | 2 + 2 A | B | c o s θ = A | 2 + B | 2 - 2 A | B | c o s θ

A | 2 + B | 2 + 2 A | B | c o s θ  = A | 2 + B | 2 - 2 A | B | c o s θ

4|A|B|cos θ =0

|A|2+|B|2cos θ =0

A=0 or B=0 so θ = 90 . so A perpendicular B

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- (i) Speed will constant throughout

(ii) Velocity will be tangential in the direction of motion

(iii) Centripetal acceleration will be a= v2/r, will always be towards centre of the circular path.

(iv) Angular momentum is constant in magnitude and direction out of the plane perpendicularly as well.

New answer posted

a year ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

To lift the disc ther must be some electrostatic force is reqired.

F=qE……. (1), E=V/d………. (2)

But the charge required during the process is - ? 0 v d ? r2, using this relation in 1 and 2

F=- ? 0 v d ? r2V/d…… (3)

Also to balance something F=mg……. (4)

From equation 3 and 4

Mg= - ? 0 v d ? r2V/d, so V = ? m g d ? 0 ? r 2 this the requird solution.

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