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New answer posted

10 months 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

10 months 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- 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

10 months 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

10 months ago

0 Follower 7 Views

New answer posted

10 months 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

10 months ago

0 Follower 2 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

10 months 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.

New answer posted

10 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, c 

Explanation – as we know average acceleration is aav= ? v ? t = v 2 - v 1 t 2 - t 1

But when acceleration is not uniform Vav is not equal to v1+v2/2

So we can write ? v = ? r ? t

? r = r 2 - r 1 = v2-v1 (t2-t1)

New answer posted

10 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation– as the given track y=x2 is a frictionless track thus total energy will be same throughout the journey.

Hence total energy at A = total energy at P . at B the particle is having only Ke but at P some KE is converted to P

Hence (KE)B = (KE)P

Total energy at A = PE= total energy at B = KE= total energy at P

= PE+KE

Potential energy at A is converted to KE and PE at P hence

(PE)P< (PE)A

Hence (height)P= (height)A

As height of p < height of A

Hence path length AB > path length BP

New answer posted

10 months ago

0 Follower 4 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 π r2, using this relation in 1 and 2

F=- 0 v d π r2V/d…… (3)

Also to balance something F=mg……. (4)

From equation 3 and 4

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

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