Physics Ncert Solutions Class 12th

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New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

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

Explanation- according to ohm's law V= IR

I= 6/6 = 1A

I= AneVd  or Vd= i/neA

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

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

Explanation – let R' be the resistance of potentiometer wire.

Effective resistance of potentiomter and variable resistor r=50ohm is 500+R'

Effective voltage across potentiometer = 10V

The current through main circuit I= V 50 + R = 10 50 + R

Potential difference across wire of potentiometer

 IR'= 10 R ' 50 + R

Since with 50 ohm resistor, null point is not obtained it is possible when

10 * R ' 50 + R < 8

10R' < 400 + 8 R '

2R'<400 or R'<200 ohm

Similarly with 10 ohm resistor, null point is obtained its is only possible when

10 * R ' 50 + R < 8

2R'>40

R'>40

10 * 3 4 R ' 10 + R ' < 8

7.5R'<80+8R'

R'>160

160

Any R' between 160 ohm and 200 ohm will achieve.

Since the null point on the last 4th segment of

...more

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

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

Explanation – power consumption in a day i.e in 5  = 10 units

Power consumption per hour = 2 units

Power consumption = 2 units =2KW= 2000J/s

Also power =V * I

2000W= 220V * l or l= 9A approx.

R= ρ l A

Power consumption in first current carrying wire

P= I2R

ρ l A l2= 1.7 * 10-8 * 10 π * 10 - 6 * 81 j/s = 4J/s approx.

Loss due to joule heating in first wire = 4 2000 * 100=0.2%

Power loss in Al wire =1.6 * 4= 6.4J/s

Fractional loss due to joule heating in second wire = 6.4 2000 * 100= 0.32%

New answer posted

4 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

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

Explanation – power consumption in a day i.e in 5  = 10 units

Power consumption per hour = 2 units

Power consumption = 2 units =2KW= 2000J/s

Also power =V * I

2000W= 220V * l or l= 9A approx.

R= ρ l A

Power consumption in first current carrying wire

P= I2R

ρ l A l2= 1.7 * 10-8 * 10 π * 10 - 6 * 81 j/s = 4J/s approx.

Loss due to joule heating in first wire = 4 2000 * 100=0.2%

Power loss in Al wire =1.6 * 4= 6.4J/s

Fractional loss due to joule heating in second wire = 6.4 2000 * 100= 0.32%

New answer posted

4 months ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- the positive charge Q is uniformly distributed at the outer surface of the enclosed sphere thus electric field inside the sphere is zero. So the effect of electric field on charge q due to positive charge Q is zero.

Now the only attractive and repulsive force between Q and q

Case 1 q>0  this creates repulsive force

Case 2 q<0 this creates attractive force

 If q is shifted from the centre then the positive charges nearer to this charge will attract it towards itself and charge q will never return to its centre.

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (a), (b), (c), (d)

Explanation- The positive charge Q is uniformly distributed along the circular ring then  electric field at the centre of ring will be zero, hence no force is experienced by the charge if it is placed at the centre of the ring.
Now the charge is displaced away from the centre in the plane of the ring. There will be net electric field opposite to displacement will push back the charge towards the centre of the ring if the charge is positive. If charge is negative, it will experience net force in the direction of displacement and the charge w

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New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, c)

Explanation- Gauss's law states that the total electric flux of an enclosed surface is given by q/? 0, where q is the charge enclosed by the surface.

So total charge inside the surface is = Q-2Q = -Q

Therefore total flux through the surface of the sphere = -Q/? 0

Now, charge 5Q is lies outside the surface, thus it makes no contribution to electric flux through the given surface. So both option a and c are true.

New answer posted

4 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (c), (d)

Explanation- electric field is not necessarily zero may it become zero by their algeabric sum. For dipole the electric field is always ∝ 1/r3 and it is also conservative.

New answer posted

4 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (b), (d)

Explanation- if we place a charge then we must experience some forces but if there would be no charge so field is continuous

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (c), (d)

Explanation- It is only possible when charges must be outside the surface or field line entering or leaving the surface are equal.

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