Physics Ncert Solutions Class 12th
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New answer posted
4 months agoContributor-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 agoContributor-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= =
Potential difference across wire of potentiometer
IR'=
Since with 50 ohm resistor, null point is not obtained it is possible when
10R'
2R'<400 or R'<200 ohm
Similarly with 10 ohm resistor, null point is obtained its is only possible when
2R'>40
R'>40
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
New answer posted
4 months agoContributor-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=
Power consumption in first current carrying wire
P= I2R
l2= 1.7 10-8 j/s = 4J/s approx.
Loss due to joule heating in first wire = 100=0.2%
Power loss in Al wire =1.6 4= 6.4J/s
Fractional loss due to joule heating in second wire = 100= 0.32%
New answer posted
4 months agoContributor-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=
Power consumption in first current carrying wire
P= I2R
l2= 1.7 10-8 j/s = 4J/s approx.
Loss due to joule heating in first wire = 100=0.2%
Power loss in Al wire =1.6 4= 6.4J/s
Fractional loss due to joule heating in second wire = 100= 0.32%
New answer posted
4 months agoContributor-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 agoContributor-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
New answer posted
4 months agoContributor-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 agoContributor-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 agoContributor-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 agoContributor-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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