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All questions of CBSE Sample Question Papers for Class 12 Exam
An electron is accelerated through a potential difference of 7 volts. The energy gained by an electron will be- a)7 erg
- b)7 watt
- c)7 eV
- d)7 volt
Correct answer is option 'C'. Can you explain this answer?
An electron is accelerated through a potential difference of 7 volts. The energy gained by an electron will be
a)
7 erg
b)
7 watt
c)
7 eV
d)
7 volt
|
Mira Joshi answered |
Potential difference = 7 V
We know that
K.E. = 1/2mv2
here, 1/2mv2 = eV
K.E. = e x 7V
K.E. = 7eV
We know that
K.E. = 1/2mv2
here, 1/2mv2 = eV
K.E. = e x 7V
K.E. = 7eV
An electric current is passed through a circuit containing two wires of same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 3 : 2 and 2 : 3, then the ratio of the current passing through the wire will be- a)2 : 3
- b)3 : 2
- c)8 : 27
- d)27 : 8
Correct answer is option 'C'. Can you explain this answer?
An electric current is passed through a circuit containing two wires of same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 3 : 2 and 2 : 3, then the ratio of the current passing through the wire will be
a)
2 : 3
b)
3 : 2
c)
8 : 27
d)
27 : 8
|
Shalini Patel answered |
I1 : I2 = 3 : 2
r1 : r2 = 2 : 3
I1 : I2 = ?
=
∴
By increasing the temperature, the specific resistance of a conductor and a semiconductor:- a)increases for both.
- b)decreases for both.
- c)increases for a conductor and decreases for a semiconductor.
- d)decreases for a conductor and increases for a semiconductor.
Correct answer is option 'C'. Can you explain this answer?
By increasing the temperature, the specific resistance of a conductor and a semiconductor:
a)
increases for both.
b)
decreases for both.
c)
increases for a conductor and decreases for a semiconductor.
d)
decreases for a conductor and increases for a semiconductor.
|
Nandini Iyer answered |
Specific resistance of a conductor increases and for a semiconductor decreases with increase in temperature because for a conductor, a temperature:
coefficient of resistivity, a = + ve
and for a semiconductor, a = – ve
coefficient of resistivity, a = + ve
and for a semiconductor, a = – ve
Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon the:- a)rate at which current change in the two coils
- b)relative position and orientation of the coils
- c)rate at which voltage induced across two coils
- d)currents in the two coils
Correct answer is option 'B'. Can you explain this answer?
Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon the:
a)
rate at which current change in the two coils
b)
relative position and orientation of the coils
c)
rate at which voltage induced across two coils
d)
currents in the two coils
|
|
Suresh Iyer answered |
Mutual inductance of a pair of two coils depends on the relative position and orientation of two coils.
Work done in moving an electron from one point to another on an equipotential surface of 10 Volt potential- a)10 eV
- b)10 V
- c)1eV
- d)zero
Correct answer is option 'D'. Can you explain this answer?
Work done in moving an electron from one point to another on an equipotential surface of 10 Volt potential
a)
10 eV
b)
10 V
c)
1eV
d)
zero
|
Mihir Yadav answered |
Explanation:
When an electron moves from one point to another on an equipotential surface, the potential difference between the two points is zero. This means that the electric field between the two points is zero, and therefore no work is done in moving the electron.
Equipotential Surfaces:
Equipotential surfaces are imaginary surfaces in an electric field where the potential at every point on the surface is the same. These surfaces are always perpendicular to the electric field lines. The potential difference between any two points on an equipotential surface is zero.
Work Done:
Work done is defined as the product of force and displacement. In the case of moving an electron, the force acting on the electron is given by the electric field. If the electron is moving along an equipotential surface, the electric field between the two points is zero. Therefore, the force on the electron is zero, and hence no work is done in moving the electron.
Mathematical Explanation:
The work done in moving a charged particle in an electric field is given by the equation W = q * V, where W is the work done, q is the charge of the particle, and V is the potential difference.
In this case, the electron has a charge of -1.6 x 10^-19 Coulombs, and the potential difference between the two points on the equipotential surface is 10 Volts. Therefore, the work done can be calculated as:
W = (-1.6 x 10^-19 C) * (10 V)
= -1.6 x 10^-18 J
Since the work done is negative, it means that work is done by an external force to move the electron against the electric field. However, in the case of an equipotential surface, the potential difference is zero, and therefore the work done is zero.
Conclusion:
In conclusion, the work done in moving an electron from one point to another on an equipotential surface is zero. This is because the potential difference between the two points is zero, and therefore the electric field between the points is zero. As a result, no force acts on the electron, and hence no work is done in moving it.
When an electron moves from one point to another on an equipotential surface, the potential difference between the two points is zero. This means that the electric field between the two points is zero, and therefore no work is done in moving the electron.
Equipotential Surfaces:
Equipotential surfaces are imaginary surfaces in an electric field where the potential at every point on the surface is the same. These surfaces are always perpendicular to the electric field lines. The potential difference between any two points on an equipotential surface is zero.
Work Done:
Work done is defined as the product of force and displacement. In the case of moving an electron, the force acting on the electron is given by the electric field. If the electron is moving along an equipotential surface, the electric field between the two points is zero. Therefore, the force on the electron is zero, and hence no work is done in moving the electron.
Mathematical Explanation:
The work done in moving a charged particle in an electric field is given by the equation W = q * V, where W is the work done, q is the charge of the particle, and V is the potential difference.
In this case, the electron has a charge of -1.6 x 10^-19 Coulombs, and the potential difference between the two points on the equipotential surface is 10 Volts. Therefore, the work done can be calculated as:
W = (-1.6 x 10^-19 C) * (10 V)
= -1.6 x 10^-18 J
Since the work done is negative, it means that work is done by an external force to move the electron against the electric field. However, in the case of an equipotential surface, the potential difference is zero, and therefore the work done is zero.
Conclusion:
In conclusion, the work done in moving an electron from one point to another on an equipotential surface is zero. This is because the potential difference between the two points is zero, and therefore the electric field between the points is zero. As a result, no force acts on the electron, and hence no work is done in moving it.
Two copper wires of same length, having different cross-sectional area in the ratio 1:3, are connected in series. The ratio of their drift velocities vd will be - a)3:1
- b)9:1
- c)1:3
- d)1:9
Correct answer is option 'A'. Can you explain this answer?
Two copper wires of same length, having different cross-sectional area in the ratio 1:3, are connected in series. The ratio of their drift velocities vd will be
a)
3:1
b)
9:1
c)
1:3
d)
1:9
|
|
Mihir Yadav answered |
Explanation:
When two copper wires of the same length but with different cross-sectional areas are connected in series, the current passing through them remains the same.
Key Concept:
Drift velocity (vd) is defined as the average velocity of charged particles, such as electrons, in a conductor when an electric field is applied.
Formula:
The drift velocity (vd) is given by the formula:
vd = I / (nAe)
where,
vd = drift velocity
I = current
n = number density of charge carriers
A = cross-sectional area
e = charge of an electron
Analysis:
In this scenario, since the wires are connected in series, the current passing through both wires is the same. Therefore, the ratio of the drift velocities (vd) of the two wires will be equal to the ratio of their respective cross-sectional areas (A).
Given that the cross-sectional areas of the two wires are in the ratio 1:3, we can assume the cross-sectional areas to be A and 3A.
Therefore, the ratio of their drift velocities (vd) will be:
vd1 / vd2 = A1 / A2
= A / 3A
= 1 / 3
Simplifying the ratio, we get:
vd1 : vd2 = 1 : 3
Therefore, the correct answer is option 'A': 3 : 1.
When two copper wires of the same length but with different cross-sectional areas are connected in series, the current passing through them remains the same.
Key Concept:
Drift velocity (vd) is defined as the average velocity of charged particles, such as electrons, in a conductor when an electric field is applied.
Formula:
The drift velocity (vd) is given by the formula:
vd = I / (nAe)
where,
vd = drift velocity
I = current
n = number density of charge carriers
A = cross-sectional area
e = charge of an electron
Analysis:
In this scenario, since the wires are connected in series, the current passing through both wires is the same. Therefore, the ratio of the drift velocities (vd) of the two wires will be equal to the ratio of their respective cross-sectional areas (A).
Given that the cross-sectional areas of the two wires are in the ratio 1:3, we can assume the cross-sectional areas to be A and 3A.
Therefore, the ratio of their drift velocities (vd) will be:
vd1 / vd2 = A1 / A2
= A / 3A
= 1 / 3
Simplifying the ratio, we get:
vd1 : vd2 = 1 : 3
Therefore, the correct answer is option 'A': 3 : 1.
The instantaneous values of emf and the current in a series ac circuit are: E = Eo sin ωt and I = Io sin (ωt + π/3) respectively, then it is- a)Necessarily a RL circuit
- b)Necessarily a RC circuit
- c)Necessarily a LCR circuit
- d)Can be RC or LCR circuit
Correct answer is option 'D'. Can you explain this answer?
The instantaneous values of emf and the current in a series ac circuit are: E = Eo sin ωt and I = Io sin (ωt + π/3) respectively, then it is
a)
Necessarily a RL circuit
b)
Necessarily a RC circuit
c)
Necessarily a LCR circuit
d)
Can be RC or LCR circuit
|
Arnab Chavan answered |
(ωt) and I = Io sin(ωt + φ), where Eo, Io, and φ are constants.
The emf E is given by Eo sin(ωt), where Eo represents the maximum value of the emf and ω is the angular frequency. The current I is given by Io sin(ωt + φ), where Io represents the maximum value of the current and φ is the phase angle.
In a series ac circuit, the emf and current are in phase with each other, meaning that the phase angle φ is 0. This means that the current reaches its maximum value Io at the same time as the emf reaches its maximum value Eo. Both the emf and current vary sinusoidally with time, following a sine function.
The angular frequency ω determines the rate at which the emf and current oscillate. It is related to the frequency f by the equation ω = 2πf, where f is the frequency of the alternating current.
Overall, the instantaneous values of the emf and current in a series ac circuit can be represented by the equations E = Eo sin(ωt) and I = Io sin(ωt), respectively.
The emf E is given by Eo sin(ωt), where Eo represents the maximum value of the emf and ω is the angular frequency. The current I is given by Io sin(ωt + φ), where Io represents the maximum value of the current and φ is the phase angle.
In a series ac circuit, the emf and current are in phase with each other, meaning that the phase angle φ is 0. This means that the current reaches its maximum value Io at the same time as the emf reaches its maximum value Eo. Both the emf and current vary sinusoidally with time, following a sine function.
The angular frequency ω determines the rate at which the emf and current oscillate. It is related to the frequency f by the equation ω = 2πf, where f is the frequency of the alternating current.
Overall, the instantaneous values of the emf and current in a series ac circuit can be represented by the equations E = Eo sin(ωt) and I = Io sin(ωt), respectively.
An electric dipole of moment p is placed parallel to the uniform electric field. The amount of work done in rotating the dipole by 90° is- a)2pE
- b)pE
- c)pE/2
- d)Zero
Correct answer is option 'B'. Can you explain this answer?
An electric dipole of moment p is placed parallel to the uniform electric field. The amount of work done in rotating the dipole by 90° is
a)
2pE
b)
pE
c)
pE/2
d)
Zero
|
Priyanka Sharma answered |
W = pE (cos θ1 – cos θ2)
θ1 = 0°
θ2 = 90°
W = pE (cos 0° – cos 90°)
= pE (1 – 0)
= PE
θ1 = 0°
θ2 = 90°
W = pE (cos 0° – cos 90°)
= pE (1 – 0)
= PE
Electric potential at a point (x, y, z) in vacuum is V = 3x2 volt. Find the electric field intensity at the point (1m, 2m, 3m) . - a)1 V/m
- b)3 V/m
- c)6 V/m
- d)2 V/m
Correct answer is option 'C'. Can you explain this answer?
Electric potential at a point (x, y, z) in vacuum is V = 3x2 volt. Find the electric field intensity at the point (1m, 2m, 3m) .
a)
1 V/m
b)
3 V/m
c)
6 V/m
d)
2 V/m
|
|
Mihir Yadav answered |
Answer:
To find the electric field intensity at a given point, you need to differentiate the electric potential with respect to each coordinate and then take the negative gradient of the potential.
Given that the electric potential at a point (x, y, z) is V = 3x^2 volts, we can find the electric field intensity at the point (1m, 2m, 3m) by taking the negative gradient of the potential.
Calculating the electric field intensity:
To find the electric field intensity, we need to calculate the partial derivatives of the potential with respect to each coordinate.
- Partial derivative with respect to x: ∂V/∂x = 6x
- Partial derivative with respect to y: ∂V/∂y = 0 (since V does not depend on y)
- Partial derivative with respect to z: ∂V/∂z = 0 (since V does not depend on z)
Now, we can calculate the electric field intensity at the given point (1m, 2m, 3m) by substituting the coordinates into the partial derivatives:
- Electric field intensity in the x-direction: Ex = -∂V/∂x = -6(1) = -6 V/m
- Electric field intensity in the y-direction: Ey = -∂V/∂y = 0
- Electric field intensity in the z-direction: Ez = -∂V/∂z = 0
The electric field intensity at the point (1m, 2m, 3m) is given by the vector sum of the components:
E = sqrt(Ex^2 + Ey^2 + Ez^2)
= sqrt((-6 V/m)^2 + (0)^2 + (0)^2)
= sqrt(36 V^2/m^2)
= 6 V/m
Therefore, the correct answer is option 'C' - 6 V/m.
To find the electric field intensity at a given point, you need to differentiate the electric potential with respect to each coordinate and then take the negative gradient of the potential.
Given that the electric potential at a point (x, y, z) is V = 3x^2 volts, we can find the electric field intensity at the point (1m, 2m, 3m) by taking the negative gradient of the potential.
Calculating the electric field intensity:
To find the electric field intensity, we need to calculate the partial derivatives of the potential with respect to each coordinate.
- Partial derivative with respect to x: ∂V/∂x = 6x
- Partial derivative with respect to y: ∂V/∂y = 0 (since V does not depend on y)
- Partial derivative with respect to z: ∂V/∂z = 0 (since V does not depend on z)
Now, we can calculate the electric field intensity at the given point (1m, 2m, 3m) by substituting the coordinates into the partial derivatives:
- Electric field intensity in the x-direction: Ex = -∂V/∂x = -6(1) = -6 V/m
- Electric field intensity in the y-direction: Ey = -∂V/∂y = 0
- Electric field intensity in the z-direction: Ez = -∂V/∂z = 0
The electric field intensity at the point (1m, 2m, 3m) is given by the vector sum of the components:
E = sqrt(Ex^2 + Ey^2 + Ez^2)
= sqrt((-6 V/m)^2 + (0)^2 + (0)^2)
= sqrt(36 V^2/m^2)
= 6 V/m
Therefore, the correct answer is option 'C' - 6 V/m.
Assertion (A): Potentiometer measures the potential difference more accurately than a voltmeter.
Reason (R): There is no current drawn from external circuit in potentiometer.- a)Both Assertion (A) and Reason (R) are true, and Reason(R) is the correct explanation of (A).
- b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- c)Assertion (A) is true, but Reason (R) is false.
- d)Assertion (A) is false, but Reason (R) is true.
Correct answer is option 'A'. Can you explain this answer?
Assertion (A): Potentiometer measures the potential difference more accurately than a voltmeter.
Reason (R): There is no current drawn from external circuit in potentiometer.
Reason (R): There is no current drawn from external circuit in potentiometer.
a)
Both Assertion (A) and Reason (R) are true, and Reason(R) is the correct explanation of (A).
b)
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
c)
Assertion (A) is true, but Reason (R) is false.
d)
Assertion (A) is false, but Reason (R) is true.
|
|
Mihir Yadav answered |
Assertion (A): Potentiometer measures the potential difference more accurately than a voltmeter.
Reason (R): There is no current drawn from the external circuit in the potentiometer.
The correct answer is option 'A' - Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Explanation:
Potentiometer:
A potentiometer is an instrument used to measure the potential difference between two points in a circuit. It consists of a long uniform resistance wire, known as the potentiometer wire, and a sliding contact called the jockey. The potentiometer wire is connected in series with a battery, while the jockey is used to make contact with the wire.
Voltmeter:
A voltmeter is a measuring instrument used to measure the potential difference (voltage) between two points in an electrical circuit. It is connected in parallel with the component or section of the circuit whose voltage is to be measured.
Accuracy of measurement:
In terms of accuracy, the potentiometer is more accurate than a voltmeter. This is because a potentiometer can measure the potential difference without any current flowing through it, whereas a voltmeter draws a small amount of current from the circuit being measured.
Reasoning:
The reason for the potentiometer's higher accuracy is that it does not draw any current from the external circuit. When a voltmeter is connected in parallel to the circuit, it creates a parallel branch that draws a small amount of current. This current can alter the potential difference being measured, leading to inaccuracies in the voltmeter reading.
On the other hand, a potentiometer does not draw any current from the circuit being measured. The potential difference is measured by adjusting the position of the sliding contact (jockey) along the potentiometer wire until the galvanometer shows zero deflection. At this point, the potential difference across the two points being measured is equal to the potential difference between the jockey and the battery.
Since no current is drawn from the circuit, the potentiometer provides a more accurate measurement of the potential difference compared to a voltmeter. Therefore, Assertion (A) is true, and Reason (R) correctly explains why the potentiometer is more accurate.
Reason (R): There is no current drawn from the external circuit in the potentiometer.
The correct answer is option 'A' - Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Explanation:
Potentiometer:
A potentiometer is an instrument used to measure the potential difference between two points in a circuit. It consists of a long uniform resistance wire, known as the potentiometer wire, and a sliding contact called the jockey. The potentiometer wire is connected in series with a battery, while the jockey is used to make contact with the wire.
Voltmeter:
A voltmeter is a measuring instrument used to measure the potential difference (voltage) between two points in an electrical circuit. It is connected in parallel with the component or section of the circuit whose voltage is to be measured.
Accuracy of measurement:
In terms of accuracy, the potentiometer is more accurate than a voltmeter. This is because a potentiometer can measure the potential difference without any current flowing through it, whereas a voltmeter draws a small amount of current from the circuit being measured.
Reasoning:
The reason for the potentiometer's higher accuracy is that it does not draw any current from the external circuit. When a voltmeter is connected in parallel to the circuit, it creates a parallel branch that draws a small amount of current. This current can alter the potential difference being measured, leading to inaccuracies in the voltmeter reading.
On the other hand, a potentiometer does not draw any current from the circuit being measured. The potential difference is measured by adjusting the position of the sliding contact (jockey) along the potentiometer wire until the galvanometer shows zero deflection. At this point, the potential difference across the two points being measured is equal to the potential difference between the jockey and the battery.
Since no current is drawn from the circuit, the potentiometer provides a more accurate measurement of the potential difference compared to a voltmeter. Therefore, Assertion (A) is true, and Reason (R) correctly explains why the potentiometer is more accurate.
A capacitor plates are charged by a battery with ‘V’ volts. After charging, battery is disconnected and a dielectric slab with dielectric constant ‘K’ is inserted between its plates, the potential across the plates of a capacitor will become:- a)Zero
- b)V/2
- c)V/K
- d)KV
Correct answer is option 'C'. Can you explain this answer?
A capacitor plates are charged by a battery with ‘V’ volts. After charging, battery is disconnected and a dielectric slab with dielectric constant ‘K’ is inserted between its plates, the potential across the plates of a capacitor will become:
a)
Zero
b)
V/2
c)
V/K
d)
KV
|
Vivek Rana answered |
Battery is disconnected.
Q = Charge remains context C′ = KC
Q′ = C′V′
Q = C′V′
Q = KCV′
V′ = 
Assertion (A): Potential difference between two points is depend on path.
Reason (R): Electric field is a conservative field.- a)Both Assertion (A) and Reason (R) are true, and Reason(R) is the correct explanation of (A).
- b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- c)Assertion (A) is true, but Reason (R) is false.
- d)Assertion (A) is false, but Reason (R) is true.
Correct answer is option 'D'. Can you explain this answer?
Assertion (A): Potential difference between two points is depend on path.
Reason (R): Electric field is a conservative field.
Reason (R): Electric field is a conservative field.
a)
Both Assertion (A) and Reason (R) are true, and Reason(R) is the correct explanation of (A).
b)
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
c)
Assertion (A) is true, but Reason (R) is false.
d)
Assertion (A) is false, but Reason (R) is true.
|
Riya Banerjee answered |
Potential difference between two points is defined as work done (W) in moving unit positive charge from one point A having potential VA to other point B having potential VB.
The work done is also explained as:
We know that F = qE
As, electric field at a point (x,y), i.e.
It concludes that whatever path S is selected on moving charge from A to B, the integral will have only one value. It means potential difference between two points is independent of path. So, assertion is not true. Electric field is a conservative field i.e. line integral of an electric field along a closed path is zero.
So, assertion is incorrect and reason is correct.
The work done is also explained as:
We know that F = qE
As, electric field at a point (x,y), i.e.
It concludes that whatever path S is selected on moving charge from A to B, the integral will have only one value. It means potential difference between two points is independent of path. So, assertion is not true. Electric field is a conservative field i.e. line integral of an electric field along a closed path is zero.
So, assertion is incorrect and reason is correct.
A 20 volt AC is applied to a circuit consisting of a resistance and a coil with negligible resistance. If the voltage across the resistance is 12 volt, the voltage across the coil is:- a)16 V
- b)10 V
- c)8 V
- d)6 V
Correct answer is option 'A'. Can you explain this answer?
A 20 volt AC is applied to a circuit consisting of a resistance and a coil with negligible resistance. If the voltage across the resistance is 12 volt, the voltage across the coil is:
a)
16 V
b)
10 V
c)
8 V
d)
6 V
|
|
Suresh Iyer answered |
VR = Effective voltage across R
∴ VR = Ieff R
VL = Effective voltage across L
VL = Ieff × L
Net V =
=
20 =
(20)2 = (12)2 + VL2
400 = 144 + VL2
VL = √400 - 144 = √266 = 16 volts
CASE STUDY
Read the following text and answer the questions
Electric potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point.
So, electric potential = work done divided by the amount of charge moved.
Negative charges move from lower to higher potential. Positive charges move from higher to lower potential.
Charges gain energy while moving through a potential difference. If an electron is accelerated from rest through a potential difference of 1V, it gains 1 eV energy.
Formula of electric potential- a) V = WQ
- b)V = W/Q
- c)W = VQ2
- d)V = WQ2
Correct answer is option 'B'. Can you explain this answer?
CASE STUDY
Read the following text and answer the questions
Electric potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point.
So, electric potential = work done divided by the amount of charge moved.
Negative charges move from lower to higher potential. Positive charges move from higher to lower potential.
Charges gain energy while moving through a potential difference. If an electron is accelerated from rest through a potential difference of 1V, it gains 1 eV energy.
Formula of electric potential
Read the following text and answer the questions
Electric potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point.
So, electric potential = work done divided by the amount of charge moved.
Negative charges move from lower to higher potential. Positive charges move from higher to lower potential.
Charges gain energy while moving through a potential difference. If an electron is accelerated from rest through a potential difference of 1V, it gains 1 eV energy.
Formula of electric potential
a)
V = WQ
b)
V = W/Q
c)
W = VQ2
d)
V = WQ2
|
|
Mira Joshi answered |
Electric potential at a point is defined using the following formula as V = W/Q, which states that the amount of work done in bringing a unit positive charge from infinity to that point is known as electric potential.
The amount of work done in bringing a charge of 10 mC from infinity to a point in an electric is 20 × 10–5 J. Electric potential at that point is- a)20 V
- b)200 V
- c)5V
- d)50V
Correct answer is option 'A'. Can you explain this answer?
The amount of work done in bringing a charge of 10 mC from infinity to a point in an electric is 20 × 10–5 J. Electric potential at that point is
a)
20 V
b)
200 V
c)
5V
d)
50V
|
|
Mira Joshi answered |
Given that
Q = 10μC = 10 x 10-6C
W = 20 x 10-5 J
Since,
V = W/Q
= 20 x 10-5 / 20 x 10-6
= 20V
Q = 10μC = 10 x 10-6C
W = 20 x 10-5 J
Since,
V = W/Q
= 20 x 10-5 / 20 x 10-6
= 20V
Which of the following is NOT the property of equipotential surface?- a)They do not cross each other.
- b)The rate of change of potential with distance on them is zero.
- c)For a uniform electric field, they are concentric spheres.
- d)They can be imaginary spheres.
Correct answer is option 'C'. Can you explain this answer?
Which of the following is NOT the property of equipotential surface?
a)
They do not cross each other.
b)
The rate of change of potential with distance on them is zero.
c)
For a uniform electric field, they are concentric spheres.
d)
They can be imaginary spheres.
|
Nabanita Pillai answered |
Equipotential surface is a surface in an electric field on which all points have the same electric potential. Here, we need to identify the property that does not belong to equipotential surfaces.
Property not belonging to Equipotential Surface:
Concentric Spheres
Equipotential surfaces are defined as surfaces on which the electric potential is constant. Hence, the rate of change of potential with distance on them is zero. They do not cross each other and can be imaginary spheres. However, for a uniform electric field, equipotential surfaces are not concentric spheres. This is because the potential difference between two points in an electric field is directly proportional to the distance between them. Therefore, equipotential surfaces for a uniform electric field are plane surfaces perpendicular to the field lines.
In summary, the property that does not belong to equipotential surfaces is that for a uniform electric field, they are not concentric spheres.
Property not belonging to Equipotential Surface:
Concentric Spheres
Equipotential surfaces are defined as surfaces on which the electric potential is constant. Hence, the rate of change of potential with distance on them is zero. They do not cross each other and can be imaginary spheres. However, for a uniform electric field, equipotential surfaces are not concentric spheres. This is because the potential difference between two points in an electric field is directly proportional to the distance between them. Therefore, equipotential surfaces for a uniform electric field are plane surfaces perpendicular to the field lines.
In summary, the property that does not belong to equipotential surfaces is that for a uniform electric field, they are not concentric spheres.
Two heaters A (500 W, 220 V) and B (1000 W, 220 V) are connected in parallel to 220 V. Which heater produces more heat- a)heater A
- b)heater B
- c)heater A and heater B produce same heat
- d)both heaters produce no heat
Correct answer is option 'B'. Can you explain this answer?
Two heaters A (500 W, 220 V) and B (1000 W, 220 V) are connected in parallel to 220 V. Which heater produces more heat
a)
heater A
b)
heater B
c)
heater A and heater B produce same heat
d)
both heaters produce no heat
|
|
Priyanka Sharma answered |
As P = V x i = i2 R = V2/R
R for heater A
R for heater A
RA = V2/P = (200)2 /500 = 96.8Ω
R for heater B
RB = V2/P = (200)2 /1000 = 48.4Ω
⇒ RA >RB
When two heaters are connected in parallel, voltage remains same for two resisters and power is calculated using the following formula:
P = V2/R
⇒ PA <PB
Therefore, power dissipation is small in heater A as compare to heater B.
Heat generated H = P × t, where t is time in seconds.
⇒ HA <HB
R for heater B
RB = V2/P = (200)2 /1000 = 48.4Ω
⇒ RA >RB
When two heaters are connected in parallel, voltage remains same for two resisters and power is calculated using the following formula:
P = V2/R
⇒ PA <PB
Therefore, power dissipation is small in heater A as compare to heater B.
Heat generated H = P × t, where t is time in seconds.
⇒ HA <HB
Two point charges placed in a medium of dielectric constant 5 are at a distance r between them, experience an electrostatic force ‘F’. The electrostatic force between them in vacuum at the same distance r will be:- a)5F
- b)F
- c)F/2
- d)F/5
Correct answer is option 'A'. Can you explain this answer?
Two point charges placed in a medium of dielectric constant 5 are at a distance r between them, experience an electrostatic force ‘F’. The electrostatic force between them in vacuum at the same distance r will be:
a)
5F
b)
F
c)
F/2
d)
F/5
|
Gaurav Kumar answered |
Q1 _______Q2
Force in the charges in the air is
= K F
= 5 F
We use alloys for making standard resistors because they have:- a)low temperature coefficient of resistivity and high specific resistance.
- b)high temperature coefficient of resistivity and low specific resistance.
- c)low temperature coefficient of resistivity and low specific resistance.
- d)high temperature coefficient of resistivity and high specific resistance.
Correct answer is option 'A'. Can you explain this answer?
We use alloys for making standard resistors because they have:
a)
low temperature coefficient of resistivity and high specific resistance.
b)
high temperature coefficient of resistivity and low specific resistance.
c)
low temperature coefficient of resistivity and low specific resistance.
d)
high temperature coefficient of resistivity and high specific resistance.
|
|
Mira Joshi answered |
Alloys have low temperature coefficient of resistivity and high specific resistance. If a = low, the value of ‘R’ with temperature will not change much and specific resistance is high, then required length of the wire will be less.
The equivalent resistance between A and B is ?
- a)3 ohms
- b)5.5 ohms
- c)7.5 ohms
- d)9.5 ohms
Correct answer is option 'C'. Can you explain this answer?
The equivalent resistance between A and B is ?
a)
3 ohms
b)
5.5 ohms
c)
7.5 ohms
d)
9.5 ohms
|
|
Riya Banerjee answered |
Redrawing the circuit, we get
3 Ω & 6 Ω are in parallel.
∴ R1 = 3 x 6 / 3 + 6 = 18/9 = 2
3 Ω & 6 Ω are in parallel.
∴ R1 = 3 x 6 / 3 + 6 = 18/9 = 2
Now R1 and 8Ω in series
∴ R2 = R1 + 8 = 2 + 8 = 10 Ω
Now R2 and 30Ωin parallel
= 300/40 = 30/4 = 15 /2
= 7.5 Ω
∴ R2 = R1 + 8 = 2 + 8 = 10 Ω
Now R2 and 30Ωin parallel
= 300/40 = 30/4 = 15 /2
= 7.5 Ω
The magnetic flux linked with the coil (in Weber) is given by the equation:
φ = 5t2 + 3t + 16 The induced EMF in the coil at time, t = 4 will be:- a)- 27 V
- b)- 43 V
- c)- 108 V
- d)210 V
Correct answer is option 'B'. Can you explain this answer?
The magnetic flux linked with the coil (in Weber) is given by the equation:
φ = 5t2 + 3t + 16 The induced EMF in the coil at time, t = 4 will be:
φ = 5t2 + 3t + 16 The induced EMF in the coil at time, t = 4 will be:
a)
- 27 V
b)
- 43 V
c)
- 108 V
d)
210 V
|
|
Tanvi Bose answered |
The magnetic flux linked with the coil (in Weber) is given by the equation:
Φ = B * A * cos(θ)
Where:
- Φ is the magnetic flux linked with the coil, measured in Weber (Wb).
- B is the magnetic field strength, measured in Tesla (T).
- A is the cross-sectional area of the coil, measured in square meters (m^2).
- θ is the angle between the magnetic field lines and the normal to the coil's surface.
This equation represents the relationship between the magnetic field strength, the cross-sectional area of the coil, and the angle at which the magnetic field lines intersect the coil. The magnetic flux is a measure of the total magnetic field passing through the coil.
Φ = B * A * cos(θ)
Where:
- Φ is the magnetic flux linked with the coil, measured in Weber (Wb).
- B is the magnetic field strength, measured in Tesla (T).
- A is the cross-sectional area of the coil, measured in square meters (m^2).
- θ is the angle between the magnetic field lines and the normal to the coil's surface.
This equation represents the relationship between the magnetic field strength, the cross-sectional area of the coil, and the angle at which the magnetic field lines intersect the coil. The magnetic flux is a measure of the total magnetic field passing through the coil.
Two magnets are held together and are allowed to oscillate in the earth’s magnetic field with like poles together. 10 oscillations per minute are made but for unlike poles together only 4 oscillations per minutes are made. The ratio of magnetic moments is- a)16:21
- b)21:16
- c)29:21
- d)21:29
Correct answer is option 'C'. Can you explain this answer?
Two magnets are held together and are allowed to oscillate in the earth’s magnetic field with like poles together. 10 oscillations per minute are made but for unlike poles together only 4 oscillations per minutes are made. The ratio of magnetic moments is
a)
16:21
b)
21:16
c)
29:21
d)
21:29
|
|
Rajesh Gupta answered |
Here,
T1 = 60/10 = 6sec,
and T2= 60/4 = 15sec
M1/M2 = T22 + T12 / T22 - T12
= 29/21
T1 = 60/10 = 6sec,
and T2= 60/4 = 15sec
M1/M2 = T22 + T12 / T22 - T12
= 29/21
The susceptibility of annealed iron at saturation is 5000. The permeability is - a)2.28 × 10–3
- b)6.28 × 10–3
- c)9.28 × 10–3
- d)1.28 × 10–3
Correct answer is option 'B'. Can you explain this answer?
The susceptibility of annealed iron at saturation is 5000. The permeability is
a)
2.28 × 10–3
b)
6.28 × 10–3
c)
9.28 × 10–3
d)
1.28 × 10–3
|
|
Mihir Yadav answered |
The permeability can be calculated using the formula:
Permeability = Susceptibility * (4π * 10^-7)
Given that the susceptibility of annealed iron at saturation is 5000, the permeability is:
Permeability = 5000 * (4π * 10^-7)
Permeability ≈ 2.257 x 10^-2
Therefore, the permeability of annealed iron at saturation is approximately 2.28.
Permeability = Susceptibility * (4π * 10^-7)
Given that the susceptibility of annealed iron at saturation is 5000, the permeability is:
Permeability = 5000 * (4π * 10^-7)
Permeability ≈ 2.257 x 10^-2
Therefore, the permeability of annealed iron at saturation is approximately 2.28.
The SI unit of magnetic field intensity is:- a)AmN–1
- b)NA–1m–1
- c)NA–2m–2
- d)NA–1m–2
Correct answer is option 'B'. Can you explain this answer?
The SI unit of magnetic field intensity is:
a)
AmN–1
b)
NA–1m–1
c)
NA–2m–2
d)
NA–1m–2
|
Kalyan Chavan answered |
SI Unit of Magnetic Field Intensity
Definition:
The SI unit of magnetic field intensity is Ampere per meter (A/m). It is denoted by the symbol NA–1m–1.
Explanation:
Ampere per meter (A/m) represents the magnetic field intensity produced by a current of one ampere flowing through a conductor per meter of length. It is a measure of the magnetic field strength at a specific point in space.
Relationship with Current:
The magnetic field intensity is directly proportional to the current flowing through the conductor. Therefore, increasing the current will result in a stronger magnetic field intensity.
Practical Applications:
The SI unit of magnetic field intensity is widely used in various applications, such as designing electromagnets, MRI machines, and magnetic sensors.
Assertion (A): Earth’s magnetic field does not affect the functioning of a moving coil galvanometer.
Reason (R): Earth’s magnetic field is too weak. - a)Both Assertion (A) and Reason (R) are true, and Reason(R) is the correct explanation of (A).
- b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- c)Assertion (A) is true, but Reason (R) is false.
- d)Assertion (A) is false, but Reason (R) is true.
Correct answer is option 'A'. Can you explain this answer?
Assertion (A): Earth’s magnetic field does not affect the functioning of a moving coil galvanometer.
Reason (R): Earth’s magnetic field is too weak.
Reason (R): Earth’s magnetic field is too weak.
a)
Both Assertion (A) and Reason (R) are true, and Reason(R) is the correct explanation of (A).
b)
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
c)
Assertion (A) is true, but Reason (R) is false.
d)
Assertion (A) is false, but Reason (R) is true.
|
|
Mira Joshi answered |
The coil of moving coil galvanometer is suspended in a very strong radial magnetic field. Earth’s magnetic field is too weak compared to that and hence its effect is negligible. So, assertion and reason both are true and the reason explains the assertion properly.
The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the center is 54 mT. What will be its value at the centre of the loop?- a)300 μT
- b)500 μT
- c)250 μT
- d)150 μT
Correct answer is option 'C'. Can you explain this answer?
The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the center is 54 mT. What will be its value at the centre of the loop?
a)
300 μT
b)
500 μT
c)
250 μT
d)
150 μT
|
|
Priyanka Sharma answered |
Magnetic field due to current carrying circular loop of radius a at a point on the axis at a distance x from the centre is given by
⇒
⇒ I = 11.94 A
Magnetic field at the centre =
⇒
⇒ I = 11.94 A
Magnetic field at the centre =
A charged particles enters into a magnetic field with an angle of 30° with respect to the direction of magnetic field. The path of the particle is- a)circular
- b)helical
- c)straight line
- d)elliptical
Correct answer is option 'B'. Can you explain this answer?
A charged particles enters into a magnetic field with an angle of 30° with respect to the direction of magnetic field. The path of the particle is
a)
circular
b)
helical
c)
straight line
d)
elliptical
|
|
Mira Joshi answered |
Charge is entered at point A making an angle 30°with the velocity v. one component of it is along the field and other component is perpendicular to it. Its perpendicular component v sin 30° will create circular motion, whereas its component along the field v cos 30° will have linear motion. Thus, resultant will have helical path.
A constant voltage is applied between the two ends of a uniform metallic wire, heat ‘H’ is developed in it. If another wire of the same material, double the radius and twice the length as compared to original wire is used, then the heat developed in it will be:- a)H/2
- b)H
- c)2H
- d)4H
Correct answer is option 'C'. Can you explain this answer?
A constant voltage is applied between the two ends of a uniform metallic wire, heat ‘H’ is developed in it. If another wire of the same material, double the radius and twice the length as compared to original wire is used, then the heat developed in it will be:
a)
H/2
b)
H
c)
2H
d)
4H
|
Jyoti Sengupta answered |
∵ V = constant
=
H′ /H = 2/1
H′ = 2H
Which statement is true for Gauss law?- a)All the charges whether inside or outside the gaussian surface contribute to the electric flux.
- b)Electric flux depends upon the geometry of the gaussian surface.
- c)Gauss theorem can be applied to non-uniform electric field.
- d)The electric field over the gaussian surface remains continuous and uniform at every point.
Correct answer is option 'D'. Can you explain this answer?
Which statement is true for Gauss law?
a)
All the charges whether inside or outside the gaussian surface contribute to the electric flux.
b)
Electric flux depends upon the geometry of the gaussian surface.
c)
Gauss theorem can be applied to non-uniform electric field.
d)
The electric field over the gaussian surface remains continuous and uniform at every point.
|
|
Jyoti Sengupta answered |
All other statements except (iv) are incorrect.
The electric field over the Gaussian surface remains continuous and uniform at every point.
The electric field over the Gaussian surface remains continuous and uniform at every point.
Two bar magnets of the same mass, same length and same breadth have magnetic moments M and 3M, respectively, are joined together pole for pole and suspended by string. The time period of assembly is 3 seconds. Now, polarity of one magnet is reversed and combination is again forced to oscillate in magnetic field, the time of oscillation is- a)3
- b)2
- c)2√2
- d)3√2
Correct answer is option 'D'. Can you explain this answer?
Two bar magnets of the same mass, same length and same breadth have magnetic moments M and 3M, respectively, are joined together pole for pole and suspended by string. The time period of assembly is 3 seconds. Now, polarity of one magnet is reversed and combination is again forced to oscillate in magnetic field, the time of oscillation is
a)
3
b)
2
c)
2√2
d)
3√2
|
|
Riya Banerjee answered |
⇒ T2 = √2T1 = 3√2
Assertion (A): A coulomb of charge is bigger than the charge on electron?
Reason (R): Electric charge is always quantized.- a)Both Assertion (A) and Reason (R) are true, and Reason(R) is the correct explanation of (A).
- b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- c)Assertion (A) is true, but Reason (R) is false.
- d)Assertion (A) is false, but Reason (R) is true.
Correct answer is option 'A'. Can you explain this answer?
Assertion (A): A coulomb of charge is bigger than the charge on electron?
Reason (R): Electric charge is always quantized.
Reason (R): Electric charge is always quantized.
a)
Both Assertion (A) and Reason (R) are true, and Reason(R) is the correct explanation of (A).
b)
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
c)
Assertion (A) is true, but Reason (R) is false.
d)
Assertion (A) is false, but Reason (R) is true.
|
|
Suresh Iyer answered |
Magnitude of charge on one electron, e = 1.6 × 10–19 Coulomb. According to the quantization of charge, charges of the particle (q) are integral multiple of charge of an electron/proton (e). i.e., q = ne. Therefore, number of charges in a coulomb,
= 0.625 x1019
Clearly shows that one coulomb of charge is bigger than the charge on electron and quantization of charge is the correct reason for it. So, both assertion and reason are true.
= 0.625 x1019
Clearly shows that one coulomb of charge is bigger than the charge on electron and quantization of charge is the correct reason for it. So, both assertion and reason are true.
The small angle between magnetic axis and geographic axis at a place is:- a)Magnetic meridian
- b)Geographic meridian
- c)Magnetic inclination
- d)Magnetic Declination
Correct answer is option 'D'. Can you explain this answer?
The small angle between magnetic axis and geographic axis at a place is:
a)
Magnetic meridian
b)
Geographic meridian
c)
Magnetic inclination
d)
Magnetic Declination
|
|
Suresh Iyer answered |
Correct option is Magnetic declination or Angle of declination. It is the small angle between geographic axis and magnetic axis.
Electrostatic force between two charged particles in air is F. If charges are placed at the same distance in brass, what will be the force?- a)zero
- b)F
- c)infinite
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Electrostatic force between two charged particles in air is F. If charges are placed at the same distance in brass, what will be the force?
a)
zero
b)
F
c)
infinite
d)
None of these
|
|
Shalini Patel answered |
Since, dielectric constant of metal is ∞. The electric force between the charges in any medium is given by
⇒ Fm = F/∞
⇒ Fm = 0
⇒ Fm = F/∞
⇒ Fm = 0
Assertion (A): If an electron is not deflected when moving through a certain region of space, then the only possibility is that no magnetic field is present in that region.
Reason (R): Force on electron is directly proportional to the strength of the magnetic field.- a)Both Assertion (A) and Reason (R) are true, and Reason(R) is the correct explanation of (A).
- b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- c)Assertion (A) is true, but Reason (R) is false.
- d)Assertion (A) is false, but Reason (R) is true.
Correct answer is option 'A'. Can you explain this answer?
Assertion (A): If an electron is not deflected when moving through a certain region of space, then the only possibility is that no magnetic field is present in that region.
Reason (R): Force on electron is directly proportional to the strength of the magnetic field.
Reason (R): Force on electron is directly proportional to the strength of the magnetic field.
a)
Both Assertion (A) and Reason (R) are true, and Reason(R) is the correct explanation of (A).
b)
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
c)
Assertion (A) is true, but Reason (R) is false.
d)
Assertion (A) is false, but Reason (R) is true.
|
Dipanjan Majumdar answered |
Assertion (A): If an electron is not deflected when moving through a certain region of space, then the only possibility is that no magnetic field is present in that region.
Reason (R): Force on electron is directly proportional to the strength of the magnetic field.
Correct answer: Option A - Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of (A).
Explanation:
The given assertion and reason both deal with the behavior of an electron in the presence of a magnetic field. Let's analyze each statement separately:
Assertion (A): If an electron is not deflected when moving through a certain region of space, then the only possibility is that no magnetic field is present in that region.
When an electron moves through a magnetic field, it experiences a force called the magnetic force. This force acts perpendicular to both the velocity of the electron and the magnetic field. The magnitude of the force is given by the equation F = qvBsinθ, where q is the charge of the electron, v is its velocity, B is the strength of the magnetic field, and θ is the angle between the velocity and the magnetic field.
If an electron is not deflected when moving through a certain region of space, it means that the magnetic force acting on it is zero. This can only happen if there is no magnetic field present in that region. Therefore, Assertion (A) is true.
Reason (R): Force on electron is directly proportional to the strength of the magnetic field.
The reason states that the force on an electron is directly proportional to the strength of the magnetic field. This is in accordance with the equation mentioned earlier, F = qvBsinθ. As the strength of the magnetic field increases, the force on the electron also increases. Therefore, Reason (R) is true.
Explanation of Correct Answer:
Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A). When an electron is not deflected in a certain region, it indicates the absence of a magnetic field in that region. The reason for this is that the force on the electron is directly proportional to the strength of the magnetic field. If there is no magnetic field present, the force on the electron will be zero, resulting in no deflection.
Reason (R): Force on electron is directly proportional to the strength of the magnetic field.
Correct answer: Option A - Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of (A).
Explanation:
The given assertion and reason both deal with the behavior of an electron in the presence of a magnetic field. Let's analyze each statement separately:
Assertion (A): If an electron is not deflected when moving through a certain region of space, then the only possibility is that no magnetic field is present in that region.
When an electron moves through a magnetic field, it experiences a force called the magnetic force. This force acts perpendicular to both the velocity of the electron and the magnetic field. The magnitude of the force is given by the equation F = qvBsinθ, where q is the charge of the electron, v is its velocity, B is the strength of the magnetic field, and θ is the angle between the velocity and the magnetic field.
If an electron is not deflected when moving through a certain region of space, it means that the magnetic force acting on it is zero. This can only happen if there is no magnetic field present in that region. Therefore, Assertion (A) is true.
Reason (R): Force on electron is directly proportional to the strength of the magnetic field.
The reason states that the force on an electron is directly proportional to the strength of the magnetic field. This is in accordance with the equation mentioned earlier, F = qvBsinθ. As the strength of the magnetic field increases, the force on the electron also increases. Therefore, Reason (R) is true.
Explanation of Correct Answer:
Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A). When an electron is not deflected in a certain region, it indicates the absence of a magnetic field in that region. The reason for this is that the force on the electron is directly proportional to the strength of the magnetic field. If there is no magnetic field present, the force on the electron will be zero, resulting in no deflection.
Two coils of self-inductances 9 mH and 4 mH are placed so close together that the effective flux in one coil is linked with other. The mutual inductance between these coils is- a)6mH
- b)36mH
- c)4mH
- d)9mH
Correct answer is option 'A'. Can you explain this answer?
Two coils of self-inductances 9 mH and 4 mH are placed so close together that the effective flux in one coil is linked with other. The mutual inductance between these coils is
a)
6mH
b)
36mH
c)
4mH
d)
9mH
|
Milan Datta answered |
Mutual inductance is a measure of the extent to which the magnetic field generated by one coil links with the other coil. It is denoted by the symbol M. In this problem, we are given two coils with self-inductances of 9 mH and 4 mH respectively, and we need to find their mutual inductance.
The formula for mutual inductance is given by:
M = √(L1 * L2)
where L1 and L2 are the self-inductances of the two coils.
Given that L1 = 9 mH and L2 = 4 mH, we can substitute these values into the formula to find the mutual inductance:
M = √(9 * 4) = √36 = 6 mH
Therefore, the mutual inductance between the two coils is 6 mH.
To summarize:
- Given coils with self-inductances of 9 mH and 4 mH.
- Mutual inductance is a measure of the extent to which the magnetic field generated by one coil links with the other coil.
- The formula for mutual inductance is M = √(L1 * L2).
- Substituting the given values, we find that the mutual inductance is 6 mH.
The formula for mutual inductance is given by:
M = √(L1 * L2)
where L1 and L2 are the self-inductances of the two coils.
Given that L1 = 9 mH and L2 = 4 mH, we can substitute these values into the formula to find the mutual inductance:
M = √(9 * 4) = √36 = 6 mH
Therefore, the mutual inductance between the two coils is 6 mH.
To summarize:
- Given coils with self-inductances of 9 mH and 4 mH.
- Mutual inductance is a measure of the extent to which the magnetic field generated by one coil links with the other coil.
- The formula for mutual inductance is M = √(L1 * L2).
- Substituting the given values, we find that the mutual inductance is 6 mH.
If the potential difference V applied across a conductor is increased to 2V with its temperature kept constant, the drift velocity of the free electrons in a conductor will:- a)remain the same.
- b)become half of its previous value.
- c)be double of its initial value.
- d)become zero.
Correct answer is option 'C'. Can you explain this answer?
If the potential difference V applied across a conductor is increased to 2V with its temperature kept constant, the drift velocity of the free electrons in a conductor will:
a)
remain the same.
b)
become half of its previous value.
c)
be double of its initial value.
d)
become zero.
|
|
Jyoti Sengupta answered |
We know
=
If temperature is kept constant, relaxation time τ - will remain constant and e, m are also constants.
Vd ∝ V
Vd ∝ 2V
=
If temperature is kept constant, relaxation time τ - will remain constant and e, m are also constants.
Vd ∝ V
Vd ∝ 2V
Two point charges + 8q and –2q are located at x = 0 and x = L respectively. The point on x axis at which net electric field is zero due to these charges is- a)8L
- b)4L
- c)2L
- d)L
Correct answer is option 'C'. Can you explain this answer?
Two point charges + 8q and –2q are located at x = 0 and x = L respectively. The point on x axis at which net electric field is zero due to these charges is
a)
8L
b)
4L
c)
2L
d)
L
|
|
Nandini Iyer answered |
Let P be the observation point at a distance r from –2q and at (L + r) from + 8q.
Given Now, Net EFI at P = 0
= EFI (Electric Field Intensity) at P due to + 8q
= EFI (Electric Field Intensity) at P due to – 2q
∴
∴
4r2 = (L + r)2
2r = L + r
r = L
∴ P is at x = L + L = 2L from origin
∴ Correct option is (C) 2L.
Given Now, Net EFI at P = 0
= EFI (Electric Field Intensity) at P due to + 8q = EFI (Electric Field Intensity) at P due to – 2q∴
∴
4r2 = (L + r)2
2r = L + r
r = L
∴ P is at x = L + L = 2L from origin
∴ Correct option is (C) 2L.
Protons move from______ to ______ potential- a)lower, higher
- b)higher, lower
- c)Random motion
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Protons move from______ to ______ potential
a)
lower, higher
b)
higher, lower
c)
Random motion
d)
None of these
|
|
Mira Joshi answered |
Proton has positive charge, in electric field positive charge moves from higher potential to lower potential.
The given figure represents the material which is
- a)ferrimagnetic
- b)ferromagnetic
- c)diamagnetic
- d)paramagnetic
Correct answer is option 'C'. Can you explain this answer?
The given figure represents the material which is
a)
ferrimagnetic
b)
ferromagnetic
c)
diamagnetic
d)
paramagnetic
|
|
Mira Joshi answered |
The given figure represents that magnetic field lines move away from the substance. The diamagnetic material has susceptibility c = 0, causing the repelling of magnetic field lines, Hence, given material is diamagnetic material
When A wire of resistance 12 R is bent to form a circle. The effective resistance across its diameter is- a)2R
- b)3R
- c)4R
- d)R
Correct answer is option 'B'. Can you explain this answer?
When A wire of resistance 12 R is bent to form a circle. The effective resistance across its diameter is
a)
2R
b)
3R
c)
4R
d)
R
|
|
Rajesh Gupta answered |
Given that resistance of wire 12 Ω. It is bent and form circle, then resistance f each semicircle is 6 Ω. Now, measuring the resistance across its diameter, these two resistances are in parallel. Their effective resistance
=3R
=3R
A current i ampere flows along an infinite long straight tube, then the magnetic induction at any point inside - a)zero
- b)infinite
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
A current i ampere flows along an infinite long straight tube, then the magnetic induction at any point inside
a)
zero
b)
infinite
c)
d)
|
|
Riya Banerjee answered |
Using Ampere’s law
Here, i is zero, for r < R, where R is radius of the tube
∴ B = 0
Here, i is zero, for r < R, where R is radius of the tube
∴ B = 0
The coil of a moving coil galvanometer is wound over a metal frame in order to:- a)reduce hysteresis
- b)increase sensitivity
- c)increase moment of inertia
- d)provide electromagnetic damping
Correct answer is option 'D'. Can you explain this answer?
The coil of a moving coil galvanometer is wound over a metal frame in order to:
a)
reduce hysteresis
b)
increase sensitivity
c)
increase moment of inertia
d)
provide electromagnetic damping
|
|
Priyanka Sharma answered |
The coil of a moving coil galvanometer is wound over metallic frame to provide electromagnetic damping so it becomes dead beat galvanometer.
A bar magnet having magnetic moment 0.4 JT–1 is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when the potential energy is- a)zero
- b)–0.64 J
- c)–0.064 J
- d)–6.40 J
Correct answer is option 'C'. Can you explain this answer?
A bar magnet having magnetic moment 0.4 JT–1 is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when the potential energy is
a)
zero
b)
–0.64 J
c)
–0.064 J
d)
–6.40 J
|
|
Riya Banerjee answered |
In stable equilibrium, Potential energy U is defined as
U = 0MB
= - (0.4)(0.16)
= - 0.064J
U = 0MB
= - (0.4)(0.16)
= - 0.064J
If the electric field is 
. Electric flux through a surface area 10 units lying in X-Y plane is- a)10 units
- b)20 units
- c)30 units
- d)40 units
Correct answer is option 'D'. Can you explain this answer?
If the electric field is . Electric flux through a surface area 10 units lying in X-Y plane is
. Electric flux through a surface area 10 units lying in X-Y plane isa)
10 units
b)
20 units
c)
30 units
d)
40 units
|
|
Preeti Iyer answered |
Direction of area element is normal to X-Y plane, i.e., z axis. Therefore, 
. Now, using the relationship between electric flux f and electric field 
= 40 units
. Now, using the relationship between electric flux f and electric field = 40 units
Two wires of the same length are shaped into a square of side 'a' and a circle with radius 'r'. If they carry same current, the ratio of their magnetic moment is:- a)2 : π
- b)π : 2
- c)π : 4
- d)4 : π
Correct answer is option 'C'. Can you explain this answer?
Two wires of the same length are shaped into a square of side 'a' and a circle with radius 'r'. If they carry same current, the ratio of their magnetic moment is:
a)
2 : π
b)
π : 2
c)
π : 4
d)
4 : π
|
Hansa Sharma answered |
l = length of wire
Area of a Square = a2
Also here l = 4a
a = l /4
∴ Area = l2 /16
A1 = l2/16
Area of a circle = πr2
Also here, 2πr = l
r = l/2π
Now area =
Area of a Square = a2
Also here l = 4a
a = l /4
∴ Area = l2 /16
A1 = l2/16
Area of a circle = πr2
Also here, 2πr = l
r = l/2π
Now area =
A2 = l2/4π
Now Magnetic moment = IA
∴ M1 = IA, & M2 = I A2
Since I (current) is same in both
∴
M1M2 = π : 4
Now Magnetic moment = IA
∴ M1 = IA, & M2 = I A2
Since I (current) is same in both
∴
M1M2 = π : 4
A conducting square loop of side 'L' and resistance 'R' moves in its plane with the uniform velocity 'v' perpendicular to one of its sides. A magnetic induction 'B' constant in time and space pointing perpendicular and into the plane of the loop exists everywhere as shown in the figure. The current induced in the loop is:
- a)BLv/R Clockwise
- b)BLv/R Anticlockwise
- c)2BLv/R Anticlockwise
- d)Zero
Correct answer is option 'D'. Can you explain this answer?
A conducting square loop of side 'L' and resistance 'R' moves in its plane with the uniform velocity 'v' perpendicular to one of its sides. A magnetic induction 'B' constant in time and space pointing perpendicular and into the plane of the loop exists everywhere as shown in the figure. The current induced in the loop is:
a)
BLv/R Clockwise
b)
BLv/R Anticlockwise
c)
2BLv/R Anticlockwise
d)
Zero
|
|
Shalini Patel answered |
Current induced is I = |e/R|
Now |e| = dφ/dt
But there is no change of flux with time, as B',A' and θ all remain constant with time.
∴ No current is induced.
But there is no change of flux with time, as B',A' and θ all remain constant with time.
∴ No current is induced.
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