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Karnataka II PUC Physics (KSEAB) study materials
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Karnataka II PUC Physics (KSEAB) practice questions

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Question 1Electric Charges and Fields; Electrostatic Potential and Capacitance

Two point charges of +2 μC and +8 μC are placed 0.3 m apart in vacuum. At what distance from the +2 μC charge (measured along the line joining them) is the net electric field zero?

  1. 0.15 m
  2. 0.06 m
  3. 0.20 m
  4. 0.10 m Correct answer.

Explanation

The zero-field point lies between two like charges, closer to the smaller charge. Setting k(2)/x² = k(8)/(0.3 − x)² gives (0.3 − x)/x = 2, so 3x = 0.3 and x = 0.10 m. The field magnitudes cancel there because the fields point in opposite directions.

Question 2Electric Charges and Fields; Electrostatic Potential and Capacitance

The SI unit of electric flux is:

  1. C m⁻²
  2. N m² C⁻¹ Correct answer.
  3. N C⁻¹
  4. V m⁻¹

Explanation

Electric flux is defined as the surface integral of the electric field over an area, so its unit is (N C⁻¹)(m²) = N m² C⁻¹. It can equivalently be written as V·m, since N m² C⁻¹ equals volt-metre.

Question 3Electric Charges and Fields; Electrostatic Potential and Capacitance

A hollow conducting sphere of radius R carries a net charge Q. Which statement about the electric field inside the cavity (r < R) is correct?

  1. It increases linearly with r from the centre.
  2. It is constant and equal to kQ/R².
  3. It is zero because the charge resides entirely on the outer surface. Correct answer.
  4. It equals kQ/r² directed radially outward.

Explanation

In electrostatic equilibrium the free charge on a conductor resides on its outer surface, and the field within the material and the enclosed cavity is zero. This follows from Gauss's law applied to a spherical surface inside the conductor, which encloses no net charge.

Question 4Electric Charges and Fields; Electrostatic Potential and Capacitance

The electric potential at a distance r from an isolated point charge Q is V. If the distance is doubled while the charge is unchanged, the potential becomes:

  1. V/2 Correct answer.
  2. 2V
  3. 4V
  4. V/4

Explanation

The potential of a point charge is V = kQ/r, which varies inversely with the first power of distance. Doubling r therefore halves the potential, giving V/2.

Question 5Electric Charges and Fields; Electrostatic Potential and Capacitance

Two capacitors of 3 μF and 6 μF are connected in series across a 12 V battery. The charge on each capacitor is:

  1. 24 μC Correct answer.
  2. 72 μC
  3. 36 μC
  4. 12 μC

Explanation

The series equivalent capacitance is (3×6)/(3+6) = 2 μF. The charge supplied is Q = CV = 2 μF × 12 V = 24 μC, and in series both capacitors carry the same charge of 24 μC.

Question 6Electric Charges and Fields; Electrostatic Potential and Capacitance

The work done in moving a test charge along an equipotential surface is:

  1. Zero Correct answer.
  2. Equal to the potential difference
  3. Infinite
  4. Maximum

Explanation

Work done is W = qΔV, and every point on an equipotential surface has the same potential, so ΔV = 0 and the work done is zero. The electric field is always perpendicular to such a surface, so the force does no work along it.

Question 7Electric Charges and Fields; Electrostatic Potential and Capacitance

An electric dipole of dipole moment p is placed in a uniform electric field E with its axis making an angle θ with the field. The magnitude of the torque acting on it is:

  1. pE tan θ
  2. pE sin θ Correct answer.
  3. pE
  4. pE cos θ

Explanation

The torque on a dipole is the cross product τ = p × E, whose magnitude is pE sin θ, where θ is the angle between p and E. The torque is maximum at θ = 90° and zero when the dipole is aligned with the field.

Question 8Electric Charges and Fields; Electrostatic Potential and Capacitance

A parallel-plate capacitor with air between the plates has capacitance C. A dielectric of relative permittivity 5 completely fills the space between the plates. The new capacitance is:

  1. 25C
  2. C
  3. 5C Correct answer.
  4. C/5

Explanation

Inserting a dielectric of relative permittivity εr multiplies the capacitance by εr because C = εr ε0 A/d. With εr = 5, the capacitance becomes 5C.

Question 9Electric Charges and Fields; Electrostatic Potential and Capacitance

A charge Q is placed at the centre of a cube of side a. The electric flux through one face of the cube is:

  1. Q/(6ε0) Correct answer.
  2. Q/(4ε0)
  3. Q/(8ε0)
  4. Q/ε0

Explanation

By Gauss's law, the total flux through the closed cube is Q/ε0. Since the charge is at the centre, symmetry distributes this flux equally over the six faces, giving Q/(6ε0) per face.

Question 10Electric Charges and Fields; Electrostatic Potential and Capacitance

Which of the following is the correct expression for the capacitance of a parallel-plate capacitor with plate area A and separation d in vacuum?

  1. C = ε0 A/d Correct answer.
  2. C = ε0 A d
  3. C = ε0 d/A
  4. C = A/(ε0 d)

Explanation

For a parallel-plate capacitor the capacitance is C = ε0 A/d, proportional to plate area and inversely proportional to separation. This follows from C = Q/V with V = Ed and E = σ/ε0.

Question 11Electric Charges and Fields; Electrostatic Potential and Capacitance

The electric potential energy of a system of two point charges q1 and q2 separated by a distance r in vacuum is:

  1. k(q1+q2)/r
  2. kq1q2/r Correct answer.
  3. kq1q2/r²
  4. kq1q2 r

Explanation

The potential energy of two point charges is U = kq1q2/r, obtained as the work done in assembling them from infinity. It is negative for unlike charges and positive for like charges.

Question 12Electric Charges and Fields; Electrostatic Potential and Capacitance

A point charge of 4 μC is located 2 m from a point P in vacuum. Taking k = 9 × 10⁹ N m² C⁻², the electric potential at P is approximately:

  1. 3.6 × 10⁴ V
  2. 1.8 × 10⁴ V Correct answer.
  3. 9 × 10³ V
  4. 4.5 × 10³ V

Explanation

The potential is V = kQ/r = (9 × 10⁹ × 4 × 10⁻⁶)/2 = 1.8 × 10⁴ V. Note that potential depends on 1/r, not on 1/r², so no distance is squared in the calculation.

Question 13Electric Charges and Fields; Electrostatic Potential and Capacitance

A capacitor of capacitance C is charged to a potential V and then disconnected from the battery. A dielectric of relative permittivity K is then inserted to fill the gap. Which pair of changes occurs?

  1. Charge remains unchanged; potential decreases to V/K. Correct answer.
  2. Charge decreases; potential remains unchanged.
  3. Charge remains unchanged; potential increases to KV.
  4. Charge increases; potential remains unchanged.

Explanation

With the battery disconnected, the charge Q on the plates cannot change. Since C rises to KC, the potential V = Q/C falls to V/K. The stored energy also decreases because the dielectric is drawn into the capacitor.

Question 14Electric Charges and Fields; Electrostatic Potential and Capacitance

The electric field inside a uniformly charged non-conducting sphere of radius R at a distance r from the centre (r < R) is proportional to:

  1. 1/r
  2. r²
  3. r Correct answer.
  4. 1/r²

Explanation

Applying Gauss's law to a sphere of radius r inside a uniformly charged sphere, the enclosed charge scales as r³ while the area scales as r², so the field is proportional to r. It rises linearly from zero at the centre to a maximum at the surface.

Question 15Electric Charges and Fields; Electrostatic Potential and Capacitance

Two charges +q and −q are separated by a distance 2a. At the midpoint of the line joining them, the electric field and potential respectively are:

  1. Zero and non-zero
  2. Non-zero and zero Correct answer.
  3. Zero and zero
  4. Non-zero and non-zero

Explanation

At the midpoint, the fields from the two charges point in the same direction, so they add to a non-zero resultant. The potentials, however, are +kq/a and −kq/a, which cancel to zero because potential is a scalar.

Question 16Electric Charges and Fields; Electrostatic Potential and Capacitance

The relation between the electric field E and the electric potential V in a region is:

  1. E = ∇V
  2. E = V²
  3. E = −∇V Correct answer.
  4. E = ∫V dr

Explanation

The electric field is the negative gradient of the potential, E = −∇V, so the field points from higher to lower potential. The minus sign reflects that moving along the field decreases potential.

Question 17Electric Charges and Fields; Electrostatic Potential and Capacitance

A charge of 2 × 10⁻⁶ C is moved between two points whose potential difference is 50 V. The work done by the external agent is:

  1. 1 × 10⁻⁷ J
  2. 2.5 × 10⁻⁸ J
  3. 1 × 10⁻⁴ J Correct answer.
  4. 4 × 10⁻⁴ J

Explanation

Work done is W = qΔV = (2 × 10⁻⁶ C)(50 V) = 1 × 10⁻⁴ J. This equals the change in potential energy of the charge between the two points.

Question 18Electric Charges and Fields; Electrostatic Potential and Capacitance

Three capacitors of 2 μF each are connected in parallel. The equivalent capacitance is:

  1. 18 μF
  2. 3/2 μF
  3. 2/3 μF
  4. 6 μF Correct answer.

Explanation

In a parallel combination the equivalent capacitance is the sum of the individual capacitances, so C = 2 + 2 + 2 = 6 μF. Each capacitor shares the same potential difference across the supply.

Question 19Electric Charges and Fields; Electrostatic Potential and Capacitance

A spherical conductor of radius 0.1 m carries a charge of 5 μC. Taking k = 9 × 10⁹ N m² C⁻², the potential at its surface is:

  1. 9 × 10⁵ V
  2. 4.5 × 10⁵ V Correct answer.
  3. 4.5 × 10⁴ V
  4. 2.25 × 10⁵ V

Explanation

For a charged conducting sphere, the surface potential is V = kQ/R = (9 × 10⁹ × 5 × 10⁻⁶)/0.1 = 4.5 × 10⁵ V. The field inside the conductor is zero, so the potential throughout the interior equals this surface value.

Question 20Electric Charges and Fields; Electrostatic Potential and Capacitance

The electric field just outside the surface of a charged conductor in equilibrium is:

  1. Parallel to the surface and equal to σ/ε0
  2. Perpendicular to the surface and equal to σ/(2ε0)
  3. Perpendicular to the surface and equal to σ/ε0 Correct answer.
  4. Zero at all points just outside the surface

Explanation

At the surface of a conductor in electrostatic equilibrium the field is normal to the surface, and Gauss's law applied to a small pillbox gives E = σ/ε0. The field has no tangential component, otherwise charges would move along the surface.

Question 21Electric Charges and Fields; Electrostatic Potential and Capacitance

An electric dipole is placed in a uniform electric field. In which orientation is its potential energy minimum?

  1. When the dipole moment is antiparallel to the field
  2. The potential energy is the same in every orientation
  3. When the dipole moment is parallel to the field Correct answer.
  4. When the dipole moment is perpendicular to the field

Explanation

The potential energy of a dipole is U = −pE cos θ, which is minimum when cos θ = 1, that is when p is parallel to E. This is the stable equilibrium orientation, and it gives U = −pE.

Question 22Electric Charges and Fields; Electrostatic Potential and Capacitance

The capacitance of an isolated spherical conductor of radius R in free space is:

  1. ε0 R
  2. 4πR/ε0
  3. 4πε0 R Correct answer.
  4. 4πε0 R²

Explanation

For an isolated sphere, the potential at the surface is V = Q/(4πε0R), so C = Q/V = 4πε0R. Capacitance therefore has the dimensions of permittivity times length.

Question 23Electric Charges and Fields; Electrostatic Potential and Capacitance

A charge of +Q is placed at the centre of a hollow conducting shell that carries no net charge. Which statement about the charge distribution on the shell is correct?

  1. A charge −Q appears on the inner surface and +Q on the outer surface. Correct answer.
  2. No charge is induced on either surface of the shell.
  3. The entire charge +Q appears on the inner surface and the outer surface is uncharged.
  4. A charge +Q appears on the inner surface and −Q on the outer surface.

Explanation

The field within the conducting material must be zero, so by Gauss's law the inner surface must carry a charge −Q that cancels the enclosed +Q. Since the shell is neutral overall, a charge +Q is left on the outer surface.

Question 24Electric Charges and Fields; Electrostatic Potential and Capacitance

Two point charges, +4 μC and −4 μC, are placed 0.2 m apart. The electric potential at the midpoint of the line joining them is:

  1. 3.6 × 10⁵ V
  2. 1.8 × 10⁵ V
  3. 7.2 × 10⁵ V
  4. Zero Correct answer.

Explanation

The midpoint is 0.1 m from each charge, so the two potentials are +k(4 μC)/0.1 and −k(4 μC)/0.1. Since potential is a scalar, these equal and opposite contributions add to zero.

Question 25Electric Charges and Fields; Electrostatic Potential and Capacitance

A parallel-plate capacitor has capacitance C and is charged to a potential V. If both the plate area and the plate separation are doubled, the new capacitance is:

  1. C/2
  2. C Correct answer.
  3. 2C
  4. 4C

Explanation

Capacitance is C = ε0A/d. Doubling both A and d multiplies the numerator and denominator by the same factor of 2, so the capacitance remains unchanged at C.

Question 26Electric Charges and Fields; Electrostatic Potential and Capacitance

A charge of 1 μC is placed at a point in a region where the electric potential is 300 V. The electrostatic potential energy of this charge in that field is:

  1. 3 × 10² J
  2. 3 × 10⁻² J
  3. 3 × 10⁻⁶ J
  4. 3 × 10⁻⁴ J Correct answer.

Explanation

The potential energy of a charge q placed at a point where the potential is V is U = qV, since potential is defined as potential energy per unit charge. Substituting q = 1 × 10⁻⁶ C and V = 300 V gives U = 3 × 10⁻⁴ J. The sign of the charge would flip the sign of U, but the magnitude follows directly from this product.

Question 27Electric Charges and Fields; Electrostatic Potential and Capacitance

Two capacitors of 3 μF and 6 μF are joined in series and the combination is connected across a 12 V battery. The charge on each capacitor is:

  1. 36 μC
  2. 12 μC
  3. 48 μC
  4. 24 μC Correct answer.

Explanation

For capacitors in series the equivalent capacitance satisfies 1/C = 1/3 + 1/6, giving C = 2 μF. The charge drawn from the battery is Q = CV = 2 μF × 12 V = 24 μC, and capacitors in series always carry equal charge. Hence each capacitor holds 24 μC regardless of its individual capacitance.

Question 28Electric Charges and Fields; Electrostatic Potential and Capacitance

A point charge Q sits at the centre of an uncharged conducting spherical shell. The shell is now connected to earth. Which statement describes the resulting charge distribution?

  1. No charge appears on either surface because the shell is earthed
  2. A charge +Q appears on the inner surface and −Q on the outer surface
  3. A charge −Q appears on the inner surface and +Q on the outer surface
  4. A charge −Q appears on the inner surface and no charge on the outer surface Correct answer.

Explanation

The field of the central charge induces a charge −Q on the inner surface, so the field within the metal is zero. Earthing lets the outer surface exchange charge with the ground, allowing the +Q that would otherwise sit there to flow away, leaving the outer surface neutral. This is why an earthed conducting shell completely screens the region outside it from the enclosed charge.

Question 29Current Electricity

A wire of resistance R is stretched uniformly so that its length becomes twice the original length. Its new resistance is

  1. R/4
  2. 4R Correct answer.
  3. R/2
  4. 2R

Explanation

Stretching keeps the volume constant, so doubling the length halves the cross-sectional area. Resistance equals resistivity times length divided by area, so doubling the length and halving the area multiplies the resistance by four.

Question 30Current Electricity

The drift velocity of free electrons in a metallic conductor carrying a steady current is of the order of

  1. 10^6 m/s
  2. 10^-15 m/s
  3. 10^-4 m/s Correct answer.
  4. 10^8 m/s

Explanation

Drift velocity equals current divided by the product of free-electron density, charge and cross-sectional area. Typical metallic values give about 10^-4 m/s, far smaller than the random thermal speed.

Question 31Current Electricity

Two resistors of 6 ohm and 3 ohm are connected in parallel. The equivalent resistance of the combination is

  1. 0.5 ohm
  2. 2 ohm Correct answer.
  3. 4.5 ohm
  4. 9 ohm

Explanation

For a parallel combination the reciprocals of the resistances add: 1/6 + 1/3 = 1/2, so the equivalent resistance is 2 ohm. The result is always smaller than the smallest individual resistance.

Question 32Current Electricity

The emf of a cell is defined as the potential difference between its terminals when

  1. the cell is short-circuited
  2. no current is drawn from the cell Correct answer.
  3. the cell is delivering maximum current
  4. the cell is connected to a load resistance

Explanation

Emf is the work done per unit charge by the source in driving charge around the complete circuit. When no current flows there is no potential drop across the internal resistance, so the terminal potential difference equals the emf.

Question 33Current Electricity

The current through a conductor is doubled while its resistance is kept constant. The heat produced per second in the conductor becomes

  1. half the original value
  2. twice the original value
  3. unchanged
  4. four times the original value Correct answer.

Explanation

Joule heating power is given by P = I^2 R. Since resistance is constant, doubling the current multiplies the power by the square of two, which is four.

Question 34Current Electricity

A potentiometer is preferred over a voltmeter for measuring the emf of a cell because

  1. its reading depends on the internal resistance of the cell
  2. it has a very low resistance and draws a large current
  3. it draws no current at the balance point and measures the true emf Correct answer.
  4. it works only with alternating current sources

Explanation

At the balance point the galvanometer shows no deflection, so no current is drawn from the cell under test. The measured potential difference therefore equals the emf rather than the reduced terminal voltage a voltmeter would show.

Question 35Current Electricity

A cell of emf 2 V and internal resistance 0.5 ohm is connected to an external resistance of 1.5 ohm. The terminal potential difference across the cell is

  1. 1.5 V Correct answer.
  2. 1.0 V
  3. 0.5 V
  4. 2.0 V

Explanation

The current is emf divided by total resistance, 2/2.0 = 1 A. The terminal voltage is the emf minus the internal drop, 2 - (1 x 0.5) = 1.5 V, which also equals the current times the external resistance.

Question 36Current Electricity

The temperature coefficient of resistance of a conductor is positive. As its temperature rises, its resistance

  1. increases Correct answer.
  2. decreases
  3. remains constant
  4. first increases and then becomes zero

Explanation

In a metallic conductor, higher temperature increases lattice vibrations, which scatter the free electrons more strongly. The resistivity and hence the resistance therefore rise with temperature, consistent with a positive temperature coefficient.

Question 37Current Electricity

Three identical cells, each of emf E and internal resistance r, are connected in parallel. The equivalent emf and internal resistance of the combination are

  1. 3E and r/3
  2. E and r/3 Correct answer.
  3. 3E and 3r
  4. E/3 and r/3

Explanation

For identical cells in parallel the emf of the combination equals that of one cell, E. The internal resistances combine as reciprocals, giving r/3, which lowers the internal drop and allows a larger current.

Question 38Current Electricity

A wire of uniform cross-section has resistance 12 ohm. It is bent to form a closed circle. The resistance between two diametrically opposite points of the circle is

  1. 24 ohm
  2. 6 ohm
  3. 3 ohm Correct answer.
  4. 12 ohm

Explanation

Bending the wire into a circle splits it into two semicircular halves of 6 ohm each, joined in parallel between the two ends. The parallel combination of 6 ohm and 6 ohm is 3 ohm.

Question 39Current Electricity

In a metre bridge experiment, the balance point is obtained at 40 cm from the left end when the unknown resistance is in the left gap. If the unknown resistance is 8 ohm, the resistance in the right gap is

  1. 20 ohm
  2. 12 ohm Correct answer.
  3. 5.33 ohm
  4. 8 ohm

Explanation

The metre bridge balances when the unknown divided by the known resistance equals the balancing length divided by the remaining length. Thus 8/R = 40/60, which gives R = 12 ohm.

Question 40Current Electricity

A galvanometer of resistance 50 ohm gives full-scale deflection for a current of 1 mA. To convert it into an ammeter of range 1 A, the shunt resistance required is approximately

  1. 0.05 ohm Correct answer.
  2. 50 ohm
  3. 0.5 ohm
  4. 50000 ohm

Explanation

A shunt is placed in parallel so that most of the current bypasses the galvanometer. Using S = Ig G /(I - Ig) with Ig = 0.001 A, G = 50 ohm and I = 1 A gives S = 0.05/0.999, about 0.05 ohm.

Question 41Current Electricity

A carbon resistor is marked with the colour bands brown, black, red and gold. Its resistance and tolerance are

  1. 1000 ohm, 5% Correct answer.
  2. 1000 ohm, 10%
  3. 100 ohm, 10%
  4. 10 ohm, 5%

Explanation

Brown stands for 1, black for 0 and red for a multiplier of 100, giving 10 x 100 = 1000 ohm. The gold fourth band indicates a tolerance of plus or minus five per cent.

Question 42Current Electricity

A conductor obeys Ohm's law. Which statement about the current density J and the applied electric field E in the conductor is correct?

  1. J is proportional to E, with the constant of proportionality equal to the conductivity Correct answer.
  2. J is independent of E
  3. J is inversely proportional to E
  4. J is proportional to the square of E

Explanation

The microscopic form of Ohm's law states that J = sigma E, where sigma is the conductivity of the material. Since the drift velocity is proportional to the electric field, the current density is directly proportional to the field.

Question 43Current Electricity

In the circuit of a Wheatstone bridge, the resistances in the four arms are P = 4 ohm, Q = 6 ohm, R = 8 ohm and S = 12 ohm, with a galvanometer between the junctions. The bridge is

  1. balanced, because P/Q equals R/S Correct answer.
  2. balanced, because P + S equals Q + R
  3. not balanced, because P/Q is less than R/S
  4. not balanced, because P/Q is greater than R/S

Explanation

A Wheatstone bridge is balanced when the ratio of the resistances in one pair of adjacent arms equals the ratio in the other pair. Here P/Q = 4/6 = 2/3 and R/S = 8/12 = 2/3, so the ratios are equal and no current flows through the galvanometer.

Question 44Current Electricity

A battery of emf E and internal resistance r delivers maximum power to an external resistance R. The value of R and the maximum power delivered are

  1. R = r/2, with power E^2/r
  2. R = 2r, with power E^2/(2r)
  3. R = r, with power E^2/(4r) Correct answer.
  4. R = 0, with power E^2/r

Explanation

The power in the external resistor is P = E^2 R/(R + r)^2, which is maximum when the external resistance equals the internal resistance. Substituting R = r gives the maximum power E^2/(4r).

Question 45Moving Charges and Magnetism; Magnetism and Matter

A straight wire of length 0.5 m carries a current of 4 A and lies at right angles to a uniform magnetic field of 0.2 T. What is the magnitude of the force on the wire?

  1. 0.04 N
  2. 0.4 N Correct answer.
  3. 4 N
  4. 0.8 N

Explanation

For a straight conductor at right angles to the field the force is F = BIL. Substituting gives F = 0.2 × 4 × 0.5 = 0.4 N, so the correct choice is 0.4 N.

Question 46Moving Charges and Magnetism; Magnetism and Matter

The SI unit of magnetic field strength, the tesla, is equivalent to which combination of base or derived units?

  1. N A⁻¹ m⁻¹ Correct answer.
  2. N A m⁻¹
  3. N A⁻¹ m
  4. N A m

Explanation

From F = BIL, the field B = F/(IL), so its unit is newton per ampere per metre, that is N A⁻¹ m⁻¹. This is the definition of one tesla: one newton per ampere per metre of conductor.

Question 47Moving Charges and Magnetism; Magnetism and Matter

A charged particle enters a uniform magnetic field with its velocity parallel to the field lines. What is the path of the particle inside the field?

  1. A circle of constant radius
  2. A helix along the field direction
  3. A parabola bending along the field
  4. A straight line with unchanged speed Correct answer.

Explanation

The magnetic force magnitude is qvB sinθ, and when the velocity is parallel to B the angle θ is zero, so sinθ vanishes and no force acts. With zero force the particle continues in a straight line at constant speed.

Question 48Moving Charges and Magnetism; Magnetism and Matter

Two long parallel wires carry currents in the same direction. What is the nature of the force between them?

  1. Alternating between attractive and repulsive
  2. Zero, because the currents are parallel
  3. Repulsive
  4. Attractive Correct answer.

Explanation

Each wire produces a magnetic field that acts on the other. For parallel currents flowing in the same direction, the resulting force is attractive, as given by Ampere's force law per unit length μ₀I₁I₂/(2πd).

Question 49Moving Charges and Magnetism; Magnetism and Matter

A bar magnet is freely suspended and comes to rest. Which statement about the Earth's magnetic field near the surface is correct?

  1. The magnet aligns east to west
  2. The south pole of the magnet points roughly towards the geographic north
  3. The magnet shows no preferred direction
  4. The north pole of the magnet points roughly towards the geographic north Correct answer.

Explanation

A freely suspended magnet aligns along the magnetic meridian with its north-seeking pole towards geographic north. This happens because the Earth behaves like a magnetic dipole whose magnetic south pole lies near the geographic north.

Question 50Moving Charges and Magnetism; Magnetism and Matter

The magnetic susceptibility of a diamagnetic material is

  1. Exactly zero
  2. Large and positive
  3. Small and negative Correct answer.
  4. Small and positive

Explanation

Diamagnetic substances are weakly repelled by a magnet and have a small negative susceptibility, typically of the order of −10⁻⁵. Their relative permeability is therefore slightly less than one.