NEET Physics Tutor Doubt 10 - NEET Physics Current Electricity and Magnetism Practice Quiz
NEET Physics Tutor Doubt 10 is a concept-based NEET Physics practice set from Kumar Physics Classes for students preparing for NEET, IIT JEE, IB and ICSE Physics.
NEET Physics Practice Quiz
NEET Physics Tutor Doubt 10 helps students revise current electricity, magnetism and mixed NEET Physics concepts with typed questions, clean diagrams and step-by-step solutions. For one-to-one online Physics support, contact Kumar Sir.
Question 1. The current in a wire bent in the form of a circular arc is as shown in the figure. The magnetic field at the centre point O is (radius of circular part: R).
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Correct answer: (2).
Solution:
Only the circular arc contributes magnetic field at O; the straight radial parts pass through O and give zero field there.
The angle of the arc is 240°, so B = μ0i2R × 240°360° = μ0i3R.
Therefore, the correct option is (2).
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Question 2+4 / -1
Question 2. A coaxial cable is made of two conductors. The inner conductor is solid and of radius R1. The outer conductor is hollow and of inner radius R2 and outer radius R3. The space between the conductors is filled with air. The two conductors carry equal currents but in opposite directions. The variation of magnetic field with distance from the common axis is best plotted as (R1 < R2 < R3).
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Correct answer: (3).
Solution:
Inside the solid inner conductor, B increases with r. Between R1 and R2, the field varies as 1/r.
Inside the material of the outer conductor, the opposite current is gradually enclosed, so B falls to zero at R3. Beyond R3, the net enclosed current is zero.
This variation matches graph (3).
Therefore, the correct option is (3).
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Question 3+4 / -1
Question 3. A thin uniform rod of length l and mass m is uniformly charged with a total charge q. The rod is rotated about a point at a distance l/3 from midpoint of the rod. The magnitude of the magnetic moment of the rod, if the angular velocity is ω, is
Correct answer: (2).
Solution:
For a rotating charged rod, magnetic moment dM = ωr22 dq.
With the axis l/3 from the midpoint, integrate r2 from -5l/6 to l/6: ∫r2dr = 7l336.
Therefore M = qω2l × 7l336 = 7qωl272.
Therefore, the correct option is (2).
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Question 4+4 / -1
Question 4. The ratio of energy density of magnetic field at centre of a current carrying loop to that at a distance R/√2 from centre of loop on its axis is (R: radius of loop).
Correct answer: (3).
Solution:
Bcentre = μ0i2R.
At x = R/√2, Baxis = μ0iR22(R2 + x2)3/2 = μ0i√27 R.
Energy density is proportional to B2, so the required ratio is 27 : 8.
Therefore, the correct option is (3).
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Question 5+4 / -1
Question 5. In which condition can wire AB perform simple harmonic motion?
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Correct answer: (3).
Solution:
For SHM, the restoring magnetic force must be proportional to displacement and directed opposite to the displacement.
In arrangement (3), F ∝ -x for wire AB, so it can perform SHM.
Therefore, the correct option is (3).
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Question 6+4 / -1
Question 6. Two charged particles a and b of same mass are accelerated through the same potential difference. On entering a region of uniform magnetic field, they describe circular paths of radii R1 and R2 respectively. The ratio of charges of a and b is
Correct answer: (1).
Solution:
After acceleration through the same potential, qV = mv2/2.
In the magnetic field, R = mv/(qB), so R ∝ 1/√q for the same m, V and B.
Hence qaqb = (R2R1)2.
Therefore, the correct option is (1).
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Question 7+4 / -1
Question 7. Two parallel, long wires carry currents i1 and i2 with i1 > i2. When the currents are in same direction, the magnetic field at a point midway between the wires is 40 μT. If the direction of i1 is reversed, the field becomes 60 μT. The ratio i1 : i2 is
Correct answer: (3).
Solution:
At the midpoint, same direction currents give opposite magnetic fields, so the field is proportional to i1 - i2.
After reversing i1, the fields add, so i1 - i2i1 + i2 = 4060 = 23.
Solving gives i1 = 5i2, so the ratio is 5 : 1.
Therefore, the correct option is (3).
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Question 8+4 / -1
Question 8. A wire carrying current i is kept in the x-y plane along the curve y = A sin(2πx/λ). A uniform magnetic field B exists in the Z-direction. The magnitude of magnetic force on the portion of the wire between x = 0 and x = λ/2 is
Correct answer: (4).
Solution:
For a wire in uniform magnetic field, the net force on a segment is i times the vector joining its endpoints crossed with B.
From x = 0 to x = λ/2, the endpoint displacement has magnitude λ/2 along x.
Therefore F = iB × λ2 = Biλ2.
Therefore, the correct option is (4).
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Question 9+4 / -1
Question 9. A circuit is composed of ten identical batteries and a resistor R = 10 Ω. Each battery has an emf of 2 V and internal resistance 0.1 Ω. The voltage difference across the resistor R is
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Correct answer: (4).
Solution:
In the shown arrangement, equal cells oppose in such a way that the net equivalent emf across R is zero.
Therefore no current is driven through R and VR = 0.
Therefore, the correct option is (4).
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Question 10+4 / -1
Question 10. Infinite number of straight wires each carrying current i are equally placed as shown in the figure. Adjacent wires have current in opposite direction. Net magnetic field at point P is
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Correct answer: (4).
Solution:
Each wire contributes a field proportional to 1/distance, with alternating signs because adjacent currents are opposite.
The resulting series is μ0i2πa (1 - 1/2 + 1/3 - 1/4 + ...).
Since 1 - 1/2 + 1/3 - 1/4 + ... = ln(2), B = μ0i2πa ln(2).
Therefore, the correct option is (4).
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Question 11+4 / -1
Question 11. Consider the following two statements: (a) For any rigid body of mass m and total charge q, rotating about an axis, the ratio of magnitude of magnetic moment to that of angular momentum is always q/2m. (b) Force experienced by a current carrying closed loop in a uniform magnetic field is zero. The correct statement(s) is/are
Correct answer: (2).
Solution:
Statement (a) is true only when mass and charge distributions are similar, not for any arbitrary rigid body.
Statement (b) is true: the net force on a closed current loop in a uniform magnetic field is zero.
Therefore, the correct option is (2).
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Question 12+4 / -1
Question 12. A long hollow cylinder carrying uniform current per unit length λ along the circumference is shown. Magnetic field inside the cylinder is
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Correct answer: (2).
Solution:
The current distribution is equivalent to a long solenoid with turns-current density λ.
The magnetic field inside is B = μ0λ.
Therefore, the correct option is (2).
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Question 13+4 / -1
Question 13. A particle of charge 0.5 C and mass 1 kg is projected from the origin with velocity v = (2i + 2j) m/s in magnetic field (4i) T. The coordinates of the particle at time t = π s are
Correct answer: (1).
Solution:
The component parallel to the magnetic field, 2 m/s along x, remains unchanged.
The perpendicular motion has period T = 2πmqB = 2π0.5 × 4 = π s.
At t = π s, the perpendicular part returns to its initial transverse position, while x = 2π m.
Therefore, the correct option is (1).
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Question 14+4 / -1
Question 14. The resultant force on the current loop ABCD due to the long current carrying conductor wire is
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Correct answer: (4).
Solution:
Only the two vertical sides of the loop give unequal horizontal forces due to different distances from the long wire.
Using B = μ0I2πr and F = iLB, the net force is proportional to 12 cm - 112 cm.
Substitution of the shown currents and length gives F = 5 × 10-4 N.
Therefore, the correct option is (4).
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Question 15+4 / -1
Question 15. A charged particle enters a magnetic field at right angle to the magnetic field. The field exists for a length equal to 1.5 times the radius of circular path of the particle. The particle will be deviated from its path by an angle
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Correct answer: (4).
Solution:
The field width is 1.5R, which is greater than the radius R of the circular path.
The particle completes a semicircular turn inside the field and exits in the opposite direction.
Hence the deviation is 180°.
Therefore, the correct option is (4).
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Question 16+4 / -1
Question 16. Rank the magnitude of ∮B.dl for closed paths shown in the figure from the smallest to the largest.
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Correct answer: (1).
Solution:
By Ampere's law, ∮B.dl = μ0Ienclosed.
Path A encloses the least current, path B encloses more, and path C encloses the maximum current.
Therefore A < B < C.
Therefore, the correct option is (1).
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Question 17+4 / -1
Question 17. An electron enters the space between the plates of a charged capacitor as shown. The surface charge density on the plate is σ. Electric field intensity in the space between the plates is E. The uniform magnetic field also exists in that space perpendicular to the direction of E. The electron moves perpendicular to both E and B without any change in direction. The time taken by the electron to travel a distance l in the space is
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Correct answer: (3).
Solution:
For no deflection, electric and magnetic forces balance: qE = qvB.
Between the plates, E = σ/ε0, so v = σ/(ε0B).
Time t = lv = ε0lBσ.
Therefore, the correct option is (3).
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Question 18+4 / -1
Question 18. A cylindrical cavity of diameter 2 m exists inside a cylinder of diameter 4 m as shown in the figure. A uniform current density 2 A/m2 exists along the length. If the magnitude of magnetic field at centre of cavity is
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Correct answer: (3).
Solution:
Use superposition: a full current-carrying cylinder plus a negative-current cylinder representing the cavity.
The field at the cavity centre is B = μ0Jd/2, where d is the separation of the two cylinder axes.
Here J = 2 A/m2 and d = 1 m, so B = μ0 = 4π × 10-7 T.
Therefore, the correct option is (3).
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Question 19+4 / -1
Question 19. The magnitude of magnetic field required to accelerate protons (mass = 1.67 × 10-27 kg) in a cyclotron that is operated at an oscillator frequency 6 MHz is approximately
Correct answer: (3).
Solution:
Cyclotron frequency is f = qB2πm, so B = 2πmfq.
Substituting proton mass, charge and f = 6 MHz gives B ≈ 0.4 T.
Therefore, the correct option is (3).
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Question 20+4 / -1
Question 20. In the figure shown, the power generated in y is maximum if y = 8 Ω. The value of resistance of resistor marked as R is
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Correct answer: (1).
Solution:
For maximum power transfer to y, y must equal the Thevenin resistance seen across its terminals.
Seen from y, the resistance is R + 2 Ω.
Given y = 8 Ω, R + 2 = 8, hence R = 6 Ω.
Therefore, the correct option is (1).
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Question 21+4 / -1
Question 21. The amount of heat generated in a resistor of resistance 2 Ω, in a time interval of 2 s, if current in it increases uniformly from zero to 4 A in the 2 s, is
Correct answer: (4).
Solution:
Since current rises uniformly from 0 to 4 A in 2 s, i = 2t.
H = ∫i2R dt = ∫02(2t)2(2)dt = 8 × 83 = 643 J.
Therefore, the correct option is (4).
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Question 22+4 / -1
Question 22. In the part of a circuit as shown in the figure, the ratio of power dissipated in 2 Ω and 12 Ω is
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Correct answer: (1).
Solution:
The 2 Ω and 4 Ω resistors are in parallel, so current divides inversely to resistance.
If the current through 4 Ω is i, then current through 2 Ω is 2i and the current through 12 Ω is 3i.
Question 23. All the bulbs shown in the figure are identical and are rated to fuse if the voltage across the bulb exceeds 200 V. The bulb(s) which will fuse in the circuit on closing the switch is
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Correct answer: (4).
Solution:
On closing the switch, the shown connections short-circuit the bulb network.
The potential difference across each bulb is zero, so no bulb fuses.
Therefore, the correct option is (4).
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Question 24+4 / -1
Question 24. A thin uniform rod is connected to a source of constant voltage supply across its ends. The heat dissipated in time t in this case is H. Now, the rod is cut into n equal parts (perpendicular to its length) and all parts are connected to the same power supply in parallel. The heat generated in second case in the same time will be
Correct answer: (1).
Solution:
Cutting the rod into n equal parts makes each part resistance R/n.
Putting n such parts in parallel gives equivalent resistance R/n2.
At the same voltage, heat in the same time is proportional to 1/R, so H' = n2H.
Therefore, the correct option is (1).
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Question 25+4 / -1
Question 25. A part of a circuit is shown in figure. The potential difference between B and C (VB - VC) is equal to 12 V. The value of current through the 4 Ω resistor is
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Correct answer: (4).
Solution:
Let the current through the 2 Ω branch from C to B be x. From the right branch, VB - VC = 10 - 2x.
Given VB - VC = 12, so x = -1 A; the current actually flows from B to C.
Combining with the 6 A current at the junction gives current through 4 Ω = 6 + 1 = 7 A.
Therefore, the correct option is (4).
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Question 26+4 / -1
Question 26. A potentiometer circuit arrangement is as shown in the figure. The potentiometer wire is 6 m long and having resistance 15r. At what distance from point A should the jockey touch the wire to get zero deflection in the galvanometer?
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Correct answer: (2).
Solution:
Current in the primary circuit is i = Er + 15r = E16r.
Potential drop across the 6 m potentiometer wire is i(15r) = 15E16, so potential gradient k = 15E96.
For balance against E/2, E2 = k l, hence l = 9630 = 3.2 m.
Therefore, the correct option is (2).
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Question 27+4 / -1
Question 27. A wire is stretched such that the length increases by 0.01%. The percentage change in the resistance is
Correct answer: (1).
Solution:
For stretching at constant volume, R = ρL/A and A ∝ 1/L, so R ∝ L2.
Therefore percentage change in R is twice the percentage change in L.
So ΔR/R = 0.02%.
Therefore, the correct option is (1).
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Question 28+4 / -1
Question 28. Six hundred identical cells of emf E and internal resistance r are connected in series in same polarity. The potential difference across any 99 cells will be
Correct answer: (1).
Solution:
In the closed series arrangement, each identical cell has terminal potential difference E - ir.
The current is such that E - ir = 0 for each cell, so the potential difference across any group of cells is zero.
Therefore, the correct option is (1).
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Question 29+4 / -1
Question 29. The equivalent resistance between A and B in the following circuit diagram is
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Correct answer: (2).
Solution:
Reduce the network step by step: the two right-side 8 Ω resistors combine in parallel to 4 Ω where applicable, followed by series and parallel combinations.
The net simplification gives 6 Ω in series with 8 Ω.
Hence RAB = 14 Ω.
Therefore, the correct option is (2).
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Question 30+4 / -1
Question 30. Uniform magnetic field of 2 T exists in a cylindrical region of radius 5 cm. The direction parallel to axis of cylinder is along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. Initially wire intersects the axis. If the wire in N-S direction is lowered from the axis by a distance 3 cm, then the magnitude and direction of the force on the wire is
Correct answer: (4).
Solution:
After lowering by 3 cm, the length of wire inside the circular field region is the chord 2√(52 - 32) cm = 8 cm.
Force F = iLB = 7 × 0.08 × 2 = 1.12 N.
Using Fleming's left-hand rule for current north to south and field east to west, the force is downward.
Therefore, the correct option is (4).
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Question 31+4 / -1
Question 31. A battery of emf E and internal resistance r is connected across a resistance R. Resistance R can be adjusted to any value greater than or equal to zero. A graph is plotted between the current (i) passing through the battery and potential difference (V) across it. Select the correct alternative.
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Correct answer: (1).
Solution:
The terminal voltage relation is V = E - ir, a straight line.
From the graph, at i = 0, V = E = 10 V. At V = 0, i = 2 A.
Thus r = Eimax = 102 = 5 Ω.
Therefore, the correct option is (1).
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Question 32+4 / -1
Question 32. The current density in a metallic conductor is plotted against electric field in a conductor at two different temperatures T1 and T2. Choose the correct statement.
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Correct answer: (2).
Solution:
The slope of the current-density versus electric-field graph is conductivity σ.
For a metal, conductivity decreases as temperature increases.
The T1 line has larger slope, so T1 is lower and T2 > T1.
Therefore, the correct option is (2).
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Question 33+4 / -1
Question 33. A metallic conductor of irregular cross-section is as shown in the figure. A constant potential difference is applied across ends A and B. Then,
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Correct answer: (3).
Solution:
In steady state, the same current crosses every section of a series conductor.
Since i = σEA, the wider cross-section P has smaller electric field than the narrower section Q.
Both statements are correct.
Therefore, the correct option is (3).
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Question 34+4 / -1
Question 34. A hemispherical network of radius a is made up using conducting wires of resistance per unit length λ. The equivalent resistance between points A and B is
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Correct answer: (1).
Solution:
By symmetry the hemispherical wire network reduces to four identical effective branches between A and B.
The equivalent reduction gives R = (2 + π)λa/8.
Therefore, the correct option is (1).
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Question 35+4 / -1
Question 35. Choose the correct statement(s).
Correct answer: (4).
Solution:
A low-voltage high-current source must have low internal resistance.
A high-tension supply should have large internal resistance to limit dangerous short-circuit current.
Ohm's law is not universal for all conducting elements, so statements (1) and (2) are correct.
Therefore, the correct option is (4).
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Question 36+4 / -1
Question 36. The given Wheatstone bridge is showing no deflection in the galvanometer joined between the points A and B. The value of resistance R as shown in the circuit is
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Correct answer: (2).
Solution:
For a balanced bridge, the ratio of upper to lower arm on the left equals the same ratio on the right.
The right upper arm is R in parallel with 200 Ω. Therefore R || 200400 = 100400.
So R || 200 = 100, which gives R = 200 Ω.
Therefore, the correct option is (2).
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Question 37+4 / -1
Question 37. Three identical circular loops each of radius a and with their centres at origin carry current i each. The planes of all three loops are perpendicular to each other. The net magnetic field at the centre origin will be
Correct answer: (2).
Solution:
Each loop produces field B = μ0i2a at the centre.
The three fields are mutually perpendicular, so the resultant is √3 times one field.
Bnet = √3 μ0i2a.
Therefore, the correct option is (2).
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Question 38+4 / -1
Question 38. A charged particle having charge 1 C is moving with velocity (2i + j) m/s. The magnetic field due to the charged particle at origin, when the charged particle is at (1 m, 1 m), is
Correct answer: (4).
Solution:
At the origin, r from charge to field point is -i - j.
v × r = (2i + j) × (-i - j) = -k.
With r = √2, B = μ04πq(v × r)r3 = -10-723/2k T.
Therefore, the correct option is (4).
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Question 39+4 / -1
Question 39. A small block of mass m and charge q is pulled (from rest) by horizontal force F on a smooth horizontal ground as shown in the figure. A uniform magnetic field B, directed into the plane exists in the region. The normal force between ground and the block would become zero at time
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Correct answer: (2).
Solution:
The block starts from rest, so horizontal acceleration is F/m and v = Ft/m.
The magnetic force qvB grows upward until it balances mg.
Set qB(Ft/m) = mg, giving t = m2gqBF.
Therefore, the correct option is (2).
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Question 40+4 / -1
Question 40. A circular loop of radius a = 1 m is placed in a uniform magnetic field B of induction 6 T. The loop carries a current 2 A. The tension developed in the loop will be (The magnetic field is perpendicular to the plane of the circular loop).
Correct answer: (4).
Solution:
For a circular current loop in a perpendicular magnetic field, tension is T = Bia.
Here T = 6 × 2 × 1 = 12 N.
Therefore, the correct option is (4).
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Question 41+4 / -1
Question 41. The average temperature coefficient of resistance of a wire is 0.00125°C-1. At 300 K, the resistance is 4 Ω. The temperature at which the resistance of the wire will be 4.5 Ω is
Correct answer: (4).
Solution:
Use R = R0[1 + α(T - T0)].
4.5 = 4[1 + 0.00125(T - 300)], so T - 300 = 100 K.
Thus T = 400 K = 127°C.
Therefore, the correct option is (4).
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Question 42+4 / -1
Question 42. Two identical heaters, when used individually take 5 minutes to boil a certain amount of water. The time taken to boil the water, when both heaters are used together in parallel, is
Correct answer: (3).
Solution:
Two identical heaters in parallel draw double the power at the same voltage.
For the same heat required, time becomes half of 5 minutes.
Therefore t = 2.5 minutes.
Therefore, the correct option is (3).
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Question 43+4 / -1
Question 43. The potential of point P in the given circuit diagram is
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Correct answer: (2).
Solution:
Let the potential of point P be x volt.
Applying KCL at P for the three branches gives (10 - x)/2 + (5 - x)/2 + (10 - x)/4 = 0 after using the indicated source polarities.
Solving gives x = 8 V.
Therefore, the correct option is (2).
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Question 44+4 / -1
Question 44. If σ1, σ2 and σ3 are conductances of three rods of identical shape and size, joined in series as shown in the figure. The equivalent conductance is
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Correct answer: (3).
Solution:
The rods are identical in shape and size, so each segment length is one-third of the total series length.
Series resistance adds; converting back to conductance gives the harmonic-type equivalent.
Question 45. For a metallic conductor, choose the correct graph.
(1)
(Ω)Resistance1 / Area of cross section(m-2)
(2)
(A)CurrentDrift speed(cm/s)
(3)
(cm/s)Drift speedArea of cross section(m2)
(4)
(cm/s)Drift speedElectric field(V/m)
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Correct answer: (4).
Solution:
For a metallic conductor, R = ρl/A, i = neAvd, and vd = eEτ/m.
The source snapshot supplied for this question is cut off after graph (3), so the graph set has been redrawn in typed HTML while preserving the original answer key.
Since vd is directly proportional to electric field E, the correct graph is graph (4).
Therefore, the correct option is (4).
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