neet physics tutor doubt 8
neet physics tutor doubt 8 is a concept-based NEET Physics practice set with instant answer checking, detailed solutions and score tracking.
neet physics tutor doubt 8 focuses on electric field, electric potential, capacitance, current electricity, circuits and graph interpretation.
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Question 1. A point charge (+ve) is brought near an isolated conducting sphere. The electric lines of field are best represented by
Solution:
For a conductor in electrostatic equilibrium, electric field lines meet the conducting surface normally. The induced charge pattern near the positive point charge is represented correctly only in figure a.
Therefore, the correct option is (1).
Question 2. Electric flux through the shown closed surfaces is
Solution:
By Gauss law, total electric flux through a closed surface is Φ = qenclosedε0.
Each surface encloses the same charge +q, so the flux does not depend on the shape or size of the surface.
Therefore, the correct option is (4).
Question 3. The dielectric constant k of an insulator cannot be
Solution:
An insulator has a finite dielectric constant. Infinite dielectric constant is associated with an ideal conductor, not an insulator.
Therefore, the correct option is (4).
Question 4. Two particles of masses m and 2m with charges 2q and q are placed in a uniform electric field E. The ratio of their accelerations will be
Solution:
Force on a charge in a uniform electric field is F = qE.
For the first particle, a1 = 2qEm. For the second particle, a2 = qE2m.
Thus a1a2 = 2qE/mqE/2m = 4.
Therefore, the correct option is (1).
Question 5. Three point charges each +q are placed at the corners of an equilateral triangle. The electric field at the centre will be
Solution:
The centre is symmetrically placed with respect to all three equal charges. The three electric field vectors have equal magnitudes and are separated by 120°.
Their vector sum is zero.
Therefore, the correct option is (4).
Question 6. A particle of mass m and charge q is placed at rest in a uniform electric field E. If gravity is neglected, its path will be
Solution:
The electric force qE is constant and acts along the field direction. Since the particle starts from rest, it accelerates along a single straight line.
Therefore, the correct option is (3).
Question 7. Eight dipoles of charges of magnitude q are placed inside a cube. The total electric flux coming out of the cube will be
Solution:
Each dipole has equal and opposite charges, so the net charge enclosed by the cube is zero.
By Gauss law, Φ = qnetε0 = 0.
Therefore, the correct option is (4).
Question 8. The flux entering and leaving a closed surface are 5 × 105 and 4 × 105 SI unit respectively. The charge inside the surface will be
Solution:
Taking outward flux as positive, Φnet = Φout - Φin = 4 × 105 - 5 × 105 = -1 × 105.
q = ε0Φ = (8.85 × 10-12)(-1 × 105) = -8.85 × 10-7 C.
Therefore, the correct option is (1).
Question 9. The points representing equal potentials in the given uniform electric field are
Solution:
In a uniform electric field, equipotential lines are perpendicular to the field lines.
S and R lie on the same equipotential line.
Therefore, the correct option is (3).
Question 10. If work done by the field in moving an electron from A to B is 6.4 × 10-19 J, then the potential difference VB - VA is
Solution:
Using the supplied answer convention, VB - VA = We.
VB - VA = 6.4 × 10-191.6 × 10-19 = 4 V.
Therefore, the correct option is (2).
Question 11. The radii of two spheres are a and b respectively. They are at equal electric potential. The ratio of their surface charge densities is
Solution:
For a sphere, σ = q4πR2 and V = 14πε0 qR.
Equal potentials give q1a = q2b, so q1q2 = ab.
Therefore σ1σ2 = q1q2 b2a2 = ba.
Therefore, the correct option is (2).
Question 12. Potential at a point x distance from the centre inside the conducting sphere of radius R and charged with Q is
Solution:
Inside a conducting sphere, potential is constant and equal to the surface potential.
Thus V = kQR.
Therefore, the correct option is (3).
Question 13. The force of attraction between two point charges at distance d apart in a medium is F. What distance apart should these point charges be kept in the same medium, so that the force between them becomes 16F?
Solution:
Coulomb force varies as F ∝ 1d2.
If the new distance is d', then 16FF = d2d'2. Hence d' = d4.
Therefore, the correct option is (2).
Question 14. Electric potential at any point is V = (-5x + 3y + √15 z) volt, then magnitude of electric field is
Solution:
Electric field is E = -∇V = 5 î - 3 ĵ - √15 k̂.
|E| = √(25 + 9 + 15) = √49 = 7 Vm-1.
Therefore, the correct option is (4).
Question 15. Which of the following statement is incorrect?
Solution:
Electrostatic field lines between like charges repel and do not produce the stated lengthwise contraction between two similar charges.
Therefore, the correct option is (3).
Question 16. Two concentric spherical shells of radii a and b are shown in figure. The outer sphere is given a charge q. The charge q' on inner sphere will be
Solution:
The inner sphere is earthed, so its potential must be zero.
Potential at the inner sphere: kqb + kq'a = 0.
Hence q' = -qab.
Therefore, the correct option is (3).
Question 17. A proton is about 1840 times heavier than an electron. When it is accelerated from rest by a potential difference of 1 kV, its kinetic energy will become
Solution:
Kinetic energy gained by a charge accelerated through potential difference V is K = qV.
For a proton through 1 kV, K = 1 keV. It does not depend on mass.
Therefore, the correct option is (3).
Question 18. The velocity v acquired by an electron starting from rest and moving through a potential difference V is shown by
Solution:
From 12mv2 = eV, velocity v = √2eVm.
Therefore v is proportional to √V, giving the concave-down graph corresponding to option (4).
Therefore, the correct option is (4).
Question 19. If the electric field is given by 5î + 4ĵ + 9k̂, the electric flux through a surface of area 20 unit lying in y-z plane will be
Solution:
For a surface in the y-z plane, the area vector is along the x-axis: A = 20î.
Φ = E · A = (5î + 4ĵ + 9k̂) · 20î = 100 unit.
Therefore, the correct option is (1).
Question 20. Dipole moment of the charge distribution shown in figure is
Solution:
Taking the -2q charge at the origin, the two +q charges at (a, 0) and (0, a) give dipole components qa along x and qa along y.
Resultant dipole moment magnitude is √[(qa)2 + (qa)2] = √2qa.
Therefore, the correct option is (1).
Question 21. If C = 6 μF, find the charge stored in capacitor C1.
Solution:
The capacitors C and 2C are in series, so Ceq = C × 2CC + 2C = 2C3.
Charge in the series combination is q = CeqV = 2C3 × 10.
With C = 6 μF, q = 40 μC. The same series charge is stored on C1.
Therefore, the correct option is (3).
Question 22. Capacity of an isolated sphere A is C. When it is enclosed by an earthed concentric sphere B, capacity of the system is increased to nC. The ratio of radii of B to A is
Solution:
For inner radius a and outer earthed radius b, capacitance is C' = 4πε0ab/(b - a).
Since isolated sphere capacitance is C = 4πε0a and C' = nC, b/(b - a) = n.
Solving gives ba = nn - 1.
Therefore, the correct option is (2).
Question 23. A parallel plate capacitor with a dielectric slab of dielectric constant k between the plates has capacity C and is charged to a potential V volt and then isolated from battery. The dielectric slab is slowly removed from the capacitor and then reinserted. The net work done by the system in this process is
Solution:
The process starts and ends in the same physical state: same capacitor, same dielectric position, and same charge.
The net change in stored energy over the full remove-and-reinsert cycle is zero, so the net work done is zero.
Therefore, the correct option is (1).
Question 24. Six identical plates of area A are arranged as shown. The distance between each two plates is d, the net capacitance is
Solution:
Using the plate connections shown, the effective arrangement reduces to one equivalent plate-pair capacitance.
Thus Ceq = ε0Ad.
Therefore, the correct option is (1).
Question 25. A wire of resistance R is stretched till its radius is half of the original value. Then the new resistance is
Solution:
R = ρlA. When radius becomes half, area becomes one-fourth.
For the same volume, length becomes four times. Hence R becomes 4/(1/4) = 16 times.
Therefore, the correct option is (4).
Question 26. The value of current I in the following circuit is
Solution:
Apply Kirchhoff's first law at each junction.
Top branch current is 15 - 8 = 7 A. Bottom branch current is 8 - 5 = 3 A.
At the right side, 7 A + 3 A enters the upper right junction and combines with the 3 A branch, giving I = 13 A.
Therefore, the correct option is (2).
Question 27. A potential difference V is applied to a conductor of length l and radius r. When potential difference is doubled, the drift velocity is
Solution:
Drift velocity is proportional to electric field, and E = V/l.
For fixed length, doubling V doubles E and hence doubles the drift velocity.
Therefore, the correct option is (3).
Question 28. The tolerance level of a resistor with the colour code red, blue, orange, gold is
Solution:
In resistor colour coding, gold as the tolerance band means ±5%.
Therefore, the correct option is (1).
Question 29. Variation of current passing through a conductor as the voltage applied across its ends varies is shown in the curve. If the resistances are determined at the points A, B, C and D, then
Solution:
Resistance at any point on a V-I graph is R = V/I, represented by the slope of the line joining that point to the origin.
The construction shows points C and D on the same line from the origin, so their V/I values are equal.
Therefore, the correct option is (1).
Question 30. Two potentiometers have uniform potential gradient across them. Two cells connected in series (i) to support each other and (ii) to oppose each other are balanced over 6 m and 2 m respectively on the potentiometer wire. The emf of the cells are in the ratio
Solution:
With uniform potential gradient, balancing length is proportional to emf.
E1 + E2E1 - E2 = 62 = 3.
Solving gives E1E2 = 21.
Therefore, the correct option is (4).
Question 31. Two wires of resistance R1 and R2 have temperature coefficients of resistance α1 and α2 respectively. These are joined in series. The effective temperature coefficient is
Solution:
For series combination, total resistance change is ΔR = α1R1ΔT + α2R2ΔT.
Since R = R1 + R2, αeff = α1R1 + α2R2R1 + R2.
Therefore, the correct option is (3).
Question 32. The total current supplied to the circuit by the battery is
Solution:
Reducing the resistor network gives an equivalent resistance of 1.5 Ω across the 6 V battery.
Therefore I = VR = 61.5 = 4 A.
Therefore, the correct option is (3).
Question 33. Two cells with the same emf ε and different internal resistances r1 and r2 are connected in series with an external resistance R. The value of R so that the potential difference across the 1st cell be zero is
Solution:
Using the given series-cell condition and setting the terminal potential difference of the first cell to zero gives R = r1 - r2.
Therefore, the correct option is (3).
Question 34. In the circuit shown, the current through 8 Ω is same before and after connecting E. The value of E is
Solution:
Before connecting E, the current in the series circuit is 12/(6 + 8 + 10) = 0.5 A.
Potential difference across the 8 Ω resistor is 0.5 × 8 = 4 V.
For the current through 8 Ω to remain the same after connecting E, E must be 4 V.
Therefore, the correct option is (3).
Question 35. The potential difference between A and B in the following figure is
Solution:
Using Kirchhoff's second law along the branch from A to B:
VA - 12 - 12 - 18 + 4 - 10 = VB.
Hence VA - VB = 48 V.
Therefore, the correct option is (2).
Question 36. Potential of point D is
Solution:
At the isolated middle point D, the algebraic sum of charges on the connected plates is zero.
C1(VD - V1) + C2(VD - V2) = 0, so VD = C1V1 + C2V2C1 + C2.
Therefore, the correct option is (3).
Question 37. Find equivalent capacitance between X and Y nearly.
C1 = C2 = C3 = 400 pF and C4 = C5 = C6 = 200 pF.
Solution:
Reducing the shown capacitance network by the series-parallel combinations gives an equivalent capacitance close to 410 pF.
Therefore, the correct option is (4).
Question 38. The electric field at R = 5 m is
Solution:
Electric field is E = -dV/dR.
At R = 5 m, the graph is on the straight segment from R = 4 to 6 where slope dV/dR = (0 - 5)/(6 - 4) = -2.5 V/m.
Therefore E = +2.5 V/m.
Therefore, the correct option is (1).
Question 39. Work done in taking a unit positive charge from P to A is W1 and from P to B is W2, then
Solution:
A and B are symmetrically placed with the same potential difference from P in the shown geometry.
Work done depends only on potential difference, hence W1 = W2.
Therefore, the correct option is (3).
Question 40. For a cell, a graph between potential difference V across the terminals of cell and the current I is plotted. The emf and internal resistance of the cell E and r are
Solution:
The emf is the terminal potential when current is zero, so E = 2 V.
For a cell, V = E - Ir. The slope magnitude of the V-I graph is r = 2/5 = 0.4 Ω.
Therefore, the correct option is (2).
Question 41. Which of the following is correct?
Solution:
Between A and C, the equivalent resistance is 4 Ω || 6 Ω = 2.4 Ω.
Between C and B, the equivalent resistance is 4 Ω || 6 Ω || 8 Ω, which is smaller.
For the same total current path comparison, the potential drop across AC is greater than across CB.
Therefore, the correct option is (2).
Question 42. Find the current I shown in figure.
Solution:
Solving the bridge network with Kirchhoff's loop and junction equations gives I = -5/12 A.
The negative sign means actual current is opposite to the arrow shown in the figure.
Therefore, the correct option is (1).
Question 43. Electric field at perpendicular distance r from infinite uniform straight linear charge distribution of charge density λ is
Solution:
Using a cylindrical Gaussian surface around the line charge, E(2πrl) = λl/ε0.
Hence E = λ2πε0r.
Therefore, the correct option is (2).
Question 44. Electric field at axial point of short electric dipole is E. Electric field at a point at the same distance on the equatorial point will be
Solution:
For a short dipole at the same distance, the equatorial field has half the axial field magnitude and is in the opposite direction.
Thus Eequatorial = -E2.
Therefore, the correct option is (1).
Question 45. An insulated sphere of radius R has a uniform volume charge density ρ. The electric field at point P inside the sphere at a distance r from the centre is
Solution:
For radius r inside the sphere, enclosed charge is ρ × 43πr3.
By Gauss law, E(4πr2) = ρ(4/3)πr3ε0, so E = rρ3ε0.
Therefore, the correct option is (2).
NEET marking is applied automatically: +4 for each correct answer, -1 for each wrong answer and 0 for unattempted questions.
