Physics is where JEE ranks are usually won or lost, since a single conceptual slip changes the whole answer. Questions here span mechanics, rotational motion, thermodynamics, waves and oscillations, electrostatics, current electricity, magnetism, optics, and modern physics. Numerical problems carry full derivations so you can see which step you skipped, not just which option was right.
For a point charge Q at the origin, if the electric potential at distance r is V(r), what is the electric field magnitude at that point?
Answer: A
The electric field is related to potential by E = -dV/dr (negative gradient of potential). The negative sign indicates field points toward lower potential.
Q.282Hard
Three point charges are arranged at the vertices of an equilateral triangle of side a. If charges are +q, +q, and -2q, what is the net electric potential at the centroid?
Answer: A
Distance from each vertex to centroid is a/√3. V = k(q + q - 2q)/(a/√3) = 0. The charges sum to zero, giving zero potential.
Q.283Hard
A charge Q is uniformly distributed on a ring of radius R. What is the electric potential at a point on the axis at distance x from the center?
Answer: A
All charge elements on the ring are equidistant from the axial point. Distance = √(R² + x²), so V = kQ/√(R² + x²).
Q.284Hard
Consider a uniformly charged disc of radius R with total charge Q. What is the electric field at the center of the disc?
Answer: B
For a uniformly charged disc, the field at the center involves integrating contributions from rings. Result: E = σ/(2ε₀) = Q/(2πε₀R²).
Q.285Medium
A parallel plate capacitor is filled with a dielectric of dielectric constant κ. How does this affect the capacitance compared to vacuum?
Answer: A
Introducing a dielectric increases capacitance by factor κ: C = κε₀A/d = κC₀. This is a fundamental property used in capacitor design.
Q.286Hard
Two point charges q₁ = 2 μC and q₂ = -2 μC are separated by 1 cm. What is the magnitude of electric field at the midpoint between them?
Answer: A
At midpoint, distance from each charge = 0.5 cm = 0.005 m. Both fields point in same direction (from +q toward -q). E_total = 2 × k × 2×10⁻⁶ / (0.005)² = 7.2 × 10⁷ V/m.
Q.287Hard
A spherical conductor of radius R is grounded and placed near an isolated point charge +Q at distance d from its center (d > R). Which statement is correct about the induced charge on the sphere?
Answer: A
The grounded sphere develops negative charge to maintain V = 0. The charge distribution is non-uniform because the near side accumulates more negative charge.
Q.288Medium
A uniformly charged infinite line with linear charge density λ = 2 × 10⁻⁸ C/m is placed along the z-axis. A point charge q = +1 μC is located at a perpendicular distance r = 0.1 m from the line. The electric field due to the line charge at the location of the point charge is perpendicular to the line. If the permittivity of free space is ε₀ = 8.85 × 10⁻¹² F/m, calculate the magnitude of the electric field at the point charge location.
Answer: A
For an infinite line charge, E = λ/(2πε₀r). Substituting: E = (2 × 10⁻⁸)/(2π × 8.85 × 10⁻¹² × 0.1) = (2 × 10⁻⁸)/(5.57 × 10⁻¹²) ≈ 3.6 × 10³ N/C
Q.289Medium
Two identical conducting spheres A and B have charges +Q and +3Q respectively. They are separated by a distance much larger than their radii. When brought into contact and then separated to the original distance, the electrostatic force between them changes by a factor of:
Answer: C
Initial force: F₁ = k(Q)(3Q)/r² = 3kQ²/r². When spheres touch, total charge = 4Q, distributed as 2Q each. Final force: F₂ = k(2Q)(2Q)/r² = 4kQ²/r². Ratio: F₂/F₁ = (4kQ²/r²)/(3kQ²/r²) = 34. The force increases by factor 34, or changes by 34 times initial. However, comparing initial to final: change factor = F₂/F₁ = 34. The force becomes (34) times, meaning it changed by multiplying with 34. If asking reduction: Answer is 32 represents the comparative analysis in different context, but correct ratio of final to initial is 34.
Q.290Easy
The SI unit of electrical conductivity is:
Answer: B
Conductivity σ = 1/ρ where ρ is resistivity. Since resistivity is in Ω·m, conductivity is in (Ω·m)⁻¹ = S/m or mho/meter.
Q.291Easy
A copper wire and an aluminum wire of the same length and cross-sectional area carry the same current. Which wire has greater power dissipation?
Answer: B
Resistivity of aluminum (2.65 × 10⁻⁸ Ω·m) is greater than copper (1.68 × 10⁻⁸ Ω·m). Since P = I²R and R = ρL/A, aluminum has higher resistance and thus higher power dissipation for same current.
Q.292Medium
In a balanced Wheatstone bridge, if the resistance of one arm is changed by 10%, by what percentage should another arm be changed to maintain balance?
Answer: A
For Wheatstone bridge: P/Q = R/S. If P increases by 10%, then Q must also increase by 10% to maintain the ratio and balance condition.
Q.293Easy
The drift velocity of electrons in a copper wire is approximately:
Answer: B
Drift velocity vd = I/(nAe) is typically 10⁻⁴ to 10⁻³ m/s for normal currents, much slower than thermal velocity (~10⁶ m/s) but much slower than light speed.
Q.294Easy
Two resistances R₁ and R₂ are connected in parallel. If R₁ = 4Ω and R₂ = 6Ω, what is the equivalent resistance?
A metallic conductor's resistance increases with temperature. This is primarily because:
Answer: B
As temperature increases, atoms vibrate more vigorously, increasing collision frequency with drifting electrons. This increases mean free path reduction and thus resistance increases.
Q.300Medium
A heating element rated 1000W, 200V is connected to a 100V supply. The heat generated will be:
Answer: A
Resistance R = V²/P = (200)²/1000 = 40Ω (constant). Heat at 100V: P = V²/R = (100)²/40 = 4010000 = 250W.
