Current in voltmeter I = V_reading/R_meter = 2V/10,000Ω = 0.0002 A. The internal resistance of the supply is very high: r = (2000-2)/0.0002 ≈ 10 MΩ. The meter reads only 2V due to the huge internal resistance of the supply.

½mv² = mCΔT (KE = heat gained). v² = 2CΔT. Therefore v = √(2CΔT). The speed of the ball-bearing is the square root of twice the product of specific heat and temperature rise.

Fringe width β = λD/d. Red light has a longer wavelength than yellow light (red ≈ 700nm, yellow ≈ 580nm). Since fringe width is directly proportional to wavelength, replacing yellow with red neon light increases the fringe width.

When wire is stretched to twice its length, volume remains constant: L×A = 2L×A'. So A' = A/2. New resistance R' = ρ(2L)/(A/2) = 4ρL/A = 4R = 4×5 = 20Ω. Stretching a wire to twice its length quadruples its resistance.

Fundamental (base) quantities in the SI system are: length, mass, time, electric current, temperature, amount of substance, and luminous intensity. Weight is a force (= mg) — a derived quantity, not a fundamental one.

In Compton scattering, a photon transfers part of its energy and momentum to a free electron. The photon loses energy, and since E = hc/λ, a lower energy photon has a longer wavelength. The wavelength of scattered photons is always greater than the incident wavelength.

Using lens formula: when object is beyond 2F (u > 2f), image forms between F and 2F on the other side, and is real, inverted, and diminished. This is exactly the condition for the image to form between F and 2F.

A voltmeter must always be connected in parallel (across) the component or source whose voltage is to be measured. A voltmeter has very high resistance so it draws minimal current and doesn't significantly disturb the circuit.

In a series circuit, the same current flows through all components regardless of their individual resistances. All three wires — copper, silver, and nichrome — carry exactly the same current. Current is the same everywhere in a series circuit.

An equipotential surface is one where all points are at the same electric potential. Since W = q×ΔV and ΔV = 0 along an equipotential surface, the work done in moving any charge along an equipotential surface is always zero, regardless of path.

E = hc/λ = (6.63×10⁻³⁴ × 3×10⁸)/(1×10⁻⁶) ≈ 1.99×10⁻¹⁹ J. Work function φ = hf₀ = 6.63×10⁻³⁴ × 4×10¹⁴ ≈ 2.65×10⁻¹⁹ J. Since E < φ, no photoelectrons are emitted. KE = E - φ < 0, so none of the given options correctly describes the situation. Answer: None.

Refractive index (and hence deviation) is greatest for shortest wavelength. In visible light, blue/violet has the shortest wavelength, so it undergoes maximum deviation. Red light has the longest wavelength and minimum deviation. Hence blue gives the maximum angle of minimum deviation.

The Balmer series involves transitions to n=2. The red line (Hα, longest wavelength/lowest energy in Balmer series) is produced by electron transition from n=3 to n=2 (λ ≈ 656 nm). This is the first and lowest energy line in the Balmer series.

Time period of a simple pendulum T = 2π√(L/g). It depends only on length L and gravitational acceleration g — NOT on the mass of the bob. Doubling the mass has absolutely no effect on the period.

Centripetal force always acts perpendicular to the velocity (toward the centre of the circular path). Since W = F·d·cosθ and θ = 90°, W = 0. Centripetal force can never do work on a particle moving in circular motion.

g = GM/R². For new planet: M' = M/2, R' = R/2. g' = G(M/2)/(R/2)² = G(M/2)/(R²/4) = 4GM/(2R²) = 2GM/R² = 2g. The surface gravity on a planet with half the mass and half the radius of Earth is twice Earth's gravity.

Distance in nth second: Sn = u + a(2n-1)/2. For n=1: S₁ = 0 + a/2 = 20m → a = 40 m/s². For n=2: S₂ = 0 + 40(3)/2 = 60m. The distance covered in the 2nd second is 60m.

A Wheatstone bridge is balanced when P/Q = R/S, which means the potential at both ends of the galvanometer are equal — so the potential difference across the galvanometer is zero. No current flows through the galvanometer when the bridge is balanced.

Angular momentum L = r × p. Computing: L = î(2×(-2)-(-1)×4) - ĵ(1×(-2)-(-1)×3) + k̂(1×4-2×3) = î(-4+4) - ĵ(-2+3) + k̂(4-6) = 0î - ĵ - 2k̂. Since L has no x-component (L·î = 0), L is perpendicular to the x-axis.

Following the decay series from U-238 step by step: α → Th-234, β → Pa-234, β → U-234, α → Th-230, α → Ra-226, α → Rn-222, α → Po-218, β → At-218, β → Rn-218, α → Po-214. Tracing this specific series (α,β,β,α,α,α,α,β,β,α), Ra-226 (⁸⁸Ra²²⁶) is NOT produced in this particular sequence.

Coherent waves must have a constant phase difference and the same frequency (hence same wavelength). Same amplitude is not a requirement for coherence. Coherence is essential for stable interference patterns to form.

An astronomical refracting telescope suffers from chromatic aberration (lens disperses different colours differently), spherical aberration (rays from edge and centre focus at different points), and produces an inverted image. All three are true for the simple astronomical telescope.

Doppler formula (both moving toward each other): f' = f×(V + v_L)/(V - v_S) = f×(V + V/10)/(V - V/10) = f×(1.1V)/(0.9V) = f×11/9 ≈ 1.22f. The listener hears a frequency approximately 1.22 times the emitted frequency.

In SHM, x = A cos(ωt). At x=A: t=0. At x=A/2: cos(ωt) = 1/2 → ωt = π/3 → t = π/(3ω) = T/6. The time from x=A to x=A/2 is T/6 minus time from start, giving T/12 for this segment.

Position of nth bright fringe: y_bright = nλD/d. 5th bright: y = 5×10⁻⁷×1/10⁻³ = 5×10⁻⁴ m. Position of nth dark fringe: y_dark = (n-½)λD/d. 10th dark: y = 9.5×10⁻⁷×1/10⁻³ = 9.5×10⁻⁴ m. Separation = 9.5×10⁻⁴ - 5×10⁻⁴ = 4.5×10⁻⁴ m.