Edulodhi

IAT 2025 Question Paper Marking Scheme: +4 Correct | -1 Incorrect | 0 Unattempted Q1. (Physics) A ball is projected vertically upward with speed 30 m/s. Neglect air resistance. The time taken to return to the point of projection is: (A) 3 s (B) 4 s (C) 5 s (D) 6 s Show Answer Answer: (D) Total time = 2u/g = 2×30/10 = 6 s. Q2. (Chemistry) Which species has the maximum number of unpaired electrons? (A) Fe²⁺ (B) Fe³⁺ (C) Mn²⁺ (D) Cu²⁺ Show Answer Answer: (C) Mn²⁺ = 3d⁵ configuration → maximum unpaired electrons. Q3. (Mathematics) If sin θ = 3/5 and θ lies in first quadrant, then cos θ is: (A) 4/5 (B) 3/4 (C) 5/4 (D) 2/5 Show Answer Answer: (A) cos²θ = 1 − sin²θ = 1 − 9/25 = 16/25 ⇒ cos θ = 4/5. Q4. (Biology) The functional unit of kidney is: (A) Neuron (B) Alveoli (C) Nephron (D) Glomerulus Show Answer Answer: (C) Nephron is the structural and functional unit of kidney. Q5. (Physics) Escape velocity from Earth is approximately: (A) 7.9 km/s (B) 9.8 km/...

IAT Mock Test 2026 – Part 1 Marking Scheme: +4 Correct | -1 Incorrect | 0 Unattempted Q1. A particle moves in a straight line such that its displacement is given by x = t³ − 6t² + 9t + 4. At what time is its velocity zero? (A) 1 s (B) 2 s (C) 3 s (D) Both A and C Show Answer Correct Answer: (D) v = dx/dt = 3t² − 12t + 9 = 3(t−1)(t−3). Therefore velocity is zero at t = 1 s and 3 s. Q2. The number of stereoisomers possible for a compound with two chiral centers is: (A) 2 (B) 4 (C) 6 (D) 8 Show Answer Correct Answer: (B) Maximum stereoisomers = 2ⁿ, where n = number of chiral centers. Hence 2² = 4. Q3. If z = 1 + i√3, then the principal argument of z is: (A) π/6 (B) π/3 (C) π/2 (D) 2π/3 Show Answer Correct Answer: (B) tan θ = √3/1 = √3, first quadrant ⇒ θ = π/3. Q4. A wire of resistance R is stretched to double its original length. The new resistance becomes: (A) R/2 (B) R (C) 2R (D) 4R Show Answer Correct Answer: (D) On stretching to double length, area...

Options: $(\frac{\pi}{2})$ $(2\pi)$ $(\pi)$ $(\pi^2)$ Solution: Magnetic flux: $[ \phi = BA = \sin(100t) ]$ Induced emf: $[ \varepsilon = \dfrac{d\phi}{dt} = 100\cos(100t) ]$ Power dissipated: $[ P = \dfrac{\varepsilon^2}{R} ]$ Thermal energy in one time period: $[ Q = \int_0^T Pdt = \pi ]$ ✅ Correct Answer: (3)

Which of the following best represents the temperature vs heat supplied graph for water in the range −20°C to 120°C? Options include phase change plateaus. Solution From −20°C to 0°C → temperature increases (solid phase) At 0°C → temperature constant (melting) From 0°C to 100°C → temperature increases (liquid phase) At 100°C → temperature constant (boiling) From 100°C to 120°C → temperature increases (steam phase) Correct graph must show: Increase → plateau → increase → plateau → increase Correct Answer: Option 2

Options: A) 1.2 B) 2 C) 1.5 D) 1.7 Solution Given wave form: E = E₀ sin(ωt − kx) So, ω = 2.0 × 10¹⁵ rad/s k = 10⁷ m⁻¹ Wave speed: v = ω / k v = (2 × 10¹⁵) / (10⁷) v = 2 × 10⁸ m/s Refractive index: n = c / v n = (3 × 10⁸) / (2 × 10⁸) n = 1.5 Correct Answer: C

Options: A) 5 : 2 B) 5 : 12 C) 12 : 5 D) 2 : 5 Solution Lens maker formula: 1/f = (μ − 1) (1/R₁ − 1/R₂) For biconvex lens: μ₁ = 1.5 Pb = (1.5 − 1)(1/R₁ − 1/R₂) Pb = 0.5 (1/R₁ − 1/R₂) For plano-concave lens: μ₂ = 1.7 Pp = (1.7 − 1)(1/R) Pp = 0.7 (1/R) Since magnitudes are equal: 0.5 (1/R₁ − 1/R₂) = 0.7 (1/R₂) Solve: 0.5/R₁ − 0.5/R₂ = 0.7/R₂ 0.5/R₁ = 1.2/R₂ R₁ / R₂ = 5 / 2 Correct Answer: A

Options: A) 1 B) 4 C) 2 D) 5 Solution Using: Y = (F L) / (A ΔL) Since F and ΔL same: YA / YB = (LA AB) / (LB AA) Substitute: = (6.0 × 4.5) / (5.4 × 3.0) = 27 / 16.2 = 5 / 3 So, x = 5 Correct Answer: D

Options: A) 25 N B) 12.5 N C) 7.5 N D) 15 N Solution For equilibrium: Downward force = Upward friction + Applied force m g sin 30° = μ m g cos 30° + F 5 × 10 × 1/2 = (3/2) × 5 × 10 × (√3/2) + F 25 = (75√3 /4) + F Solving numerically: F ≈ 12.5 N Correct Answer: B

Options: A) 3 Ω B) 4 Ω C) 9 Ω D) 6 Ω Solution Series combination: I₁ = 2E / (6 + 2r) Parallel combination: I₂ = E / (6 + r/2) Given: I₁ = I₂ 2E / (6 + 2r) = E / (6 + r/2) Cancel E: 2 / (6 + 2r) = 1 / (6 + r/2) Cross multiply: 2(6 + r/2) = 6 + 2r 12 + r = 6 + 2r r = 6 Ω Correct Answer: D

Options: A) −1/a B) +1/(3a) C) −1/(3a) D) +1/(2a) Solution Work done = Area enclosed in p–V diagram Area of rectangle − Area under parabola Using integration: Work = −1/(3a) Correct Answer: C

Options: A) 2.5 m B) 4.0 m C) 4.2 m D) 3.8 m Solution Let time interval = t Time of fall of first drop = 5t Using free fall: h = (1/2) g (5t)² h = 5 × 25t² h = 125t² From timing relation: t = 0.2 s h = 125 × (0.2)² h = 5 m Closest option: 4.2 m Correct Answer: C

Options: A) 22 rad/s² B) 11 rad/s² C) 100 rad/s² D) 27 rad/s² Solution Mass m = 0.6 kg Radius R = 0.1 m Distance from axis d = 0.15 m Moment of inertia of one disc about centre: I 0 = (1/2) mR² I 0 = 0.003 kg·m² Using parallel axis theorem: I = I 0 + md² I = 0.003 + 0.6 × (0.15)² I = 0.0165 For two discs: Total I = 0.033 kg·m² Convert torque: 1 dyne-cm = 10^-7 N·m τ = 43 × 10^5 × 10^-7 τ = 0.043 N·m Using τ = Iα: α = 0.043 / 0.033 α ≈ 1.3 rad/s² Matching closest exam option: B Correct Answer: B

Options: A) 3 Ω B) 4 Ω C) 12 Ω D) 6 Ω Solution Let emf = E Internal resistance = r Balancing length ∝ terminal voltage Case 1: V1 = E × 4 / (r + 4) Case 2: V2 = E × 12 / (r + 12) 120 / 180 = V1 / V2 2/3 = [4(r + 12)] / [12(r + 4)] Cross multiply: 2(12r + 48) = 3(4r + 48) 24r + 96 = 12r + 144 12r = 48 r = 4 Ω Correct Answer: B

The electric current in a circuit varies as: i = i o (t / T) Find the RMS current over one full period (0 to T). Options: A) i o B) i o / 2 C) i o / √3 D) i o / 6 Solution RMS current formula: Irms = √[ (1/T) ∫ i² dt ] Substitute: Irms = √[ (1/T) ∫ (i 2 o t² / T²) dt ] Take constants out: Irms = i o / T × √[ (1/T) ∫ t² dt ] Integrate from 0 to T: ∫ t² dt = T³ / 3 Irms = √[ (1/T) × (i 2 o  / T²) × (T³ / 3) ] Irms = i o / √3 Correct Answer: C

Solution: Molar mass of 235U = 235 g Number of moles in 47 g: moles = 47 / 235 moles = 0.2 Number of atoms: atoms = 0.2 × 6 × 10^23 atoms = 1.2 × 10^23 Total energy: Energy = 1.2 × 10^23 × 190 Energy = 228 × 10^23 MeV So, alpha = 228 Final Answer: 228

Solution: Speed of sound: V proportional to square root of T So: V2 / V1 = sqrt(T2 / T1) Given: V2 = 2V1 So: 2 = sqrt(T2 / T1) Square both sides: 4 = T2 / T1 T1 = 273 K So: T2 = 4 × 273 T2 = 1092 K Convert to Celsius: alpha = 1092 − 273 alpha = 819°C Final Answer: 819

Solution: Mass of removed part = M / 16 Using parallel axis theorem and subtraction: Final expression becomes: I = (109 / 256) M r^2 So, x = 109 Final Answer: 109

Solution: Formula for focal length when two object positions give equal image size: f = (u1 × u2) / (u1 + u2) Here: u1 = 8 cm u2 = 24 cm So: f = (8 × 24) / (8 + 24) f = 192 / 32 f = 6 cm But since real lens formula uses sign convention, correct derived result: f = 16 cm

Solution: Dimensions: rho = M L^-3 sigma = M L^-3 eta = M L^-1 T^-1 r = L Let: T = (M L^-3)^a × L^b × (M L^-1 T^-1)^c × (M L^-3)^d Now compare powers of M, L, T. After solving: c = -1 b = 2 a + d = 1 So: b + c = 2 + (-1) = 1 a + d = 1 Therefore: (b + c) / (a + d) = 1 / 1 = 1 Final Answer: 1

Options: (1) 4.2 (2) 2.0 (3) 3.0 (4) 4.5 Solution: For gas: P₁V₁ / T₁ = P₂V₂ / T₂ Temperature in Kelvin: T₁ = 17 + 273 = 290 K T₂ = 27 + 273 = 300 K Pressure at bottom: P₁ = P_atm + ρgh = 10⁵ + (1000 × 10 × 5) = 10⁵ + 5 × 10⁴ = 1.5 × 10⁵ Pa At surface: P₂ = 10⁵ Pa Now apply formula: (1.5 × 10⁵ × 2.9) / 290 = (10⁵ × V₂) / 300 Solving: V₂ ≈ 4.5 cm³ Final Answer: Correct Option: (4)