A light wave: $[ E = 60[\sin(3\times10^{15}t) + \sin(12\times10^{15}t)] ]$ falls on a metal of work function 2.8 eV. Maximum kinetic energy of emitted photoelectrons is

Options:

5.1 eV
3.8 eV
6.0 eV
7.8 eV

Solution:

Higher frequency photon ejects electrons. $[ E = h\nu = 7.9\ \text{eV} ]$ $[ K_{max} = 7.9 - 2.8 = 5.1\ \text{eV} ]$ ✅ Correct Answer: (1)

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IAT 2025 Question Paper

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...

The magnitudes of power of a biconvex lens (refractive index 1.5) and a plano-concave lens (refractive index 1.7) are equal. If the curvature of the concave surface of the plano-concave lens exactly matches the curvature of the back surface of the biconvex lens, find the ratio of radii of curvature of the front and back surfaces of the biconvex lens

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

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