In an experiment with photoelectric effect, the stopping potential.
Answer key (C) & worked solution
Correct answer: Option C
Updated 19 Oct 2026
JEE Main 2025, 29 Jan Shift 2 question paper with solutions
JEE Main 2025, 29 Jan Shift 2 was held on 29 January 2025. The bank holds 67 of its questions (Physics 22, Chemistry 24, Mathematics 21). A sample with answer keys and worked solutions is below, or sit the complete shift paper in the official NTA CBT interface (180 minutes, 300 marks, +4 / −1). Real PYQ papers are never filtered by your covered syllabus.
2025 Exam session year
300 Maximum marks (+4 / −1)
180 min Official duration
67 Questions in bank for this shift
Try questions from JEE Main 2025, 29 Jan Shift 2
The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have
Answer key (C) & worked solution
Correct answer: Option C
See-back effect Low thermal conductivity High electrical conductivity
Identify the essential amino acids from below : (A) Valine (B) Proline (C) Lysine (D) Threonine (E) Tyrosine Choose the correct answer from the options given below :
Answer key (A) & worked solution
Correct answer: Option A
Valine, Lysine and Threonine are essential amino acids.
Drug X becomes ineffective after 50% decomposition. The original concentration of drug in a bottle was 16 mg/mL which becomes 4 mg/mL in 12 months. The expiry time of the drug in months is _____. Assume that the decomposition of the drug follows first order kinetics.
Answer key (D) & worked solution
Correct answer: Option D
C = 16 mg/mLC = 4 mg/mLt = 12 monthsUsing the first-order kinetics equation: ln(C) = ln(C) - ktln(4) = ln(16) - k(12) k(12) = ln(16) - ln(4) = ln(16/4) = ln(4) k = ln(4)/12 ≈ 0.2888 months-1 t1/2 = 0.693/k
t1/2 = 0.693 / (ln(4)/12) = (0.693 * 12) / ln(4) ≈ 6 months
t1/2 = 0.693 / (ln(4)/12) = (0.693 * 12) / ln(4) ≈ 6 months
Let a straight line L pass through the point P(2,–1,3) and be perpendicular to the lines . If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is :
Answer key (C) & worked solution
Correct answer: Option C
Vector parallel to ‘L’ \left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k} \\
2&1&{ - 2} \\
1&3&4
\end{array}} \right| = 10\hat i - 10\hat j + 5\hat k Equation of ‘L’ Let Q(2λ + 2, –2λ – 1, λ + 3) ⇒ 2λ + 2 = 0 ⇒ λ = –1 ⇒ Q (0, 1, 2) d(P, Q) = 3
Let the function f(x) = (x – 1)|x – ax + 2| + cos|x| be not differentiable at the two points x = α = 2 and x = β. Then the distance of the point (α,β) from the line 12x + 5y + 10 = 0 is equal to :
Answer key (A) & worked solution
Correct answer: Option A
cos|x| is always differentiable ∴ we have to check only for |x – ax + 2| ∴ Not differentiable at x – ax + 2 = 0 One root is given, α = 2 ∴ 4 – 2a + 2 = 0 a = 3 ∴ other root β = 1 but for x = 1 f(x) is differentiable (Drop)