About this paper
This is the actual Maharashtra State Board (MSBSHSE) HSC (Class 12) Science Physics (54) board paper of the February–March 2026 examination (Day 06). The paper code printed on the question paper is J-137. The questions are transcribed in full from the board’s own printed question paper as published in the AglaSem Docs archive (‘MH 12th Question Paper March 2026 Physics’); figures and diagrams asked for are described in words. The model answers are original Doubtly answers, worked out and checked independently, not the board’s official answer key.
Paper pattern
Maximum marks 70, time 3 hours, four sections. Section A: Q.1 has ten MCQs (1 mark each) and Q.2 has eight very short answer questions (1 mark each) — 18 marks, all compulsory. Section B: Q.3 to Q.14, twelve 2-mark questions, attempt any eight (16 marks). Section C: Q.15 to Q.26, twelve 3-mark questions, attempt any eight (24 marks). Section D: Q.27 to Q.31, five 4-mark questions, attempt any three (12 marks). Log tables are allowed, calculators are not. The physical constants printed on the paper are \(h = 6.63\times10^{-34}\) J s, \(c = 3\times10^8\) m/s, \(\pi = 3.142\), \(g = 9.8\) m/s\(^2\), \(\varepsilon_0 = 8.85\times10^{-12}\) C\(^2\)/N m\(^2\) and \(\mu_0 = 4\pi\times10^{-7}\) Wb/A m.
How to use it
Attempt the paper in one 3-hour sitting without a calculator, choosing questions the way the board allows (any eight in B and C, any three in D). Then compare with the model answers: check the formula, the substitution with units and the final answer for each numerical, and the steps and labelled diagram for each derivation.
FAQs
Is this the official board paper?
Yes, the questions are from the board’s March 2026 Physics paper (J-137). The solutions are Doubtly’s own model answers; the board does not publish a full answer key.
How many questions must I answer in Section D?
Any three of the five 4-mark questions (Q.27 to Q.31), for 12 marks.
Why is Q.7 answered as 12.57 W?
The fan reaches 60 r.p.m. in \(2\pi\) seconds, so \(\alpha = 1\) rad/s\(^2\) and \(\tau = 2\) N m. Power rating is taken as torque × angular speed at the working speed of the fan, \(P = \tau\omega = I\alpha\omega\), which gives \(4\pi \approx 12.57\) W. The note in the solution explains the other value some students get.