PhysicsCBSE Class 12 · Board exam 2025, set 55/7/1 · 1 mark · MCQ

CBSE Class 12 Physics 2025: question 4

Set in the CBSE Class 12 board exam of 2025, paper 55/7/1, section A, question 4, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q4 · 1 mark · MCQAlternating Current

An alternating current is given by I = I $_{0}$ cos $(100 π)$ t. The least time the current takes to decrease from its maximum value to zero will be :

(A) $(\frac{1}{200})$ s
(B) $(\frac{1}{150})$ s
(C) $(\frac{1}{100})$ s
(D) $(\frac{1}{50})$ s

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 55/7/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Find the time period from ω.

Hint 2

Max to zero of a cosine takes what fraction of one cycle?

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

$ω = 100 π$ , so $T = \frac{2 π}{ω} = \frac{1}{50}$ s.
½ mark
The current falls from maximum to zero in a quarter cycle: $t = \frac{T}{4} = \frac{1}{200}$ s.
½ mark

Answer: 1/200 s, option (A)

Where the marks are

Finds T = 1/50 s and takes T/4 = 1/200 s (1 mark)

Revise the chapter

Chapter: Alternating Current (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Alternating Current, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 55/7/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Physics 2025: question 4

Hints

Step-by-step solution

Revise the chapter

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