A spherical ball has a variable diameter 5/2(3x + 1). The rate of change of its volume w.r.t. x, when x = 1, is : (A) 225π (B) 300π (C) 375π (D) 125π

MathsCBSE Class 12 · Board exam 2025, set 65/7/1 · 1 mark · MCQ

CBSE Class 12 Mathematics 2025: question 10

Set in the CBSE Class 12 board exam of 2025, paper 65/7/1, section A, question 10, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q10 · 1 mark · MCQApplication of Derivatives

A spherical ball has a variable diameter $\frac{5}{2} (3 x + 1)$ . The rate of change of its volume w.r.t. x, when $x = 1$ , is :

(A) $225 π$
(B) $300 π$
(C) $375 π$
(D) $125 π$

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 65/7/1.

Hints

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

Hint 1

The radius is half the diameter.

Hint 2

Write $V = \frac{4}{3} π r^{3}$ in terms of x and use the chain rule.

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.

Radius $r = \frac{5}{4} (3 x + 1)$ , so $V = \frac{4}{3} π \cdot \frac{125}{64} (3 x + 1)^{3}$ .
$\frac{d V}{d x} = \frac{125 π}{48} \cdot 3 (3 x + 1)^{2} \cdot 3 = \frac{375 π}{16} (3 x + 1)^{2}$ . At $x = 1$ : $\frac{375 π}{16} \times 16 = 375 π$ .
1 mark

Answer: (C) $375 π$

Where the marks are

Writes V in terms of x, differentiates and substitutes $x = 1$ to get $375 π$ (1 mark)

Revise the chapter

Chapter: Application of Derivatives (notes, formulas, where marks are lost and every past question)
Topic: Rate of Change of Quantities

Practise this chapterBoard-style questions from Application of Derivatives, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 65/7/1: every question, by section.

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CBSE Class 12 Mathematics 2025: question 10

Hints

Step-by-step solution

Revise the chapter

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