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Solution for If an Error of K% is Made in Measuring the Radius of a Sphere, Then Percentage Error in Its Volume is (A) K% (B) 3k% (C) 2k% (D) K/3% - CBSE (Science) Class 12 - Mathematics

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Question

If an error of k% is made in measuring the radius of a sphere, then percentage error in its volume is
(a) k%
(b) 3k%
(c) 2k%
(d) k/3%

Solution

(b) 3k%
Let x be the radius of the sphere and y be its volume.
Then,

\[\frac{∆ x}{x} \times 100 = k\]

\[\text { Also }, y = \frac{4}{3}\pi x^3 \]

\[ \Rightarrow \frac{dy}{dx} = 4\pi x^2 \]

\[ \Rightarrow \frac{∆ y}{y} = \frac{4\pi x^2}{y}dx = \frac{4\pi x^2}{\frac{4}{3}\pi x^3} \times \frac{kx}{100}\]

\[ \Rightarrow \frac{∆ y}{y} \times 100 = 3k\]

\[\text { Hence, the error in the volume is } 3k % .\]

  Is there an error in this question or solution?

APPEARS IN

 RD Sharma Mathematics for Class 12 by R D Sharma (Set of 2 Volume) (2018-19 Session) (with solutions)
Chapter 14: Differentials, Errors and Approximations
Q: 3 | Page no. 13

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Solution for question: If an Error of K% is Made in Measuring the Radius of a Sphere, Then Percentage Error in Its Volume is (A) K% (B) 3k% (C) 2k% (D) K/3% concept: Approximations. For the courses CBSE (Science), PUC Karnataka Science, CBSE (Arts), CBSE (Commerce)
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