#### Question

If the ratio of base radius and height of a cone is 1 : 2 and percentage error in radius is λ %, then the error in its volume is

##### Options

λ %

2 λ %

3 λ %

none of these

#### Solution

3 *λ* %

Let the radius of the cone be *x, *the height be 2*x* and the volume be y.

\[\frac{∆ x}{x} = \lambda \] %

\[ \Rightarrow y = \frac{1}{3}\pi x^2 \times 2x = \frac{2}{3}\pi x^3 \]

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

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

\[ \Rightarrow \frac{∆ y}{y} = 3\lambda\%\]

Is there an error in this question or solution?

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If the Ratio of Base Radius and Height of a Cone is 1 : 2 and Percentage Error in Radius is λ %, Then the Error in Its Volume is Concept: Approximations.

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