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  • Explain solution RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 8 maths

Explain solution RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 8 maths

Answers (1)

Answer: 3k %

Hint: volume of a sphere of radius x is given by v=\frac{4}{3} \pi x^{3}

Given: we have \Delta x=0.01 k x

Solution: Suppose the error in measuring the radius of a sphere be k

⇒ Suppose x be the radius of the sphere and \Delta x be the error in the value of x

⇒ Thus, we have \Delta x=\left(\frac{k}{100}\right) \times(x)

So, \Delta x=0.01 k x

⇒ Differentiate v with respect to x

\begin{aligned} &\frac{d v}{d x}=\frac{d}{d x}\left(\frac{4}{3}\right) \pi x^{3} \\\\ &\Rightarrow \frac{d v}{d x}=\frac{4 \pi}{3} \frac{d}{d x}\left(x^{3}\right) \Rightarrow \frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}

\begin{aligned} &\frac{d v}{d x}=\frac{4 \pi}{3}\left(3 x^{2}\right) \frac{d v}{d x}=4 \pi x^{2} \\\\ &\Rightarrow \Delta y=\frac{d y}{d x} \Delta x, \text { Here } \frac{d v}{d x}=4 \pi x^{2} \Delta x=0.01 k x \end{aligned}

\therefore \Delta x=0.04 k \pi x^{3}

⇒ percentage of error is,

\begin{aligned} &\text { Error }=\frac{0.04 k \pi x^{3}}{\frac{4}{3} \pi x^{3}} \times 100 \% \\\\ \end{aligned}

              =\frac{0.04 k \pi x^{3}}{4 \pi x^{3}} \times 100 \times 3=3 k \%

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