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Please solve RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 5 maths textbook solution

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Answer: 0.3%

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

Given: let  x be the radius and \Delta x be the error in the value x

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

⇒ Thus, we have \Delta x=\left(\frac{0.1}{100}\right) \times(x)

So,   \Delta x=0.001 x

⇒ Volume of a sphere,  v=\frac{4}{3} \pi x^{3}

So, Differentiate v with respect x

\begin{aligned} &\Rightarrow \frac{d v}{d x}=\frac{d}{d x}\left(\frac{4}{3} \pi x^{3}\right) \Rightarrow \frac{d v}{d x}=\frac{4 \pi}{3} \frac{d}{d x}\left(x^{3}\right) \\\\ \\\\ &\Rightarrow \frac{d v}{d x}=\frac{4 \pi}{3}\left(4 x^{2}\right)=4 \pi x^{2} \end{aligned}

\Rightarrow As we know, y=f(x)  and \Delta x  is a smaller increment,

\begin{aligned} &\Delta y=\left(\frac{d v}{d x}\right) \Delta x=0.001 x \\\\ &\Delta x=0.004 \pi x^{3} \\\\ &\text { Here, } \frac{d v}{d x}=4 \pi x^{2} \text { and } \Delta x=0.001 x \end{aligned}


\Rightarrow Percentage error    =\frac{0.004 \pi x^{3}}{\frac{4}{3} \pi x^{3}} \times 100=0.003 \times 100=0.3 \%

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