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explain solution RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 4 maths

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Answer: 2% Error

Hint: area of a cubical box of radius x is given by s=6 x^{2}

Given: \Delta x is the error in the value of x \; \; \Delta x=0.01 x

Solution: suppose x be the of cubical box

So,\Delta x=0.01 x

Differentiate  s with respect to x 

\begin{aligned} &\frac{d s}{d x}=\frac{d}{d x}\left(6 x^{2}\right) \\\\ &\Rightarrow \frac{d s}{d x}=6 \times \frac{d}{d x}\left(x^{2}\right) \\\\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \Rightarrow \frac{d s}{d x}=6 x 2 x=12 x \end{aligned}

⇒ As we know that if   y=F(x) \Delta x  is a small increment in x , then the corresponding increment in  y

\Delta x=F(x+\Delta y)-F(x)  , is approximately given as  \Delta y=\left(\frac{d y}{d x}\right) \Delta x

\begin{aligned} &\Rightarrow \text { Here, } \frac{d s}{d x}=12 x \Delta x=0.01 x \\\\ &\Delta s=12 x \times 0.01 x \\\\ &\Delta s=0.12 x^{2} \end{aligned}

\Rightarrow  Percentage of error is

\text { Error }=\frac{0.12 x^{2}}{6 x^{2}} \times 100 \%=0.02 \times 100 \%=2 \%


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