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Need solution for RD Sharma maths class 12 chapter 13 Differentials Errors and Approximations exercise multiple choice question 9

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Answer:    \frac{1}{8}%

 Hint:  Here, we use the concept of  P=d P=\frac{d P}{d V} \Delta V


     P V^{\frac{1}{4}}= Constant

Percentage error in V=\frac{-1}{2} \%


Let k be a constant

\begin{aligned} &\quad P V^{\frac{1}{4}}=k \\\\ &\text { So, } \quad P=\frac{k}{V^{\frac{1}{4}}} \end{aligned}

Let’s differentiate

        \frac{d P}{d V}=\frac{-k}{4} \times V^{\frac{-5}{4}}

Here, percentage error in   V=\frac{-1}{2} \%

        \begin{aligned} &\frac{\Delta V}{V}=\frac{-1}{200} \\\\ &\Delta V=\frac{-V}{200} \end{aligned}

Approximate change in P

        P=d P=\left(\frac{d P}{d V}\right) \Delta V

            =\frac{1}{800} k V^{\frac{-1}{4}}            [From equation (i)]

            =\frac{1}{8} \% \text { of } P

Percentage increase in V=\frac{1}{8} \%


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