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Please solve RD Sharma class 12 chapter 13 Differentials Errors and Approximations exercise fill in the blanks question 2 maths textbook solution

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Answer:  0.06 x^{3} m^{3}

Hint:  Here we use the formula of volume of cube,


Given:  Side of the cube =x \; meter

Also,increase in side =2% of side


Increase in side =2% of side =0.02x

Hence,  \Delta x=0.02 x

And volume of the cube, V=x^{3} m^{3}

We need to find approximate change in volume of the cube

Now,     \Delta V=\frac{d V}{d x} \Delta x

                  \begin{aligned} &=\left(\frac{d V}{d x}\right) \times \Delta x \\\\ &\Rightarrow 3 x^{2} \times 0.02 x \end{aligned}                  [because 2% of  0.02x is=0.02x]

                  =0.06 x^{3}

Hence, the approximate change in the volume of the cube is 0.06 x^{3} m^{3}

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