The approximate change in the volume of a cube of side $x$ metres caused by increasing the side by $3$ % is :

0.06 x^{3} m^{3}

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0.6 x^{3} m^{3}

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0.09 x^{3} m^{3}

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0.9 x^{3} m^{3}

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Solution

The correct option is B $0.09 x^{3} m^{3}$
Volume $(V)$ of a cube with side $x$ is given by, $V = x^{3}$
$\Rightarrow \frac{d V}{d x} = 3 x^{2} \equiv Δ V = 3 x^{2} Δ x$ ,
Now it is given that $Δ x = .03 x$
$∴ Δ V = .09 x^{3}$ m $^{3}$

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Similar questions

The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is

A. 0.06 x³ m³ B. 0.6 x³ m³ C. 0.09 x³ m³ D. 0.9 x³ m³

The approximate change in the volume of a cube of side x metre caused by increasing the side by 3% is;

(a) $0.06 x^{3} m^{3}$ (b) $0.6 x^{3} m^{3}$
(c) $0.09 x^{3} m^{3}$ (d) $0.9 x^{3} m^{3}$

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of a cube of side

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The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is :

The correct option is B 0.09x3m3Volume (V) of a cube with side x is given by, V=x3⇒dVdx=3x2≡ΔV=3x2Δx, Now it is given that Δx=.03x∴ΔV=.09x3 m3

The approximate change in the volume of a cube of side $x$ metres caused by increasing the side by $3$ % is :

The correct option is B $0.09 x^{3} m^{3}$
Volume $(V)$ of a cube with side $x$ is given by, $V = x^{3}$
$\Rightarrow \frac{d V}{d x} = 3 x^{2} \equiv Δ V = 3 x^{2} Δ x$ ,
Now it is given that $Δ x = .03 x$
$∴ Δ V = .09 x^{3}$ m $^{3}$