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Consider the following statements $A$ is relative error in the area of a square when the relative error in its side is 0.4 $B$ is relative error in the volume of a sphere when the relative error in its radius is 0.3 $C$ is relative error in the surface area of a closed cylinder whose height is equal to its radius, when the relative error in its height is 0.2 $D$ is approximate error in $y=x^2+x-3$ when $x=2$ and $\delta x=0.1$
The ascending order of the values of errors in these statements is
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The ascending order of the values of errors in these statements is
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Verified Answer
The correct answer is:
$C, D, A, B$
Calculation of $A$
Let $S$ be the area of square of side $a$
$$
\begin{aligned}
& \therefore \quad S=a^2 \\
& \Rightarrow \quad \Delta S=\frac{d s}{d a} \times \Delta a \Rightarrow \Delta S=2 a \times \Delta a \\
& \Rightarrow \quad \frac{\Delta S}{S}=\frac{2 a \times \Delta a}{S} \Rightarrow \frac{\Delta S}{S}=\frac{2 a \times \Delta a}{a^2} \\
& =\quad \frac{\Delta a}{a}=2 \times 0.4=0.8 \quad\left[\therefore \frac{\Delta a}{a}=0.4\right] \\
& \therefore \quad A=0.8 \\
&
\end{aligned}
$$
Calculation of $B$
Let $V$ be the volume of the sphere of radius $r$.
$$
\therefore \quad V=\frac{4}{3} \pi r^3
$$
$$
\begin{aligned}
& \Rightarrow \quad \Delta V=\frac{d V}{d r} \times \Delta r \Rightarrow \Delta V=4 \pi r^2 \times \Delta r \\
& \Rightarrow \quad \frac{\Delta V}{V}=\frac{4 \pi r^2 \times \Delta r}{V} \\
& \begin{array}{r}
\Rightarrow \quad \frac{\Delta V}{V}=\frac{4 \pi r^2 \times \Delta r}{\frac{4}{3} \pi r^3}=\frac{3 \Delta r}{r}=3 \times 0.3=0.9 \\
{[\because \Delta r}
\end{array} \\
& {\left[\because \frac{\Delta r}{r}=0.3\right]} \\
& \therefore \quad B=0.9 \\
&
\end{aligned}
$$
Calculation of $C$
Let $S^{\prime}$ be the surface area of closed cylinder with radius $r$ and height $h$
$$
\begin{aligned}
& \therefore \quad S^{\prime}=2 \pi r h+2 \pi r^2 \\
& \Rightarrow \quad S^{\prime}=2 \pi h^2+2 \pi h^2 \\
& {[\because r=h]} \\
& \Rightarrow \quad S^{\prime}=4 \pi h^2 \\
& \Rightarrow \quad \Delta S^{\prime}=\frac{d S^{\prime}}{d h} \cdot \Delta h \\
& \Rightarrow \quad \Delta S^{\prime}=8 \pi h \times \Delta h \\
& \Rightarrow \quad \frac{\Delta S^{\prime}}{S}=\frac{8 \pi h \times \Delta h}{S} \\
& \Rightarrow \quad \frac{\Delta S^{\prime}}{S}=\frac{8 \pi h \times \Delta h}{4 \pi h^2} \\
& =\frac{2 \Delta h}{h}=2 \times 0.2=0.4 \quad\left[\because \frac{\Delta h}{h}=0.2\right] \\
& \therefore \quad c=0.4 \\
&
\end{aligned}
$$
Calculation of $D$
We have, $y=x^2+x-3 \Rightarrow \frac{d y}{d x}=2 x+1$
Now, $\quad \Delta y=\frac{d y}{d x} \Delta x \quad \Rightarrow \Delta y=(2 x+1) \Delta x$
$$
\begin{array}{lcc}
\Rightarrow & \Delta y=(2 \times 2+1) \times 01 & {[\because x=2 \Delta x=01]} \\
\Rightarrow & \Delta y=5 \times 01=0.5 \\
\therefore & \quad D=0.5 \\
\therefore & C < D < A < B .
\end{array}
$$
Let $S$ be the area of square of side $a$
$$
\begin{aligned}
& \therefore \quad S=a^2 \\
& \Rightarrow \quad \Delta S=\frac{d s}{d a} \times \Delta a \Rightarrow \Delta S=2 a \times \Delta a \\
& \Rightarrow \quad \frac{\Delta S}{S}=\frac{2 a \times \Delta a}{S} \Rightarrow \frac{\Delta S}{S}=\frac{2 a \times \Delta a}{a^2} \\
& =\quad \frac{\Delta a}{a}=2 \times 0.4=0.8 \quad\left[\therefore \frac{\Delta a}{a}=0.4\right] \\
& \therefore \quad A=0.8 \\
&
\end{aligned}
$$
Calculation of $B$
Let $V$ be the volume of the sphere of radius $r$.
$$
\therefore \quad V=\frac{4}{3} \pi r^3
$$
$$
\begin{aligned}
& \Rightarrow \quad \Delta V=\frac{d V}{d r} \times \Delta r \Rightarrow \Delta V=4 \pi r^2 \times \Delta r \\
& \Rightarrow \quad \frac{\Delta V}{V}=\frac{4 \pi r^2 \times \Delta r}{V} \\
& \begin{array}{r}
\Rightarrow \quad \frac{\Delta V}{V}=\frac{4 \pi r^2 \times \Delta r}{\frac{4}{3} \pi r^3}=\frac{3 \Delta r}{r}=3 \times 0.3=0.9 \\
{[\because \Delta r}
\end{array} \\
& {\left[\because \frac{\Delta r}{r}=0.3\right]} \\
& \therefore \quad B=0.9 \\
&
\end{aligned}
$$
Calculation of $C$
Let $S^{\prime}$ be the surface area of closed cylinder with radius $r$ and height $h$
$$
\begin{aligned}
& \therefore \quad S^{\prime}=2 \pi r h+2 \pi r^2 \\
& \Rightarrow \quad S^{\prime}=2 \pi h^2+2 \pi h^2 \\
& {[\because r=h]} \\
& \Rightarrow \quad S^{\prime}=4 \pi h^2 \\
& \Rightarrow \quad \Delta S^{\prime}=\frac{d S^{\prime}}{d h} \cdot \Delta h \\
& \Rightarrow \quad \Delta S^{\prime}=8 \pi h \times \Delta h \\
& \Rightarrow \quad \frac{\Delta S^{\prime}}{S}=\frac{8 \pi h \times \Delta h}{S} \\
& \Rightarrow \quad \frac{\Delta S^{\prime}}{S}=\frac{8 \pi h \times \Delta h}{4 \pi h^2} \\
& =\frac{2 \Delta h}{h}=2 \times 0.2=0.4 \quad\left[\because \frac{\Delta h}{h}=0.2\right] \\
& \therefore \quad c=0.4 \\
&
\end{aligned}
$$
Calculation of $D$
We have, $y=x^2+x-3 \Rightarrow \frac{d y}{d x}=2 x+1$
Now, $\quad \Delta y=\frac{d y}{d x} \Delta x \quad \Rightarrow \Delta y=(2 x+1) \Delta x$
$$
\begin{array}{lcc}
\Rightarrow & \Delta y=(2 \times 2+1) \times 01 & {[\because x=2 \Delta x=01]} \\
\Rightarrow & \Delta y=5 \times 01=0.5 \\
\therefore & \quad D=0.5 \\
\therefore & C < D < A < B .
\end{array}
$$
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