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The image formed by a convex mirror of focal length $30 \mathrm{~cm}$ is a quarter of the size of the object. The distance of the object from the mirror is
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The correct answer is:
$90 \mathrm{~cm}$
Focal length, $f=30 \mathrm{~cm}$
$\begin{aligned}
& \mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}=\frac{1}{4} \\
& \mathrm{u}=-4 \mathrm{v} \\
& \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{v}}-\frac{1}{4 \mathrm{v}}=\frac{4-1}{4 \mathrm{v}} \\
& \frac{1}{\mathrm{f}}=\frac{3}{4 \mathrm{v}} \\
& \mathrm{f}=\frac{4 \mathrm{v}}{3} ; \frac{30 \times 3}{4}=\mathrm{v} \\
& \mathrm{u}=\frac{-4 \times 30 \times 3}{4}=-90 \mathrm{~cm}
\end{aligned}$
$\begin{aligned}
& \mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}=\frac{1}{4} \\
& \mathrm{u}=-4 \mathrm{v} \\
& \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{v}}-\frac{1}{4 \mathrm{v}}=\frac{4-1}{4 \mathrm{v}} \\
& \frac{1}{\mathrm{f}}=\frac{3}{4 \mathrm{v}} \\
& \mathrm{f}=\frac{4 \mathrm{v}}{3} ; \frac{30 \times 3}{4}=\mathrm{v} \\
& \mathrm{u}=\frac{-4 \times 30 \times 3}{4}=-90 \mathrm{~cm}
\end{aligned}$
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