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(B) For a convex mirror,the focal length $f = +10\,cm$. The object $AB$ has length $5\,cm$. Point $B$ is at $u_B = -20\,cm$. Since $AB = 5\,cm$,point

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$\frac{10}{21}\,cm$

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(B) For a convex mirror,the focal length $f = +10\,cm$. The object $AB$ has length $5\,cm$. Point $B$ is at $u_B = -20\,cm$. Since $AB = 5\,cm$,point $A$ is at $u_A = -20 - 5 = -25\,cm$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we get $v = \frac{uf}{u-f}$.For point $A$ $(u_A = -25\,cm)$:$v_A = \frac{(-25) 	imes 10}{-25 - 10} = \frac{-250}{-35} = \frac{50}{7}\,cm$.For point $B$ $(u_B = -20\,cm)$:$v_B = \frac{(-20) 	imes 10}{-20 - 10} = \frac{-200}{-30} = \frac{20}{3}\,cm$.The size of the image is the difference between the image positions: $|v_A - v_B| = |\frac{50}{7} - \frac{20}{3}| = |\frac{150 - 140}{21}| = \frac{10}{21}\,cm$.

$A$ convex mirror of focal length $10\,cm$ is shown in the figure. $A$ linear object $AB = 5\,cm$ is placed along the optical axis. Point $B$ is at a distance of $20\,cm$ from the pole of the mirror. Then,the size of the image of $AB$ will be:

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