(A) The magnification $m$ for a mirror is given by $m = -v/u$. $1$. $m = -2$: Negative sign indicates a real and inverted image. Since $|m| > 1$,it is

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$1-(b, c), 2-(b, c), 3-(b, d), 4-(a, d)$

(A) The magnification $m$ for a mirror is given by $m = -v/u$. $1$. $m = -2$: Negative sign indicates a real and inverted image. Since $|m| > 1$,it is a concave mirror. So,$(b, c)$. $2$. $m = -1/2$: Negative sign indicates a real and inverted image. Since $|m| $3$. $m = +2$: Positive sign indicates a virtual and erect image. Since $|m| > 1$,it is a concave mirror. So,$(b, d)$. $4$. $m = +1/2$: Positive sign indicates a virtual and erect image. Since $|m| < 1$,it is a convex mirror. So,$(a, d)$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

Match the corresponding entries of Column-$I$ with Column-$II$. (Where $m$ is the magnification produced by the mirror)

A
$1-(b, c), 2-(b, c), 3-(b, d), 4-(a, d)$
B
$1-(b, c), 2-(b, c), 3-(b, d), 4-(a, d)$
C
$1-(b, c), 2-(b, c), 3-(b, d), 4-(a, d)$
D
$1-(b, c), 2-(b, c), 3-(b, d), 4-(a, d)$

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$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y + 5 = 0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x = 1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x + 8y = 27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x = 9$

Match the corresponding entries of Column-$I$ with Column-$II$. (Where $m$ is the magnification produced by the mirror)

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In a series $LCR$ circuit,$R=300 \Omega, L=0.9 \text{ H}, C=2.0 \mu\text{F}$ and $\omega=1000 \text{ rad/s}$,then the impedance of the circuit is: (in $\Omega$)

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