Using the magnification formula for mirrors: \[ m = \frac{f}{u-f} \] For the concave mirror, the object distance is \( u = -18 \, \text{cm} \), and the focal length is \( f = \frac{R}{2} = 6 \, \text{cm} \), where \( R = 12 \, \text{cm} \): \[ m_1 = \frac{6}{18 - 6} = \frac{1}{2} \]

For the convex mirror, the object distance is the same, and the focal length is positive: \[ m_2 = \frac{6}{18 + 6} = \frac{1}{4} \] Hence, the ratio of the sizes of the images formed by the convex mirror and the concave mirror is: \[ \frac{m_2}{m_1} = \frac{1/4}{1/2} = \frac{1}{2} \]

Thus, the correct answer is: \[ \frac{1}{2} \]
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
An organic compound (X) with molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ is not readily oxidised. On reduction it gives $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}(\mathrm{Y})\right.$ which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis give 2,3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are: