We use the formula for a camera to relate the size of the image, the size of the landscape, the height of the camera, and the focal length: \[ \frac{\text{Size of image}}{\text{Size of landscape}} = \frac{\text{Focal length}}{\text{Height of camera}}. \] Here, the size of the image is \( 2 \times 2 \) cm (so the area is 4 cm\(^2\)), the size of the landscape is 400 km\(^2\), and the height of the camera is 18 km. Substituting these values into the equation and solving for the focal length, we find that the focal length is \( 0.9 \, \text{cm} \).
Final Answer: \( 0.9 \, \text{cm} \).
For a particular ideal gas, which of the following graphs represents the variation of mean squared velocity of the gas molecules with temperature?
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: