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The solar cell converts light from the sun to electricity.

No expanded working has been attached for this question yet. The verified answer is Option D.

Latent heat is thermal energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process.

This is as a result of the Hydrogen bonding in water. Due to the high Kinetic Energy of water, the hydrogen bonds can't hold for too long but from 4^\circC, there's an increase in the hydrogen bonds hence strenghtening it. From the point of production of more hydrogen bonds, the bonding is no longer easily broken. Generally, these and a few factors based on these cause the density of water to be maximum at 4^\circC.

A saturated vapour is one which is in a dynamic equilibrium with its own liquid.

This is a type of constructive interference. When two waves of equal amplitude are interfering constructively, the resulting amplitude is twice as large as the amplitude of an individual wave. = A + A = 2A

Dispersion occurs when different wavelengths of light are refracted by different amounts as they pass through the water droplets, causing the light to spread out into its constituent colors, thereby creating a rainbow.

The 3rd overtone for an open pipe has 5 antinodes and 4 nodes.

Comparing the given equation with the wave equation: \(Asin(kx - \omega t)\), we have that A = 10 which is the amplitude of the wave.

Diffraction refers to the various things that occur when a wave encounters an obstacle. This bending of the wave after it passes a small gap is an example of diffraction.

\(\eta = \frac{1}{sin\theta}\) \(1.5 = \frac{1}{sin\theta}\) \(sin\theta = \frac{1}{1.5} = 0.667\) \(\theta = sin^{-1}(0.667) = 41.835^\circ\) \(\approxeq 41.8^\circ\)

For the 4th harmonic, \(L = 2\lambda\) \(4m = 2\lambda \implies \lambda = \frac{4m}{2}\) =\(2m\) Recall, \(v = f\lambda\) \(1 = f \times 2 \implies f = \frac{1}{2}\) = 0.50Hz

The number of images formed = \(\frac{360}{n}\) where the value gotten is odd else, = \(\frac{360}{n} -1\). Solving, we have that \(\frac{360}{20} = 18\) Hence, the no of images formed is 18 - 1 = 17.

Given, image size = 3U object size = U focal length = 10cm lens to object distance = \(\frac{focal length \times object size}{image size}\) = \(\frac{10 \times u}{3u} = 3.33\) distance between image and object = 10 + 3.33 = 13.33cm

The time required to complete one full cycle is the period; the inverse of the period gives the frequency.