የፊዚክስ መግቢያ ፈተና፣ 2005 ዓ.ም. (2013)

Light comes in packets of energy called photons, and one photon carries the energy \(E = hf\). A photon has no charge and no rest mass, so it is a packet of electromagnetic energy rather than a piece of matter. That makes D correct. Option A is tempting because the word quantum is used elsewhere for matter as well, but the smallest unit of matter is a particle such as an electron, and a photon is not made of matter.

The time constant of an RC circuit is \(\tau = RC\), so \(\tau = (1.0\times10^{6}\,\Omega)(5\times10^{-6}\,\mathrm{F}) = 5\) s. Once the capacitor is fully charged no current flows and the whole 30 V sits across it, so \(Q_{max} = C\varepsilon = (5\times10^{-6})(30) = 1.5\times10^{-4}\) C, which is 150 microcoulombs. Both numbers in option A come out of those two rules. Option C is what you reach if the prefixes are not tracked: the \(10^{6}\) of the megohm and the \(10^{-6}\) of the microfarad cancel exactly, so the time constant is 5 s and not 10 s.

Newton's third law says forces come in pairs of equal size and opposite direction. The fins push on the water, and the water pushes back on the fish with the same force the other way. To be pushed east, the fish has to push the water west, so C is correct. Option A is the usual trap: pushing the water east would send the fish west.

Writing \(\vec{A} = c\vec{B}\) means \(\vec{A}\) lies along the same line as \(\vec{B}\). A positive \(c\) keeps the direction and only changes the length, while a negative \(c\) turns the vector around. Opposite directions therefore force \(c\) to be negative. Option D fails because \(c = 0\) would make \(\vec{A}\) the zero vector, and the question says both vectors are non zero.

Faraday's law gives the induced emf as \(\varepsilon = -N\dfrac{\Delta\Phi}{\Delta t}\). That expression contains the number of turns, how fast the flux changes (so the speed of the magnet) and how strong the magnet is, but resistance appears nowhere in it. Resistance decides the induced current through \(I = \varepsilon/R\), not the emf itself, so A is the one the emf does not depend on. Option C looks like a safe thing to rule out, yet a faster magnet changes the flux more quickly and does raise the emf.

Use \(v = f\lambda\), so \(f = \dfrac{v}{\lambda} = \dfrac{50}{25\,000} = 2.0\times10^{-3}\) Hz. A wavelength of 25 km is enormous next to a speed of 50 m/s, so one whole wave takes about 500 seconds to pass and the frequency has to be a very small number. Options C and D can be dropped on that reasoning alone, since a frequency of 1 Hz or more would mean at least one full wave every second.