This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Welcome back forti — been a while! Let's pick up where you left off. SECTION I 1. a) Distinguish (by defining) between an intrinsic and an extrinsic semiconductor. State an example of each. An intrinsic semiconductor* is a pure semiconductor material, such as silicon or germanium, that has not been doped with impurities. Its electrical conductivity is solely due to thermally generated electron-hole pairs. Example: Pure Silicon or Pure Germanium*. An extrinsic semiconductor* is a semiconductor material that has been doped with impurities to increase its conductivity. Doping creates either an excess of free electrons (n-type) or an excess of holes (p-type). Example: n-type Silicon (doped with phosphorus) or p-type Germanium* (doped with boron). 1. b) A current of 3A flows past a point in a circuit in 20 s. i) Determine the charge that has passed the point. Step 1: Identify the given values and the formula for charge. Given: Current I = 3\, A, Time t = 20\, s. The formula for charge is Q = I × t. Step 2: Substitute the values into the formula and calculate the charge. Q = 3\, A × 20\, s Q = 60\, C The charge that has passed the point is 60 C. ii) Calculate the number of electrons that have passed the point. Step 1: Identify the elementary charge and the formula relating charge and number of electrons. The elementary charge e = 1.6 × 10^-19\, C. The formula is Q = n × e, where n is the number of electrons. Step 2: Rearrange the formula to solve for n and substitute the values. n = (Q)/(e) n = 60\, C1.6 × 10^-19\, C/electron n = 3.75 × 10^20\, electrons The number of electrons that have passed the point is 3.75 × 10^20. 2. Three 40 bulbs are connected to a 240V power source through an ammeter as shown in the figure below. 2. a) What property of the ammeter enables it to be connected as shown? An ammeter has a very low internal resistance, which allows it to be connected in series in a circuit without significantly affecting the total resistance or the current flow. 2. b) Give two advantages of connecting bulbs in parallel in our homes. Each appliance receives the full supply voltage*, allowing them to operate at their intended power. If one appliance fails or is switched off*, the others continue to operate independently. 2. c) Calculate; i) The combined resistance of the above circuit. Step 1: Calculate the equivalent resistance of the two bulbs connected in parallel. The two parallel bulbs each have a resistance of R_1 = 40\, and R_2 = 40\, . The formula for two parallel resistors is R_p = (R_1 × R_2)/(R_1 + R_2). R_p = (40\, × 40\, )/(40\, + 40\, ) R_p = (1600\, ^2)/(80\, ) R_p = 20\, Step 2: Calculate the total resistance of the circuit. The parallel combination (R_p) is in series with the third bulb (R_3 = 40\, ). The total resistance R_total = R_p + R_3. R_total = 20\, + 40\, R_total = 60\, The combined resistance of the circuit is 60\, . ii) The current flowing through the ammeter. Step 1: Use Ohm's Law to calculate the total current. Given: Voltage V = 240\, V, Total resistance R_total = 60\, . The formula for current is I = (V)/(R_total). I = 240\, V60\, I = 4\, A The current flowing through the ammeter is 4\, A. iii) The power consumed when all the bulbs are in use. Step 1: Use the formula for power consumed in the circuit. Given: Voltage V = 240\, V, Total current I = 4\, A. The formula for power is P = V × I. P = 240\, V × 4\, A P = 960\, W The power consumed when all the bulbs are in use is 960\, W. 3. a) Explain why the specific latent heat of vaporization of a substance is much greater than its specific latent heat of fusion. The specific latent heat of vaporization is much greater than the specific latent heat of fusion because more energy is required to overcome the intermolecular forces and completely separate the molecules in a liquid to form a gas. In vaporization, molecules must break free from the liquid's surface and move far apart, doing work against atmospheric pressure and intermolecular attraction. In fusion, molecules in a solid only need enough energy to break free from their fixed positions and move past each other in the liquid state, where intermolecular forces are still significant. 3. b) i) Give two differences between boiling and evaporation. Boiling occurs at a specific temperature (boiling point) throughout the entire liquid, forming bubbles, while evaporation* occurs at any temperature from the surface of the liquid. Boiling is a rapid process, whereas evaporation* is a slow and gradual process. 3. b) ii) State one way by which evaporation can be increased. One way evaporation can be increased is by increasing the surface area of the liquid. 4. a) i) Define the principal focus and the focal length of a convex lens. The principal focus* (F) of a convex lens is the point on the principal axis where rays of light parallel to the principal axis converge after passing through the lens. The focal length* (f) of a convex lens is the distance between the optical center of the lens and its principal focus. 4. a) ii) Explain what is meant by a real image. A real image is an image formed when actual light rays converge at a point after reflection or refraction. Real images can be projected onto a screen. 4. a) iii) Draw a labeled diagram to show how such an image is produced by a convex lens. I cannot draw diagrams directly. However, to produce a real magnified image with a convex lens, the object must be placed between the principal focus (F) and twice the focal length (2F) from the optical center of the lens. Here's how the diagram would be drawn: 1. Draw a principal axis and a convex lens at its center. 2. Mark the principal focus (F) and twice the focal length (2F) on both sides of the lens. 3. Place an object (e.g., an arrow) upright between F and 2F on one side. 4. Draw two principal rays from the top of the object: Ray 1: Parallel to the principal axis, refracts through the lens and passes through F on the other side. Ray 2: Passes through the optical center of the lens without deviation. 5. The point where these two refracted rays intersect on the other side of the lens forms the top of the image. 6. Draw the image as an inverted arrow from the principal axis to this intersection point. The image will be real, inverted, and magnified, located beyond 2F on the opposite side of the lens. 4. b) A ray of light travels at a speed of 3 × 10^8 m/s in air and enters a glass of refractive index 1.5. Calculate the speed of light in the glass. Step 1: Identify the given values and the formula for refractive index. Given: Speed of light in air c = 3 × 10^8\, m/s, Refractive index of glass n = 1.5. The formula for refractive index is n = (c)/(v), where v is the speed of light in the medium. Step 2: Rearrange the formula to solve for v and substitute the values. v = (c)/(n) v = 3 × 10^8\, m/s1.5 v = 2 × 10^8\, m/s The speed of light in the glass is 2 × 10^8\, m/s. Send me the next one 📸