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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Here are the solutions to the questions: 6. The change in potential energy of an electron is -1.28 × 10^-4 J when it transverse between two points A and B. Find the change in potential between these points. (q = 1.602 × 10^-19 C) Step 1: Identify the given values and the formula. The change in potential energy ( U) is related to the charge (q) and the change in potential ( V) by the formula U = q V. Given: U = -1.28 × 10^-4 J The charge of an electron is q = -1.602 × 10^-19 C. (The problem states q = 1.602 × 10^-19 C, which is the magnitude of the elementary charge. For an electron, the charge is negative.) Step 2: Rearrange the formula and substitute the values. V = ( U)/(q) V = -1.28 × 10^-4 J-1.602 × 10^-19 C Step 3: Calculate the change in potential. V ≈ 0.7989 × 10^15 V V ≈ 7.989 × 10^14 V There is a significant discrepancy between this calculated value and the provided options. If we assume a typo in the question and that the potential energy was -1.28 × 10^-14 J instead of -1.28 × 10^-4 J, then: V = -1.28 × 10^-14 J-1.602 × 10^-19 C V ≈ 0.7989 × 10^5 V V ≈ 7.989 × 10^4 V This value is approximately 8 × 10^4 V, which matches option (d). Assuming this typo: The correct option is (d). 8 × 10^4 V 7. Two charges 3.21 × 10^-9 C and 5.4 × 10^-9 C are situated at 3m and 6m respectively on the positive x-axis of a co-ordinate system. Find the total electric potential at a point 9m away from the origin due to these charges. Step 1: Identify the given values and the formula for electric potential. The electric potential (V) due to a point charge (Q) at a distance (r) is given by V = (kQ)/(r), where k = 9 × 10^9 N·m^2/C^2. Given: Q_1 = 3.21 × 10^-9 C at x_1 = 3m Q_2 = 5.4 × 10^-9 C at x_2 = 6m Point of interest is at x_P = 9m. Step 2: Calculate the distance from each charge to the point of interest. Distance from Q_1 to P: r_1 = x_P - x_1 = 9m - 3m = 6m Distance from Q_2 to P: r_2 = x_P - x_2 = 9m - 6m = 3m Step 3: Calculate the potential due to each charge and sum them. V_1 = (9 × 10^9 N·m^2/C^2) × (3.21 × 10^-9 C)6m = (9 × 3.21)/(6) V = 4.815 V V_2 = (9 × 10^9 N·m^2/C^2) × (5.4 × 10^-9 C)3m = (9 × 5.4)/(3) V = 16.2 V Total potential V_total = V_1 + V_2 = 4.815 V + 16.2 V = 21.015 V Step 4: Compare with options. The calculated value 21.015 V is approximately 21 V. The correct option is (b). 21 V 8. Copper contains 8.4 × 10^28 free electrons/m^3. A copper wire of cross-sectional area 7.4 × 10^-7m^2 carries a current of 1 A. The electron drift speed is approximately: Step 1: Identify the given values and the formula for drift speed. The current (I) in a conductor is related to the number density of charge carriers (n), the cross-sectional area (A), the drift speed (v_d), and the charge of each carrier (q) by the formula I = n A v_d q. Given: n = 8.4 × 10^28 m^-3 A = 7.4 × 10^-7m^2 I = 1 A Charge of an electron q = 1.602 × 10^-19 C Step 2: Rearrange the formula to solve for drift speed. v_d = (I)/(n A q) Step 3: Substitute the values and calculate the drift speed. v_d = 1 A(8.4 × 10^28 m^-3) × (7.4 × 10^-7 m^2) × (1.602 × 10^-19 C) v_d = (1)/(8.4 × 7.4 × 1.602 × 10^(28 - 7 - 19)) m/s v_d = (1)/(99.58608 × 10^2) m/s v_d = (1)/(9958.608) m/s v_d ≈ 0.0001004 m/s v_d ≈ 1.0 × 10^-4 m/s The correct option is (d). 10^-4 m/s 9. A given wire of resistance 10 has a length of 5m and a cross-sectional area of 4.0 × 10^-3m^2. Calculate the conductivity of the wire. Step 1: Identify the given values and the formulas for resistivity and conductivity. Resistance (R) is given by R = (L)/(A), where is resistivity, L is length, and A is cross-sectional area. Conductivity () is the reciprocal of resistivity, = (1)/(). Given: R = 10 L = 5m A = 4.0 × 10^-3m^2 Step 2: Calculate the resistivity (). Rearrange the resistance formula: = (R × A)/(L) = 10 × 4.0 × 10^-3m^25m = 40 × 10^-3 · m5 = 8 × 10^-3 · m Step 3: Calculate the conductivity (). = (1)/() = (1)/(8 × 10^-3) · m = (1000)/(8)^-1m^-1 = 125^-1m^-1 Step 4: Convert the conductivity to ^-1cm^-1 and compare with options. Since 1m = 100cm, then 1m^-1 = 10^-2cm^-1. = 125^-1m^-1 = 125 × 10^-2^-1cm^-1 = 1.25^-1cm^-1 There is a significant discrepancy between this calculated value and the provided options, which are in the range of 10^6 to 10^8. If we assume a typo in the cross-sectional area, and it was 4.0 × 10^-10m^2 instead of 4.0 × 10^-3m^2, then: = 10 × 4.0 × 10^-10m^25m = 8 × 10^-10 · m = (1)/(8 × 10^-10) · m = 1.25 × 10^9^-1m^-1 Converting to ^-1cm^-1: = 1.25 × 10^9^-1m^-1 = 1.25 × 10^9 × 10^-2^-1cm^-1 = 1.25 × 10^7^-1cm^-1 This matches option (b). Assuming this typo: The correct option is (b). 1.25 × 10^7 ^-1cm^-1 10. A wire of diameter 1mm carries a current of 20A. What is the current density in the wire? Step 1: Identify the given values and the formula for current density. Current density (J) is defined as current (I) per unit cross-sectional area (A), J = (I)/(A). Given: Diameter d = 1mm Current I = 20A Step 2: Convert diameter to meters and calculate the radius. d = 1mm = 1 × 10^-3m r = (d)/(2) = 1 × 10^-3m2 = 0.5 × 10^-3m = 5 × 10^-4m Step 3: Calculate the cross-sectional area (A). A = r^2 = (5 × 10^-4m)^2 A = (25 × 10^-8)m^2 A ≈ 3.14159 × 25 × 10^-8m^2 A ≈ 78.53975 × 10^-8m^2 A ≈ 7.854 × 10^-7m^2 Step 4: Calculate the current density (J). J = (I)/(A) = 20A7.854 × 10^-7m^2 J ≈ 2.546 × 10^7 A/m^2 Step 5: Compare with options. The calculated value 2.546 × 10^7 A/m^2 is approximately 2.55 × 10^7 A/m^2. The correct option is (a). 2.55 × 10^7 A/m^2 11. An electric bulb is rated 60W, 220V. Calculate the resistance of its Filament when it is operating normally. Step 1: Identify the given values and the formula for power, voltage, and resistance. The power (P) dissipated by a resistor is related to the voltage (V) across it and its resistance (R) by the formula P = (V^2)/(R). Given: Power P = 60W Voltage V = 220V Step 2: Rearrange the formula to solve for resistance. R = (V^2)/(P) Step 3: Substitute the values and calculate the resistance. R = (220V)^260W R = 48400V^260W R = 806.666... R ≈ 806.7 The options are not visible for this question, but the calculation is straightforward. 806.7 That's 2 down. 3 left today — send the next one.