dataset_name: electrical_engineering
description: "\nThe following are multiple choice questions (with answers) about electrical\
  \ engineering.\n\nQ: A point pole has a strength of 4\u03C0 * 10^-4 weber. The force\
  \ in newtons on a point pole of 4\u03C0 * 1.5 * 10^-4 weber placed at a distance\
  \ of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's\
  \ think step by step. The force between two point poles is given by m_1m_2/(mu_0\
  \ 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in\
  \ the question, we calculate that the force is approximately 15 N. The answer is\
  \ (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30\
  \ mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density\
  \ is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's\
  \ think step by step. The torque on a coil in a uniform magnetic field is given\
  \ by BANI, where B is the magnetic flux density, A is the area of the coil, N is\
  \ the number of turns, and I is the current. So we have that I = (Torque)/(BAN),\
  \ or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch\
  \ built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1\
  \ (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset\
  \ latch; in the case where S=1 and R=1, the circuit has no stable state; instead\
  \ a race condition will be produced within the circuit, so the device will be in\
  \ an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ:\
  \ Two long parallel conductors carry 100 A. If the conductors are separated by 20\
  \ mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1\
  \ N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length\
  \ between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r),\
  \ where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging\
  \ in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\n\
  Q: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03A9\
  \ and total number of conductors is 100. Find the total resistance\n(A) 200\u03A9\
  \ (B) 100\u03A9 (C) 50\u03A9 (D) 10\u03A9\nA: Let's think step by step. In lap winding,\
  \ effectively two resistors are connected in parallel, so the actual resistance\
  \ of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50\
  \ Ohms. The answer is (C)."
include: _mmlu_flan_cot_fewshot_template_yaml
task: mmlu_flan_cot_fewshot_electrical_engineering
