dataset_name: engineering
description: The following are multiple choice questions (with answers) about engineering.
fewshot_config:
  sampler: first_n
  samples:
  - question: "A point pole has a strength of 4π * 10^-4 weber. The force in newtons on a point pole of 4π * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be
      (A) 25 N. (B) 12.5 N. (C) 15 N. (D) 10 N. (E) 7.5 N. (F) 20 N. (G) 5 N. (H) 3.75 N. (I) N/A (J) N/A"
    target: "Let's think step by step. The force between two point poles is given by \\( \\frac{m_1 m_2}{\\mu_0 4 \\pi r^2} \\), in analogy to Coulomb’s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (C)."
  - question: "The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is \\( 240 \\times 10^{-6} \\) N-m on full scale. If magnetic flux density is 1 Wb/m² range of meter is
      (A) 2 mA. (B) 5 mA. (C) 1.5 mA. (D) 0.5 mA. (E) 6 mA. (F) 4 mA. (G) 3 mA. (H) 1 mA. (I) 2.5 mA. (J) 3.5 mA."
    target: "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 = \\frac{\\text{Torque}}{BAN} \\), or \\( \\frac{240 \\times 10^{-6}}{1200 \\times 10^{-6} \\times 100 \\times 1} = 2 \\times 10^{-3} \\). The answer is (A)."
  - question: "In an SR latch built from NOR gates, which condition is not allowed
      (A) S=0, R=2 (B) S=2, R=2 (C) S=1, R=1 (D) S=1, R=-1 (E) S=1, R=2 (F) S=0, R=0 (G) S=2, R=0 (H) S=1, R=0 (I) S=2, R=1 (J) S=0, R=1"
    target: "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 (C)."
  - question: "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
      (A) 100 N. (B) 0.01 N. (C) 10 N. (D) 0.5 N. (E) 0.001 N. (F) 5 N. (G) 1 N. (H) 50 N. (I) 0.1 N. (J) 0.0001 N."
    target: "Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\( \\frac{\\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 (I)."
  - question: "In a 2 pole lap winding dc machine, the resistance of one conductor is 2Ω and the total number of conductors is 100. Find the total resistance
      (A) 50Ω (B) 1Ω (C) 25Ω (D) 200Ω (E) 10Ω (F) 100Ω (G) 500Ω (H) 150Ω (I) 75Ω (J) 20Ω"
    target: "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 (A)."
group: mmlu_pro_flan_cot_fewshot_stem
include: _mmlu_pro_flan_cot_fewshot_template_yaml
task: mmlu_pro_flan_cot_fewshot_engineering
