1. The potentiometer compares two emf sources by the sum and difference method. To calculate E₁/E₂, the formula used is:, a) (l₁ + l₂) / (l₁ – l₂), b) (l₁ – l₂) / (l₁ + l₂), c) l₁ / l₂, d) (l₂ – l₁) / (l₂ + l₁), 2. In the sum and difference method, the null point balancing lengths for the two cells are measured:, a) Separately, b) When cells are connected in series and opposition, c) Only when cells are connected in parallel, d) Only with a single cell at a time, 3. The sum balancing length corresponds to:, a) When both cells assist each other, b) When both oppose each other, c) When only E₁ is used, d) When only E₂ is used, 4. The difference balancing length corresponds to:, a) Two cells assist, b) Two cells oppose, c) Only stronger cell, d) Only weaker cell, 5. The potentiometer works on which principle?, a) Ohm’s law, b) Kirchhoff's law, c) Null deflection method, d) Wheatstone bridge concept, 6. The ratio E₁/E₂ = (l_sum + l_diff) / (l_sum – l_diff) is valid when:, a) E₁ > E₂, b) E₁ < E₂, c) Both cells are identical, d) For any two cells, 7. If the balancing length for sum is 180 cm and for difference is 20 cm, then E₁/E₂ is:, a) 10, b) 5, c) 2, d) 1.29, 8. A potentiometer is preferred over a voltmeter for comparing EMFs because:, a) It draws no current from the cell, b) It is cheaper, c) It uses fewer components, d) It is more compact, 9. For accurate readings, the potentiometer wire must have:, a) High resistance & short length, b) Low resistance & uniform cross-section, c) High resistance & non-uniform cross-section, d) Variable resistance, 10. If the null point shifts towards the right end of the wire, it indicates:, a) Driving cell emf decreased, b) Test cell emf increased, c) Resistance increased, d) Potentiometer wire heated up.

JKC E1/E2 Using Potentiometer

oleh
Cetakan
Benamkan

Papan mata

Gaya visual

Pilihan

Tukar templat

Pulihkan autosimpan: ?