NEET Practice Questions (MCQs) with Answers & Solutions

The induced emf (electromotive force) in a dynamo is directly proportional to the rotational velocity of the armature. According to Faraday's Law:

\[ \text{emf} \propto \text{rotational velocity} \]

If the rotational velocity of the dynamo armature is doubled, the induced emf will also double. Therefore, if the original emf is \(E\), then the new emf when the rotational velocity is doubled will be \(2E\). Thus, the induced emf will become two times the original value.

fact

The transformer works on the principle of mutual induction and the voltage transformation ratio is given by the formula: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). Here, \(V_p\) is the primary voltage, \(V_s\) is the secondary voltage, \(N_p\) is the number of primary turns, and \(N_s\) is the number of secondary turns. Given \(V_p = 20V\), \(N_p = 500\), and \(N_s = 5000\), the secondary voltage \(V_s\) can be calculated as: \[ V_s = V_p \times \frac{N_s}{N_p} = 20V \times \frac{5000}{500} = 200V \] The frequency remains the same at 50 Hz. Hence, the correct answer is

200V, 50Hz

To find the current in the primary coil, we first need to determine the current in the secondary coil. The power consumed by the bulb is 30W and the voltage across it is 6V. Using the formula \( P = V \times I \), we get: \[ I_s = \frac{P}{V} = \frac{30W}{6V} = 5A \] The transformer is step-down, so the primary voltage is higher than the secondary. The current transformation ratio is given by \( \frac{I_s}{I_p} = \frac{V_p}{V_s} \). Given \(V_p = 200V\) and \(V_s = 6V\), we can calculate the primary current \(I_p\) as: \[ I_p = I_s \times \frac{V_s}{V_p} = 5A \times \frac{6V}{200V} = 0.15A \] Hence, the current in the primary is 0.15 A.

$ I = I_0 (1 - e^{-Rt \over L}) $ $ I_\alpha = I _0 (1 -e ^{- \alpha} ) = 1.5 $ $ I1 = 1.5 (1- e -2) \Rightarrow { I_\alpha \over I_1 } = { 1 \over 1 -e^{-2} } $

$ \varepsilon = M { dI \over dt} \Rightarrow 0.005 { d \over dt } ( 10 sin 100 \pi t ) = \varepsilon _ { max } = 5 \pi $

$ I_0 = 12/6 = 2A . $ Current becomes half in time t = 0.693 L/R = 1ms

A DC ammeter measures the average value of current. For an alternating current (AC), the average value over a complete cycle is zero because AC alternates in direction and spends equal time in positive and negative half-cycles. Hence, a DC ammeter cannot measure AC.

In AC, the resistance of a coil is influenced by both its inherent resistance and reactance (inductive or capacitive). The reactance adds to the resistance, resulting in a higher effective resistance when compared to DC. This is why the resistance will increase in AC.