NEET Practice Questions (MCQs) with Answers & Solutions

The context specifies: 'Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst (first row transition metals utilise 3d and 4s electrons for bonding).'

The context states: 'They are usually non stoichiometric and are neither typically ionic nor covalent... They have high melting points, higher than those of pure metals.' The core definition also specifies small atoms trapped, not large ones.

According to the provided text, 'In fact, the I or rms current is the equivalent dc current that would produce the same average power loss as the alternating current.' This highlights the fundamental definition and significance of RMS current in terms of power dissipation.

The text states: '$I = \frac{I_m}{\sqrt{2}} = 0.707 I_m$'. This is the standard definition of the RMS value for a sinusoidal alternating current.

The context explicitly mentions, 'It is customary to measure and specify rms values for ac quantities. For example, the household line voltage of 220 V is an rms value'. This confirms that such general ratings refer to RMS values.

The text provides the calculation: '$V_m = \sqrt{2} V = (1.414)(220 V) = 311 V'$. This shows how to convert from RMS voltage to peak voltage.

The text states: 'In terms of I, the average power, denoted by P is $P = I_m^2 R/2 = I^2 R$'. It also mentions, 'In terms of rms values, the equation for power [Eq. (7.7)] and relation between current and voltage in ac circuits are essentially the same as those for the dc case.' This supports $P=I^2R$ as the equivalent expression.

As stated in 'POINTS TO PONDER 1.', 'When a value is given for ac voltage or current, it is ordinarily the rms value. The voltage across the terminals of an outlet in your room is normally 240 V. This refers to the rms value of the voltage.' Similarly, the 220V rating for the bulb in example 7.1 is an RMS value.

From Example 7.1(a): 'We are given P = 100 W and V = 220 V. The resistance of the bulb is $R = V^2 / P = (220 V)^2 / 100 W = 48400 / 100 \Omega = 484 \Omega$'.

From Example 7.1(c): 'Since, $P = I V$, then $I = P / V = 100 W / 220 V = 0.454 A$'.