NEET Practice Questions (MCQs) with Answers & Solutions

At resonance, $X_L = X_C$, so the impedance formula $Z = \sqrt{R^2 + (X_L - X_C)^2}$ simplifies to $Z = \sqrt{R^2 + 0^2} = R$.

The NCERT text describes, 'Resonant circuits have a variety of applications, for example, in the tuning mechanism of a radio or a TV set. ... In tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. When this happens, the amplitude of the current with the frequency of the signal of the particular radio station in the circuit is maximum.'

At resonance, 'the voltages across L and C cancel each other... and the current amplitude is $v_m/R$, the total source voltage appearing across R.'

The text explains resonance as, 'If such a system is driven by an energy source at a frequency that is near the natural frequency, the amplitude of oscillation is found to be large.'

The resonant angular frequency is given by $\omega_0 = 1/\sqrt{LC}$. Given $L = 1.00 \text{ mH} = 1.00 \times 10^{-3} \text{ H}$ and $C = 1.00 \text{ nF} = 1.00 \times 10^{-9} \text{ F}$. So, $\omega_0 = 1/\sqrt{(1.00 \times 10^{-3}) \times (1.00 \times 10^{-9})} = 1/\sqrt{1.00 \times 10^{-12}} = 1/(1.00 \times 10^{-6}) = 1.00 \times 10^6 \text{ rad/s}$.

At resonance, the peak current $i_m = v_m/R$. Given $v_m = 100 \text{ V}$ and $R = 100 \Omega$. So, $i_m = 100 \text{ V} / 100 \Omega = 1 \text{ A}$. The provided text also specifically mentions this for the given values: 'Since $i_m = v_m/R$ at resonance, the current amplitude for case (i) is twice to that for case (ii).'

At resonance, $X_L = X_C$, which means $(X_L - X_C) = 0$. From $\tan \phi = (X_L - X_C)/R$, we get $\tan \phi = 0$, so $\phi = 0$. This implies current and voltage are in phase.

According to the NCERT text, 'Halogen atoms are more electronegative than carbon, therefore, carbon-halogen bond of alkyl halide is polarised; the carbon atom bears a partial positive charge whereas the halogen atom bears a partial negative charge.'

As we go down the group in the periodic table, the size of the halogen atom increases. Fluorine atom is the smallest and iodine atom is the largest. Consequently, the carbon-halogen bond length also increases from C-F to C-I.

The NCERT states, 'In order to dissolve haloalkane in water, energy is required to overcome the attractions between the haloalkane molecules and break the hydrogen bonds between water molecules. Less energy is released when new attractions are set up between the haloalkane and the water molecules as these are not as strong as the original hydrogen bonds in water. As a result, the solubility of haloalkanes in water is low.'