What will be the reading of the voltmeter in the given circuit (in $V$)?

The tuning circuit of a radio receiver has a resistance of $50\,\Omega$,an inductor of $10\,mH$ and a variable capacitor. $A$ $1\,MHz$ radio wave produces a potential difference of $0.1\,mV$. The value of the capacitor to produce resonance is.......$pF$ (Take $\pi^2 = 10$).

$A$ series $LCR$ circuit is connected across a source of alternating emf of changing frequency and resonates at frequency $f_0$. Keeping capacitance constant,if the inductance $(L)$ is increased by $\sqrt{3}$ times and resistance is increased $(R)$ by $1.4$ times,the resonant frequency now is

The resonance point in the $X_L - f$ and $X_C - f$ curves is

$A$ series $L-C-R$ circuit containing a resistance of $120 \Omega$ has an angular frequency of $4 \times 10^5 \ rad \ s^{-1}$. At resonance,the voltage across the resistance and the inductor are $60 \ V$ and $40 \ V$ respectively. The value of the inductance is: (in $mH$)

$A$ resistor of $R=6 \Omega$,an inductor of $L=1 \text{ H}$,and a capacitor of $C=17.36 \mu \text{F}$ are connected in series with an $AC$ source. Find the $Q$ factor.