The correct option of this question will be (b).
Solution —
As $V_{R}=V_{L}=V_{C}=10 V$
$\therefore \quad R=X_{L}=X_{C}$ and $Z=R$
and $V=I R=10 V$
When the capacitor is short circuited, the impedance of the circuit is
$Z^{\prime}=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{R^{2}+R^{2}}=\sqrt{2} R$
and the current in the circuit is
$I^{\prime}=\frac{V}{Z^{\prime}}=\frac{10 V }{\sqrt{2 R}}$
$\therefore$ The voltage across the inductance is
$V_{L}^{\prime}=I^{\prime} X_{L}=\left(\frac{10 V }{\sqrt{2} R}\right) R=\frac{10}{\sqrt{2}} V$