It is given in the fig that the AC changes as #2sin(100t)#. So the angular frequency of the AC is #omega=100"rad/s"#
Now Capacitive reactance of the circuit #X_C=1/(omega C)#,, where C is the unknown capacitance in the given RLC series circuit.
The inductive reactance of the circuit #X_L=omega L#,, where # L=0.1H# is the inductance in the given RLC series circuit.
Again it is also given that resistance connected in RLC series circuit is #R=10Omega# and the power factor is #cos phi=1/sqrt2#, where #phi# is the angle subtended by the phasor with #R # in the impedance triangle.
As #cos phi=1/sqrt2# then #phi = cos ^-1(1/sqrt2)=45^@#
#tanphi= abs(X_L-X_C)/R#
#=>tan45^@= abs(omegaL-1/(omegaC))/R#
#=>1= abs(100xx0.1-1/(100C))/10#
#=>10= abs(10-1/(100C))#
If #10>1/(100C)# then value of C will be impossible one. So inthis case #10<1/(100C)# and the overall circuit reactance is capacitive giving a leading phase angle #45^@#
So the equation becomes
#=>10= -10+1/(100C)#
#=>1/(100C)=20#
#=>C=1/2000F=500/10^6F=500xx10^-6F=500muF# which is option (3)
