A parallel plate capacitor of capacitance 90pF is connected to a battery of emf 20V. If a dielectric material of dielectric constant K=53 is inserted between the plates, the magnitude of the induced charge will be.

For parallel plate capacitor
Q=CV
Initial charge Qi=90pF × 20V = 1.8nC
When dielectric material is inserted
C'=KC
Final charge Qf=(KC)V = 53 × 90pF × 20V = 95400pC = 95.4nC
Induced charge = Qf - Qi = 95.4nC - 1.8nC = 93.6nC
However, the induced charge is given by Q_induced = Q_f - Q_i = C_fV - C_iV = (K-1)CV = (53-1) * 90pF * 20V = 52 * 1800pC = 93600pC = 93.6nC This is different from the provided answer. Let's re-examine the calculation.
Initial charge Qi = 90pF * 20V = 1.8nC
When the dielectric is inserted, the capacitance becomes C' = KC = 53 * 90pF = 4770pF
The final charge Qf = C'V = 4770pF * 20V = 95400pC = 95.4nC
The induced charge is the difference between the final and initial charges: Induced charge = Qf - Qi = 95.4nC - 1.8nC = 93.6nC
This calculation is consistent. There appears to be an error in the provided solution or options. The correct induced charge is 93.6nC which is not among the given options.