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(D) Initial energy stored in the capacitor is given by $U_i = \frac{1}{2} C V^2$.Substituting the values: $U_i = \frac{1}{2} 	imes 14 \, pF 	imes (1

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$864$

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(D) Initial energy stored in the capacitor is given by $U_i = \frac{1}{2} C V^2$.Substituting the values: $U_i = \frac{1}{2} 	imes 14 \, pF 	imes (12 \, V)^2 = \frac{1}{2} 	imes 14 	imes 144 = 1008 \, pJ$.When the battery is disconnected,the charge $Q$ on the plates remains constant.After inserting the dielectric slab of constant $k = 7$,the new capacitance becomes $C' = kC = 7 	imes 14 \, pF = 98 \, pF$.The new energy stored in the capacitor is $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2(kC)} = \frac{U_i}{k}$.$U_f = \frac{1008 \, pJ}{7} = 144 \, pJ$.The loss in electrical potential energy is converted into the mechanical energy of the slab.Mechanical energy $= U_i - U_f = 1008 \, pJ - 144 \, pJ = 864 \, pJ$.

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