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(A) Given that point $A$ is earthed,its potential is $V_A = 0 \ V$.From the circuit,we can find the potentials of other points relative to $A$:$1$. Mo

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$A$

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(A) Given that point $A$ is earthed,its potential is $V_A = 0 \ V$.From the circuit,we can find the potentials of other points relative to $A$:$1$. Moving from $A$ to $B$ through the $2 \ V$ cell,the potential increases by $2 \ V$ (assuming the positive terminal is towards $B$). Thus,$V_B = 2 \ V$.$2$. Moving from $A$ to $D$ through the $3 \ \Omega$ resistor,the current flows from $D$ to $A$ (since $A$ is at $0 \ V$ and $D$ is at a higher potential). Let the current be $I$. The potential at $D$ is $V_D = V_A + I 	imes 3 = 3I$.$3$. Moving from $B$ to $D$ through the $8 \ V$ battery,$V_D - V_B = 8 \ V$,so $V_D = V_B + 8 = 2 + 8 = 10 \ V$.$4$. Now,for point $C$,moving from $D$ to $C$ through the $4 \ V$ cell,$V_C = V_D - 4 = 10 - 4 = 6 \ V$.Comparing the potentials: $V_A = 0 \ V$,$V_B = 2 \ V$,$V_C = 6 \ V$,and $V_D = 10 \ V$.The point with the least potential is $A$.

The figure shows a network of resistances and cells in which the point $A$ is earthed. The point which has the least potential is

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