Charge redistribution on connected conductors

Steps:

• Sphere X starts uncharged (0 charge); sphere Y starts positively charged (+Q total).
• The metal rod connects X and Y, making them a single conductor where charges can flow freely.
• Positive charge from Y spreads equally to both spheres (assuming identical size).
• Each ends with +Q/2: X positive, Y positive but less than initial.

Why A is correct:

• Connected conductors share charge equally per electrostatic equilibrium, conserving total charge (+Q split between two).

Why the others are wrong:

• B and C: No charge transfer occurs; ignores conduction via rod.
• D: Total charge conserved as positive, not neutralized.

Final answer: A