The potential difference across resistor R2 after two time constants, is :

2 V

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7.6 V

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10 V

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12 V

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Solution

The correct option is A $2 V$
When switch S is closed, the capacitor will start charging.
Potential across C at instant time t is $V_{c} (t) = V_{0} (1 - e^{- t / C R_{2}})$
At $t = 2 C R_{2}, V_{c} = V_{0} (1 - e^{- 2}) = 12 (1 - 1 / e^{2}) = 10 V$
When capacitor is charging, no current pass through resistor $R_{1}$ . So

Potential across $R_{1}$ is zero.
Thus the potential across $R_{2} = V_{0} - V_{c} = 12 - 10 = 2 V$

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