A capacitor is connected to a $20\, {V}$ battery through a resistance of $10\, \Omega .$ It is found that the potential difference across the capacitor rises to $2\, {V}$ in $1\, \mu {s}$. The capacitance of the capacitor is $....\,\mu {F}$ Given : $\ln \left(\frac{10}{9}\right)=0.105$
  • A$9.52$
  • B$0.95$
  • C$0.105$
  • D$1.85$
JEE MAIN 2021, Diffcult
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