QNEET-2024-Ph-154: NEET 2024 Physics - Class 12 NEET-2024-Ph-154 Solution: The amplitude of the charge oscillating in a circuit decreases…

The given equation for the amplitude of the charge oscillating in a circuit is: Q = Q 0 e _ R t 2 L We need to find the time, t , at which charge amplitude decreases to 0.50 Q 0 . So, we set Q = 0.50 Q 0 : 0.50 Q 0 = Q 0 e _ R t 2 L Divide both sides by Q 0 : 0.50 = e _ R t 2 L Take the natural logarithm on both sides to solve for t : ln _ ( 0.50 ) = _ R t 2 L Recall that ln _ ( 0.50 ) = _ ln _ ( 2 ) . Substituting the given value ln _ ( 2 ) = 0.693 , we get: _ 0.693 = _ R t 2 L Remove the negative signs from both sides: 0.693 = R t 2 L Now, solve for t : t = 2 L _ 0.693 R Substituting the given values R = 1.5 � and L = 12   mH = 12 _ 10 _ 3   H , we have: t = 2 _ 12 _ 10 _ 3 _ 0.693 1.5 Calculate the numerator and denominator: t = 16.632 _ 10 _ 3 1.5 Finally, compute the value of t : t = 11.09 _ 10 _ 3   s = 11.09   ms Therefore, the time at which the charge amplitude decreases to 0.50 Q 0 is nearly 11.09 ms. Thus, the correct answer is: Option B: 11.09 ms