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Question

(A) Let the principal be $P$. Given that the principal increases continuously at a rate of $6 \%$ per year,we have the differential equation: $\frac{d

Vedclass · Accepted Answer

$\frac{50}{3} \log 2$

Vedclass · Answer

(A) Let the principal be $P$. Given that the principal increases continuously at a rate of $6 \%$ per year,we have the differential equation: $\frac{dP}{dt} = \frac{6}{100} P = 0.06 P$ Separating the variables,we get: $\frac{dP}{P} = 0.06 dt$ Integrating both sides: $\int \frac{dP}{P} = \int 0.06 dt$ $\log P = 0.06 t + C$ Initially,at $t = 0$,$P = 6000$. So,$\log 6000 = C$. Thus,$\log P = 0.06 t + \log 6000$,which implies $\log(\frac{P}{6000}) = 0.06 t$. We want to find the time $t$ when the principal doubles,i.e.,$P = 12000$. $\log(\frac{12000}{6000}) = 0.06 t$ $\log 2 = \frac{6}{100} t$ $t = \frac{100}{6} \log 2 = \frac{50}{3} \log 2$ Therefore,the time required is $\frac{50}{3} \log 2$ years. Option $(A)$ is correct.

In a bank,the principal increases continuously at the rate of $6 \%$ per year. Then the time required to double $₹ 6000$ is (in years)

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