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(B) Let $P$ be the principal at any time $t$. Given that the principal increases continuously at the rate of $10 \%$ per year,we have the differential

Vedclass · Accepted Answer

$3296$

Vedclass · Answer

(B) Let $P$ be the principal at any time $t$. Given that the principal increases continuously at the rate of $10 \%$ per year,we have the differential equation: $\frac{dP}{dt} = 0.10 P$ Separating the variables,we get: $\frac{dP}{P} = 0.10 dt$ Integrating both sides: $\int \frac{dP}{P} = \int 0.10 dt$ $\ln(P) = 0.10 t + C$ $P(t) = e^{0.10 t + C} = Ae^{0.10 t}$ At $t = 0$,$P = 2000$. Substituting these values: $2000 = Ae^{0.10(0)} \implies A = 2000$ So,the equation for the principal is $P(t) = 2000 e^{0.10 t}$. After $t = 5$ years: $P(5) = 2000 e^{0.10(5)} = 2000 e^{0.5}$ Given $e^{0.5} = 1.648$: $P(5) = 2000 	imes 1.648 = 3296$ Thus,the amount after $5$ years will be Rs. $3296$.

The principal increases continuously in a newly opened bank at the rate of $10 \%$ per year. An amount of Rs. $2000$ is deposited with this bank. How much will it become after $5$ years? $(e^{0.5} = 1.648)$

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