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In a bank, the principal increases continuously at the rate of \(6 \%\) per year. Then the time required to double \(₹ 6000\) rupees is (in years)
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Verified Answer
The correct answer is:
\(\frac{50}{3} \log 2\)
According to given information, let the principal \(P=₹ 6000\) getting double in time ' \(t\) ' with rate of \(6 \%\) per year, so
\(\frac{d P}{d t}=\frac{6}{100} P \Rightarrow \int_{6000}^{12000} \frac{d p}{P}=\frac{3}{50} \int_0^t d t\)
\(\begin{array}{ll}
\Rightarrow & {[\log P]_{6000}^{12000}=\frac{3}{50} t} \\
\Rightarrow & \log 12000-\log 6000=\frac{3}{50} t \\
\Rightarrow & \log 2=\frac{3}{50} t \Rightarrow t=\frac{50}{3} \log 2
\end{array}\)
Hence, option (a) is correct.
\(\frac{d P}{d t}=\frac{6}{100} P \Rightarrow \int_{6000}^{12000} \frac{d p}{P}=\frac{3}{50} \int_0^t d t\)
\(\begin{array}{ll}
\Rightarrow & {[\log P]_{6000}^{12000}=\frac{3}{50} t} \\
\Rightarrow & \log 12000-\log 6000=\frac{3}{50} t \\
\Rightarrow & \log 2=\frac{3}{50} t \Rightarrow t=\frac{50}{3} \log 2
\end{array}\)
Hence, option (a) is correct.
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