MHT CET · Maths · Differential Equations
Radium decompose at the rate proportional to the amount present at any time. If \(\mathrm{P} \%\) of amount disappears in one year, then amount of radium left after 2 years is
- A \(\left(10-\frac{P}{10}\right)^2\)
- B \(x_0\left[1+\frac{P}{100}\right]^2\)
- C \(x_0\left[1-\frac{P}{100}\right]^2\)
- D \(x_0\left[10-\frac{P}{100}\right]^2\)
Answer & Solution
Correct Answer
(C) \(x_0\left[1-\frac{P}{100}\right]^2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{P} \%\) amount disappears in one year.
Let initial amount of radium \(=\mathrm{x}_0\)
\(\therefore\) Amount left after 1 year \(=\mathrm{x}_0-\frac{\mathrm{P}}{100} \times \mathrm{x}_0=\mathrm{x}_0\left(1-\frac{\mathrm{P}}{100}\right)\)
Amount left after 2 years
\(
\begin{aligned}
& =x_0\left(1-\frac{P}{100}\right)-\frac{P}{100} \times x_0\left(1-\frac{P}{100}\right) \\
& =x_0\left(1-\frac{P}{100}\right)\left(1-\frac{P}{100}\right)=x_0\left(1-\frac{P}{100}\right)^2
\end{aligned}
\)
Let initial amount of radium \(=\mathrm{x}_0\)
\(\therefore\) Amount left after 1 year \(=\mathrm{x}_0-\frac{\mathrm{P}}{100} \times \mathrm{x}_0=\mathrm{x}_0\left(1-\frac{\mathrm{P}}{100}\right)\)
Amount left after 2 years
\(
\begin{aligned}
& =x_0\left(1-\frac{P}{100}\right)-\frac{P}{100} \times x_0\left(1-\frac{P}{100}\right) \\
& =x_0\left(1-\frac{P}{100}\right)\left(1-\frac{P}{100}\right)=x_0\left(1-\frac{P}{100}\right)^2
\end{aligned}
\)
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- A random variable X assumes values \(1,2,3, \ldots \ldots ., \mathrm{n}\) with equal probabilities. If \(\operatorname{var}(X): E(X)=4: 1\), then \(n\) is equal toMHT CET 2024 Hard
- \(\operatorname{cosec} 2 \theta-\cot 2 \theta=\)MHT CET 2020 Medium
- The equation of the line passing through the point \((1,2,3)\) and perpendicular to the
lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\bar{r}=\lambda(-3 \hat{\imath}+2 \hat{\jmath}+5 \hat{k})\) isMHT CET 2020 Easy - Two cards are drawn successively with replacement from a well shuffled pack of 52 cards, then mean of number of queens isMHT CET 2023 Medium
- \(\int_1^2 \frac{\mathrm{d} x}{\left(x^2-2 x+4\right)^{\frac{3}{2}}}=\frac{\mathrm{k}}{\mathrm{k}+5}\), then \(\mathrm{k}\) has the valueMHT CET 2023 Hard
- If \(\overline{\mathrm{p}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\) and \(\overline{\mathrm{q}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\). Then a vector of magnitude \(5 \sqrt{3}\) units perpendicular to the vector \(\overline{\mathrm{q}}\) and coplanar with \(\overline{\mathrm{p}}\) and \(\overline{\mathrm{q}}\) isMHT CET 2023 Easy
More PYQs from MHT CET
- Identify the product ' \(\mathrm{B}\) ' in the following reaction. Dry ice \(\underset{\text { Dryether }}{\stackrel{\mathrm{CH}_3 \mathrm{MgBr}}{\longrightarrow}} \mathrm{A} \underset{\text { dil. } \mathrm{HCl}}{\stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow}} \mathrm{B}\)MHT CET 2023 Medium
- Which among the following is NOT a pair of isomers?MHT CET 2024 Medium
- Let the inductance and resistance be denoted by 'L' and 'R' respectively. The dimensions of \(\left(\frac{\mathrm{L}}{\mathrm{R}}\right)\) areMHT CET 2020 Easy
- Formation of p-hydroxyazobenzene from benzene diazonium chloride and phenol in mild alkaline medium is aMHT CET 2020 Hard
- The force required to take away a flat circular plate of radius 2 \(\mathrm{cm}\) from the surface of water is \(\left[\right.\) Surface tension of water \(\left.=70 \times 10^{-3} \mathrm{Nm}^{-1}, \pi=\frac{22}{7}\right]\)MHT CET 2021 Medium
- The value of is ________MHT CET 2019 Hard