MHT CET 2021 20th September Morning Shift | Differential Equations Question 128 | Mathematics

An ice ball melts at the rate which is proportional to the amount of ice at that instant. Half the quantity of ice melts in 20 minutes, $$x_0$$ is the initial quantity of ice. If after 40 minutes the amount of ice left is $$\mathrm{Kx}_0$$, then $$\mathrm{K}=$$

$$\frac{1}{2}$$

$$\frac{1}{8}$$

$$\frac{1}{4}$$

$$\frac{1}{3}$$

MHT CET 2020 19th October Evening Shift

MCQ (Single Correct Answer)

-0

The integrating factor of the differential equation $x \frac{d y}{d x}+y \log x=x^2$ is

$x^{\log (\sqrt{x})}$

$(\log x)^2$

$x^{\log x}$

$(\log x)^x$

MHT CET 2020 19th October Evening Shift

MCQ (Single Correct Answer)

-0

The rate of disintegration of a radio active element at time $t$ is proportional to its mass, at the time. Then the time during which the original mass of 1.5 gm . Will disintegrate into its mass of 0.5 gm . is proportional to

$\log 5$

$\log 3$

$\log 2$

$\log 4$

MHT CET 2020 19th October Evening Shift

MCQ (Single Correct Answer)

-0

The general solution of the differential equation $\left(1+y^2\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$ is

$x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} y}\right)^2}{2}+C$

$e^{\tan ^{-1} y}=\left(e^{\tan ^{-1} x}\right)^2+C$

$x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} x} x\right)^2}{2}+C$

$e^{\tan ^{-1} y}=\left(e^{\tan ^{-1} y}\right)^2+C$

PREVIOUS NEXT

MHT CET Subjects