1
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation, having general solution as $A x^2+B y^2=1$, where $A$ and $B$ are arbitrary constants, is

A
$x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}-x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2-y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
B
$x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}-x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2+y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
C
$x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2+y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
D
$x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2-y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
2
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

A radio active substance has half-life of $h$ days, then its initial decay rate is given by Note that at $\mathrm{t}=0, \mathrm{M}=\mathrm{m}_{\mathrm{o}}$

A
$\frac{\mathrm{m}_{\mathrm{o}}}{\mathrm{h}}(\log 2)$
B
$\left(\mathrm{m}_{\mathrm{o}} \mathrm{h}\right)(\log 2)$
C
$-\frac{\mathrm{m}_{\mathrm{o}}}{\mathrm{h}}(\log 2)$
D
$\left(-\mathrm{m}_{\mathrm{o}} \mathrm{h}\right)(\log 2)$
3
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation of $y=\mathrm{e}^x\left(\mathrm{a}+\mathrm{bx}+x^2\right)$ is

A
$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-2 y=0$
B
$\frac{\mathrm{d}^2 y}{\mathrm{dx}^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=0$
C
$\frac{\mathrm{d}^2 y}{\mathrm{dx}^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-2 \mathrm{e}^x+y=0$
D
$\frac{\mathrm{d}^2 y}{\mathrm{dx}}+2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-\mathrm{e}^x+2 y=0$
4
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

An ice ball melts at the rate which is proportional to the amount of ice at that instant. Half of the quantity of ice melts in 15 minutes. $x_0$ is the initial quantity of ice. If after 30 minutes the amount of ice left is $\mathrm{kx}_0$, then the value of $k$ is

A
$\frac{1}{2}$
B
$\frac{1}{3}$
C
$\frac{1}{4}$
D
$\frac{1}{8}$
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