1
MHT CET 2024 3rd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation $\left[\frac{1+\left(\frac{d y}{d x}\right)^2}{\left(\frac{d^2 y}{d x^2}\right)}\right]^{\frac{3}{2}}=\mathrm{kx}$ is of

A
order $=2$, degree $=3$
B
order $=3$, degree $=2$
C
order $=2$, degree $=2$
D
order $=3$, degree $=3$
2
MHT CET 2024 3rd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation of family of circles, whose centres are on the X -axis and also touch the Y -axis is

A
$4\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2=\left(x^2+y^2\right)^2$
B
$\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2=\left(x^2+y^2\right)^2$
C
$2\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2=\left(x^2+y^2\right)^2$
D
$\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2=4\left(x^2+y^2\right)^2$
3
MHT CET 2024 2nd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A body cools according to Newton's law of cooling from $100^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in 15 minutes. If the temperature of the surrounding is $20^{\circ} \mathrm{C}$, then the temperature of the body after cooling down for one hour is

A
$30^{\circ} \mathrm{C}$
B
$25^{\circ} \mathrm{C}$
C
$35^{\circ} \mathrm{C}$
D
$40^{\circ} \mathrm{C}$
4
MHT CET 2024 2nd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $y=y(x)$ is the solution of the differential equation $\left(\frac{5+\mathrm{e}^x}{2+y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{e}^x=0$ satisfying $y(0)=1$, then a value of $y(\log 13)$ is

A
$-$1
B
0
C
1
D
2
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