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
MHT CET Subjects
EXAM MAP
Medical
NEETAIIMS
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
Defence
NDA
Staff Selection Commission
SSC CGL Tier I
CBSE
Class 12