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

If $A=\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$, then $A^{-1}=$

A
$-\frac{1}{2}\left[\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right]$
B
$\frac{1}{14}\left[\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right]$
C
$\frac{1}{14}\left[\begin{array}{cc}-3 & -2 \\ 4 & -2\end{array}\right]$
D
$-\frac{1}{14}\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]$
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 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

A ball ' $A$ ' is projected vertically upwards with certain initial speed. Another ball 'B' of same mass is projected at an angle of $30^{\circ}$ with vertical with the same initial speed. At the highest point, the ratio of potential energy of ball A to that of ball B will be

$$\left(\sin 90^{\circ}=1, \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\right)$$

A
$4: 3$
B
$3: 4$
C
$4: 1$
D
$3: 2$
4
MHT CET 2024 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is 1.6 m . The period of oscillation of the sphere in second is (acceleration due to gravity, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

A
$0.8 \pi$
B
$0.6 \pi$
C
$0.4 \pi$
D
$0 \cdot 2 \pi$
MHT CET Papers
EXAM MAP