1
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The statement $\sim(p \leftrightarrow \sim q)$ is

A
equivalent to $\mathrm{p} \leftrightarrow \mathrm{q}$
B
a fallacy
C
a tautology
D
equivalent to $\sim \mathrm{p} \leftrightarrow \mathrm{q}$
2
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^2+x$, then

A
$\alpha=-6, \beta=\frac{1}{2}$
B
$\alpha=-6, \beta=-\frac{1}{2}$
C
$\alpha=2, \beta=-\frac{1}{2}$
D
$\alpha=2, \beta=\frac{1}{2}$
3
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

A circular coil of resistance ' $R$ ', area ' $A$ ', number of turns ' N ' is rotated about its vertical diameter with angular speed ' $\omega$ ' in a uniform magnetic field of magnitude ' $B$ '. The average power dissipated in a complete cycle is

A
$\frac{\mathrm{N}^2 \mathrm{~A}^2 \mathrm{~B}^2 \omega^2}{2 \mathrm{R}}$
B
$\frac{\mathrm{BNA} \omega}{\mathrm{R}}$
C
$\frac{\mathrm{BNA} \omega}{2 \mathrm{R}}$
D
$\frac{N^2 A^2 B^2 \omega^2}{R}$
4
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The linear speed of a particle at the equator of the earth due to its spin motion is ' V '. The linear speed of the particle at latitude $30^{\circ}$ is

$$\left[\begin{array}{l} \sin 30^{\circ}=\cos 60^{\circ}=1 / 2 \\ \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2 \end{array}\right]$$

A
$\frac{\mathrm{V}}{\sqrt{2}}$
B
$\frac{\mathrm{V}}{2}$
C
$\frac{\sqrt{3}}{2} \mathrm{v}$
D
$\mathrm{V}$
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