1
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

A uniformly charged conducting sphere of diameter 3.5 cm has a surface charge density of $20 \mu \mathrm{Cm}^{-2}$. The total electric flux leaving the surface of the sphere is nearly [permittivity of free space, $\varepsilon_0=8.85 \times 10^{-12} \mathrm{SI}$ unit]

A
$57 \times 10^2 \mathrm{~Wb}$
B
$70 \times 10^2 \mathrm{~Wb}$
C
$87 \times 10^2 \mathrm{~Wb}$
D
$35 \times 10^3 \mathrm{~Wb}$
2
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

A stone is projected with kinetic energy E, making an angle $\theta$ with the horizontal. When it reaches a highest point, its kinetic energy is

A
$\mathrm{E}^2 \sin ^2 \theta$
B
$\mathrm{E} \sin \theta$
C
$E \cos ^2 \theta$
D
$\mathrm{E} \cos \theta$
3
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

In Young's double slit experiment, the intensity on screen at a point where path difference is $\frac{\lambda}{4}$ is $\frac{K}{2}$. The intensity at a point when path difference is ' $\lambda$ ' will be

A
4 K
B
2 K
C
K
D
$\frac{\mathrm{K}}{4}$
4
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $M$ is the magnetisation induced in the material, H is the magnetic field intensity, B is the net magnetic field inside the material then the correct relation between them is ( $\mu_0=$ permeability of free space)

A
$\quad \mathrm{B}=\frac{\mu_0}{(\mathrm{H}+\mathrm{M})}$
B
$\quad B=\mu_0(H-M)$
C
$\quad \mathrm{B}=\frac{\mu_0}{(\mathrm{H}-\mathrm{M})}$
D
$\quad B=\mu_0(H+M)$

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