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

The electrostatic potential inside a charged spherical ball is given by $\mathrm{V}=\mathrm{ar}^2+\mathrm{b}$ where ' r ' is the distance from its centre and ' $a$ ' and ' $b$ ' are constants. The volume charge density of the ball is [ $\varepsilon_0=$ permittivity of free space $]$

A
$-24 \pi \mathrm{a} \varepsilon_0 \mathrm{r}$
B
$-6 \mathrm{a} \varepsilon_0 \mathrm{r}$
C
$-24 \pi \mathrm{a} \varepsilon_0$
D
$-6 \mathrm{a} \varepsilon_0$
2
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

A charge $$17.7 \times 10^{-4} \mathrm{C}$$ is distributed uniformly over a large sheet of area $$200 \mathrm{~m}^2$$. The electric field intensity at a distance $$20 \mathrm{~cm}$$ from it in air will be $$\left[\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2\right]$$

A
$$5 \times 10^5 \mathrm{~N} / \mathrm{C}$$
B
$$6 \times 10^5 \mathrm{~N} / \mathrm{C}$$
C
$$7 \times 10^5 \mathrm{~N} / \mathrm{C}$$
D
$$8 \times 10^5 \mathrm{~N} / \mathrm{C}$$
3
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $$\mathrm{E}_{\mathrm{a}}$$ and $$\mathrm{E}_{\mathrm{q}}$$ represent the electric field intensity due to a short dipole at a point on its axial line and on the equatorial line at the same distance '$$r$$' from the centre of the dipole, then

A
$$\mathrm{E}_{\mathrm{a}}=\mathrm{E}_{\mathrm{q}}$$
B
$$\mathrm{E}_{\mathrm{a}}=\frac{1}{2} \mathrm{E}_{\mathrm{q}}$$
C
$$\mathrm{E}_{\mathrm{a}}=\frac{1}{\sqrt{2}} \mathrm{E}_{\mathrm{q}}$$
D
$$\mathrm{E}_{\mathrm{a}}=2 \mathrm{E}_{\mathrm{q}}$$
4
MHT CET 2023 14th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The electric field intensity on the surface of a solid charged sphere of radius '$$r$$' and volume charge density '$$\rho$$' is ($$\varepsilon_0=$$ permittivity of free space)

A
$$\frac{\rho r}{3 \varepsilon_0}$$
B
$$\frac{\rho}{4 \pi \varepsilon_0 \mathrm{r}}$$
C
zero
D
$$\frac{5 \rho \mathrm{r}}{6 \varepsilon_0}$$
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