1
JEE Main 2023 (Online) 1st February Evening Shift
+4
-1 Given below are two statements: One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one.

Reason R : Capacitance of metallic spheres depend on the radii of spheres

In light of the above statements, choose the correct answer from the options given below.

A
Both $$\mathbf{A}$$ and $$\mathbf{R}$$ are true but $$\mathbf{R}$$ is not the correct explanation of $$\mathbf{A}$$
B
Both $$\mathbf{A}$$ and $$\mathbf{R}$$ are true and $$\mathbf{R}$$ is the correct explanation of $$\mathbf{A}$$
C
$$\mathbf{A}$$ is false but $$\mathbf{R}$$ is true
D
$$\mathbf{A}$$ is true but $$\mathbf{R}$$ is false
2
JEE Main 2023 (Online) 25th January Morning Shift
+4
-1 A parallel plate capacitor has plate area 40 cm$$^2$$ and plates separation 2 mm. The space between the plates is filled with a dielectric medium of a thickness 1 mm and dielectric constant 5. The capacitance of the system is :

A
$$\mathrm{10\varepsilon_0~F}$$
B
$$\mathrm{24\varepsilon_0~F}$$
C
$$\mathrm{\frac{3}{10}\varepsilon_0~F}$$
D
$$\mathrm{\frac{10}{3}\varepsilon_0~F}$$
3
JEE Main 2022 (Online) 29th July Evening Shift
+4
-1 Two identical thin metal plates has charge $$q_{1}$$ and $$q_{2}$$ respectively such that $$q_{1}>q_{2}$$. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :

A
$$\frac{\left(q_{1}+q_{2}\right)}{C}$$
B
$$\frac{\left(q_{1}-q_{2}\right)}{C}$$
C
$$\frac{\left(q_{1}-q_{2}\right)}{2 C}$$
D
$$\frac{2\left(q_{1}-q_{2}\right)}{C}$$
4
JEE Main 2022 (Online) 28th July Evening Shift
+4
-1 A slab of dielectric constant $$\mathrm{K}$$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $$\frac{3}{4} \mathrm{~d}$$, where $$\mathrm{d}$$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :

(Given $$\mathrm{C}_{0}$$ = capacitance of capacitor with air as medium between plates.)

A
$$\frac{4 K C_{0}}{3+K}$$
B
$$\frac{3 K C_{0}}{3+K}$$
C
$$\frac{3+K}{4 K C_{0}}$$
D
$$\frac{K}{4+K}$$
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