1
JEE Main 2020 (Online) 8th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A capacitor is made of two square plates each of side 'a' making a very small angle $$\alpha $$ between them, as shown in figure. The capacitance will be close to : JEE Main 2020 (Online) 8th January Evening Slot Physics - Capacitor Question 93 English
A
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 + {{\alpha a} \over {d}}} \right)$$
B
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {4d}}} \right)$$
C
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {2d}}} \right)$$
D
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{3\alpha a} \over {2d}}} \right)$$
2
JEE Main 2020 (Online) 8th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In the given circuit, value of Y is : JEE Main 2020 (Online) 8th January Evening Slot Physics - Semiconductor Question 135 English
A
toggles between 0 and 1
B
1
C
will not execute
D
0
3
JEE Main 2020 (Online) 8th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A particle moves such that its position vector $$\overrightarrow r \left( t \right) = \cos \omega t\widehat i + \sin \omega t\widehat j$$ where $$\omega $$ is a constant and t is time. Then which of the following statements is true for the velocity $$\overrightarrow v \left( t \right)$$ and acceleration $$\overrightarrow a \left( t \right)$$ of the particle :
A
$$\overrightarrow v $$ and $$\overrightarrow a $$ both are perpendicular to $$\overrightarrow r $$
B
$$\overrightarrow v $$ and $$\overrightarrow a $$ both are parallel to $$\overrightarrow r $$
C
$$\overrightarrow v $$ is perpendicular to $$\overrightarrow r $$ and $$\overrightarrow a $$ is directed towards the origin
D
$$\overrightarrow v $$ is perpendicular to $$\overrightarrow r $$ and $$\overrightarrow a $$ is directed away from the origin
4
JEE Main 2020 (Online) 8th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Consider two charged metallic spheres S1 and S2 of radii R1 and R2, respectively. The electric fields E1 (on S1) and E2 (on S2) on their surfaces are such that E1/E2 = R1/R2. Then the ratio V1 (on S1) / V2 (on S2) of the electrostatic potentials on each sphere is :
A
(R1/R2)2
B
(R2/R1)
C
(R1/R2)3
D
R1/R2
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