1
JEE Main 2021 (Online) 26th August Evening Shift
+4
-1
A parallel plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K1 and K2 of same area A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the capacitor will be given by:

A
$${{{\varepsilon _0}A} \over d}\left( {{1 \over 2} + {{{K_1}{K_2}} \over {{K_1} + {K_2}}}} \right)$$
B
$${{{\varepsilon _0}A} \over d}\left( {{1 \over 2} + {{{K_1}{K_2}} \over {2({K_1} + {K_2})}}} \right)$$
C
$${{{\varepsilon _0}A} \over d}\left( {{1 \over 2} + {{{K_1} + {K_2}} \over {{K_1}{K_2}}}} \right)$$
D
$${{{\varepsilon _0}A} \over d}\left( {{1 \over 2} + {{2({K_1} + {K_2})} \over {{K_1}{K_2}}}} \right)$$
2
JEE Main 2021 (Online) 26th August Morning Shift
+4
-1
The material filled between the plates of a parallel plate capacitor has resistivity 200 $$\Omega$$m. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is : (given the value of relative permittivity of material is 50)
A
9.0 $$\mu$$A
B
9.0 mA
C
0.9 mA
D
0.9 $$\mu$$A
3
JEE Main 2021 (Online) 27th July Evening Shift
+4
-1
Match List I with List II.

List - I List - II
(a) Capacitance, C (i) $${M^1}{L^1}{T^{ - 3}}{A^{ - 1}}$$
(b) Permittivity of free space, $${\varepsilon _0}$$ (ii) $${M^{ - 1}}{L^{ - 3}}{T^4}{A^2}$$
(c) Permeability of free space, $${\mu _0}$$ (iii) $${M^{ - 1}}{L^{ - 2}}{T^4}{A^2}$$
(d) Electric field, E (iv) $${M^1}{L^1}{T^{ - 2}}{A^{ - 2}}$$

Choose the correct answer from the options given below
A
(a) $$\to$$ (iii), (b) $$\to$$ (ii), (c) $$\to$$ (iv), (d) $$\to$$ (i)
B
(a) $$\to$$ (iii), (b) $$\to$$ (iv), (c) $$\to$$ (ii), (d) $$\to$$ (i)
C
(a) $$\to$$ (iv), (b) $$\to$$ (ii), (c) $$\to$$ (iii), (d) $$\to$$ (i)
D
(a) $$\to$$ (iv), (b) $$\to$$ (iii), (c) $$\to$$ (ii), (d) $$\to$$ (i)
4
JEE Main 2021 (Online) 27th July Evening Shift
+4
-1
A simple pendulum of mass 'm', length 'l' and charge '+ q' suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be

A
$${\tan ^{ - 1}}\left[ {{q \over {mg}} \times {{{C_1}({V_2} - {V_1})} \over {({C_1} + {C_2})(d - t)}}} \right]$$
B
$${\tan ^{ - 1}}\left[ {{q \over {mg}} \times {{{C_2}({V_2} - {V_1})} \over {({C_1} + {C_2})(d - t)}}} \right]$$
C
$${\tan ^{ - 1}}\left[ {{q \over {mg}} \times {{{C_2}({V_1} + {V_2})} \over {({C_1} + {C_2})(d - t)}}} \right]$$
D
$${\tan ^{ - 1}}\left[ {{q \over {mg}} \times {{{C_1}({V_1} + {V_2})} \over {({C_1} + {C_2})(d - t)}}} \right]$$
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