1
JEE Main 2020 (Online) 7th January Morning Slot
+4
-1
A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1 + $$\alpha$$x) where 'x' is the distance measured from one of the plates. If (ad) << 1, the total capacitance of the system is best given by the expression :
A
$${{A{ \in _0}K} \over d}\left( {1 + {{\left( {{{\alpha d} \over 2}} \right)}^2}} \right)$$
B
$${{A{ \in _0}K} \over d}\left( {1 + {{\alpha d} \over 2}} \right)$$
C
$${{A{ \in _0}K} \over d}\left( {1 + {{{\alpha ^2}{d^2}} \over 2}} \right)$$
D
$${{A{ \in _0}K} \over d}\left( {1 + \alpha d} \right)$$
2
JEE Main 2019 (Online) 12th April Evening Slot
+4
-1
In the given circuit, the charge on 4 $$\mu$$F capacitor will be :
A
5.4 $$\mu$$C
B
9.6 $$\mu$$C
C
13.4 $$\mu$$C
D
24 $$\mu$$C
3
JEE Main 2019 (Online) 12th April Morning Slot
+4
-1
Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K1, K2 and K3. The first capacitor is filled as shown in fig.I, and the second one is filled as shown in fig II. If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be (E1 refers to capacitor (I) and E2 to capacitor (II)):
A
$${{{E_1}} \over {{E_2}}} = {{\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)} \over {{K_1}{K_2}{K_3}}}$$
B
$${{{E_1}} \over {{E_2}}} = {{{K_1}{K_2}{K_3}} \over {\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}$$
C
$${{{E_1}} \over {{E_2}}} = {{\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)} \over {9{K_1}{K_2}{K_3}}}$$
D
$${{{E_1}} \over {{E_2}}} = {{9{K_1}{K_2}{K_3}} \over {\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}$$
4
JEE Main 2019 (Online) 10th April Evening Slot
+4
-1
A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E, as shown in figure. Its bob has mass m and charge q. The time period of the pendulum is given by :
A
$$2\pi \sqrt {{L \over {\sqrt {{g^2} - {{{q^2}{E^2}} \over {{m^2}}}} }}}$$
B
$$2\pi \sqrt {{L \over {\left( {g + {{qE} \over m}} \right)}}}$$
C
$$2\pi \sqrt {{L \over {\sqrt {{g^2} + {{{q^2}{E^2}} \over {{m^2}}}} }}}$$
D
$$2\pi \sqrt {{L \over {\left( {g - {{qE} \over m}} \right)}}}$$
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