1
JEE Main 2021 (Online) 16th March Morning Shift
+4
-1
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $${3 \over 4}$$d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation :
A
$$C' = {{3 + K} \over {4K}}{C_0}$$
B
$$C' = {{4 + K} \over {3}}{C_0}$$
C
$$C' = {{4K} \over {K + 3}}{C_0}$$
D
$$C' = {{4} \over {3 + K}}{C_0}$$
2
JEE Main 2021 (Online) 26th February Morning Shift
+4
-1
Consider the combination of 2 capacitors C1 and C2 with C2 > C1, when connected in parallel, the equivalent capacitance is $${{15} \over 4}$$ times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $${{{C_2}} \over {{C_1}}}$$.
A
$${{15} \over {11}}$$
B
No Solutions
C
$${{29} \over {15}}$$
D
$${{15} \over {4}}$$
3
JEE Main 2021 (Online) 25th February Evening Shift
+4
-1
An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle '$$\alpha$$' with the plates. It leaves the plates at angle '$$\beta$$' with kinetic energy K2. Then the ratio of kinetic energies K1 : K2 will be :
A
$${{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}$$
B
$${{\cos \beta } \over {\cos \alpha }}$$
C
$${{{{\sin }^2}\beta } \over {{{\cos }^2}\alpha }}$$
D
$${{\cos \beta } \over {\sin \alpha }}$$
4
JEE Main 2021 (Online) 24th February Morning Shift
+4
-1
Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be :
A
4 : 1
B
1 : 2
C
2 : 1
D
1 : 4
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