Test: Electric Potential and Capacitance (November 23) - NEET MCQ

# Test: Electric Potential and Capacitance (November 23) - NEET MCQ

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## 15 Questions MCQ Test Daily Test for NEET Preparation - Test: Electric Potential and Capacitance (November 23)

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Test: Electric Potential and Capacitance (November 23) - Question 1

### The four capacitors, each of 25μ F are connected as shown in fig. The dc voltmeter reads 200 V. The charge on each plate of capacitor is    [1994]

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 1

Charge on each plate of each capacitor

Q = CV =  25 x10-6 x 200
= 5 x10-3C

Test: Electric Potential and Capacitance (November 23) - Question 2

### A parallel plate condenser with oil between the plates (dielectric constant of oil K = 2) has a capacitance C. If the oil is removed, then capacitance of the capacitor becomes

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 2

When oil is placed between space of plates

When oil is removed   ... (2)

On comparing both equations, we get
C ' = C/2

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Test: Electric Potential and Capacitance (November 23) - Question 3

### A capacitor is charged to store an energy U.The charging battery is disconnected. Anidentical capacitor is now connected to the firstcapacitor in parallel. The energy in each of thecapacitor is [2000]

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 3

Energy stored in a capacitor is

As the battery is disconnected, total charge Q is shared equally by two capacitors.
So energy of each capacitor

Test: Electric Potential and Capacitance (November 23) - Question 4

The capacity of a parallel plate condenser is 10μF when the distance between its plates is 8cm. If the distance between the plates is reducedto 4 cm then the capacity of this parallel platecondenser will be [2001]

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 4

C = 10 μF          d = 8 cm
C ' = ?                d ' = 4 cm

If d is halved then C will be doubled.

Hence, C ' = 2C = 2 × 10μF = 20μF

Test: Electric Potential and Capacitance (November 23) - Question 5

A capacitor C1 is charged to a potential difference V. The charging battery is then removed and the capacitor is connected to an uncharged capacitor C2. The potential difference across the combination is

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 5

Charge Q = C1V
Total capacity of combination (parallel)

C = C1+ C2

Test: Electric Potential and Capacitance (November 23) - Question 6

Each corner of a cube of side l has a negative charge, –q. The electrostatic potential energy of a charge q at the centre of the cube is    [2002]

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 6

Length of body diagonal =

∴ Distance of centre of cube from each
corner

P.E at centre
= 8 × Potential Energy due to A

Test: Electric Potential and Capacitance (November 23) - Question 7

Three capacitors each of capacity 4μF are to be connected in such a way that the effective capacitance is 6 μF. This can be done by       [2003]

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 7

For series,

For parallel, Ceq = C '+C3 = 2 + 4 = 6μF

Test: Electric Potential and Capacitance (November 23) - Question 8

As per the diagram, a point charge +q is placed at the origin O. Work done in taking another point charge –Q from the point A [coordinates (0, a)] to another point B [coordinates (a, 0)] along the straight path AB is:

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 8

We know that potential energy of two charge system is given by

According to question,

ΔU= UB–UA = 0
We know that for conservative force,
W = –ΔU = 0

Test: Electric Potential and Capacitance (November 23) - Question 9

Two charges q1 and q2 are placed 30 cm apart, as shown in the figure. A third charge q3 is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is  where k is

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 9

We know that potential energy of discrete system of charges is given by

According to question,

Test: Electric Potential and Capacitance (November 23) - Question 10

A parallel plate air capacitor is charged to apotential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates    [2006]

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 10

If we increase the distance between the plates its capacity decreases resulting in higher potential as we know Q = CV. Since Q is constant (battery has been disconnected), on decreasing C, V will increase

Test: Electric Potential and Capacitance (November 23) - Question 11

Two condensers, one of capacity C and other of capacity C/2 are connected to a V-volt battery, as shown. [2007]

The work done in charging fully both the  condensers is

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 11

Work done = Change in energy

Test: Electric Potential and Capacitance (November 23) - Question 12

The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is [2008]

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 12

The energy required to charge a parallel
plate condenser is given by

As  and V = E.d

Test: Electric Potential and Capacitance (November 23) - Question 13

Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be        [2009]

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 13

In series combination of capacitors

Thus, the capacitance and breakdown
voltage of the combination will be C/3 and 3V

Test: Electric Potential and Capacitance (November 23) - Question 14

The electric potential at a point (x, y, z) is given by V = – x2y – xz3 + 4. The electric field  at that point is   [2009]

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 14

The electric field at a point is equal to
negative of potential gradient at that point.

Test: Electric Potential and Capacitance (November 23) - Question 15

A condenser of capacity C is charged to a potential difference of V1. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V2 is

Detailed Solution for Test: Electric Potential and Capacitance (November 23) - Question 15

q = CV1 cos ωt

Also,   and V = V1 cosωt

At t = t1 , V = V2 and i = -ωCV1 sinωt1

(–ve sign gives direction)

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