Electrodynamics Test

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Electrodynamics Test - 6

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Weekly Quiz Competition

Question 1
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Directions: The following question has four choices out of which ONLY ONE is correct.
Seven conductors each of resistance R are arranged as shown in the figure. (i), fig. (ii) and fig. (iii). If R₁, R₂ and R₃ are the respective equivalent resistances between A and B, then

A
maximum (R₁, R₂, R₃) = R₁

B
R₃ > R₂ < R₁

C
minimum (R₁, R₂, R₃) = R₂

D
R₂ < R₁ < R₃

Solution

For the first case
All the resistances are in series so

R₁ = 7R

For the second case ,we have

R₂ = R + 2R | | 2R + R
R₂ = 3R,

max (R₁, R_2, R₃) = R₁
Min (R₁, R₂, R₃) = R₃
R₃ < R₂ < R₁
Question 2
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Directions: The following question has four choices out of which ONLY ONE is correct.

Following question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. Any given statement in Column I can match with ONE OR MORE statements (s) in column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example:

If the correct matches are A - p, s and t; B - q and r, C - p and q; and D - s and t: the correct darkening of bubbles will look the following.

A
(A) (p, q), (B) (q, r), (C) r, (D) q

B
(A) (q, p), (B) (s, q), (C) q, (D) s

C
(A) (r, s), (B) (q, r), (C) s, (D) p

D
(A) (s, r), (B) (r, s), (C) q, (D) r

Solution

V_AB = 2 - 5
V_AB - 10 = - 4
Or 2 - 5 = 10 - 4
15 = 6
= 2.5 A
V_A - V_B = 0V
(Q) Using krischoffs laws in loop (1) and (2) we get
For loop (1), 4 = 2, + 2 (+ )
4, + 2 = 4
2, + = 2 â¦â¦.(1)
For loop (2) - 2 = - = 2 ( + )
2 = 3 + 2
2, + 3 = 2 â¦â¦.(2)
= 0.
V_A - V_B = 0
(R) =
= 2.5A
V_A -V_B - 12 = - 3 2.5
V_A - V_B - 12 - 7.5 = 4.5
V_A - V_B = 4.5 V
(s) applying Kirschoff`s voltage laws in loop (1) and (2) we get
For (1) 3 = 1 + â¦â¦â¦â¦.(1)
For (2) 4.5 = - - +
4.5 = - 2 + â¦â¦â¦â¦.(2)
(1) - (2) will give
- 1.5 = 3
= - 0.5
V_A - V_B = 0.5 V
(t) V_A - V_C = 4.5 V = 4
= = .125
V_A - V_B = 2.25 V
Question 3
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Directions: The following question has four choices out of which ONLY ONE is correct.

An unknown resistance R is connected to two cells E₁ and E₂ of negligible internal resistance across AB as shown below:

The emf of cell E₁ is

A
36 V

B
18 V

C
54 V

D
45 V

Solution

In loop CBDC using kirchoff`s voltage law we get
E₁ = 4 x 3 + 8 x 3
E₁ = 12 + 24 = 36V
Question 4
1 / -0

Directions: The following question has four choices out of which ONLY ONE is correct.

An unknown resistance R is connected to two cells E₁ and E₂ of negligible internal resistance across AB as shown below:

The potential drop across A and B must be

A
54 V

B
90 V

C
45 V

D
18 V

Solution

Using Kirschoff`s voltage laws in loop ACDA we get
E₂ = 8 3 + 6 5
E₂ = 54 V
V_A - E_Z + E₁ - VB = 0
V_AB = E₂ - E₁ = 54 - 36
V_AB = 18 V
Question 5
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Directions: The following question has four choices out of which ONLY ONE is correct.

An unknown resistance R is connected to two cells E₁ and E₂ of negligible internal resistance across AB as shown below:

The unknown resistance R is found to be

A
18

B
9

C
27

D
36

Solution

V_AB = 2 R
V_AB = 18 = 2 R
R = 9
Question 6
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Four capacitors are connected in a circuit, as shown in the figure. The effective capacitance (in ÂµF) between P and Q will be

A
5 ÂµF

B
10 ÂµF

C
2 ÂµF

D
7.5 ÂµF

Solution
Question 7
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Two conducting spheres A and B of respective radii a and b are at the same potential. The ratio of the surface charge densities of A and B is

A

B

C

D

Solution
Question 8
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Three charges 2q, -q and -q are located at the vertices of an equilateral triangle. At the circumcentre of the triangle,

A
the field is zero, but the potential is non-zero

B
the potential is zero and the field is infinity

C
both the field and the potential are zero

D
the field is non-zero, but the potential is zero

Solution

At the point P , the resultant electric field due all the charges will be the vector sum of individual electric fields hence resultant electric field will not be zero.

Electric potential at point P will be zero as the distance of all the charges from point P is same and sum of the all the charges is zero.
Question 9
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A particle of mass m carrying charge q is kept at rest in a uniform electric field E and then released. The kinetic energy gained by the particle, when it moves through a distance y, is

A
qEy²

B
qEy

C
qEy²

D
qE²y

Solution

Using conservation of energy
Increase in kinetic energy = Decrease in potential energy

, where q(-dV) is decrease in potential energy. As displacement and electric field are along the same direction, dV = -Ey.
Question 10
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Directions: The following question has four choices out of which ONLY ONE is correct.

C, V, U and Q are capacitance, potential difference, energy stored and charge of a parallel plate capacitor respectively. The quantities that increase when a dielectric slab is introduced between the plates without disconnecting the battery are

A
V and C

B
V and U

C
U and Q

D
V and Q

Solution

As the battery is connected, V will remain the same. With the insertion of a dielectric slab, the capacitance of the capacitor will increase; hence charge will also increase.
Question 11
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Directions: The following question has four choices out of which ONLY ONE is correct.

A heater of 220 V heats a volume of water in 5 minutes. The same heater when connected to 110 V heats the same volume of water in

A
2 minutes

B
20 minutes

C
10 minutes

D
2.5 minutes

Solution

Heat produced

i.e. when voltage is halved, heat produced becomes one-fourth. Hence time taken to heat the water becomes four times.

So the time taken will be 20 minutes
Question 12
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Directions: The following question has four choices out of which ONLY ONE is correct.

Two different conductors have the same resistance at 0^oC. It is found that the resistance of the first conductor at t₁^oC is equal to the resistance of the second conductor at t₂°C. The ratio of the temperature coefficients of resistance of the conductors, is

A

B

C

D
`

Solution

_{As the variation of the resistance with temperature is given as}
For the first conductor

For the second conductor

As per given condition, we have
Question 13
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Directions: The following question has four choices out of which ONLY ONE is correct.

In the given circuit diagram the current through the battery and the charge on the capacitor respectively in steady state are

A
1 A and 3 μC

B
17 A and 0 μC

C
A and μC

D
11 A and 3 μC

Solution

Current = = 11 A

Q = CV = 0.5 x 6 = 3C
Question 14
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Directions: The following question has four choices out of which ONLY ONE is correct.

A potentiometer wire of length 10 m and resistance 20 is connected in series with a 15 V battery and an external resistance of 40. A secondary cell of emf E in the secondary circuit is balanced by 240 cm long potentiometer wire. The emf E of the cell is

A
2.4 V

B
1.2 V

C
2.0 V

D
3 V

Solution

I = = 0.25 A
Potential gradient = = 0.5 V/m

PD across 240 cm = 0.5 2.4 = 1.2 V
Question 15
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Two parallel, large and thin metal plates have equal surface charge densities ( = 26.4 x 10^-12 C/m²) of opposite signs. The electric field between these plates is

A
1.5 N/C

B
1.5 x 10^-10 N/C

C
3 N/C

D
3 x 10^-10 N/C

Solution
Question 16
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Three charged capacitors of capacitance , respectively, having charge each are connected with resistor of resistance through switch S. S is closed at time t = 0. For this given situation, the initial current in the circuit is

A
20 A

B
10 A

C
2 A

D
0.1 A

Solution

The current in the RC circuit is given by

Where is the RC time constant and on just closing the circuit , t= 0 so

Where V₁, V₂and V₃ are potential differences across C₁, C₂ and C₃, respectively and V = Q / C
Question 17
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In the given figure, . Block A is neutral while charge on B is . Block B is released from rest at a distance of 1.8 m from A on the smooth horizontal surface. Initially, the spring has its free length and assume the blocks to be point masses.

The amplitude of oscillation of the combined mass will be

A

B

C

D

Solution

From
At
Question 18
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In the given circuit, it is observed that current I is independent of the value of the resistance R₆. The resistance value must satisfy

A
R₁R₂R₅ = R₃ R₄ R₆

B

C
R₁ R₄ = R₂ R₃

D
R₁ R₃ = R₂ R₄ = R₅ R₆

Solution

or R₁ R₄ = R₂ R₃
Question 19
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Directions: The following question has four choices out of which ONE or MORE is/are correct.

A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles

A
the charge on the capacitor increases

B
the voltage across the plates increases

C
the capacitance increases

D
the energy stored in the capacitor increases

Solution

The charging battery is removed. Therefore, q = constant
The distance between the plates is increased. Therefore, C decreases.
Now, V = , q is constant and C is decreasing.
Therefore, V should increase.
U = again q is constant and C is decreasing.
Therefore U should increase.
Question 20
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An elliptical cavity is carved within a perfect conductor. A positive charge q is placed at the centre of the cavity. The points A and B are on the cavity surface as shown in the figure. Then the

A
electric field near A in the cavity = electric field near B in the cavity

B
charge density at A = charge density at B

C
potential at A = potential at B

D
total electric field flux through the surface of the cavity is q/

Solution

Under electrostatic condition, all points lying on the conductor are at same potential. Therefore, potential at A = potential at B. Hence, option (3) is correct. From Gauss theorem, total flux through the surface of the cavity will be q/.