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NEET Minor Test - 5 - NEET MCQ

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30 Questions MCQ Test - NEET Minor Test - 5

NEET Minor Test - 5 for NEET 2024 is part of NEET preparation. The NEET Minor Test - 5 questions and answers have been prepared according to the NEET exam syllabus.The NEET Minor Test - 5 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for NEET Minor Test - 5 below.

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NEET Minor Test - 5 - Question 1

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The angle between the electric lines of force and the equipotential surface is

A. 0^∘
B. 45^∘
C. 90^∘
D. 180^∘

Detailed Solution for NEET Minor Test - 5 - Question 1

dV = −Edrcosθ

For equipotential surface,

dV = 0

cosθ = 0

⇒ θ = 90∘

NEET Minor Test - 5 - Question 2

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Two point charges −q and +q are placed at a distance of L, as shown in the figure.
The magnitude of electric field intensity at a distance R(R≫L) varies as:

A. 1/R²
B. 1/R³
C. 1/R⁴
D. 1/R⁶

Detailed Solution for NEET Minor Test - 5 - Question 2

For R≫L, arrangement is an electric dipole

where p = qL

E ∝ 1/R³

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NEET Minor Test - 5 - Question 3

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The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If C is its initial capacitance, its final capacitance is equal to:

A. C/4
B. 2C
C. C/2
D. 4C

Detailed Solution for NEET Minor Test - 5 - Question 3

NEET Minor Test - 5 - Question 4

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Six charges +q, −q, +q, −q, +q and − q are fixed at the corners of a hexagon of side d as shown in the figure. The work done in bringing a charge q₀ to the centre of the hexagon from infinity is: (ε₀. - permittivity of free space )

A.
B. Zero
C.
D.

Detailed Solution for NEET Minor Test - 5 - Question 4

Potential at the centre of hexagon is zero.

V0 = 0

∴ w_∞→0= q∆v = q0(0)

= 0

NEET Minor Test - 5 - Question 5

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The equivalent capacitance of the combination shown in the figure is

A. 3C
B. 2C
C. C/2
D. 3C/2

Detailed Solution for NEET Minor Test - 5 - Question 5

Given circuit is

Points 1, 2, 3 are at same potential (as they are connected by conducting wire)

So the capacitor is short circuited. It does not store any charge.

The circuit can be redrawn as

C_AB = C + C = 2C (Parallel combination)

NEET Minor Test - 5 - Question 6

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Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop.

Calculate the potential of the bigger drop.

A. 660 V
B. 1320 V
C. 1520 V
D. 1980 V

Detailed Solution for NEET Minor Test - 5 - Question 6

Electric potential due to a charged sphere = kQ/R

k = 9 × 10⁹N−m²/C²

Q: charge on sphere

R: Radius of sphere

Let charge and radius of smaller drop is q and r respectively

For smaller drop, V = kq/r = 220V

Let R be radius of bigger drop,

As volume remains the same

Now, using charge conservation,

⇒ Q = 27q

= 9 × 220 = 1980 V

NEET Minor Test - 5 - Question 7

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In a certain region of space with volume 0.2 m³, the electric potential is found to be 5V throughout. The magnitude of electric field in this region is:

A. 0.5N/C
B. 1N/C
C. 5N/C
D. zero

Detailed Solution for NEET Minor Test - 5 - Question 7

since, electric potential is constant throughout the volume, hence electric field,

E = −dV/dr

= 0

NEET Minor Test - 5 - Question 8

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The capacitance of a parallel plate capacitor with air as medium is 6µF. With the introduction of a dielectric medium, the capacitance becomes 30µF. The permittivity of the medium is:

(E₀ = 8.85 × 10⁻¹²C²N⁻¹m⁻²)

A. 1.77 × 10⁻¹²C²N⁻¹m⁻²
B. 0.44 × 10⁻¹⁰C²N⁻¹m⁻²
C. 5.00C²N⁻¹m⁻²
D. 0.44 × 10⁻¹³C²N⁻¹m⁻²

Detailed Solution for NEET Minor Test - 5 - Question 8

Capacitance of a parallel plate capacitor with air is

Here, A = area of plates of capacitor, d = distance between the plates Capacitance of a same parallel plate capacitor with introduction of dielectric medium of

dielectric constant K is

Dividing (ii) by(i)

NEET Minor Test - 5 - Question 9

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Two point charges A and B, having charges +Q and −Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes

A. 4F / 3
B. F
C. 9F / 16
D. 16F / 9

Detailed Solution for NEET Minor Test - 5 - Question 9

In case I :

In case II :

∴

From equations (i) and (ii),

NEET Minor Test - 5 - Question 10

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An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is

A. smaller
B. 5 times greater
C. 10 times greater
D. equal

Detailed Solution for NEET Minor Test - 5 - Question 10

Force experienced by a charged particle in an electric field, F = qE

As F = ma

As electron and proton both fall from same height at rest. Then initial velocity =0

From

∴

∴ Electron has smaller mass so it will take smaller time.

NEET Minor Test - 5 - Question 11

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The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is

A. Independent of the distance between the plates
B. Linearly proportional to the distance between the plates
C. Proportional to the square root of the distance between the plates
D. Inversely proportional to the distance between the plates

Detailed Solution for NEET Minor Test - 5 - Question 11

For isolated capacitor, charge Q= constant.

Electrostatic force,

NEET Minor Test - 5 - Question 12

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Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e+Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of

[given ma of hydrogen mh = 1.67×10⁻²⁷kg]

A. 10⁻²³C
B. 10⁻³⁷C
C. 10⁻⁴⁷C
D. 10⁻²⁰C

Detailed Solution for NEET Minor Test - 5 - Question 12

A hydrogen atom consists of an electron and a proton.

∴ Charge on one hydrogen atom

= qe + qp = −e + (e + Δe) = Δe

Since a hydrogen atom carry a net charge Δe

∴ Electrostatic force

will act between two hydrogen atoms.

The gravitational force between two hydrogen atoms is given as

Since, the net force on the system is zero,F_e = F_g

Using eqns. (i) and (ii), we get

NEET Minor Test - 5 - Question 13

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A capacitor of 2 μF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is

A. 75%
B. 80%
C. 0%
D. 20%

Detailed Solution for NEET Minor Test - 5 - Question 13

Initially, the energy stored in 2 μF capacitor is

Initially, the charge stored in 2μF capacitor is Q_i = CV = (2 × 10⁻⁶)V = 2V × 10⁻⁶ coulomb.

When switch S is turned to position 2, the charge flows and both the capacitors share charges till a common potential V_C is reached

Finally, the energy stored in both the capacitors

% loss of energy,

= 80%

NEET Minor Test - 5 - Question 14

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An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 105NC⁻¹.It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is

A. 8 mC
B. 2 mC
C. 5 mC
D. 7 mC

Detailed Solution for NEET Minor Test - 5 - Question 14

Here θ = 30°,E = 2 × 10⁵NC⁻¹

τ = 4Nm,l = 2cm = 0.02m,q= ?

τ = pEsinθ = (ql)Esinθ

∴

= 2mc

NEET Minor Test - 5 - Question 15

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People living at sea level have around 5 million RBC per cubic millimeter of their blood whereas those living at an altitude to 5400 metres have around 8 million. This is because at high altitude

A.
People eat more nutritive food, therefore more RBCs are formed
B.
People get pollution-free air to breath and more oxygen is available
C.
Atmospheric O₂ level is less and hence more RBCs are needed to absorb the required amount of O₂ to survive
D.
There is more UV radiation which enhances RBC production.

Detailed Solution for NEET Minor Test - 5 - Question 15

At high altitudes composition of air remains almost same as at sea level, but density (barometric pressure) of air gradually decreases due to which arterial pO₂ is also decreased (hypoxemia). High altitudes has complex conditions to which human body has to acclimatize. Number of red blood cells per unit volume of blood is likely to be higher in a person living at high altitudes. This is in response to the air being less dense at high altitude. More number of red blood cells are needed to trap O₂ from rarefied air having low pO₂ (partial pressure of oxygen).

NEET Minor Test - 5 - Question 16

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An electric bulb is rated 220 V–100 W. The power consumed by it when operated on 110 V will be

A. 50 W
B. 75 W
C. 40 W
D. 25 W

Detailed Solution for NEET Minor Test - 5 - Question 16

Resistance of the bulb

R = V² / P = (220)² / 100 = 484 Ω

Power when connected across 110 V source

∴ Power = 110 × 110 / 484 = 25 W

NEET Minor Test - 5 - Question 17

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A (100 W, 200 V) bulb is connected to a 160 V power supply. The power consumption would be

A. 125 W
B. 100 W
C. 80 W
D. 64 W

Detailed Solution for NEET Minor Test - 5 - Question 17

Power = 100 W, Voltage = 200 V

The resistance of bulb = V² / P = 200 x 200 / 100 = 400Ω

When the bulb is connected across 160 V

Current in bulb = 160 / 400 A,

Power consumption = VI = 160 × 160 / 400 = 64 W

NEET Minor Test - 5 - Question 18

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The relation between current and drift velocity is

A.
I = nev_d
B.
I/A = nev_d
C.
IA=nev_d
D.
I = nv_dA / e

Detailed Solution for NEET Minor Test - 5 - Question 18

As we know, I = neAv_d

Therefore, I/A = nev_d

NEET Minor Test - 5 - Question 19

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A conductor wire having 10²⁹ free electrons carries a current of 20 A. If the cross-section of the wire is 1 mm², then the drift velocity of electrons will be

A. 6.25 × 10⁻³ m s⁻¹
B. 1.25 × 10⁻⁵ m s⁻¹
C. 1.25 × 10⁻³ m s⁻¹
D. 1.25 × 10⁻⁴ m s⁻¹

Detailed Solution for NEET Minor Test - 5 - Question 19

= 1.25 × 10⁻³m/s

NEET Minor Test - 5 - Question 20

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The current passing through the ideal ammeter, in the circuit given below, is

A. 1.25 A
B. 0.5 A
C. 0.75 A
D. 1 A

Detailed Solution for NEET Minor Test - 5 - Question 20

To calculate the equivalent resistance for the ammeter, the resistors of 2 Ω and 2 Ω are in parallel with each other and then the whole circuit has 3 resistances in series.

So for parallel combination,

∴

Now, internal resistance of the battery, that is 1 Ω, resistance of 2 Ω and resistance R = 1 Ω, are all in series.

R_eq for series is the sum of all the resistances.

∴ R_eq = 1 Ω + 2 Ω + 1 Ω = 4 Ω

Hence current, I = V / R

I = 4 / 4 = 1 A

NEET Minor Test - 5 - Question 21

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Shown in the figure is a meter-bridge set up with null deflection in the galvanometer. The value of the unknown resistor R is

A. 13.75 Ω
B. 220 Ω
C. 110 Ω
D. 55 Ω

Detailed Solution for NEET Minor Test - 5 - Question 21

Meter bridge works on the principle of balanced Wheatstone's bridge. So, using that concept, balanced condition for a meter bridge is,

⇒

⇒ R = 55 × 4 = 220 Ω

NEET Minor Test - 5 - Question 22

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Constant current is flowing through a linear conductor of non-uniform area of cross-section. The charge flowing per second through the area of conductor at any cross-section is

A. Proportional to the area of cross-section
B. Inversely proportional to the area of cross-section
C. Independent of the area of cross-section
D. Dependent on the length of conductor

Detailed Solution for NEET Minor Test - 5 - Question 22

The quantity of charge flowing through any cross-section per second is defined as Current.

And as it is given that, Current through the conductor is constant so it will be same throughout irrespective of the cross-section area.

Hence, the charge flowing per second will be independent of the area of cross-section.

NEET Minor Test - 5 - Question 23

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A meter bridge is used to determine the resistance of an unknown wire by measuring the balance point length l. If the wire is replaced by another wire of same material but with double the length and half the thickness, the balancing point is expected to be

A. 1 / 8l
B. 1 / 4l
C. 8l
D. 16l

Detailed Solution for NEET Minor Test - 5 - Question 23

In a meter bridge the ratio of two resistances is

Where l and l′ are balancing lengths.

Resistance

In material remains same ρ = ρ′

Given, l′ = 2l

r′ = r / 2

∴

R′ = 8R

Therefore, the new balancing point is expected to be 8l.

NEET Minor Test - 5 - Question 24

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Potential difference between points A and B (i.e. V_A - V_B) is

A.
4 V
B.
2 V
C.
6 V
D.
8 V

Detailed Solution for NEET Minor Test - 5 - Question 24

along wire ACDB

V_A - 2 - 6 + 4 = V_B

V_A - V_B = 4 V

NEET Minor Test - 5 - Question 25

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When the switches are arranged so that the current through the battery is maximum, what is the voltage across points A and B-

A. 4 V
B. 0 V
C. 1 V
D. 2 V

Detailed Solution for NEET Minor Test - 5 - Question 25

When S₁ is closed

current through the battery is maximum

I = 24 / 1 = 24 Ampere

V_A− V_B= E – ir

= 24− 24 × 1 = 0

NEET Minor Test - 5 - Question 26

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The dimensions [MLT⁻²A⁻²] belong to the

A. Magnetic flux
B. Self inductance
C. Magnetic permeability
D. Electric permittivity

Detailed Solution for NEET Minor Test - 5 - Question 26

Dimensional formula of magnetic permeability is [MLT⁻²A⁻²]

NEET Minor Test - 5 - Question 27

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A uniform conducting wire of length 12a and resistance 'R' is wound up as a current carrying coil in the shape of,

(i) an equilateral triangle of side 'a'.

(ii) a square of side 'a'.

The magnetic dipole moments of the coil in each case respectively are

A. 3 Ia² and 3 Ia²
B. 3 Ia² and Ia²
C. 3 Ia² and 4 Ia²
D. 4 Ia² and 3 Ia²

Detailed Solution for NEET Minor Test - 5 - Question 27

Current in the loop will be V/R = l which is same for both loops.

Now magnetic moment of Triangle loop

and magnetic moment of square loop = N′IA′

M₂v = 3la²

NEET Minor Test - 5 - Question 28

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The magnetic field B on the axis of a circular coil at distance x far away from its centre are related as:

A.
Bαx^-1
B.
Bαx^-4
C.
Bαx^-3
D.
Bαx^-2

Detailed Solution for NEET Minor Test - 5 - Question 28

dB=μ_oidlsin90/4πr²=μ_oidl/4πr²
B_net= ∫dBsinθ
= ∫μ_oidlR/4πr²r== (μ_oiR/4πr³) ∫dl
= (μ_oiR/4πr³)2πR= μ_oiR²/2(R²+x²)^3/2
B= μ_oiR²/2(R2+x²)^3/2
If x>>>R
B= μ_oiR²/2(x²)^3/2

[=B ∝ x^-3]

NEET Minor Test - 5 - Question 29

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The magnetic susceptibility is negative for

A. ferromagnetic material only
B. paramagnetic and ferromagnetic materials
C. diamagnetic material only
D. paramagnetic material only

Detailed Solution for NEET Minor Test - 5 - Question 29

Magnetic susceptibility is negative for diamagnetic material only.

NEET Minor Test - 5 - Question 30

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A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2Weber/m². The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be

A. 0.24 Nm
B. 0.12 Nm
C. 0.15 Nm
D. 0.20 Nm

Detailed Solution for NEET Minor Test - 5 - Question 30

The required torque is τ = NIABsinθ

where N is the number of turns in the coil, I is the current through the coil, B is the uniform magnetic field, A is the area of the coil and 0 is the angle between the direction of the magnetic field and normal to the plane of the coil.

Here,

N = 50, I = 2A, A = 0.12m × 0.1m = 0.012m²

∴

= 0.20Nm

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