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NEET Part Test - 4 - NEET MCQ


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30 Questions MCQ Test - NEET Part Test - 4

NEET Part Test - 4 for NEET 2024 is part of NEET preparation. The NEET Part Test - 4 questions and answers have been prepared according to the NEET exam syllabus.The NEET Part Test - 4 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for NEET Part Test - 4 below.
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NEET Part Test - 4 - Question 1

A and C are concentric conducting spherical shells of radius a and c respectively. A  is surrounded by a concentric dielectric medium of inner radius a, outer radius b and dielectric constant K as shown in the figure. If sphere A is given a charge Q, the potential at the outer surface of the dielectric is.

Detailed Solution for NEET Part Test - 4 - Question 1

Charge on outer surface of C = - charge on inner surface of C
Hence potential at B due to charge on conductor C = 0 charge on outer surface of dielectric = - charge on inner surface of dielectric
∴ Potential at B due to charge on dielectric = 0 
Potential at B due to charge on A 
∴ net potential at B 

NEET Part Test - 4 - Question 2

In the given circuit the rms value of voltage across the capacitor C, inductor L and resister R1 are 12V, 10V and 5V respectively. Then the peak voltage across R2 is

Detailed Solution for NEET Part Test - 4 - Question 2

In parallel combination each branch should have equal potential but in AC circuit in a branch 




∴ 

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NEET Part Test - 4 - Question 3

Two identical capacitors C1 and C2 are connected in series with a battery. They are fully charged. Now a dielectric slab is inserted between the plates of C2. The potential difference across C1 will :

Detailed Solution for NEET Part Test - 4 - Question 3

Potential difference across ‘C1

When dielectric is inserted, C2 will increase. So V1 will increase.

NEET Part Test - 4 - Question 4

A solenoid having an iron core has its terminals connected across an ideal DC source. If the iron core is removed, the current flowing through the solenoid

Detailed Solution for NEET Part Test - 4 - Question 4

Current will have to increase in order to oppose the cause (decrease in flux). 

NEET Part Test - 4 - Question 5

An electric field ‘E’ whose direction is radially outward varies as distance from origin ‘r’ as shown in the graph. E is taken as positive if its direction is away from the origin. Then the work done by electric field on a 2 C charge if it is taken from (1, 1, 0) to (3, 0, 0) is 

Detailed Solution for NEET Part Test - 4 - Question 5

work done on 2C charge 
[∵ r for (1, 1, 0) = √2 & r for (3,0, 0) = 3] 
= 2 x (area of E-r graph from r = √2 m to r = 3) 

NEET Part Test - 4 - Question 6

A negative test charge is moving near a long straight current carrying wire. A force will act on the test charge in a direction parallel to the direction of the current, if the motion of the charge is in a direction:

Detailed Solution for NEET Part Test - 4 - Question 6

The magnetic force shall act on the negatively charged test particle in direction parallel to current carrying wire, if the velocity  of the negative charge is as shown. 

NEET Part Test - 4 - Question 7

Potential difference between centre and the surface of sphere of radius R and having uniform volume charge density ρ within it will be :

Detailed Solution for NEET Part Test - 4 - Question 7


NEET Part Test - 4 - Question 8

The magnetic field at the origin due to the current flowing in the wire is 

Detailed Solution for NEET Part Test - 4 - Question 8

NEET Part Test - 4 - Question 9

There exists a uniform electric field in the space as shown. Four points A, B, C and D are marked which are equidistant from the origin. If VA, VB, VC and VD are their potentials respectively, then   

Detailed Solution for NEET Part Test - 4 - Question 9

Four lines, perpendicular to lines of electric field and passing through A, B, C and D are drawn. These are equipotential lines. As potential decreases in the direction of electric field, therefore VA > VB > VD > VC 

NEET Part Test - 4 - Question 10

A current carrying wire AB of the length L is turned along a circle, as shown in figure. The magnetic field at the centre O. 

   

Detailed Solution for NEET Part Test - 4 - Question 10

Let r be the radius of new arc 
∴ r(2π - θ) = L or 
∴ Magnetic field at centre O is

NEET Part Test - 4 - Question 11

Total electric force on an electric dipole placed in an electric field of a point charge is:

Detailed Solution for NEET Part Test - 4 - Question 11

Electric field of a point charge is non-uniform hence net force can never be zero. 

NEET Part Test - 4 - Question 12

The magnitude of magnetic field at O (centre of the circular part) due to the current carrying coil as shown is :

Detailed Solution for NEET Part Test - 4 - Question 12

Magnetic field due to circular segment 
Due to one straight wire segment (sin 45° + sin 0°) = 
Net field 

NEET Part Test - 4 - Question 13

Eight point charges (can be assumed as small spheres uniformly charged and their centres at the corner of the cube) having values q each are fixed at vertices of a cube. The electric flux through square surface ABCD of the cube is 

Detailed Solution for NEET Part Test - 4 - Question 13

Net charge enclosed inside the cube 
net electric flux through cube 
∵ charges are symmetrically placed so the flux through each face 

NEET Part Test - 4 - Question 14

A charged particle is moving in the region around a long current carrying wire. Due to wire, it may experience:

Detailed Solution for NEET Part Test - 4 - Question 14

The charge particle will experience magnetic as will as gravitational force due to wire. Gravitation force is very small.

NEET Part Test - 4 - Question 15

The figure shows a charge q placed  inside a cavity in an uncharged conductor. 
Now if an external electric field is switched on :    

Detailed Solution for NEET Part Test - 4 - Question 15

The distribution of charge on the outer surface, depends only on the charges outside, and it distributes itself such that the net, electric field inside the outer surface due to the charge on outer surface and all the outer charges is zero. Similarly the distribution of charge on the inner surface, depends only on the charges inside the inner surface, and it distributes itself such that the net, electric field outside the inner surface due to the charge on inner surface and all the inner charges is zero.
Also the force on charge inside the cavity is due to the charge on the inner surface. Hence only induced charge on outer surface will redistribute 

NEET Part Test - 4 - Question 16

Two infinite sheets carrying current in same direction  (of equal current per unit length K) are separated by a distance ‘d’. A proton is released from a point between the plates with a velocity parallel to the sheets but perpendicular to the direction of current in the sheets. Then the path of the proton is 

Detailed Solution for NEET Part Test - 4 - Question 16

As the net force acting on the proton is zero, it will move on straight line

NEET Part Test - 4 - Question 17

Three large parallel charged conducting plates are placed at a small distance d apart from each other. The surface charge density on the face B of the conductor is +5.0 εo coulomb/m2 and that on face E is + 6.0 εo columb/m2. Charges on other faces are not known. The electric field intensity at P (shown in the figure) between the plates is     

Detailed Solution for NEET Part Test - 4 - Question 17

From Gauss theorem surface charge density on C face is – σ . 

NEET Part Test - 4 - Question 18

A horizontal metallic rod of mass 'm'  and length  ' l '  is supported by two vertical identical springs of spring constant 'K'  each and natural length  l​0. A current  ' i ' is flowing in the rod in the direction shown. If the rod is in equilibrium then the length of each spring in this state is:

Detailed Solution for NEET Part Test - 4 - Question 18

The force on the rod due to magnetic field and gravity is i ℓ B–mg (upwards)
Hence the extension in the springs is

(Note that effective spring constant is 2k)
Therefore the length of the spring is 

NEET Part Test - 4 - Question 19

A 150 m long metal wire connects points A and B. The electric potential at point B is 50 V less than that at point A. If the conductivity of the metal is 60 × 106 mho/m, then magnitude of the current density in the wire is equal to :

Detailed Solution for NEET Part Test - 4 - Question 19


(or as J = σ E)
using values j = 20 x 106 A/m2

NEET Part Test - 4 - Question 20

A toroid of mean radius ' a ' , cross section radius ' r ' and total number of turns N. It carries a current ' i '. The torque experienced by the toroid if a uniform magnetic field of strength B is applied :

Detailed Solution for NEET Part Test - 4 - Question 20

The resultant magnetic dipole moment of toroid is zero. of small parts of toroid turn along a circle and hence there resultant is zero.
∴ Torque acting on it is zero. 

NEET Part Test - 4 - Question 21

The equivalent resistance between A and B will be (in ?) 

Detailed Solution for NEET Part Test - 4 - Question 21


As C & D are at same potential by symmetry of circuit ° 
It is balanced wheat-stone bridge Hence the circuit has equivalant resistance 7/3 Ω 

NEET Part Test - 4 - Question 22

As shown in figure, a permanent magnet and current carrying coil are placed. If the coil is moved towards magnet, then current in coil (Magnet is symmetrical) :

Detailed Solution for NEET Part Test - 4 - Question 22

No change in the flux occurs due to the described motion of the magnet. Hence no current will be induced in the coil. 

NEET Part Test - 4 - Question 23

In the figure shown:

Detailed Solution for NEET Part Test - 4 - Question 23

Current through resistance woll be from
A to B if
20 – ε > 2 ⇒ ε < 18
and from
B to A if  20 – ε < 2

NEET Part Test - 4 - Question 24

A super conducting loop having an inductance 'L' is kept in a magnetic field which is varying with respect to time. If is the total flux, e = total induced emf, then:

Detailed Solution for NEET Part Test - 4 - Question 24

For any loop εtotal , = iR, For superconductor, R = 0 

∴ εtotal = 0 ⇒  ⇒  φ = constant

NEET Part Test - 4 - Question 25

A battery of internal resistance ' r ' and e.m.f. ε is connected to a variable external resistor AB. If the sliding contact is moved from A to B, then terminal potential difference of battery will :

Detailed Solution for NEET Part Test - 4 - Question 25

V = E - i r. Now, if we go from A → B, then resistance of the circuit increases, ' i ' decreases
∴ V increases 

NEET Part Test - 4 - Question 26

Figure shows a conducting  horizontal rod of resistance r  is made to oscillate simple harmonically with a fixed amplitude in a uniform and constant magnetic field B, directed inwards. The ends of rod always touch two parallel fixed vertical  conducting rails. The ends of rails are joined by an inductor and a capacitor having self inductance and capacitance 1/π Henry and 1/π farad respectively. The amplitude of current in the circuit depends on the frequency of oscillation  of rod. The amplitude of the current will be maximum when the time period of rod is : (do not consider self inductance anywhere other than in the inductor) 

Detailed Solution for NEET Part Test - 4 - Question 26

Here oscillating rod is an AC source because emf induced in it is (vBℓ) ; which varies sinusoidally because v varies sinusoidally.
Maximum current will flow through the circuit under resonance condition. Therefore time period of oscillation of rod  is 

NEET Part Test - 4 - Question 27

In the figure shown the equivalent capacitance between 'A' and 'B' is :

Detailed Solution for NEET Part Test - 4 - Question 27

Equivalent circuit is

NEET Part Test - 4 - Question 28

A very small circular loop of area 5 × 10–4 m2 and resistance 2 ohm  is initially concentric and coplanar with a stationary loop of radius 0.1 m. If one ampere constant current is passed through the bigger loop and the smaller loop is rotated about its diameter with constant angular velocity ω. The current induced (in ampere) in the smaller loop will be :

Detailed Solution for NEET Part Test - 4 - Question 28


NEET Part Test - 4 - Question 29

The capacitor shown in figure 1 is charged completely by connecting switch S to contact a. If switch S is thrown to contact b at time t = 0, which of the curves in figure 2 above represents the magnitude of the current through the resistor R as function of time t ?

Detailed Solution for NEET Part Test - 4 - Question 29

The potential difference across completely charged capacitor is V. As the switch is pushed to b, the initial current in the resistor R is V/R. Hence J is the correct curve. 

NEET Part Test - 4 - Question 30

At a given instant the current and self induced emf in an inductor are directed as shown in figure. If the induced emf is 17 volt and rate of change of current is 25 kA/s the correct statement is :

Detailed Solution for NEET Part Test - 4 - Question 30

Current is decreasing and 

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