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30 Questions MCQ Test NEET Past Year Papers - Test: NEET 2021 Past Year Paper

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Test: NEET 2021 Past Year Paper - Question 1

An infinitely long straight conductor carries a current of 5A as shown. An electron is moving with a speed of 105 m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 1

M.F due to long current carrying wire at 20 cm,

B = 0.5 × 10−5 T
Now,
Force experienced by electron = qVB
Fm = eVB
Fm = 1.6 × 10−19 × 105 × 0.5 × 10−5
= 8 × 10−20 N

Test: NEET 2021 Past Year Paper - Question 2

A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 2

Given, frequency of SIHM = n
as we know,
∵ In one time period PE become maximum & minimum two time 
So, It means frequency of PE is 2n.

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Test: NEET 2021 Past Year Paper - Question 3

A radioactive nucleus undergoes spontaneous decay in the sequencewhere Z is the atomic number of element X. The possible decay particles in the sequence are:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 3

As we know due to α-decay atomic number decreases by 2 and mass number decreases by 4. Due to β− decay mass number remains unchange and atomic number increases by 1.
Due to  β+ decay mass number remains unchange but atomic number decreares by 1.
So, option (3) is correct option.

Test: NEET 2021 Past Year Paper - Question 4

The escape velocity from the Earth's surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is :

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 4

As we know escape velocity is given by,

Escape velocity from Earth’s surface = v
Escape velocity from surface of another planet = ?
Rplanet = 4Rearth
Splanet = Searth
So we can write,

It means,

Test: NEET 2021 Past Year Paper - Question 5

The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 5

Given,
T1/2 = 100 hours
t = 150 hours
∵ no. of nuclide remain after n half life, N = N0/2n
∵ 100 hours = 1 half life
150 hours = 1/100 x 150
= 3/2 half life
So, Fraction Remain after 3/2 half life

Test: NEET 2021 Past Year Paper - Question 6

A convex lens 'A' of focal length 20cm and a concave lens 'B' of focal length 5cm are kept along the same axis with a distance 'd' between them. If a parallel beam of light falling on 'A' leaves 'B' as a parallel beam, then the distance 'd' in cm will be :

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 6


Total distance = F1 + F2
= 25 cm

Test: NEET 2021 Past Year Paper - Question 7

A capacitor of capacitance 'C', is connected across an ac source of voltage V, given by V = V0 sin ωt The displacement current between the plates of the capacitor, would then be given by:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 7

∵ q = CV
And  V = V0 sin ωt
Id = ?
As we know,

Test: NEET 2021 Past Year Paper - Question 8

A small block slides down on a smooth inclined plane, starting from rest at time t = 0 Let Sn be the distance travelled by the block in the interval t = n − 1 to t = n Then, the ratio is:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 8


Now,
eq (1) divide by (2)

Test: NEET 2021 Past Year Paper - Question 9

A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 9

∵ At every pt.
KE + PE =constant
so, (Kε)A + (Pε)A = (kε)B + (Pε)B
0 + mgS = 3(Pε)B + (Pε)B
mgS = 4mgh
h = S/4
Now,
∵ At pt.B -
Kε = 3Pε

Test: NEET 2021 Past Year Paper - Question 10

In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36cm length of wire. If another cell of EMF 2.5V replaces the first cell, then at what length of the wire, the balance point occurs?

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 10

Given,
E1 = 1.5 V
l1 = 36 cm
E2 = 2.5v
l2 = ?
AS we know,

l2 = 60 cm

Test: NEET 2021 Past Year Paper - Question 11

For a plane electromagnetic wave propagating in x −direction, which one of the following combination gives the correct possible directions for electric field (E) and magnetic field (B) respectively?

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 11

Test: NEET 2021 Past Year Paper - Question 12

Polar molecules are the molecules:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 12

A polar molecule is a molecule in which one end of the molecule is slightly positive, while the other end is slightly negative. Therefore, these molecules have permanent electric dipole moment.

Test: NEET 2021 Past Year Paper - Question 13

The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is d/2, then the viscous force acting on the ball will be :

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 13


Fr + Fb = Mg
Fr = Mg-  Fb

Test: NEET 2021 Past Year Paper - Question 14

Match Column - I and Column - II and choose the correct match from the given choices.

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 14

Standard formula for each physical quantity needs to be used. 
(A → Q), (B → P), (C → S), (D → R)

Test: NEET 2021 Past Year Paper - Question 15

Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10 \% of the input energy. How much power is generated by the turbine? (g = 10m/s2)

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 15

Change in potential energy of water per second = 15 × g × 60
= 9000 J⁄s
Energy remaining per second after loss = 90% of 9000 J⁄s
= (90/100) x 9000
= 8100 J⁄s = 8.1 kW

Test: NEET 2021 Past Year Paper - Question 16

A lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since :

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 16

The larger the objective, the more light telescope collects and increases the brightness of image and large focal length enhances the magnifying power of the telescope.

Test: NEET 2021 Past Year Paper - Question 17

The electron concentration in an: n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 17

Electrons effective mass is smaller than holes therefore mobility of electrons is higher than holes and for equal electric field, drift velocity of the electron will be greater compared to holes.
As concentration is also same for both the cases, hence magnitude of current due to electron will be greater compared to that of holes as I = neAv.

Test: NEET 2021 Past Year Paper - Question 18

A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6MeV while that of fragments is 8.5MeV. The total gain in the Binding Energy in the process is:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 18


BE gain = (120 × 8.5) + (120 × 8.5) − (240 × 7.6)
= 2040 − 1824
= 216 MeV

Test: NEET 2021 Past Year Paper - Question 19

A thick current carrying cable of radius 'R' carries current 'I' uniformly distributed across its cross-section. The variation of magneticfield B(r) due to the cable with the distance 'r' from the ax is of the cable is represented by:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 19

For inside part
J = I/πR2


Test: NEET 2021 Past Year Paper - Question 20

Two charged spherical conductors of radius R1 and R2 are connected by a wire. Then the ratio of surface charge densities of the spheres (σ12) is :

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 20


V1 = V2

Test: NEET 2021 Past Year Paper - Question 21

If E and G respectively denote energy and gravitational constant, then E/G has the dimensions of:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 21

Dimensional Formulas, Energy,
[E] = M1 L2 T−2
Gravitational constant

Now, 

Test: NEET 2021 Past Year Paper - Question 22

A spring is stretched by 5cm by a force 10N. The time period of the oscillations when a mass of 2kg is suspended by it is:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 22

Spring force, F = kx
Spring constant, k = F/x = 10/0.5 = 200 N⁄m
Time period, T = 2π
T = 2π= 0.2 π = 0.628 s

Test: NEET 2021 Past Year Paper - Question 23

Column - I gives certain physical terms associated with flow of current through a metallic conductor. Column - II gives some mathematical relations involving electrical quantities. Match Column - I and Column - II with appropriate relations.

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 23


A − R, B − S, C − P, D − Q

Test: NEET 2021 Past Year Paper - Question 24

A dipole is placed in an electric field as shown. In which direction will it move? 

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 24


F1 = qE1, F2 = qE2
E1 > E2 ⇒ F1 > F2
Hence, net force is towards right and its potential energy will decrease.

Test: NEET 2021 Past Year Paper - Question 25

Consider the following statements (A) and (B) and identify the correct answer.
(A) A zener diode is connected in reverse bias, when used as a voltage regulator.
(B) The potential barrier of p − n junction lies between 0.1V to 0.3V

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 25

(A) is correct while (B) is incorrect because Si diode has barrier potential of 0.7V.

Test: NEET 2021 Past Year Paper - Question 26

A screw gauge gives the following readings when used to measure the diameter of a wire
Main scale reading : 0mm
Circular scale reading : 52 divisions
Given that 1mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 26

L.C. = 1/100 mm
= 0.01 mm
Now,
Diameter of wire = 52 × 0.01 = 0.52 mm = 0.052 cm

Test: NEET 2021 Past Year Paper - Question 27

An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance ' R ' are connected in series to an ac source of potential difference ' V ' volts as shown in figure. Potential difference across L,C and R is 40V, 10V and 40V, respectively. The amplitude of current flowing through LCR series circuit is 10√2A. The impedance of the circuit is:

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 27




⇒ z = 50 Ω

Test: NEET 2021 Past Year Paper - Question 28

A parallel plate capacitor has a uniform electric field 'in the space between the plates. If the distance between the plates is 'd' and the area of each plate is 'A', the energy stored in the capacitor is : ∈0 = permittivity of free space)

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 28

Electric energy density u = 1/2 ∈0E2
Volume of the space between plates of Capacitor is Ad.
∴ Energy Stored = Energy density × Volume
 1/2 ∈​​​​​​​0E2 Ad

Test: NEET 2021 Past Year Paper - Question 29

An electromagnetic wave of wavelength 'λ' is incident on a photosensitive surface of negligible work function. If ' m ' mass is of photoelectron emitted from the surface has de-Broglie wavelength λd, then :

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 29

From Einstein’s equation,
Eincident = ϕ ± KEmax

Test: NEET 2021 Past Year Paper - Question 30

Find the value of the angle of emergence from the prism. Refractive index of the glass is √3.

Detailed Solution for Test: NEET 2021 Past Year Paper - Question 30


No refraction at face AC.
Apply Snell's law at face BC, we get
μ1sini = μ2sinr
√3sin30= 1 × sine
√3/2 = sine
⇒ e = 60

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