NEET 2021 Question paper & Solutions PDF by Embibe
PHYSICS (Code -M5)
1. An infinitely long straight conductor carries a current of 5π΄ as shown. An electron is moving with a speed of 105m/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.
(1) 4 Γ 10β20N
(2) 8π Γ 10β20N
(3) 4π Γ 10β20N
(4) 8 Γ 10β20N
Correct Answer: (4)
Q.1. Solution:
M.F due to long current carrying wire at 20 cm,
π΅ =π0πΌ
2ππ
π΅ =2 Γ 10β7 Γ 5
20 Γ 10β2
π΅ = 0.5 Γ 10β5 T
Now,
Force experienced by electron = πVπ΅
πΉπ = πππ΅
πΉπ = 1.6 Γ 10β19 Γ 105 Γ 0.5 Γ 10β5
= 8 Γ 10β20 N
2. A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is:
(1) n
(2) 2π
(3) 3n
(4) 4n
Correct Answer: (2)
Q.2. Solution:
Given,
frequency of SIHM = π
as we know,
β΅ In one time period ππΈ become maximum & minimum two time
So,
It means frequency of ππΈ is 2π.
3. A radioactive nucleus πAX undergoes spontaneous decay in the sequence Z
AX β Zβ1B β Zβ3C β Zβ2D where
Z is the atomic number of element X. The possible decay particles in the sequence are:
(1) πΌ, π½β, π½+
(2) πΌ, π½+, π½β
(3) π½+, πΌ, π½β
(4) π½β, πΌ, π½+
Correct Answer: (3)
Q.3. Solution:
As we know due to πΌ-decay atomic number decreases by 2 and mass number decreases by 4. Due to π½βdecay
mass number remains unchange and atomic number increases by 1.
Due to π½+ decay mass number remains unchange but atomic number decreares by 1.
So, option (3) is correct option.
4. The escape velocity from the Earth's surface is π£. The escape velocity from the surface of another planet having a
radius, four times that of Earth and same mass density is :
(1) π£
(2) 2π£
(3) 3π£
(4) 4π£
Correct Answer: (4)
Q.4. Solution:
As we know escape velocity is given by,
π£π = β2ππ = β2πΊπ
43ππ 3
4
3ππ 2 = β
2πΊπ
π
4
3ππ 2
ππ β βπ 2 β π
Escape velocity from Earthβs surface = π£
Escape velocity from surface of another planet =?
π ππππππ‘ = 4π ππππ‘β
πππππππ‘ = πππππ‘β
So we can write ,
(π£π)earth
(π£π) planet =
π ππππ‘β
π ππππππ‘=
π
4π =
1
4
It means,
(π£π )earth
= (π£π)Planet
(π£π)Planet = 4π£
5. The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours
would be :
(1) 1/2
(2) 1
2β2
(3) 2
3
(4) 2
3β2
Correct Answer: (2)
Q.5. Solution:
Given,
π12= 100 hours
π‘ = 150 hours
β΅ no. of nuclide remain after π half life, π =π0
2π
β΅ 100 hours = 1 half life
150 hours =1
100Γ 150
=3
2 half life
So, Fraction Remain after 3
2 half life
π
π0= (
1
2)3/2
=1
2β2
6. A convex lens 'A' of focal length 20cm and a concave lens ' π΅ ' 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 :
(1) 25
(2) 15
(3) 50
(4) 30
Correct Answer: (1)
Q.6. Solution:
Total distance = π1 + π2
= 25 ππ
7. A capacitor of capacitance 'πΆ', is connected across an ac source of voltage π, given by V = V0sin πt The displacement
current between the plates of the capacitor, would then be given by:
(1) Id = V0πCcosπt
(2) Id =V0
πCcos πt
(3) πΌπ =π0
ππΆsin ππ‘
(4) Id = V0πCsinπt
Correct Answer: (1)
Q.7. Solution:
β΅ π = πΆπ
And π = π0 sinππ‘
πΌπ =?
As we know,
β΅ πΌπ = π0πβ
ππ‘
= π0ππΈπ΄
ππ‘ (β΅ πΈ =
π
π)
=π0π΄ ππ
π ππ‘
= (π0π΄
π)
π
ππ‘ π0 sinππ‘
πΌπ = πΆπ0π cosππ‘
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 β 1 to t = n Then, the ratio Sn
Sn+1 is:
(1) 2πβ1
2π
(2) 2πβ1
2π+1
(3) 2π+1
2πβ1
(4) 2π
2πβ1
Correct Answer: (2)
Q.8. Solution:
β΅ πππ‘β = π’ + (2πβ1
2) π
but π’ = 0
β΄ π π = (2π β 1
2) π β¦ (1)
π π+1 = [2(π + 1) β 1
2] π β¦(2)
Now,
eq (1) divide by (2)
π π
π π+1=
2π β 1
2π + 1
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:
(1) S
4,3gS
2
(2) S
4,β3gS
2
(3) S
2,β3gS
2
(4) S
4, β
3gS
2
Correct Answer: (4)
Q.9. Solution:
β΅ At every ππ‘.
πΎπΈ + ππΈ =constant
so,(πΎπ)π΄ + (ππ)π΄ = (ππ)π΅ + (ππ)π΅
0 + πππ = 3(ππ)π΅ + (ππ)π΅
πππ = 4ππβ
β = π/4
Now,
β΅At ππ‘. π΅ β
πΎπ = 3ππ
1
2ππ£2 = 3
πππ
4
π£ = β3
2ππ
10. In a potentiometer circuit a cell of EMF 1.5 π 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?
(1) 60cm
(2) 21.6cm
(3) 64cm
(4) 62cm
Correct Answer: (1)
Q.10. Solution:
Given,
πΈ1 = 1.5 V
π1 = 36 cm
πΈ2 = 2.5v
π2 =?
AS we know,
πΈ1
πΈ2=
π1π2
1 β 5
2 β 5=
36
π2
3
5=
36
π2
π2 = 60 cm
11. For a plane electromagnetic wave propagating in π₯ βdirection, which one of the following combination gives the
correct possible directions for electric field (E) and magnetic field (B) respectively?
(1) π^+ π
^
, π^+ π
^
(2) βπ^+ π
^
, βπ^β π
^
(3) π^+ π
^
, βπ^β π
^
(4) βπ^+ π
^
, βπ^+ π
^
Correct Answer: (2)
Q.11. Solution:
Direction = οΏ½βοΏ½ Γ οΏ½βοΏ½
(i) (πΜ + οΏ½ΜοΏ½) Γ (πΜ + οΏ½ΜοΏ½) = 0
(ii) (βπΜ + οΏ½ΜοΏ½) Γ (βπΜ β οΏ½ΜοΏ½)
= 0 + πΜ β (βπΜ) + 0
= 2πΜ
(iii) (πΜ + οΏ½ΜοΏ½) Γ (βπΜ β οΏ½ΜοΏ½) = 0
(iv) (βπΜ + οΏ½ΜοΏ½) Γ (βπΜ + οΏ½ΜοΏ½)
= 0
12. Polar molecules are the molecules:
(1) having zero dipole moment.
(2) acquire a dipole moment only in the presence of electric field due to displacement of charges.
(3) acquire a dipole moment only when magnetic field is absent.
(4) having a permanent electric dipole moment.
Correct Answer: (4)
Q.12. Solution:
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.
13. The velocity of a small ball of mass π and density d, when dropped in a container filled with glycerine becomes
constant after some time. If the density of glycerine is π
2, then the viscous force acting on the ball will be :
(1) Mg
2
(2) Mg
(3) 3
2Mg
(4) 2Mg
Correct Answer: (1)
Q.13. Solution:
πΉπ + πΉπ = ππ
πΉπ = ππ β πΉπ
= ππππ β ππππ
= ππ [π βπ
2]
=πππ
2
=(ππ)π
2=
ππ
2
14. Match Column - I and Column - II and choose the correct match from the given choices.
Column - I Column - II
(A) Root mean square speed of gas molecules (P) 1
3nmπ£
Β― 2
(B) Pressure exerted by ideal gas (Q) β
3RT
M
(C) Average kinetic energy of a molecule (R) 5
2RT
(D) Total internal energy of 1 mole of a diatomic
gas (S)
3
2kBT
(1) (A) β (R), (B) β (P), (C) β (S), (D) β (Q)
(2) (A) β (Q), (B) β (R), (C) β (S), (D) β (P)
(3) (A) β (Q), (B) β (P), (C) β (S), (D) β (R)
(4) (A) β (R), (B) β (Q), (C) β (P), (D) β (S)
Correct Answer: (3)
Q.14. Solution:
Standard formula for each physical quantity needs to be used. (π΄ β π), (π΅ β π), (πΆ β π) & (π· β π )
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)
(1) 10.2kW
(2) 8.1kW
(3) 12.3kW
(4) 7.0kW
Correct Answer: (3)
Q.15. Solution:
Change in potential energy of water per second = 15 Γ π Γ 60
= 9000 π½ π β
Energy remaining per second after loss = 90% of 9000 π½ π β
= (90
100) Γ 9000
= 8100 π½ π β = 8.1 ππ
16. A lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since :
(1) a large aperture contributes to the quality and visibility of the images.
(2) a large area of the objective ensures better light gathering power.
(3) a large aperture provides a better resolution.
(4) all of the above.
Correct Answer: (4)
Q.16. Solution:
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.
17. The electron concentration in an : π-type semiconductor is the same as hole concentration in a π-type semiconductor.
An external field (electric) is applied across each of them. Compare the currents in them.
(1) current in n βtype = current in p βtype.
(2) current in n βtype > current in p βtype.
(3) current in n βtype > current in p βtype.
(4) No current will flow in p-type, current will only flow in π-type.
Correct Answer: (3)
Q.17. Solution:
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 πΌ = πππ΄π£.
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 :
(1) 0.9MeV
(2) 9.4MeV
(3) 804MeV
(4) 216MeV
Correct Answer: (4)
Q.18. Solution:
π΅πΈ gain = (120 Γ 8.5) + (120 Γ 8.5) β (240 Γ 7.6)
= 2040 β 1824
= 216 πππ
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 axis of the cable is represented by:
(1)
(2)
(3)
(4)
Correct Answer: (3)
Solution:
For inside part
π½ =πΌ
ππ 2
β« οΏ½βοΏ½ β ππ = ΞΌ0π β¦(i)
π΅ 2ππ = ΞΌ0 (πΌ
ππ 2) Γ ππ2
π΅ 2ππ = ΞΌ0πΌπ2
π 2
π΅ = πΆπ where πΆ = constant
For outside part
β« οΏ½βοΏ½ β ππ = ΞΌ0πΌ
π΅(2ππ) = ΞΌ0πΌ
β΄ π΅ =πΆβ²
π , where πΆβ² = constant
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 (π1/π2) is :
(1) π 1
π 2
(2) R2
R1
(3) β(π 1
π 2)
(4) β(π 1
π 2)
Correct Answer: (2)
Solution:
π1 = π2
ππ1
π 1=
ππ2
π 2
π14π(π 1)2
π 1=
π2 4π(π 2)2
π 2
π1
π2=
π 2
π 1
21. If πΈ and πΊ respectively denote energy and gravitational constant, then E
G has the dimensions of:
(1) [M2][Lβ1][T0]
(2) [M][Lβ1][Tβ1]
(3) [M][L0][T0]
(4) [M2][Lβ2][Tβ1]
Correct Answer: (1)
Solution:
Dimensional Formulas,
Energy, [πΈ] = π1 πΏ2 πβ2
Gravitational constant
[πΊ] = [πΉπ2
π2] =
π1 πΏ1πβ2ΓπΏ2
π2
[πΊ] = πβ1 πΏ3 πβ2
Now,
[πΈ
πΊ] =
π πΏ2 πβ2
πβ1 πΏ3 πβ2 = π2 πΏβ1 ππ
Correct option (1)
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:
(1) 0.0628 s
(2) 6.28 s
(3) 3.14 s
(4) 0.628 s
Correct Answer: (2)
Solution:
Spring force, πΉ = ππ₯
Spring constant, π =πΉ
π₯=
10
0.05= 200 π πβ
Time period, π = 2πβπ
π
π = 2πβ2
200= 0.2 π = 6.285 s
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.
Column - I Column - II
(A) Drift Velocity (P) m
ne2π
(B) Electrical Resistivity (Q) neπ£d
(C) Relaxation Period (R) eE
mπ
(D) Current Density (S) πΈ
π½
(1) (A) β (R), (B) β (S), (C) β (P), (D) β (Q)
(2) (A) β (R), (B) β (S), (C) β (Q), (D) β (P)
(3) (A) β (R), (B) β (P), (C) β (S), (D) β (Q)
(4) (A) β (R), (B) β (Q), (C) β (S), (D) β (P)
Correct Answer: (1)
Solution:
Drift velocity, π = (ππΈ
π) π
Electrical resistivity, π =πΈ
π½
Relaxation period, π =π
ππ2π
Current Density, π½ = ππππ
π΄ β π , π΅ β π, πΆ β π,π· β π
24. A dipole is placed in an electric field as shown. In which direction will it move?
(1) towards the left as its potential energy will increase.
(2) towards the right as its potential energy will decrease.
(3) toward, the left as its potential energy will decrease.
(4) towards the right as its potential energy will increase.
Correct Answer: (2)
Solution:
πΉ1 = ππΈ1 , πΉ2 = ππΈ2
πΈ1 > πΈ2 β πΉ1 > πΉ2
Hence, net force is towards right and its potential energy will decrease.
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 β n junction lies between 0.1V to 0.3V
(1) (A) and (B) both are correct.
(2) (A) and (B) both are incorrect.
(3) (A) is correct and (B) is incorrect.
(4) (A) is incorrect but (B) is correct.
Correct Answer: (3)
Solution:
(π΄)is correct while (π΅) is incorrect because ππ diode has barrier potential of 0.7V.
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:
(1) 0.52cm
(2) 0.026cm
(3) 0.26cm
(4) 0.052cm
Correct Answer: (3)
Solution:
πΏ β πΆ. =1
100 ππ
= 0.01 ππ
Now,
Diameter of wire = 52 Γ 0.01 = 0.52 mm = 0.052 cm
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 10β2A. The impedance of the circuit is :
(1) 4β2Ξ©
(2) 5/β2Ξ©
(3) 4Ξ©
(4) 5Ξ©
Correct Answer: (4)
Solution:
π = βππ 2 + (ππ β ππΏ)2
= β402 + (10 β 40)2 = 50 volt
πrms = π£πππ
π§
β Impedance, π§ =ππππ
ππππ
β π§ =ππππ Γ β2
10
β =β2 Γ 50
10β2
β π§ = 50 Ξ©
28. A parallel plate capacitor has a uniform electric field ' Eβ
' in the space between the plates. If the distance between the
plates is 'd' and the area of each plate is ' π΄', the energy stored in the capacitor is : π0 = permittivity of free space)
(1) 1
2π0E
2
(2) π0EAd
(3) 1
2π0πΈ
2π΄π
(4) E2Ad
π0
Correct Answer: (3)
Solution:
Electric energy density π’ =1
2π0πΈ
2
Volume of the space between plates of Capacitor is π΄π.
β΄ Energy Stored = Energy density ΓVolume
=1
2π0πΈ
2π΄π
29. An electromagnetic wave of wavelength ' π ' is incident on a photosensitive surface of negligible work function. If ' π
' mass is of photoelectron emitted from the surface has de-Broglie wavelength πd, then :
(1) π = (2m
hc) πd
2
(2) πd = (2mc
h)π2
(3) π = (2mc
h)πd
2
(4) π = (2h
mc) πd2
Correct Answer: (3)
Solution:
From Einsteinβs equation,
πΈincident = π Β± πΎπΈmax
βπ = 0 +1
2ππ£2
βπ
π=
π2
2πβ π = β
2πβπ
π,
ππ =β
π=
β
β2πβππ
= βπβ
2ππ
β π = (2ππ
β) ππ
2
30. Find the value of the angle of emergence from the prism. Refractive index of the glass is β3.
(1) 60β
(2) 30β
(3) 45β
(4) 90β
Correct Answer: (1)
Solution:
No refraction at face π΄πΆ.
Apply Snell's law at face π΅πΆ, we get
π1sinπ = π2sinr
β3sin30β = 1 Γ sinπ
β3
2= sinπ
β π = 60β
31. The equivalent capacitance of the combination shown in the figure is :
(1) 3πΆ
(2) 2C
(3) C/2
(4) 3C/2
Correct Answer: (2)
Solution:
Redrawing the circuit between π΄ & π΅
Capacitor 3 can be eliminated and Capacitor1 & 2 are in parallel.
πΆπ΄π΅ = πΆ1 + πΆ2
= πΆ + πΆ
= 2πΆ
32. If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of
energy.
(1) [F][A][T]
(2) [F][A][T2]
(3) [F][A][Tβ1]
(4) [F][Aβ1][T]
Correct Answer: (2)
Solution:
[πΈπππππ¦] = [πΉ]π [π΄]π [π]π
[ππΏ2πβ2] = [ππΏπβ2]π [πΏπβ2]π [π]π
[ππΏ2πβ2] = [πππΏπ+ππβ2πβ2π+π]
On comparing powers of [π]
π = 1
On comparing powers of [πΏ]
π + π = 2
β 1 + π = 2
β π = 1
On comparing powers of [π]
β2π β 2π + π = β2
β β2 β 2 + π = β2
β π = 2
[πΈπππππ¦] = [πΉ] [π΄] [π2]
33. A cup of coffee cools from 90βC to 80βC in t minutes, when the room temperature is 20βC. The time taken by a similar
cup of coffee to cool from 80βC to 60βC at a room temperature same at 20βC is :
(1) 13
10t
(2) 13
5t
(3) 10
13t
(4) 5
13t
Correct Answer: (2)
Solution:
From Newtonβs Law of cooling,
Ξπ
Ξπ‘= π(π β ππ )
βπ2βπ1
π‘β π (
π1+π2
2β πs) πs
where, π1 =Initial temperature of body
π2 = Final temperature of body
ππ = Surrounding temperature
π‘ =time taken
For case1,
90 β 80
π‘= π (
90 + 80
2β 20)
β10
π‘= 65π
β π =10
65π‘β¦ (1)
For case 2,
80 β 60
tβ²= π (
80 + 60
2β 20)
β20
tβ²= 50k
Putting the value k from eq. (π),
β20
tβ²= 50 Γ
10
65π‘
β tβ² =20 Γ 65
500π‘
β π‘β² =13
5π‘
34. The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section
and same material is 0.25 Ξ©. What will be the effective resistance if they are connected in series?
(1) 0.25Ξ©
(2) 0.5Ξ©
(3) 1Ξ©
(4) 4Ξ©
Correct Answer: (4)
Solution:
Let the resistance of each curve be π
Effective resistance in parallel combination,
π π =π
4
β π = 4π π
β π = 4 Γ 0.25 = 1Ξ
Effective resistance is series combination,
π π = 4π
β π π = 4 Γ 1 = 4
35. The number of photons per second on an average emitted by the source of monochromatic light of wavelength
600 nm, when it delivers the power of 3.3 Γ 10β3 watt will be : (h = 6.6 Γ 10β34 Js)
(1) 1018
(2) 1017
(3) 1016
(4) 1015
Correct Answer: (3)
Solution:
Energy of each photon, πΈ =βπ
π
β πΈ =6.6Γ10β34Γ3Γ108
600Γ10β9
β πΈ = 3.3 Γ 10β19 π½
Number of photons emitted per second, π =π
πΈ
β π =3.3Γ10β3
3.3Γ10β19
β π = 1016
36. Three resistors having resistances r1, r2 and r3 are connected as shown in the given circuit. The ratio π3
π1 of currents in
terms of resistances used in the circuit is :
(1) r1
r2+r3
(2) π2
π2+π3
(3) π1
π1+π2
(4) π1
π1+π2
Correct Answer: (2)
Solution:
Potential difference between π΅ and πΆ,
ππ΅πΆ = π3π3 = π1 Γπ2π3
π2 + π3
β π3 = π1 Γπ2
π2 + π3
37. A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put
perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a
distance of :
(1) 20cm from the lens, it would be a real image.
(2) 30cm from the lens, it would be a real image.
(3) 30cm from the plane mirror, it would be a virtual image.
(4) 20cm from the plane mirror, it would be a virtual image.
Correct Answer: (4)
Solution:
Using lens formula,
1
π£β
1
π’=
1
π
β1
π£1β
1
(β60)=
1
(+30)
β1
π£1=
1
30β
1
60
β1
π£1=
1
60
β π£1 = +60 cm
Distance of image forwed by lens from Mirror,
π = π£1 β 40 = 60 β 40 = 20 cm
The final image will be formed at a distance of 20 cm from Mirror.
Distance of image from lens in 20 cm.
Again, light will refract from lens
Using lens formula,
1
π£β
1
π’=
1
π
β1
π£1β
1
(β20)=
1
(+30)
β1
π£1=
1
30β
1
20
β1
π£1=
β10
600
β π£1 = β60 cm
Hence, the image will be virtual and formed at a distance of 60 from lens or 20 from mirror.
38. For the given circuit, the input digital signals are applied at the terminals A, B and C. What would be the output at the
terminal π¦ ?
Truth table of AND gate for output π¦1
π΄ π΅ π¦1
0 0 0
0 1 0
1 0 0
1 1 1
Truth table of NAND gate for output π¦2
π΅ πΆ π¦2
0 0 1
0 1 1
1 0 1
1 1 0
Truth table of OR gate for output π¦
π¦1 π¦2 π¦
0 0 0
0 1 1
1 0 1
1 1 1
39. A step down transformer connected to an ac mains supply of 220V is made to operate at 11 V, 44 W lamp. Ignoring
power losses in the transformer, what is the current in the primary circuit?
(1) 0.2A
(2) 0.4A
(3) 2A
(4) 4A
Correct Answer: (1)
Solution:
Power loss is zero. Hence, output Power is equal to input Power.
So, Input Power, π1 = 44 W
β π1π1 = π1
β π1 =π1
π1=
44
220= 0.2 A
40. 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 :
(1) β3Ia2 and 3Ia2
(2) 3Ia2 and Ia 2
(3) 3Ia2 and 4Ia2
(4) 4Ia2 and 3Ia2
Correct Answer: (1)
Solution:
Area of equilateral triangle of side π,
π΄ =1
2Γ π Γ
β3π
2=
β3
4π2
Magnetic moment of each triangle,
π = πΌπ΄ =β3
4π2πΌ
12π length of wire will form 4 loops of triangle of side π.
Magnetic moment of triangle,
π1 = ππ = 4 Γβ3
4π2πΌ = β3πΌπ2
Similarly, magnetic moment of square
π2 = 3 Γ (πΌπ2)
= 3πΌπ2
41. In the product
Fβ
= q(π£β
Γ Bβ)
= qπ£β
Γ (Bπ^+ B
^π^+ B0π
^
)
For q = 1 and π£β
= 2π^+ 4π
^+ 6π
^
and Fβ
= 4π^β 20π
^+ 12π
^
What will be the complete expression for Bβ
?
(1) β8π^β 8π
^β 6π
^
(2) β6π^β 6π
^β 8π
^
(3) 8π^+ 8π
^β 6π
^
(4) 6π^+ 6π
^β 8π
^
Correct Answer: (2)
Solution:
4π β 20π + 12π = (2π + 4π + 6π) Γ (π΅πΜ + π΅π + π΅ποΏ½ΜοΏ½)
4π β 20π + 12π = |+π βπ +π2 4 6π΅ π΅ π΅0
|
= (4π΅0 β 6π΅)π β (2π΅0 β 6π΅)π + (2π΅ β 4π΅)π
4π΅0 β 6π΅ = 4
2π΅0 β 6π΅ = 20
2π΅ β 4π΅ = 12 β π΅ = 6
S0 2π΅0 β 6(β6) = 20
2π΅0 = 20 β 36 β π΅0 = β8
β΄ οΏ½βοΏ½ = β6π β 6π β 8οΏ½ΜοΏ½
42. A particle moving in a circle of radius R with a uniform speed takes a time 'I' to complete one revolution. If this
particle were projected with the same speed at an angle ' π ' to the horizontal, the maximum height attained by it equals
4R. The angle of projection, $\theta$, is then given by :
(1) π = cosβ1 (gT2
π2R)1/2
(2) π = cosβ1 (π2π
gT2)1/2
(3) π = sinβ1 (π2π
ππ2)1/2
(4) π = sinβ1 (2gT2
π2R)1/2
Correct Answer: (4)
Solution:
π =2π
π=
2ππ
π£β π£ =
2ππ
π
π» =π£2sin2 π
2πβ 4π =
4π2π 2sin2 π
2ππ2
β sin2π =2ππ2
π2π
β sinπ = (2ππ2
π2π )
1/2
β π = sinβ1 (2ππ2
π2π )2
43. A series LCR circuit containing 5.0 H inductor, 80 πF capacitor and 40 Ξ© resistor is connected to 230 V variable
frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at
the resonant angular frequency are likely to be:
(1) 25rad/s and 75rad/s
(2) 50rad/s and 25rad/s
(3) 46rad/s and 54rad/s
(4) 42rad/s and 58rad/s
Correct Answer: (3)
Solution:
πΏ = 5π», πΆ = 80ππΉ, π = 40π; ππππ = 230 π
Half power frequencies,
π1 = βπ πΆ+βπ 2πΆ2+4πΏπΆ
2πΏπΆ and π2 =
π πΆ+βπ 2πΆ2+4πΏπΆ
2πΏπΆ
π1 =(β40Γ80Γ10β6)+β(40Γ80Γ10β6)2+(4Γ5Γ80Γ10β6)
2Γ5Γ80Γ10β6
Here, (40 Γ 80 Γ 10β6)2 is negligible.
We get,
π1 = 46 πππ π β
And π2 = 54 πππ π β
44. From a circular ring of mass 'M' and radius 'R'an arc corresponding to a 90β sector is removed. The moment of inertia
of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the
ring is 'K' times β²MR 2β². Then the value of 'K' is :
(1) 3
4
(2) 7
8
(3) 1
4
(4) 1
8
Correct Answer: (1)
Solution:
mass per unit length of the ring is
π =π
2ππ
β΄ Mass of remaining ring is πβ² = π Γ3
4(2ππ )
β΄ πβ² =3
4π.
Moment of inertia of remaining part is
πΌβ² = πβ²π 2 =3
4ππ 2 = πΎππ 2
β΄ πΎ =3
4
45. A uniform rod of length 200cm and mass 500g is balanced on a wedge placed at 40cm mark. A mass of 2kg is
suspended from the rod at 20cm and another unknown mass ' m ' is suspended from the rod at 160cm mark as shown in
the figure. Find the value of 'm' such that the rod is in equilibrium. (g = 10m/s2)
(1) 1
2kg
(2) 1
3kg
(3) 1
6kg
(4) 1
12kg
Correct Answer: (4)
Solution:
Torque balance about β²πβ²
2π Γ 20 = 0.5 π Γ 60 + ππ Γ 120
β 2 = 1.5 + 6π
β π =0.5
6=
1
12 ππ
46. Twenty-seven drops of same size are charged at 220V each. They combine to form a bigger drop. Calculate the
potential of the bigger drop.
(1) 660V
(2) 1320V
(3) 1520V
(4) 1980V
Correct Answer: (4)
Solution:
ππππ = π2 3β ππ ππππ
= (27)2 3β Γ 220
= 1980 π
47. A car starts from rest and accelerates at 5m/s2 At t = 4s, a ball is dropped out of a window by a person sitting in the
car. What is the velocity and acceleration of the ball at π‘ = 6π ?
(Take g = 10m/s2 )
(1) 20m/s, 5m/s2
(2) 20m/s, 0
(3) 20β2m/s β 0
(4) 20β2m/s, 10m/s2
Correct Answer: (4)
Solution:
Car velocity at π‘ = 4 π is π£ = π’ + ππ‘
π£ = 0 + 5 Γ 4 = 20 ππ β1
For an observe on ground, the ball follows parabolic path same as horizontal projectile.
Velocity at 6π is π£1 = βπ£π₯2 + π£π¦
2
π£1 = βπ£2 + π2(βπ‘)2
= β202 + 102 Γ 22
= 20β2 ππ β1
And as the body is under free fall, its acceleration at π‘ = 6π is 10 ππ β2
(20β2 ππ β1, 10 ππ β2)
48. A particle of mass ' m ' is projected with a velocity π£ = kVe(k < 1) from the surface of the earth. (Ve =escape
velocity) The maximum height above the surface reached by the particle is:
(1) π (π
1βπ)2
(2) π (π
1+π)2
(3) R2k
1+k
(4) Rk2
1βk2
Correct Answer: (4)
Solution:
π£ = πππ(π < 1)
1
2ππ£2 β
πΊππ
π = 0 β
πΊππ
π + β
1
2ππ2 Γ
2πΊπ
π β
πΊππ
π = β
πΊππ
π + β
πΊπππ2
π = βπΊππ(
1
π + ββ
1
π )
ππ = β2πΊπ
π
π2
π =
β
π (π + β)β 1 +
π
β=
1
π2
βπ
β=
1
π2β 1
β β =π πΎ2
1 β π2
49. A ball of mass 0.15kg is dropped from a height 10m strikes the ground and rebounds to the same height. The
magnitude of impulse imparted to the ball is (g = 10m/s2) nearly:
(1) 0kgm/s
(2) 4.2kgm/s
(3) 2.1kgm/s
(4) 1.4kgm/s
Correct Answer: (2)
Solution:
π = 0.15 kg; β = 10 m.
π£ = β2πβ = β2 Γ 9.8 Γ 10 = 14 π/π
Impulse
= βππ£ β ππ£ = β2ππ£
= 2 Γ 0.15 Γ 14
= β4.2 ππ π/π
(To raise back to same height ,rebound velocity is same as the hitting velocity)
50. Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres coinciding. If R1 >>
R2, the mutual inductance M between them will be directly proportional to:
(1) R1
R2
(2) R2
R1
(3) R1
2
R2
(4) R2
2
R1
Correct Answer: (4)
Solution:
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