NEET-XII-Physics
P1 year:2019
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- Qstn #16A body weighs 200 N on the surface of the
earth. How much will it weigh half way down
to the centre of the earth ?
(1) 150 N
(2) 200 N
(3) 250 N
(4) 100 NdigAnsr: 4Ans : ( 4 )
Sol.
d
g'
R
C
g
Acceleration due to gravity at a depth d from
surface of earth
' =
d
g g 1 ...(1)
R
Where g = acceleration due to gravity at
earth's surface
Multiplying by mass 'm' on both sides of (1)5
' =
d
mg mg 1
R
=
R
d
2
= = =
R 200
200 1 100N
2R 2
- Qstn #17Six similar bulbs are connected as shown in
the figure with a DC source of emf E and zero
internal resistance.

The ratio of power consumption by the bulbs
when
(i ) all are glowing and
(ii) in the situation
when two from section A and one from section
B are glowing, will be :
(1) 4 : 9
(2) 9 : 4
(3) 1 : 2
(4) 2 : 1digAnsr: 2Ans : ( 2 )
Sol. (i) All bulbs are glowing
RR
RR
RR
E
= + =
eq
R R 2R
R
3 3 3
= =
2 2
i
eq
E 3E
Power (P) ...(1)
R 2R
(ii) Two from section A and one from section
B are glowing.
E
R
R
R
= + =
eq
R 3R
R R
2 2
=
2
f
2E
Power (P ) ...(2)
3R
= =
2
i
2
f
P 3E 3R
9 : 4
P 2R 2E
- Qstn #18For a p-type semiconductor, which of the
following statements is true ?
(1) Electrons are the majority carriers and
trivalent atoms are the dopants.
(2) Holes are the majority carriers and
trivalent atoms are the dopants.
(3) Holes are the majority carriers and
pentavalent atoms are the dopants.
(4) Electrons are the majority carriers and
pentavalent atoms are the dopants.digAnsr: 2Ans : ( 2 )
Sol. In p-type semiconductor, an intrinsic
semiconductor is doped with trivalent
impurities, that creates deficiencies of
valence electrons called holes which are
majority charge carriers.
- Qstn #19Average velocity of a particle executing SHM in
one complete vibration is :
(1)``\frac{ωA}{2}``
(2) Aω
(3)``\frac{ω2A}{2}``
(4) ZerodigAnsr: 4Ans : ( 4 )
Sol. In one complete vibration, displacement is
zero. So, average velocity in one complete
vibration
f i
y yDisplacement
0
Time interval T
= = =
- Qstn #20The unit of thermal conductivity is :
(1) ``J m K^{-1}``
(2) ``J m^{-1} K^{-1}``
(3) ``W m K^{-1}``
(4) ``W m^{-1} K^{-1}``digAnsr: 4Ans : ( 4 )
Sol. The heat current related to difference of
temperature across the length l of a
conductor of area A is
= ▵
dH KA
T
dt
(K = coefficient of thermal
conductivity)
∴
dH
K
A dt T
=
▵
Unit of K = Wm-1 K-1
6
- Qstn #21A solid cylinder of mass 2 kg and radius 4 cm
rotating about its axis at the rate of 3 rpm.
The torque required to stop after 2
revolutions is
(1) ``2 × 10^{{-6}}`` N m
(2) ``2 × 10^{-3}`` N m
(3) ``12 × 10^{-4}`` N m
(4) ``2 × 10^6`` N mdigAnsr: 1Ans : ( 1 )
Sol. Work energy theorem.
2 2
f i
1
W I( )
2
= ω ω = 2 revolution
= 2 × 2 = 42 rad
i
2
W 3 rad / s
60
=
2 2 2
i
1 1
- mr (0 )
2 2
&implies; = ω
&implies; =
2
2
2
1 1 2
2 (4 10 ) 3
2 2 60
-
4
6
2 10 Nm
&implies; =
- Qstn #22A force F = 20 + 10 y acts on a particle in y-
direction where F is in newton and y in meter.
Work done by this force to move the particle
from y = 0 to y = 1 m is
(1) 30 J
(2) 5 J
(3) 25 J
(4) 20 JdigAnsr: 3Ans : ( 3 )
Sol. Work done by variable force is
f
i
y
y
W Fdy=
Here, y
i
= 0, y
f
= 1 m
∴
1
1 2
0
0
10y
W (20 10y)dy 20y
2
= + = +
= 25 J
- Qstn #23Which of the following acts as a circuit
protecting device?
(1) Conductor
(2) Inductor
(3) Switch
(4) FusedigAnsr: 4Ans : ( 4 )
Sol. Fuse wire has less melting point so when
excess current flows, due to heat produced
in it, it melts.
- Qstn #24In the circuits shown below, the readings of
voltmeters and the ammeters will be

Circuit 2
(1) ``V_2 > V_1 `` and `` i_1 = i_2``
(2) ``V_2 = V_1 `` and ``i_1 > i_2``
(3) ``V_2 = V_1`` and ``i_1 = i_2``
(4) ``V_2 > V_1`` and ``i_1 > i_2``digAnsr: 3Ans : ( 3 )
Sol. For ideal voltmeter, resistance is infinite and
for the ideal ammeter, resistance is zero.
1 1
10
V i 10 10 10 volt
10
= = =
2 2
10
V i 10 10 10 volt
10
= = =
V
1
= V
2
1 2
10 V
i i 1 A
10
= = =
- Qstn #25A hollow metal sphere of radius R is uniformly
charged. The electric field due to the sphere
at a distance r from the centre
(1) Increases as r increases for `` r < R`` and for
``r > R``
(2) Zero as r increases for `` r < R``, decreases as
r increases for `` r > R``
(3) Zero as r increases for `` r < R``, increases as
r increases for ``r > R``
(4) Decreases as r increases for `` r < R`` and for
``r > R``digAnsr: 2Ans : ( 2 )
Sol.
C
R
++
+
+
+
+ +
+
r
Q
Charge Q will be distributed over the surface
of hollow metal sphere.
(i) For r < R (inside)
By Gauss law,
en
in
0
q
E · dS 0= =
&implies; E
in
= 0 =∵
en
( q 0)
(ii) For r > R (outside)
++
+
+
+
+ +
+
r
0 2
1
E
r
E = 0
in
E
R
r
O
en
0
0
q
E · dS =
Here, q
en
= Q =∵
en
( q Q)
∴ 2
0
0
Q
E 4 r =
∴
0 2
1
E
r
- Qstn #26At a point A on the earth’s surface the
angle of dip, ``δ`` = +25°. At a point
B on the earth’s surface the angle of dip, ``δ``
= -25°. We can interpret that:
(1) A and B are both located in the northern
hemisphere.
(2) A is located in the southern hemisphere
and B is located in the northern
hemisphere.
(3) A is located in the northern hemisphere
and B is located in the southern
hemisphere.
(4) A and B are both located in the southern
hemisphere.digAnsr: 3Ans : ( 3 )
Sol. Angle of dip is the angle between earth's
resultant magnetic field from horizontal. Dip
is zero at equator and positive in northern
hemisphere.
BV
BH
δ = (+) ve
B
B
V B
B
H
δ = (-) ve
In southern hemisphere dip angle is
considered as negative.
- Qstn #27The total energy of an electron in an atom in
an orbit is -3.4 eV. Its kinetic and potential
energies are, respectively:
(1) -3.4 eV, -3.4 eV
(2) -3.4 eV, -6.8 eV
(3) 3.4 eV, -6.8 eV
(4) 3.4 eV, 3.4 eVdigAnsr: 3Ans : ( 3 )
Sol. In Bohr's model of H atom
∴
U
K.E. TE
2
= =
∴ K.E. = 3.4 eV
U = -6.8 eV
- Qstn #28In total internal reflection when the angle of
incidence is equal to the critical angle for the
pair of media in contact, what will be angle of
refraction?
(1) 180°
(2) 0°
(3) Equal to angle of incidence
(4) 90°digAnsr: 4Ans : ( 4 )
Sol.
90°
iC
At i = i
c
, refracted ray grazes with the
surface.
So angle of refraction is 90°.
8
- Qstn #29The work done to raise a mass m from the
surface of the earth to a height h, which is
equal to the radius of the earth, is:
(1) mgR
(2) 2mgR
(3) ½mgR
(4)``\frac32``mgRdigAnsr: 3Ans : ( 3 )
Sol.
h = R
RM
Initial potential energy at earths surface is
=
i
GMm
U
R
Final potential energy at height h = R
=
f
GMm
U
2R
As work done = Change in PE
∴ W = U
f
- U
i
= = =
2
GMm gR m mgR
2R 2R 2
(∵GM = gR
2)
- Qstn #30When an object is shot from the bottom of a
long smooth inclined plane kept at an angle 60°
with horizontal, it can travel a distance ``x_1``
along the plane. But when the inclination is
decreased to 30° and the same object is shot
with the same velocity, it can travel ``x_2``
distance. Then ``x_1 : x_2`` will be:
(1) 1 : ``\sqrt2``
(2) ``\sqrt2`` : 1
(3) 1 : ``\sqrt3``
(4) 1 : 2``\sqrt3``digAnsr: 3Ans : ( 3 )
Sol.
60°
g sin60°
30°
g sin30°
u
u
(Stopping distance) =
2
1
u
x
2gsin60
(Stopping distance) =
2
2
u
x
2gsin30
&implies;
= = =
1
2
x sin30 1 2
1: 3
x sin60 2 3