NEET-XII-Physics
exam-1 year:2016
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- Qstn #15A black body is at a temperature of 5760 K. The
energy of radiation emitted by the body at
wavelength 250 nm is ``U_1``, at wavelength 500 nm
is ``U_2`` and that at 1000 nm is ``U_3``. Wien's constant,
b = 2.88 × ``10^6`` nmK. Which of the following is
correct?
(1) ``U_1`` = 0
(2) ``U_3`` = 0
(3) ``U_1`` > ``U_2``
(4) ``U_2`` > ``U_1``digAnsr: 4Ans : 4
Sol. Maximum amount of emitted radiation
corresponding to m =
b
T
m =
62.88 10 nmK
5760K
= 500 nm
U
U2
Emitted
(radiation)
250nm
500nm 1000nm
From the graph U1 < U2 > U3
- Qstn #16Coefficient of linear expansion of brass and steel
rods are ``\alpha_1`` and ``\alpha_2``. Lengths of brass and steel rods
are ``l_1`` and ``l_2`` respectively. If (``l_2`` - ``l_1``) is maintained
same at all temperatures, which one of the following
relations holds good ?
(1) ``\alpha_1`` ``l_2`` = ``\alpha_2````l_1``
(2)``\alpha_1````l_2^2``= ``\alpha_2````l_1^2``
(3) ``\alpha_1^2````l_2`` = ``\alpha_2^2````l_1``
(4) ``\alpha_1````l_1`` = ``\alpha_2````l_2``digAnsr: 4Ans : 4
Sol. Change in length for both rods should be same
▵1 = ▵2
11▵T = 22▵T
11 = 22
- Qstn #17A npn transistor is connected in common emitter
configuration in a given amplifier. A load resistance
of 800 ``\Omega`` is connected in the collector circuit and
the voltage drop across it is 0.8 V. If the current
amplification factor is 0.96 and the input resistance
of the circuit is 192``\Omega``, the voltage gain and the
power gain of the amplifier will respectively be :
(1) 4, 3.84
(2) 3.69, 3.84
(3) 4, 4
(4) 4, 3.69digAnsr: 1Ans : 1
Sol. Given = 0.96
so, =
=
0.96
1 0.04
&implies; = 24
Voltage gain for common emitter configuration
Av = .
L
i
R
R = 24 ×
=
800
100
192
Power gain for common emitter configuration
Pv = Av = 24 × 100 = 2400
Voltage gain for common base configuration
Av = .
L
P
R
R = 0.96 ×
800
192
= 4
Power gain for common base configuration
Pv = Av = 4 × 0.96 = 3.84
*In the question it is asked about common
emitter configuration but we got above
answer for common base configuration.
- Qstn #18The intensity at the maximum in a Young's double
slit experiment is ``I_0``. Distance between two slits is
d = 5 ``\lambda``, where ``\lambda`` is the wavelength of light used
in the experiment. What will be the intensity in front
of one of the slits on the screen placed at a distance
D = 10 d ?
(1) ``I_0``
(2)``I_0``/4
(3) 3/4 ``I_0``
(4)``I_0``/2digAnsr: 4Ans : 4
Sol. Path difference
= S2P - S1P
= + 2 2D d D
S1
S2
O B
P
D
d
= D
+
2
2
1 d
1 D
2 D
= D
+ =
2 2
2
d d
1 1
2D2D
▵x =
= = =
2d d 5
2 10d 20 20 4
▵ =
=
2
.
4 2
So, intensity at the desired point is
I = I0cos2
2
= I0cos2
4
= 0
I
2
- Qstn #19A uniform circular disc of radius 50 cm at rest is
free to turn about an axis which is perpendicular
to its plane and passes through its centre. It is
subjected to a torque which produces a constant
angular acceleration of 2.0 rad ``s^{-2}``. Its net
acceleration in m ``s^{-2}`` at the end of 2.0 s is
approximately :
(1) 8.0
(2) 7.0
(3) 6.0
(4) 3.0digAnsr: 1Ans : 1
Sol. Particle at periphery will have both radial and
tangential acceleration
at = R= 0.5 × 2 = 1 m/s2
= 0 + t
= 0 + 2 × 2 = 4 rad/sec
ac = 2R = (4)2 × 0.5 = 16 × 0.5 = 8 m/s2
atotal = + = +
2 2 2 2 2
p ca a 1 8 8m/s
*In this question we have assumed the point
to be located at periphery of the disc.
- Qstn #20An electron of mass m and a photon have same
energy E. The ratio of de-Broglie wavelengths
associated with them is :
(1)``\frac 1c(\frac{E}{2m})^\frac 12``
(2)``(\frac{E}{2m})^\frac 12``
(3) ``c(2mE)^\frac 12``
(4) ``\frac {1}{xc}(\frac{2m}{E})^\frac 12``
(c being velocity of light)digAnsr: 1Ans : 1
Sol. For electron e =
h
2mE
for Photon E = pc &implies; =Ph
hc
E
&implies;
= =
1/2
e
Ph
h E E 1
hc 2m c2mE
- Qstn #21A disk and a sphere of same radius but different
masses roll off on two inclined planes of the same
altitude and length. Which one of the two objects
gets to the bottom of the plane first ?
(1) Disk
(2) Sphere
(3) Both reach at the same time
(4) Depends on their massesdigAnsr: 2Ans : 2
Sol. a =
+
2
2
gsin
K
1
R
for disc ; = =
2
2
K 1
0.5
2R
for sphere ; = =
2
2
K 2
0.4
5R
a(sphere) > a(disc)
∴ sphere reaches first
- Qstn #22The angle of incidence for a ray of light at a
refracting surface of a prism is 45° . The angle of
prism is 60°. If the ray suffers minimum deviation
through the prism, the angle of minimum deviation
and refractive index of the material of the prism
respectively, are :
(1) 45°, ``\frac {1}{\sqrt 2}``
(2) 30°, ``\sqrt 2``
(3) 45°, ``\sqrt 2``
(4) 30°,``\frac {1}{\sqrt 2}``digAnsr: 2Ans : 2
Sol. i = 45°; A = 60°; m = 2i - A = 30°
µ =
+
mAsin
2
sin A /2
=
= =
sin45 1 2
. 2
sin30 12
- Qstn #23When an ``\alpha``-particle of mass 'm' moving
with velocity 'v' bombards on a heavy nucleus of
charge 'Ze', its distance of closest approach from the nucleus
depends on m as :
(1) 1/m
(2) ``\frac {1}{\sqrt m}``
(3) ``\frac {1}{m^2}``
(4) mdigAnsr: 1Ans : 1
Sol. At closest distance of approach, the kinetic energy
of the particle will convert completely into
electrostatic potential energy.
&implies; =
21 KQqmv
2 d
&implies; d
1
m
- Qstn #24A particle of mass 10 g moves along a circle of
radius 6.4 cm with a constant tangential
acceleration. What is the magnitude of this
acceleration if the kinetic energy of the particle
becomes equal to 8 × ``10^{-4}`` J by the end of the
second revolution after the beginning of the
motion?
(1) 0.1 ``\frac {m}{s^2}``(2) 0.15 ``\frac {m}{s^2}``
(3) 0.18 ``\frac {m}{s^2}`` (4) 0.2 ``\frac {m}{s^2}``digAnsr: 1Ans : 1
Sol. 2
1
mv
2
= E &implies;
21 10 v
2 1000
= 8 × 10-4
&implies; v2 = 16 × 10-2 &implies; v = 4 × 10-1 = 0.4 m/s
Now,
v2 = u2 + 2ats (s = 4R)
&implies;
= +
2
t
16 22 6.4
0 2a 4
100 7 100
&implies; at =
16 7 100
100 8 22 6.4
= 0.1 m/s2
- Qstn #25The molecules of a given mass of a gas
have r.m.s. velocity of 200 ``\frac {m}{s}`` at 27°C and
1.0 × ``10^5 Nm^{-2}`` pressure. When the temperature
and pressure of the gas are respectively, 127°C and
0.05 × 105 ``10^5 Nm^{-2}``, the r.m.s. velocity of
its molecules in ``ms^{-1}`` is :
(1) 100 ``\sqrt 2``
(2)``\frac {400}{\sqrt 3}``
(3)``\frac {100\sqrt 2 }{3}``
(4) 100/3digAnsr: 2Ans : 2
Sol. v T &implies; =
v 400
200 300
&implies; v =
200 2
3
m/s
&implies; v =
400
3
m/s
- Qstn #26A long straight wire of radius a carries a steady
current I. The current is uniformly distributed over
its cross-section. The ratio of the magnetic fields
B and B', at radial distances a/2 and 2a respectively,
from the axis of the wire is :
(1)``\frac {1}{4}``
(2)``\frac {1}{2}``
(3) 1
(4) 4digAnsr: 3Ans : 3
Sol. For points inside the wire
=
0
2
µ Ir
B (r R)
2 R
For points outside the wire
= ≥
0µ IB (r R)
2 r
according to the question
=
0
2
0
I(a / 2)
B 2 a
IB '
2 (2a)
= 1 : 1
- Qstn #27A particle moves so that its position vector is given
by ``\vec{r}`` = ``cos\omega``t ``\hat{x}`` + ``sin\omega``t ``\hat{y}`` .
Where ``\omega`` is a constant.
Which of the following is true ?
(1) Velocity and acceleration both are perpendicular
to ``\vec{r}``.
(2) Velocity and acceleration both are parallel to r``\vec{r}``
(3) Velocity is perpendicular to r``\vec{r}`` and acceleration
is directed towards the origin
(4) Velocity is perpendicular to r``\vec{r}`` and acceleration
is directed away from the origin.digAnsr: 3Ans : 3
Sol. = +
ˆ ˆr cos t x sin t y
= +
ˆ ˆv sin t x cos t y
= + =
2 2ˆ ˆa cos t x sin t y r
=
r.v 0 hence
r v
a is directed towards the origin.
- Qstn #28What is the minimum velocity with which a body
of mass m must enter a vertical loop of radius R
so that it can complete the loop ?
(1) ``\sqrt {gR}``
(2) ``\sqrt {2gR}``
(3) ``\sqrt {3gR}``
(4) ``\sqrt {5gR}``digAnsr: 4Ans : 4
Sol. When minimum speed of body is 5gR , then no
matter from where it enters the loop, it will
complete full vertical loop.
- Qstn #29When a metallic surface is illuminated with radiation
of wavelength ``\lambda``, the stopping potential is V. If the
same surface is illuminated with radiation of
wavelength 2``\lambda``, the stopping potential is V/4. The
threshold wavelength for the metallic surface is :-
(1) 4``\lambda``
(2) 5``\lambda``
(3) 5/2 ``\lambda``
(4) 3 ``\lambda``digAnsr: 4Ans : 4
Sol. eV =
0
hc hc
...(i)
eV/4 =
0
hc hc
2
...(ii)
From equation (i) and (ii)
&implies; 4 =
0
0
1 1
1 1
2
On solving 0 = 3