NEET 2018 Past Year Paper with Detailed Solutions | NEET Mock Test Series 2025 PDF Download

 Page 2


  
 
  
  
 
    
   
 
  
 
 
    
   
      
 
 
 
 
 
 
  
 
  
 
 
 
 
 
 
  
  
  
  
  
  
  
 
  
  
 
    
Solution: 
From the graph is clear that this is an isobaric process.  
P= constant 
At constant pressure 
W = nR?T            … (1) 
?Q=nC p?T (for constant pressure) 
5
Q n R T ...(1)
2
??
? ? ?
??
??
 
Equation (1)/(2) 
nR T 2
5 5
n R T
2
?
??
??
?
??
??
 
 
Q 2. The fundamental frequency in an open organ pipe is equal to the third harmonic of a 
closed organ pipe if the length of the closed organ pipe is 20 cm, the length of the open organ 
pipe is 
Option A 12.5 cm 
Option B 8 cm 
Option C 13.2 cm 
Option D 16 cm 
Correct Option C 
Solution: 
For open organ pipe 
 
open
2
2
nv
General formula f=
2l
vv
f
2
?
?
? ? ?
??
?
 
For pipe closed at one end 
 
General formula 
nv
f=
4l
 
Fundamental 
v
f
4
? 
Page 3


  
 
  
  
 
    
   
 
  
 
 
    
   
      
 
 
 
 
 
 
  
 
  
 
 
 
 
 
 
  
  
  
  
  
  
  
 
  
  
 
    
Solution: 
From the graph is clear that this is an isobaric process.  
P= constant 
At constant pressure 
W = nR?T            … (1) 
?Q=nC p?T (for constant pressure) 
5
Q n R T ...(1)
2
??
? ? ?
??
??
 
Equation (1)/(2) 
nR T 2
5 5
n R T
2
?
??
??
?
??
??
 
 
Q 2. The fundamental frequency in an open organ pipe is equal to the third harmonic of a 
closed organ pipe if the length of the closed organ pipe is 20 cm, the length of the open organ 
pipe is 
Option A 12.5 cm 
Option B 8 cm 
Option C 13.2 cm 
Option D 16 cm 
Correct Option C 
Solution: 
For open organ pipe 
 
open
2
2
nv
General formula f=
2l
vv
f
2
?
?
? ? ?
??
?
 
For pipe closed at one end 
 
General formula 
nv
f=
4l
 
Fundamental 
v
f
4
? 
  
 
  
  
 
    
As given: 
1 2 2
3v v 3 1
4l 2l 4 20 2l
13.2
? ? ?
?
?
 
 
Q 3. The efficiency of an ideal heat engine working between the freezing point and boiling 
point of water, is 
Option A 6.25% 
Option B 20% 
Option C 26.8% 
Option D 12.5% 
Correct Option C 
Solution: Efficiency of heat engine 
12
12
1
2
1
Q W Q
work done
Efficiency
Energy available
QQ
Q
T
h1
T
273
1
373
1 0.731
26.8%
??
?
?
??
??
??
??
 
 
Q 4. At what temperature will the rms speed of oxygen molecules become just sufficient for 
escape from the Earth's atmosphere? (Given: Mass of oxygen molecule (m) = 2.76×10
-26
kg 
Option A 5.016×10
4
 K 
Option B 8.360×10
4
 K 
Option C 2.508×10
4
 K 
Option D 1.254×10
4
 K 
Correct Option B 
Solution: 
Escape velocity=11.2km/s=11200m/s 
0
3
rms
RT
V
M
? 
? ?
23
2
3
26
30 1.38 10 T
11.2 10
2.76 10
?
?
?
? ? ?
??
?
 
T = 8.360 ×10
4
 
 
  
Page 4


  
 
  
  
 
    
   
 
  
 
 
    
   
      
 
 
 
 
 
 
  
 
  
 
 
 
 
 
 
  
  
  
  
  
  
  
 
  
  
 
    
Solution: 
From the graph is clear that this is an isobaric process.  
P= constant 
At constant pressure 
W = nR?T            … (1) 
?Q=nC p?T (for constant pressure) 
5
Q n R T ...(1)
2
??
? ? ?
??
??
 
Equation (1)/(2) 
nR T 2
5 5
n R T
2
?
??
??
?
??
??
 
 
Q 2. The fundamental frequency in an open organ pipe is equal to the third harmonic of a 
closed organ pipe if the length of the closed organ pipe is 20 cm, the length of the open organ 
pipe is 
Option A 12.5 cm 
Option B 8 cm 
Option C 13.2 cm 
Option D 16 cm 
Correct Option C 
Solution: 
For open organ pipe 
 
open
2
2
nv
General formula f=
2l
vv
f
2
?
?
? ? ?
??
?
 
For pipe closed at one end 
 
General formula 
nv
f=
4l
 
Fundamental 
v
f
4
? 
  
 
  
  
 
    
As given: 
1 2 2
3v v 3 1
4l 2l 4 20 2l
13.2
? ? ?
?
?
 
 
Q 3. The efficiency of an ideal heat engine working between the freezing point and boiling 
point of water, is 
Option A 6.25% 
Option B 20% 
Option C 26.8% 
Option D 12.5% 
Correct Option C 
Solution: Efficiency of heat engine 
12
12
1
2
1
Q W Q
work done
Efficiency
Energy available
QQ
Q
T
h1
T
273
1
373
1 0.731
26.8%
??
?
?
??
??
??
??
 
 
Q 4. At what temperature will the rms speed of oxygen molecules become just sufficient for 
escape from the Earth's atmosphere? (Given: Mass of oxygen molecule (m) = 2.76×10
-26
kg 
Option A 5.016×10
4
 K 
Option B 8.360×10
4
 K 
Option C 2.508×10
4
 K 
Option D 1.254×10
4
 K 
Correct Option B 
Solution: 
Escape velocity=11.2km/s=11200m/s 
0
3
rms
RT
V
M
? 
? ?
23
2
3
26
30 1.38 10 T
11.2 10
2.76 10
?
?
?
? ? ?
??
?
 
T = 8.360 ×10
4
 
 
  
  
 
  
  
 
    
Q 5. Unpolarised light is incident from air on a plane surface of a material of refractive index 
'µ'. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are 
perpendicular to each other, which of the following options is correct for this situation? 
Option A i = sin
-1
 
1 ??
??
?
??
 
Option B Reflected light is polarised with its electric vector perpendicular to the plane of 
incidence 
Option C Reflected light is polarised with its electric vector parallel to the plane of incidence 
Option D i = tan
-1
 
1 ??
??
?
??
 
Correct Option B 
Solution: 
 
This is the condition of polarisation in which light is incident on an interface at Brewster's angle. 
Thus reflected light is polarised with its E perpendicular to the plane of incidence. 
 
Q 6. In Young's double slit experiment the separation d between the slits is 2mm, the 
wavelength ? of the light used is 5896 Å and distance D between the screen and slits is 100 
cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular 
width to 0.21° (with same ? and D) the separation between the slits needs to be changed to 
Option A 2.1 mm 
Option B 1.9 mm 
Option C 1.8 mm 
Option D 1.7 mm 
Correct Option B 
Solution: 
Angular width =
d
d
For new seperation
'
d'
?
?
?
??
?
??
 
Page 5


  
 
  
  
 
    
   
 
  
 
 
    
   
      
 
 
 
 
 
 
  
 
  
 
 
 
 
 
 
  
  
  
  
  
  
  
 
  
  
 
    
Solution: 
From the graph is clear that this is an isobaric process.  
P= constant 
At constant pressure 
W = nR?T            … (1) 
?Q=nC p?T (for constant pressure) 
5
Q n R T ...(1)
2
??
? ? ?
??
??
 
Equation (1)/(2) 
nR T 2
5 5
n R T
2
?
??
??
?
??
??
 
 
Q 2. The fundamental frequency in an open organ pipe is equal to the third harmonic of a 
closed organ pipe if the length of the closed organ pipe is 20 cm, the length of the open organ 
pipe is 
Option A 12.5 cm 
Option B 8 cm 
Option C 13.2 cm 
Option D 16 cm 
Correct Option C 
Solution: 
For open organ pipe 
 
open
2
2
nv
General formula f=
2l
vv
f
2
?
?
? ? ?
??
?
 
For pipe closed at one end 
 
General formula 
nv
f=
4l
 
Fundamental 
v
f
4
? 
  
 
  
  
 
    
As given: 
1 2 2
3v v 3 1
4l 2l 4 20 2l
13.2
? ? ?
?
?
 
 
Q 3. The efficiency of an ideal heat engine working between the freezing point and boiling 
point of water, is 
Option A 6.25% 
Option B 20% 
Option C 26.8% 
Option D 12.5% 
Correct Option C 
Solution: Efficiency of heat engine 
12
12
1
2
1
Q W Q
work done
Efficiency
Energy available
QQ
Q
T
h1
T
273
1
373
1 0.731
26.8%
??
?
?
??
??
??
??
 
 
Q 4. At what temperature will the rms speed of oxygen molecules become just sufficient for 
escape from the Earth's atmosphere? (Given: Mass of oxygen molecule (m) = 2.76×10
-26
kg 
Option A 5.016×10
4
 K 
Option B 8.360×10
4
 K 
Option C 2.508×10
4
 K 
Option D 1.254×10
4
 K 
Correct Option B 
Solution: 
Escape velocity=11.2km/s=11200m/s 
0
3
rms
RT
V
M
? 
? ?
23
2
3
26
30 1.38 10 T
11.2 10
2.76 10
?
?
?
? ? ?
??
?
 
T = 8.360 ×10
4
 
 
  
  
 
  
  
 
    
Q 5. Unpolarised light is incident from air on a plane surface of a material of refractive index 
'µ'. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are 
perpendicular to each other, which of the following options is correct for this situation? 
Option A i = sin
-1
 
1 ??
??
?
??
 
Option B Reflected light is polarised with its electric vector perpendicular to the plane of 
incidence 
Option C Reflected light is polarised with its electric vector parallel to the plane of incidence 
Option D i = tan
-1
 
1 ??
??
?
??
 
Correct Option B 
Solution: 
 
This is the condition of polarisation in which light is incident on an interface at Brewster's angle. 
Thus reflected light is polarised with its E perpendicular to the plane of incidence. 
 
Q 6. In Young's double slit experiment the separation d between the slits is 2mm, the 
wavelength ? of the light used is 5896 Å and distance D between the screen and slits is 100 
cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular 
width to 0.21° (with same ? and D) the separation between the slits needs to be changed to 
Option A 2.1 mm 
Option B 1.9 mm 
Option C 1.8 mm 
Option D 1.7 mm 
Correct Option B 
Solution: 
Angular width =
d
d
For new seperation
'
d'
?
?
?
??
?
??
 
  
 
  
  
 
    
0.20 d'
0.21 d
0.20 d'
0.21 2
d' 1.904
??
??
??
 
 
Q 7. An astronomical refracting telescope will have large angular magnification and high 
angular resolution, when it has an objective lens of  
Option A large focal length and large diameter 
Option B large focal length and small diameter 
Option C small focal length and large diameter 
Option D small focal length and small diameter 
Correct Option A 
Solution: 
R. P ? d 
0
e
f
angular magnification=
f
And
Aperture of objective
Resolving power
wavelength
?
 
? Greater the focal length of the objective, more is the magnification. 
Also for a telescope to have high resolution the aperture of objective lens should be large. 
 
Q 8. The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and 
C are K A, K B and K C, respectively. AC is the major axis and SB is perpendicular to AC at the 
position of the Sun S as shown in the figure. Then 
 
 
 
Option A K B<K A<K C 
Option B K A>K B>K C 
Option C K A<K B<K C 
Option D K B>K A>K C 
Correct Option B 
Solution: We know that angular momentum remains conserved In motion of a planet around the 
sun 
? ?
m r×v =constant 
The nearest point in perigee which is point A. Next nearest is B and then C 
Therefore  
V
A B C
VV ?? 
? K A>K B>K C