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 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 
 
  
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FAQs on NEET 2018 Past Year Paper with Detailed Solutions - NEET Mock Test Series 2024

1. What is NEET 2018?
Ans. NEET 2018 is the National Eligibility cum Entrance Test conducted by the Central Board of Secondary Education (CBSE) for admission to undergraduate medical and dental courses in government and private medical colleges in India.
2. What is the pattern of NEET 2018 exam?
Ans. NEET 2018 exam consists of a single paper with 180 multiple-choice questions from Physics, Chemistry, and Biology. The duration of the exam is 3 hours.
3. How can I access past year papers for NEET 2018?
Ans. Past year papers for NEET 2018 can be accessed online from various websites such as the official website of CBSE, NEET-UG, and other educational portals.
4. Are detailed solutions available for NEET 2018 past year papers?
Ans. Yes, detailed solutions for NEET 2018 past year papers are available online on various educational portals as well as in NEET-UG guidebooks.
5. Can practicing NEET 2018 past year papers help in preparation for the exam?
Ans. Yes, practicing NEET 2018 past year papers can help in preparation for the exam as it gives a clear idea of the exam pattern, level of difficulty, and types of questions asked in the exam. It also helps in identifying weak areas and improving time management skills.
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