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# Test Paper 9 Solution MBBS Notes | EduRev

## MBBS : Test Paper 9 Solution MBBS Notes | EduRev

``` Page 1

HINT – SHEET
1. Temperature of interface
1 1 22
12
KK
T
KK
q+q
=
+
300 100 200 0
60C
300 200
´ +´
= =°
+
2. As the temperature of water is increased from 2°C
to 3°C the density of water increases (rememeber
anomalous behaviour of water), also the volume
of sphere increases. Therefore buoyant force on
sphere due to water shall increase.
3. Note the volume has been plotted along Y-axis.
W = SPDV
= (0.6 – 0.8) × 10
3
+ (0.4 – 0.6) × 2 × 10
3
+ (0.2–0.4) ×3×10
–3
= – 200 – 400 – 600 = –1200 J
(0999DMD310319010) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 09
20-10-2019
LTS/HS-1/7
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
+91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
Test Type : Unit Test # 07
4. Rate of cooling (R) =
( )
44
0
A TT
t mc
Îs-
Dq
=
Þ
2
3
A Area r1
R
m volume r r
µµ µµ
Þ ()
1/3
11
RateR
rm
µµ
3 1/3
4
m =  × r r m
3
éù
r p Þµ
êú
ëû
Q
Þ
1/3
1/3
12
21
Rm 1
R m3
æö
æö
==
ç÷ ç÷
èø
èø
5. (i) Thermal energy
Q 300
TC mc 15 J/°C
T 45 25
====
D-
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.4 1 4 2 1 4 4 1 1 3 3 4 2 3 3 2 3 3 2 4
Que.21222324 2526 272829303132333435 3637 383940
Ans.2 4 4 3 3 3 3 3 2 4 2 1 3 4 2 1 1 4 2 1
Que.41424344 4546 474849505152535455 5657 585960
Ans.2 3 3 2 3 2 1 2 2 3 3 3 2 1 1 4 4 2 1 4
Que.61626364 6566 676869707172737475 7677 787980
Ans.4 1 3 2 1 1 3 4 3 3 2 4 2 2 4 4 1 2 2 3
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.3 2 4 4 1 4 2 4 1 4 1 2 4 3 1 1 3 4 4 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.2 2 3 3 1 2 1 1 2 1 4 3 3 2 1 4 3 3 3 3
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 4 2 2 4 2 2 3 2 3 4 1 2 2 3 4 3 3 2 1
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.1 1 2 1 4 2 4 3 2 3 2 4 1 2 4 2 4 3 2 2
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 4 4 3 4 2 3 2 2 2 3 4 4 3 3 2 2 3 1 4
Page 2

HINT – SHEET
1. Temperature of interface
1 1 22
12
KK
T
KK
q+q
=
+
300 100 200 0
60C
300 200
´ +´
= =°
+
2. As the temperature of water is increased from 2°C
to 3°C the density of water increases (rememeber
anomalous behaviour of water), also the volume
of sphere increases. Therefore buoyant force on
sphere due to water shall increase.
3. Note the volume has been plotted along Y-axis.
W = SPDV
= (0.6 – 0.8) × 10
3
+ (0.4 – 0.6) × 2 × 10
3
+ (0.2–0.4) ×3×10
–3
= – 200 – 400 – 600 = –1200 J
(0999DMD310319010) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 09
20-10-2019
LTS/HS-1/7
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
+91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
Test Type : Unit Test # 07
4. Rate of cooling (R) =
( )
44
0
A TT
t mc
Îs-
Dq
=
Þ
2
3
A Area r1
R
m volume r r
µµ µµ
Þ ()
1/3
11
RateR
rm
µµ
3 1/3
4
m =  × r r m
3
éù
r p Þµ
êú
ëû
Q
Þ
1/3
1/3
12
21
Rm 1
R m3
æö
æö
==
ç÷ ç÷
èø
èø
5. (i) Thermal energy
Q 300
TC mc 15 J/°C
T 45 25
====
D-
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.4 1 4 2 1 4 4 1 1 3 3 4 2 3 3 2 3 3 2 4
Que.21222324 2526 272829303132333435 3637 383940
Ans.2 4 4 3 3 3 3 3 2 4 2 1 3 4 2 1 1 4 2 1
Que.41424344 4546 474849505152535455 5657 585960
Ans.2 3 3 2 3 2 1 2 2 3 3 3 2 1 1 4 4 2 1 4
Que.61626364 6566 676869707172737475 7677 787980
Ans.4 1 3 2 1 1 3 4 3 3 2 4 2 2 4 4 1 2 2 3
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.3 2 4 4 1 4 2 4 1 4 1 2 4 3 1 1 3 4 4 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.2 2 3 3 1 2 1 1 2 1 4 3 3 2 1 4 3 3 3 3
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 4 2 2 4 2 2 3 2 3 4 1 2 2 3 4 3 3 2 1
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.1 1 2 1 4 2 4 3 2 3 2 4 1 2 4 2 4 3 2 2
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 4 4 3 4 2 3 2 2 2 3 4 4 3 3 2 2 3 1 4
LTS/HS-2/7 0999DMD310319010
Target : Pre-Medical 2020/NEET-UG/20-10-2019
AL LEN
(ii) Specific heat is nothing but thermal capacity per
unit mass
3
mc 15
C 600 J/kg-°C
m 25 10
-
===
´
6. The image on the screen is real and inverted.
Triangle
aperture
Screen
20 cm 10 cm
The size of the image on the screen has aperture size
given by
10
Size 1.0 0.5 cm
20
æö
==
ç÷
èø
7. For mixture of gases,
12
1 v 2v
v
12
nC nC
C
nn
+
=
+
where
f
CR
2
=
, f is degree of freedom
and
12
1 p 2p
p
12
nC nC
C
nn
+
=
+
where     p
f
C 1R
2
æö
=+
ç÷
èø
For helium, n
1
= 4, f = 3
For oxygen,
2
1
n , f = 5
2
=
\
p
v
5R 1 7R
4
C
47
2 22
3R 1 5R
C 29
4
2 22
´ +´
==
´ +´
= 1.62
8. The efficiency of reversible engine is always greater
than that of irreversible engine. In case of irreversible
engine, a part of the energy may be dissipated against
friction, etc.
9.
t = 20 cm m = 3/2
curvature = 20 cm
object
Considering refraction at the curved surface,
u = –20;     m
2
= 1
m
1
= 3/2;    R = +20
Applying
2 1 21
v uR
mmm-m
-=
1 3/2 1 3/2
v 10
v 20 20
-
- = Þ =-
-
i.e. 10 cm below the curved surface or 10 cm above
the actual position of flower.
10.
( ) ( ) X 125 Y 70
500 40
- - --
=
ForY = 50
X = 1375.0°X
11. for lens u = wants to see = ¥; v = can see = –5 m
\ From
1 11
f vu
=-
Þ
1 11
f5
=-
-¥
Þ f = –5 m.
12. Absolute temperatures of the black body
corresponding to curve P and Q are in the inverse
ratio of l
m
(Wein's displacement law).
i.e.,
P
Q
T 1987
T 2980
=
Page 3

HINT – SHEET
1. Temperature of interface
1 1 22
12
KK
T
KK
q+q
=
+
300 100 200 0
60C
300 200
´ +´
= =°
+
2. As the temperature of water is increased from 2°C
to 3°C the density of water increases (rememeber
anomalous behaviour of water), also the volume
of sphere increases. Therefore buoyant force on
sphere due to water shall increase.
3. Note the volume has been plotted along Y-axis.
W = SPDV
= (0.6 – 0.8) × 10
3
+ (0.4 – 0.6) × 2 × 10
3
+ (0.2–0.4) ×3×10
–3
= – 200 – 400 – 600 = –1200 J
(0999DMD310319010) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 09
20-10-2019
LTS/HS-1/7
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
+91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
Test Type : Unit Test # 07
4. Rate of cooling (R) =
( )
44
0
A TT
t mc
Îs-
Dq
=
Þ
2
3
A Area r1
R
m volume r r
µµ µµ
Þ ()
1/3
11
RateR
rm
µµ
3 1/3
4
m =  × r r m
3
éù
r p Þµ
êú
ëû
Q
Þ
1/3
1/3
12
21
Rm 1
R m3
æö
æö
==
ç÷ ç÷
èø
èø
5. (i) Thermal energy
Q 300
TC mc 15 J/°C
T 45 25
====
D-
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.4 1 4 2 1 4 4 1 1 3 3 4 2 3 3 2 3 3 2 4
Que.21222324 2526 272829303132333435 3637 383940
Ans.2 4 4 3 3 3 3 3 2 4 2 1 3 4 2 1 1 4 2 1
Que.41424344 4546 474849505152535455 5657 585960
Ans.2 3 3 2 3 2 1 2 2 3 3 3 2 1 1 4 4 2 1 4
Que.61626364 6566 676869707172737475 7677 787980
Ans.4 1 3 2 1 1 3 4 3 3 2 4 2 2 4 4 1 2 2 3
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.3 2 4 4 1 4 2 4 1 4 1 2 4 3 1 1 3 4 4 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.2 2 3 3 1 2 1 1 2 1 4 3 3 2 1 4 3 3 3 3
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 4 2 2 4 2 2 3 2 3 4 1 2 2 3 4 3 3 2 1
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.1 1 2 1 4 2 4 3 2 3 2 4 1 2 4 2 4 3 2 2
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 4 4 3 4 2 3 2 2 2 3 4 4 3 3 2 2 3 1 4
LTS/HS-2/7 0999DMD310319010
Target : Pre-Medical 2020/NEET-UG/20-10-2019
AL LEN
(ii) Specific heat is nothing but thermal capacity per
unit mass
3
mc 15
C 600 J/kg-°C
m 25 10
-
===
´
6. The image on the screen is real and inverted.
Triangle
aperture
Screen
20 cm 10 cm
The size of the image on the screen has aperture size
given by
10
Size 1.0 0.5 cm
20
æö
==
ç÷
èø
7. For mixture of gases,
12
1 v 2v
v
12
nC nC
C
nn
+
=
+
where
f
CR
2
=
, f is degree of freedom
and
12
1 p 2p
p
12
nC nC
C
nn
+
=
+
where     p
f
C 1R
2
æö
=+
ç÷
èø
For helium, n
1
= 4, f = 3
For oxygen,
2
1
n , f = 5
2
=
\
p
v
5R 1 7R
4
C
47
2 22
3R 1 5R
C 29
4
2 22
´ +´
==
´ +´
= 1.62
8. The efficiency of reversible engine is always greater
than that of irreversible engine. In case of irreversible
engine, a part of the energy may be dissipated against
friction, etc.
9.
t = 20 cm m = 3/2
curvature = 20 cm
object
Considering refraction at the curved surface,
u = –20;     m
2
= 1
m
1
= 3/2;    R = +20
Applying
2 1 21
v uR
mmm-m
-=
1 3/2 1 3/2
v 10
v 20 20
-
- = Þ =-
-
i.e. 10 cm below the curved surface or 10 cm above
the actual position of flower.
10.
( ) ( ) X 125 Y 70
500 40
- - --
=
ForY = 50
X = 1375.0°X
11. for lens u = wants to see = ¥; v = can see = –5 m
\ From
1 11
f vu
=-
Þ
1 11
f5
=-
-¥
Þ f = –5 m.
12. Absolute temperatures of the black body
corresponding to curve P and Q are in the inverse
ratio of l
m
(Wein's displacement law).
i.e.,
P
Q
T 1987
T 2980
=
AL LEN
0999DMD310319010 LTS/HS-3/7
Area under curves represent the total power radiated
by a body and is proportional to the fourth power of
absolute temperature (Stefan's law)
\
4
PP
QQ
AT 16
A T 81
æö
== ç÷
ç÷
èø
13.
L
H
T 7002
11
T 2100 3
h=- =-=
2
% 100 66%
3
h=´=
Actual efficiency is 40% which is 60% of the
theoretical efficiency.
14. In the first case, the temperature difference is greater
as compared to the second case. So, the rate of loss
of heat quicker.
15.
( )
()
rms
11
rms2
2
v
T
vT
=
Given, (V
rms
)
1
= 100 m/s
T
1
= 27°C

= 27 + 273 = 300 K
T
2
= 127°C = 127 + 273 = 400 K
\ From Eq. (i)
rms2
100 300 3
(v ) 400 2
==
Þ( )
rms
2
2 100 200
v m/s
33
´
==
16.
( )
( )
3
3
A AA
AA A AA
3
BB B BB B BB
4/3 rC mC rC
mC 4/3rC rC
pr æö r
= =´
ç÷
prr
èø
3
1 2 11
2 1 3 12
æ ö æ ö æö
= ´ ´=
ç ÷ ç ÷ ç÷
è ø è ø èø
17. This is the statement of the second law of
thermodynamics.
18. Process AB is isochoric,
\ W
AB
= PDV = 0
Process BC is isothermal
\ W
BC
= RT
2
· ln
2
1
V
V
æö
ç÷
èø
Process CA is isobaric
\ W
CA
= –PDV = –RDT = –R(T
1
–T
2
) = R(T
2
–T
1
)
(Negative sign is taken because of compression)
19. Change in L
A
= change in L
B
i.e., DL
A
= DL
B
Þa
A
DTL
A
= a
B
DTL
B
or a
A
L
A
= a
B
L
B
20. As the ray moves toward the normal while entering
medium 2 from 1, we have n
2
> n
1
For total internal reflection at interface of 2 and 3,
n
2
> n
3
.
Besides n
3
should also be less than n
1
or else ray
would have emerged in medium 3, parallel to its path
in medium 1.
Hence, n
3
< n
1
< n
2
is the correct order.
21. According to Wien's law
m
1
T
lµ and from the figure
(lm)
1
< (lm)
3
< (lm)
2
therefore T
1
> T
3
> T
2
.
22.
2N
PK
3V
= from the kinetic-theory account for
pressure.
3 PV
N
2K
=
AA
N 3 PV
n
N 2 KN
==
23. By symmetry
I
AB
= I
BC
and I
= I
DC
\ No current in BO and OD
\ T
B
= T
O
= T
D
24. For an adiabatic process, PV
g
= constant
TV
g – 1

= constant ;
and
1
T P constant
-g
g
=
g = 5/3 (argon being a monoatomic gas).
Page 4

HINT – SHEET
1. Temperature of interface
1 1 22
12
KK
T
KK
q+q
=
+
300 100 200 0
60C
300 200
´ +´
= =°
+
2. As the temperature of water is increased from 2°C
to 3°C the density of water increases (rememeber
anomalous behaviour of water), also the volume
of sphere increases. Therefore buoyant force on
sphere due to water shall increase.
3. Note the volume has been plotted along Y-axis.
W = SPDV
= (0.6 – 0.8) × 10
3
+ (0.4 – 0.6) × 2 × 10
3
+ (0.2–0.4) ×3×10
–3
= – 200 – 400 – 600 = –1200 J
(0999DMD310319010) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 09
20-10-2019
LTS/HS-1/7
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
+91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
Test Type : Unit Test # 07
4. Rate of cooling (R) =
( )
44
0
A TT
t mc
Îs-
Dq
=
Þ
2
3
A Area r1
R
m volume r r
µµ µµ
Þ ()
1/3
11
RateR
rm
µµ
3 1/3
4
m =  × r r m
3
éù
r p Þµ
êú
ëû
Q
Þ
1/3
1/3
12
21
Rm 1
R m3
æö
æö
==
ç÷ ç÷
èø
èø
5. (i) Thermal energy
Q 300
TC mc 15 J/°C
T 45 25
====
D-
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.4 1 4 2 1 4 4 1 1 3 3 4 2 3 3 2 3 3 2 4
Que.21222324 2526 272829303132333435 3637 383940
Ans.2 4 4 3 3 3 3 3 2 4 2 1 3 4 2 1 1 4 2 1
Que.41424344 4546 474849505152535455 5657 585960
Ans.2 3 3 2 3 2 1 2 2 3 3 3 2 1 1 4 4 2 1 4
Que.61626364 6566 676869707172737475 7677 787980
Ans.4 1 3 2 1 1 3 4 3 3 2 4 2 2 4 4 1 2 2 3
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.3 2 4 4 1 4 2 4 1 4 1 2 4 3 1 1 3 4 4 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.2 2 3 3 1 2 1 1 2 1 4 3 3 2 1 4 3 3 3 3
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 4 2 2 4 2 2 3 2 3 4 1 2 2 3 4 3 3 2 1
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.1 1 2 1 4 2 4 3 2 3 2 4 1 2 4 2 4 3 2 2
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 4 4 3 4 2 3 2 2 2 3 4 4 3 3 2 2 3 1 4
LTS/HS-2/7 0999DMD310319010
Target : Pre-Medical 2020/NEET-UG/20-10-2019
AL LEN
(ii) Specific heat is nothing but thermal capacity per
unit mass
3
mc 15
C 600 J/kg-°C
m 25 10
-
===
´
6. The image on the screen is real and inverted.
Triangle
aperture
Screen
20 cm 10 cm
The size of the image on the screen has aperture size
given by
10
Size 1.0 0.5 cm
20
æö
==
ç÷
èø
7. For mixture of gases,
12
1 v 2v
v
12
nC nC
C
nn
+
=
+
where
f
CR
2
=
, f is degree of freedom
and
12
1 p 2p
p
12
nC nC
C
nn
+
=
+
where     p
f
C 1R
2
æö
=+
ç÷
èø
For helium, n
1
= 4, f = 3
For oxygen,
2
1
n , f = 5
2
=
\
p
v
5R 1 7R
4
C
47
2 22
3R 1 5R
C 29
4
2 22
´ +´
==
´ +´
= 1.62
8. The efficiency of reversible engine is always greater
than that of irreversible engine. In case of irreversible
engine, a part of the energy may be dissipated against
friction, etc.
9.
t = 20 cm m = 3/2
curvature = 20 cm
object
Considering refraction at the curved surface,
u = –20;     m
2
= 1
m
1
= 3/2;    R = +20
Applying
2 1 21
v uR
mmm-m
-=
1 3/2 1 3/2
v 10
v 20 20
-
- = Þ =-
-
i.e. 10 cm below the curved surface or 10 cm above
the actual position of flower.
10.
( ) ( ) X 125 Y 70
500 40
- - --
=
ForY = 50
X = 1375.0°X
11. for lens u = wants to see = ¥; v = can see = –5 m
\ From
1 11
f vu
=-
Þ
1 11
f5
=-
-¥
Þ f = –5 m.
12. Absolute temperatures of the black body
corresponding to curve P and Q are in the inverse
ratio of l
m
(Wein's displacement law).
i.e.,
P
Q
T 1987
T 2980
=
AL LEN
0999DMD310319010 LTS/HS-3/7
Area under curves represent the total power radiated
by a body and is proportional to the fourth power of
absolute temperature (Stefan's law)
\
4
PP
QQ
AT 16
A T 81
æö
== ç÷
ç÷
èø
13.
L
H
T 7002
11
T 2100 3
h=- =-=
2
% 100 66%
3
h=´=
Actual efficiency is 40% which is 60% of the
theoretical efficiency.
14. In the first case, the temperature difference is greater
as compared to the second case. So, the rate of loss
of heat quicker.
15.
( )
()
rms
11
rms2
2
v
T
vT
=
Given, (V
rms
)
1
= 100 m/s
T
1
= 27°C

= 27 + 273 = 300 K
T
2
= 127°C = 127 + 273 = 400 K
\ From Eq. (i)
rms2
100 300 3
(v ) 400 2
==
Þ( )
rms
2
2 100 200
v m/s
33
´
==
16.
( )
( )
3
3
A AA
AA A AA
3
BB B BB B BB
4/3 rC mC rC
mC 4/3rC rC
pr æö r
= =´
ç÷
prr
èø
3
1 2 11
2 1 3 12
æ ö æ ö æö
= ´ ´=
ç ÷ ç ÷ ç÷
è ø è ø èø
17. This is the statement of the second law of
thermodynamics.
18. Process AB is isochoric,
\ W
AB
= PDV = 0
Process BC is isothermal
\ W
BC
= RT
2
· ln
2
1
V
V
æö
ç÷
èø
Process CA is isobaric
\ W
CA
= –PDV = –RDT = –R(T
1
–T
2
) = R(T
2
–T
1
)
(Negative sign is taken because of compression)
19. Change in L
A
= change in L
B
i.e., DL
A
= DL
B
Þa
A
DTL
A
= a
B
DTL
B
or a
A
L
A
= a
B
L
B
20. As the ray moves toward the normal while entering
medium 2 from 1, we have n
2
> n
1
For total internal reflection at interface of 2 and 3,
n
2
> n
3
.
Besides n
3
should also be less than n
1
or else ray
would have emerged in medium 3, parallel to its path
in medium 1.
Hence, n
3
< n
1
< n
2
is the correct order.
21. According to Wien's law
m
1
T
lµ and from the figure
(lm)
1
< (lm)
3
< (lm)
2
therefore T
1
> T
3
> T
2
.
22.
2N
PK
3V
= from the kinetic-theory account for
pressure.
3 PV
N
2K
=
AA
N 3 PV
n
N 2 KN
==
23. By symmetry
I
AB
= I
BC
and I
= I
DC
\ No current in BO and OD
\ T
B
= T
O
= T
D
24. For an adiabatic process, PV
g
= constant
TV
g – 1

= constant ;
and
1
T P constant
-g
g
=
g = 5/3 (argon being a monoatomic gas).
LTS/HS-4/7 0999DMD310319010
Target : Pre-Medical 2020/NEET-UG/20-10-2019
AL LEN
25. E = e sAT
4
= es4pR
2
T
4
( )( )
( )( )
24
1
24
2
e 1 4000 E
E
e 4 2000
s
=
s
=
( )
( )
4
4
4
2 . 2000
1
16 2000
=
26.
( )
12
1 11
1
f RR
æö
=m--
ç÷
èø
R = 10 cm
f = 20 f = 20
f'
( )
11
f' 31
10 10
æö
=--
ç÷
-
èø
10
f'
4
-
=
eq
1 141 24
f 20 10 20 20 10
=- + =-
; f
eq
= –10/3 cm
27.
ii
WW
W
mix
iW
mL
m
c
mm
q-
q=
+
Q
m
i
= m
w
Þ
i
W
W
mix
L
336
800
c
4.2
0C
22
q-
+-
q = = =°
or
1 gm ice ® 80 cal ® 0°C
1 gm of 80°C water ® 0°C ® 80 cal
so final temperature will be 0°C
28. For a given pressure, V is small for T
1
. Since V µ T,
therefore, T
1
< T
2
.
29. L = f
0
+ f
e
= 44 and
0
e
f
m 10
f
==
This gives f
0
= 40 cm
30.
eq 12
2 11
K KK
=+
1
1
K A
Q
L
Dq
=
and
2
2
K A
Q
L
Dq
=
eq
eq
A 1 A
QK.
1
L 2L
K
Dq Dq
==
12
1 A
.
2L 1 11
2KK
Dq
=
æö
+
ç÷
èø
12
1 A Q
·
A Q A Q
L
Q · L Q · L
D
=
DD
+
12
12
12
Q Q 1
=
11
QQ
QQ
=
+
+
31. m
1
sin a
1
= m
2
sin a
2
a
1
a
2
m
1
m
2
12
12
cc
sin sin
vv
a=a
12
12
sin sin
ff
aa
=
ll
Þ
2
21
1
sin
sin
a
l =l
a
Page 5

HINT – SHEET
1. Temperature of interface
1 1 22
12
KK
T
KK
q+q
=
+
300 100 200 0
60C
300 200
´ +´
= =°
+
2. As the temperature of water is increased from 2°C
to 3°C the density of water increases (rememeber
anomalous behaviour of water), also the volume
of sphere increases. Therefore buoyant force on
sphere due to water shall increase.
3. Note the volume has been plotted along Y-axis.
W = SPDV
= (0.6 – 0.8) × 10
3
+ (0.4 – 0.6) × 2 × 10
3
+ (0.2–0.4) ×3×10
–3
= – 200 – 400 – 600 = –1200 J
(0999DMD310319010) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 09
20-10-2019
LTS/HS-1/7
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
+91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
Test Type : Unit Test # 07
4. Rate of cooling (R) =
( )
44
0
A TT
t mc
Îs-
Dq
=
Þ
2
3
A Area r1
R
m volume r r
µµ µµ
Þ ()
1/3
11
RateR
rm
µµ
3 1/3
4
m =  × r r m
3
éù
r p Þµ
êú
ëû
Q
Þ
1/3
1/3
12
21
Rm 1
R m3
æö
æö
==
ç÷ ç÷
èø
èø
5. (i) Thermal energy
Q 300
TC mc 15 J/°C
T 45 25
====
D-
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.4 1 4 2 1 4 4 1 1 3 3 4 2 3 3 2 3 3 2 4
Que.21222324 2526 272829303132333435 3637 383940
Ans.2 4 4 3 3 3 3 3 2 4 2 1 3 4 2 1 1 4 2 1
Que.41424344 4546 474849505152535455 5657 585960
Ans.2 3 3 2 3 2 1 2 2 3 3 3 2 1 1 4 4 2 1 4
Que.61626364 6566 676869707172737475 7677 787980
Ans.4 1 3 2 1 1 3 4 3 3 2 4 2 2 4 4 1 2 2 3
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.3 2 4 4 1 4 2 4 1 4 1 2 4 3 1 1 3 4 4 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.2 2 3 3 1 2 1 1 2 1 4 3 3 2 1 4 3 3 3 3
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 4 2 2 4 2 2 3 2 3 4 1 2 2 3 4 3 3 2 1
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.1 1 2 1 4 2 4 3 2 3 2 4 1 2 4 2 4 3 2 2
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 4 4 3 4 2 3 2 2 2 3 4 4 3 3 2 2 3 1 4
LTS/HS-2/7 0999DMD310319010
Target : Pre-Medical 2020/NEET-UG/20-10-2019
AL LEN
(ii) Specific heat is nothing but thermal capacity per
unit mass
3
mc 15
C 600 J/kg-°C
m 25 10
-
===
´
6. The image on the screen is real and inverted.
Triangle
aperture
Screen
20 cm 10 cm
The size of the image on the screen has aperture size
given by
10
Size 1.0 0.5 cm
20
æö
==
ç÷
èø
7. For mixture of gases,
12
1 v 2v
v
12
nC nC
C
nn
+
=
+
where
f
CR
2
=
, f is degree of freedom
and
12
1 p 2p
p
12
nC nC
C
nn
+
=
+
where     p
f
C 1R
2
æö
=+
ç÷
èø
For helium, n
1
= 4, f = 3
For oxygen,
2
1
n , f = 5
2
=
\
p
v
5R 1 7R
4
C
47
2 22
3R 1 5R
C 29
4
2 22
´ +´
==
´ +´
= 1.62
8. The efficiency of reversible engine is always greater
than that of irreversible engine. In case of irreversible
engine, a part of the energy may be dissipated against
friction, etc.
9.
t = 20 cm m = 3/2
curvature = 20 cm
object
Considering refraction at the curved surface,
u = –20;     m
2
= 1
m
1
= 3/2;    R = +20
Applying
2 1 21
v uR
mmm-m
-=
1 3/2 1 3/2
v 10
v 20 20
-
- = Þ =-
-
i.e. 10 cm below the curved surface or 10 cm above
the actual position of flower.
10.
( ) ( ) X 125 Y 70
500 40
- - --
=
ForY = 50
X = 1375.0°X
11. for lens u = wants to see = ¥; v = can see = –5 m
\ From
1 11
f vu
=-
Þ
1 11
f5
=-
-¥
Þ f = –5 m.
12. Absolute temperatures of the black body
corresponding to curve P and Q are in the inverse
ratio of l
m
(Wein's displacement law).
i.e.,
P
Q
T 1987
T 2980
=
AL LEN
0999DMD310319010 LTS/HS-3/7
Area under curves represent the total power radiated
by a body and is proportional to the fourth power of
absolute temperature (Stefan's law)
\
4
PP
QQ
AT 16
A T 81
æö
== ç÷
ç÷
èø
13.
L
H
T 7002
11
T 2100 3
h=- =-=
2
% 100 66%
3
h=´=
Actual efficiency is 40% which is 60% of the
theoretical efficiency.
14. In the first case, the temperature difference is greater
as compared to the second case. So, the rate of loss
of heat quicker.
15.
( )
()
rms
11
rms2
2
v
T
vT
=
Given, (V
rms
)
1
= 100 m/s
T
1
= 27°C

= 27 + 273 = 300 K
T
2
= 127°C = 127 + 273 = 400 K
\ From Eq. (i)
rms2
100 300 3
(v ) 400 2
==
Þ( )
rms
2
2 100 200
v m/s
33
´
==
16.
( )
( )
3
3
A AA
AA A AA
3
BB B BB B BB
4/3 rC mC rC
mC 4/3rC rC
pr æö r
= =´
ç÷
prr
èø
3
1 2 11
2 1 3 12
æ ö æ ö æö
= ´ ´=
ç ÷ ç ÷ ç÷
è ø è ø èø
17. This is the statement of the second law of
thermodynamics.
18. Process AB is isochoric,
\ W
AB
= PDV = 0
Process BC is isothermal
\ W
BC
= RT
2
· ln
2
1
V
V
æö
ç÷
èø
Process CA is isobaric
\ W
CA
= –PDV = –RDT = –R(T
1
–T
2
) = R(T
2
–T
1
)
(Negative sign is taken because of compression)
19. Change in L
A
= change in L
B
i.e., DL
A
= DL
B
Þa
A
DTL
A
= a
B
DTL
B
or a
A
L
A
= a
B
L
B
20. As the ray moves toward the normal while entering
medium 2 from 1, we have n
2
> n
1
For total internal reflection at interface of 2 and 3,
n
2
> n
3
.
Besides n
3
should also be less than n
1
or else ray
would have emerged in medium 3, parallel to its path
in medium 1.
Hence, n
3
< n
1
< n
2
is the correct order.
21. According to Wien's law
m
1
T
lµ and from the figure
(lm)
1
< (lm)
3
< (lm)
2
therefore T
1
> T
3
> T
2
.
22.
2N
PK
3V
= from the kinetic-theory account for
pressure.
3 PV
N
2K
=
AA
N 3 PV
n
N 2 KN
==
23. By symmetry
I
AB
= I
BC
and I
= I
DC
\ No current in BO and OD
\ T
B
= T
O
= T
D
24. For an adiabatic process, PV
g
= constant
TV
g – 1

= constant ;
and
1
T P constant
-g
g
=
g = 5/3 (argon being a monoatomic gas).
LTS/HS-4/7 0999DMD310319010
Target : Pre-Medical 2020/NEET-UG/20-10-2019
AL LEN
25. E = e sAT
4
= es4pR
2
T
4
( )( )
( )( )
24
1
24
2
e 1 4000 E
E
e 4 2000
s
=
s
=
( )
( )
4
4
4
2 . 2000
1
16 2000
=
26.
( )
12
1 11
1
f RR
æö
=m--
ç÷
èø
R = 10 cm
f = 20 f = 20
f'
( )
11
f' 31
10 10
æö
=--
ç÷
-
èø
10
f'
4
-
=
eq
1 141 24
f 20 10 20 20 10
=- + =-
; f
eq
= –10/3 cm
27.
ii
WW
W
mix
iW
mL
m
c
mm
q-
q=
+
Q
m
i
= m
w
Þ
i
W
W
mix
L
336
800
c
4.2
0C
22
q-
+-
q = = =°
or
1 gm ice ® 80 cal ® 0°C
1 gm of 80°C water ® 0°C ® 80 cal
so final temperature will be 0°C
28. For a given pressure, V is small for T
1
. Since V µ T,
therefore, T
1
< T
2
.
29. L = f
0
+ f
e
= 44 and
0
e
f
m 10
f
==
This gives f
0
= 40 cm
30.
eq 12
2 11
K KK
=+
1
1
K A
Q
L
Dq
=
and
2
2
K A
Q
L
Dq
=
eq
eq
A 1 A
QK.
1
L 2L
K
Dq Dq
==
12
1 A
.
2L 1 11
2KK
Dq
=
æö
+
ç÷
èø
12
1 A Q
·
A Q A Q
L
Q · L Q · L
D
=
DD
+
12
12
12
Q Q 1
=
11
QQ
QQ
=
+
+
31. m
1
sin a
1
= m
2
sin a
2
a
1
a
2
m
1
m
2
12
12
cc
sin sin
vv
a=a
12
12
sin sin
ff
aa
=
ll
Þ
2
21
1
sin
sin
a
l =l
a
AL LEN
0999DMD310319010 LTS/HS-5/7
32.
app.
Mass expelled
Mass remained × T
g=
D
4
x /100 1
1.25 10 / C
x 80 8000
-
= = = ´°
´
33.
3PV
C
M
=
5
1
3 24 10 10 100
cms
20
-
´ ´ ´´
=
= 6 × 10
4
cm s
–1
= 600 ms
–1
34. According to Newton's law
12
t
q -q
12
0
k
2
q +q éù
= -q
êú
ëû
Initially,
( )
0
80 64 80 64
K
52
- +æö
= -q
ç÷
èø
Þ 3.2 = K[72 – q
0
] ....(i)
Finally,
( )
0
64 52 64 52
K
102
- +éù
= -q
êú
ëû
Þ 1.2 = K[58 – q
0
] ....(ii)
On solving equation (i) and (ii) q
0
= 49°C s
35. Since
0
e
f
m
f
=
Also m =
Angle subtended by the image
Angle subtended by the object
\
0
e
f
f
a
=
b
Þ
0
e
f 602
a 24
f5
´b ´
= = =°
36. The volume of matter in portional AB of the curve
is almost constant and pressure is decreasing. These
are the characteristics of liquid state.
37. (P = constant)
PP
nC TC Q5
WnRT R2
D D
= ==
DD
38. Ideal gas equation PV = mRT =
A
N
RT
N
æö
ç÷
èø
where N
= Number of molecule, N
A
\
1 1 12
2 2 21
N P VT
N P VT
æ öæ öæ ö
=
ç ÷ç ÷ç ÷
è øè øè ø

P V 2T4
2P V/4 T1
æ öæ öæö
==
ç ÷ç ÷ç÷
è øè øèø
39. Given u = (f + x
1
) and u = (f + x
2
), v = (f + x
2
)
the focal length
( )( )
( )( )
12
12
fx fx uv
f
u v fx fx
++
==
+ + ++
On solving, we get f
2
= x
1
x
2
or
12
f xx =
40.
dQ KA
dt
Dq
=
l
, For both rods K, A and Dq are same
Þ
dQ1
dt
µ
l
So
( )
( )
semi circular
straight
dQ / dt
dQ / dt
straight
semi circular
2r2
r
= ==
pp
l
l
41.
21
12
TT 1
1,
T T1
h=-=
-h
( )
2
1 2 12
T 1
T T T/T1
w==
--

( )
11
1/11
-h
==
h éù -h-
ëû
As
1 0.1
10% 0.1,  = 9
0.1
-
h= =w=
```
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