Test Paper 10 Solution MBBS Notes | EduRev

MBBS : Test Paper 10 Solution MBBS Notes | EduRev

 Page 1


HINT – SHEET
ANSWER KEY
(0999DMD310319012) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 11
17-11-2019
LTS/HS-1/4
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 # 09
1. Energy is released in a process when total binding
energy (BE) of the nucleus is increased or we can
say when total BE of products is more than the
reactants. By calculation we can see that only in
case of option (3), this happens.
Given, W ® 2Y
BE of reactants = 120 × 7.5 = 900 MeV
and BE of products = 90 × 8 + 60 × 8.5 = 1230 MeV
i.e., Be of products > BE of reactants.
2. v
0
 = 4 × 10
15
8
0 15
0
c 3 10
m
4 10
´
l==
n´
= 750Å
0
12400
eV f=
l
 = 16.5 eV
3. E
total
 = 160 J, P.E.
max
 = E
total
 = 160 J
KE
max
 = 
( )
2
26
11
Ka 2 10 0.01 100 J
22
=´´´=
4. 2pr = nl Þ  n =
2r p
l
=
-
-
´ ´´
´
11
10
2 3.14 5.3 10
1.1 10
=3
5. Minimum wavelength of continuous X-ray
spectrum is
l
min
 (Å) = 3
12375 12375
E(eV) 80 10
=
´
 
0.155 »
here energy of the incident electrons 80 KeV is
more than ionization energy of k-shell electrons
is 72.5 KeV. So characteristics X-ray spectrum
will also be obtained because energy of incident
electron is enough to knock out the electron from
K or L shells.
6. Activity of a radioactive substance,
R = lN \ l = 
R
N
Here R = N
2
 particles per second, N = N
1
\
2
1
N
N
l=
7.
2
0 max
hc1
W mv
2
=+
l
Assuming W
0
 to be negligible in comparison to 
hc
l
i.e. 
2
max
1
mv µ
l
 Þ max
1
v µ
l
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 3 2 3 4 2 4 3 2 1 2 1 3 3 1 3 3 1 4 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.3 1 1 1 2 2 2 2 4 1 2 2 4 3 4 3 3 2 3 1
Que.41424344 4546 474849505152535455 5657 585960
Ans.1 4 4 3 3 4 3 4 4 1 4 2 2 4 2 3 2 2 4 2
Que.61626364 6566 676869707172737475 7677 787980
Ans.1 2 2 3 3 2 3 1 3 2 1 1 3 4 2 2 4 2 3 4
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.2 2 2 1 1 3 1 2 1 2 2 1 2 2 2 3 2 3 3 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.1 2 1 2 1 1 4 3 2 1 3 3 4 2 3 2 4 2 2 1
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.4 1 4 4 1 1 4 3 4 2 2 2 2 1 2 3 1 1 1 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 2 3 4 2 2 4 1 2 4 3 2 2 3 2 4 4 2 1 1
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 3 3 1 2 2 2 2 2 4 3 3 4 2 4 2 1 1 4 2
Page 2


HINT – SHEET
ANSWER KEY
(0999DMD310319012) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 11
17-11-2019
LTS/HS-1/4
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 # 09
1. Energy is released in a process when total binding
energy (BE) of the nucleus is increased or we can
say when total BE of products is more than the
reactants. By calculation we can see that only in
case of option (3), this happens.
Given, W ® 2Y
BE of reactants = 120 × 7.5 = 900 MeV
and BE of products = 90 × 8 + 60 × 8.5 = 1230 MeV
i.e., Be of products > BE of reactants.
2. v
0
 = 4 × 10
15
8
0 15
0
c 3 10
m
4 10
´
l==
n´
= 750Å
0
12400
eV f=
l
 = 16.5 eV
3. E
total
 = 160 J, P.E.
max
 = E
total
 = 160 J
KE
max
 = 
( )
2
26
11
Ka 2 10 0.01 100 J
22
=´´´=
4. 2pr = nl Þ  n =
2r p
l
=
-
-
´ ´´
´
11
10
2 3.14 5.3 10
1.1 10
=3
5. Minimum wavelength of continuous X-ray
spectrum is
l
min
 (Å) = 3
12375 12375
E(eV) 80 10
=
´
 
0.155 »
here energy of the incident electrons 80 KeV is
more than ionization energy of k-shell electrons
is 72.5 KeV. So characteristics X-ray spectrum
will also be obtained because energy of incident
electron is enough to knock out the electron from
K or L shells.
6. Activity of a radioactive substance,
R = lN \ l = 
R
N
Here R = N
2
 particles per second, N = N
1
\
2
1
N
N
l=
7.
2
0 max
hc1
W mv
2
=+
l
Assuming W
0
 to be negligible in comparison to 
hc
l
i.e. 
2
max
1
mv µ
l
 Þ max
1
v µ
l
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 3 2 3 4 2 4 3 2 1 2 1 3 3 1 3 3 1 4 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.3 1 1 1 2 2 2 2 4 1 2 2 4 3 4 3 3 2 3 1
Que.41424344 4546 474849505152535455 5657 585960
Ans.1 4 4 3 3 4 3 4 4 1 4 2 2 4 2 3 2 2 4 2
Que.61626364 6566 676869707172737475 7677 787980
Ans.1 2 2 3 3 2 3 1 3 2 1 1 3 4 2 2 4 2 3 4
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.2 2 2 1 1 3 1 2 1 2 2 1 2 2 2 3 2 3 3 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.1 2 1 2 1 1 4 3 2 1 3 3 4 2 3 2 4 2 2 1
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.4 1 4 4 1 1 4 3 4 2 2 2 2 1 2 3 1 1 1 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 2 3 4 2 2 4 1 2 4 3 2 2 3 2 4 4 2 1 1
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 3 3 1 2 2 2 2 2 4 3 3 4 2 4 2 1 1 4 2
LTS/HS-2/4 0999DMD310319012
Target : Pre-Medical 2020/NEET-UG/17-11-2019
AL LEN
8.
2R 2R
2R
M
M M
22
R 22
GM 3 GM
F 3F3
(2R) 4R
éù
===
êú
ëû
9. Df  = 10pt + 8t
34
pp æö
- p+
ç÷
èø
 = 2t
12
p
p+
t = 0.5    Þ  
13
12 12
p
Df=p+ =p
10. Let the nucleus is 
z
X
A
. b
+
 decay is represented as
z
X
A
 ® 
z?1
X
A 
+ 
1
e
o 
+ n + Q
2
\     Q
2
 = [m
n
(
z
X
A
)]?m
n 
(
z-1
y
A
)?m
e
]c
2
= [m
n
(
z
X
A
) ? zm
e 
? m
n 
(
z-1
Y
A
) ? (z-1) me ? 2m
e
]c
2
=
 
[m(
z
X
A
) ? m(
z-1
Y
A
) ? 2m
e
]c
2
 = (M
x
?M
y
?2m
e
)c
2
b- decay is represented as
= 
z
X
A
 ® 
z+1
A
Y 
+ 
-1
e
0 
+ 
v
 + a
1
Q
1
 = [m
n
(
z
X
A
)] ?m
n 
(
z+1
Y
A
) ? m
e
]c
2
= [m
n
(
z
X
A
) ? zm
e 
? m
n 
(
z+1
Y
A
) ? (z+1) me]c
2
= [m(
z
X
A
) ? m(
z-1
Y
A
)]c
2
 = (M
x
?M
y
)c
2
11.
22
R 22
Gm Gm
??
F ij
(2a)
a
2
=+
æö
ç÷
èø
r
  
m
(0, 0)
m
(0, 2a)
æ
ç
è
a
2
ö
÷
ø
, 0
m
   
2
2
Gm1
??
4ij
a4
æö
=+
ç÷
èø
12.
2
23
13
1 11
RZ
nn
æö
=-
ç÷
l
èø
Þ 
2
22
1 11
R(1)
6561 2 3
éù
=-
êú
ëû
and 
2
22
1 11
R(2)
24
éù
=-
êú
l
ëû
Therefore l = 1215Å
13. 1
0
1
hc
eV = -f
l
     Þ  1
0
2
hc
3eV = -f
l
2
1
hc
3
hc
1
-f
l
=
-f
l
        Þ   
12
3hc hc
3 - f = -f
ll
12
3hc hc
2 - =f
ll
    Þ   
( )
21
12
hc
3
2
f = l -l
ll
14.
m m d
F = 
2
2
Gm
d
        
50% mass transfer
m ? 
m
2
d + 
d
2
m + 
m
2
m
2
   =
3
2
d    
3m
2
F'=
2
m 3m
G
22
3
d
2
æ öæö
ç ÷ç÷
è øèø
æö
ç÷
èø
 Þ F' = 
F
3
16. x = a sin
3
 wt
= 
3 sin3t
a sint
44
w éù
w-
êú
ëû
x = 
3a
4
 sin wt ? 
a
4
sin 3wt
So periodic but no SHM
17. Number of lines in absorption spectrum
   = (n ?1)  Þ 5 = n ? 1 Þ n = 6
\ Number of bright lines in the emission
spectrum
= 
n(n – 1) 6(6 – 1)
15
22
==
18.
235 231 231
90
92 91
X XX
-
ab
¾¾ ® ¾¾®
Therefore, one alpha and one electron are emitted.
19. K = 2
T = 
m
2
K
p
 Þ  T = 
2
2
2
p
 Þ T = 2p
20. Active fraction
No. of active nuclei
(No. of active No. of decay)nuclei +
t/Th 3
1 1 11
1 78 22
= ==
+
t = 3 Th = 3 × 1.4 × 9 yr = 4.2 × 10
9
 yr
21. By Moseley's low 
n
 = a (Z ? b)
comparing with eqn of parabola, y
2
 = 4ax
22. By using l = 
h
2mE
E = 10
?32
 J = constant for both particles.
Hence l µ 
1
m
Since m
p
 > m
e
 so l
p
 < l
e
Page 3


HINT – SHEET
ANSWER KEY
(0999DMD310319012) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 11
17-11-2019
LTS/HS-1/4
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 # 09
1. Energy is released in a process when total binding
energy (BE) of the nucleus is increased or we can
say when total BE of products is more than the
reactants. By calculation we can see that only in
case of option (3), this happens.
Given, W ® 2Y
BE of reactants = 120 × 7.5 = 900 MeV
and BE of products = 90 × 8 + 60 × 8.5 = 1230 MeV
i.e., Be of products > BE of reactants.
2. v
0
 = 4 × 10
15
8
0 15
0
c 3 10
m
4 10
´
l==
n´
= 750Å
0
12400
eV f=
l
 = 16.5 eV
3. E
total
 = 160 J, P.E.
max
 = E
total
 = 160 J
KE
max
 = 
( )
2
26
11
Ka 2 10 0.01 100 J
22
=´´´=
4. 2pr = nl Þ  n =
2r p
l
=
-
-
´ ´´
´
11
10
2 3.14 5.3 10
1.1 10
=3
5. Minimum wavelength of continuous X-ray
spectrum is
l
min
 (Å) = 3
12375 12375
E(eV) 80 10
=
´
 
0.155 »
here energy of the incident electrons 80 KeV is
more than ionization energy of k-shell electrons
is 72.5 KeV. So characteristics X-ray spectrum
will also be obtained because energy of incident
electron is enough to knock out the electron from
K or L shells.
6. Activity of a radioactive substance,
R = lN \ l = 
R
N
Here R = N
2
 particles per second, N = N
1
\
2
1
N
N
l=
7.
2
0 max
hc1
W mv
2
=+
l
Assuming W
0
 to be negligible in comparison to 
hc
l
i.e. 
2
max
1
mv µ
l
 Þ max
1
v µ
l
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 3 2 3 4 2 4 3 2 1 2 1 3 3 1 3 3 1 4 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.3 1 1 1 2 2 2 2 4 1 2 2 4 3 4 3 3 2 3 1
Que.41424344 4546 474849505152535455 5657 585960
Ans.1 4 4 3 3 4 3 4 4 1 4 2 2 4 2 3 2 2 4 2
Que.61626364 6566 676869707172737475 7677 787980
Ans.1 2 2 3 3 2 3 1 3 2 1 1 3 4 2 2 4 2 3 4
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.2 2 2 1 1 3 1 2 1 2 2 1 2 2 2 3 2 3 3 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.1 2 1 2 1 1 4 3 2 1 3 3 4 2 3 2 4 2 2 1
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.4 1 4 4 1 1 4 3 4 2 2 2 2 1 2 3 1 1 1 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 2 3 4 2 2 4 1 2 4 3 2 2 3 2 4 4 2 1 1
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 3 3 1 2 2 2 2 2 4 3 3 4 2 4 2 1 1 4 2
LTS/HS-2/4 0999DMD310319012
Target : Pre-Medical 2020/NEET-UG/17-11-2019
AL LEN
8.
2R 2R
2R
M
M M
22
R 22
GM 3 GM
F 3F3
(2R) 4R
éù
===
êú
ëû
9. Df  = 10pt + 8t
34
pp æö
- p+
ç÷
èø
 = 2t
12
p
p+
t = 0.5    Þ  
13
12 12
p
Df=p+ =p
10. Let the nucleus is 
z
X
A
. b
+
 decay is represented as
z
X
A
 ® 
z?1
X
A 
+ 
1
e
o 
+ n + Q
2
\     Q
2
 = [m
n
(
z
X
A
)]?m
n 
(
z-1
y
A
)?m
e
]c
2
= [m
n
(
z
X
A
) ? zm
e 
? m
n 
(
z-1
Y
A
) ? (z-1) me ? 2m
e
]c
2
=
 
[m(
z
X
A
) ? m(
z-1
Y
A
) ? 2m
e
]c
2
 = (M
x
?M
y
?2m
e
)c
2
b- decay is represented as
= 
z
X
A
 ® 
z+1
A
Y 
+ 
-1
e
0 
+ 
v
 + a
1
Q
1
 = [m
n
(
z
X
A
)] ?m
n 
(
z+1
Y
A
) ? m
e
]c
2
= [m
n
(
z
X
A
) ? zm
e 
? m
n 
(
z+1
Y
A
) ? (z+1) me]c
2
= [m(
z
X
A
) ? m(
z-1
Y
A
)]c
2
 = (M
x
?M
y
)c
2
11.
22
R 22
Gm Gm
??
F ij
(2a)
a
2
=+
æö
ç÷
èø
r
  
m
(0, 0)
m
(0, 2a)
æ
ç
è
a
2
ö
÷
ø
, 0
m
   
2
2
Gm1
??
4ij
a4
æö
=+
ç÷
èø
12.
2
23
13
1 11
RZ
nn
æö
=-
ç÷
l
èø
Þ 
2
22
1 11
R(1)
6561 2 3
éù
=-
êú
ëû
and 
2
22
1 11
R(2)
24
éù
=-
êú
l
ëû
Therefore l = 1215Å
13. 1
0
1
hc
eV = -f
l
     Þ  1
0
2
hc
3eV = -f
l
2
1
hc
3
hc
1
-f
l
=
-f
l
        Þ   
12
3hc hc
3 - f = -f
ll
12
3hc hc
2 - =f
ll
    Þ   
( )
21
12
hc
3
2
f = l -l
ll
14.
m m d
F = 
2
2
Gm
d
        
50% mass transfer
m ? 
m
2
d + 
d
2
m + 
m
2
m
2
   =
3
2
d    
3m
2
F'=
2
m 3m
G
22
3
d
2
æ öæö
ç ÷ç÷
è øèø
æö
ç÷
èø
 Þ F' = 
F
3
16. x = a sin
3
 wt
= 
3 sin3t
a sint
44
w éù
w-
êú
ëû
x = 
3a
4
 sin wt ? 
a
4
sin 3wt
So periodic but no SHM
17. Number of lines in absorption spectrum
   = (n ?1)  Þ 5 = n ? 1 Þ n = 6
\ Number of bright lines in the emission
spectrum
= 
n(n – 1) 6(6 – 1)
15
22
==
18.
235 231 231
90
92 91
X XX
-
ab
¾¾ ® ¾¾®
Therefore, one alpha and one electron are emitted.
19. K = 2
T = 
m
2
K
p
 Þ  T = 
2
2
2
p
 Þ T = 2p
20. Active fraction
No. of active nuclei
(No. of active No. of decay)nuclei +
t/Th 3
1 1 11
1 78 22
= ==
+
t = 3 Th = 3 × 1.4 × 9 yr = 4.2 × 10
9
 yr
21. By Moseley's low 
n
 = a (Z ? b)
comparing with eqn of parabola, y
2
 = 4ax
22. By using l = 
h
2mE
E = 10
?32
 J = constant for both particles.
Hence l µ 
1
m
Since m
p
 > m
e
 so l
p
 < l
e
Leader Test Series/Joint Package Course/NEET-UG/17-11-2019
AL LEN
0999DMD310319012 LTS/HS-3/4
23. Potential due to ring at its axis point
x
P
M,R
 Total potential at P
 
P
22
GM
V
Rx
=-
+
2 2 22
GM G(2M)
V
R R 4RR
--
=+
++
 
GM G(2M)
2R 5R
-
=-
 
GM12
R 25
æö
=-+
ç÷
èø
24. l
neutron
 µ
1
T
 Þ  
12
21
T
T
l
=
l
Þ  
2
(273 927) 1200
2
(273 27) 300
l+
= ==
l+
 Þ  l
2
 =
2
l
26. The electrostatic P.E. is zero when the electron
and proton are far apart from each other. Work
done in pulling electronand proton far away from
each other
W = E
?
 ? E
i
 = 0 ? E
i
 = ?
2
13.6
– eV
n
æö
ç÷
èø
Þ W = 
–19
2
13.6
1.6 10
(2)
´´ J = 3.4 × 1.6 × 10
?19
 J
27. b-particle carries one unit of negative charge and
a-particle carries 2 units of positive charge and
g-photon carries no charge, therefore electronic
energy levels of the atom changes for a and
b decay, but not for g-decay.
28. On applying constant force there is no changing
in time period force can change MP only
29. After removing one electron from helium atom it
will become hydrogen like atom.
\ E = 24.6 + (13.6) (2)
2
 = 79 eV
30. K.E acquired by the electron
K = eV = 20 × 10
3
 eV
& the energy of photon E = eV
= 0.05 × 20 × 10
3
 eV = 10
3
 eV
thus, 
hc
l
 = 10
3
 eV
33
hc 1240
10 eV 10
l==
 nm = 1.24 nm
31. Energy of the skylab in the first orbit is
GMm GMm
2(2R) 4R
- =-
Total energy required to place the skylab into the
orbit of radius 2R from the surface of earth is
GMm GMm 3GMm
4R R 4R
æö
- --=
ç÷
èø
= 
2
3gRm3
mgR
4R4
=
Energy of the skylab in the second orbit
= ?(GMm)/6R. Energy needed to shift the skylab
from the first orbit to the second orbit is
GMm GMm GMm 2 mgR
4R 6R R 24 12
- - = ´=
32. Stopping potential equals to maximum kinetic
energy. Since stopping potential is varying
linearly with the frequency. Therefore max. KE
for both the metals also vary linearly with
frequency.
33.
2 22
11
Energy KA m A
22
= =w
2
2
1g
mA
2L
ìü
ïï
=
íý
ïï
îþ
2 2g
TL L
2
g
ìü
pp
w===
ïï
íý
p
ïï
îþ
Energy = 
2
mgA
E
2L
=
(a)
()
2
mgAE
E'
22L2
==
(b)
()
2
2
mg 2A mgA
E' 4 4E
2L 2L
éù
= ==
êú
ëû
35. By conservation of momentum both particles
must have same magnetic of momentum and
Q 
D
h
P
l=
 so same l
D
36.
m
R
3m
V
at
 Surface = V
due
 to point mass + V
due
 to shell
V
surface
 = 
Gm G3m
RR
-- æ öæö
+
ç ÷ç÷
è øèø
 4Gm
R
-
=
Page 4


HINT – SHEET
ANSWER KEY
(0999DMD310319012) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 11
17-11-2019
LTS/HS-1/4
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 # 09
1. Energy is released in a process when total binding
energy (BE) of the nucleus is increased or we can
say when total BE of products is more than the
reactants. By calculation we can see that only in
case of option (3), this happens.
Given, W ® 2Y
BE of reactants = 120 × 7.5 = 900 MeV
and BE of products = 90 × 8 + 60 × 8.5 = 1230 MeV
i.e., Be of products > BE of reactants.
2. v
0
 = 4 × 10
15
8
0 15
0
c 3 10
m
4 10
´
l==
n´
= 750Å
0
12400
eV f=
l
 = 16.5 eV
3. E
total
 = 160 J, P.E.
max
 = E
total
 = 160 J
KE
max
 = 
( )
2
26
11
Ka 2 10 0.01 100 J
22
=´´´=
4. 2pr = nl Þ  n =
2r p
l
=
-
-
´ ´´
´
11
10
2 3.14 5.3 10
1.1 10
=3
5. Minimum wavelength of continuous X-ray
spectrum is
l
min
 (Å) = 3
12375 12375
E(eV) 80 10
=
´
 
0.155 »
here energy of the incident electrons 80 KeV is
more than ionization energy of k-shell electrons
is 72.5 KeV. So characteristics X-ray spectrum
will also be obtained because energy of incident
electron is enough to knock out the electron from
K or L shells.
6. Activity of a radioactive substance,
R = lN \ l = 
R
N
Here R = N
2
 particles per second, N = N
1
\
2
1
N
N
l=
7.
2
0 max
hc1
W mv
2
=+
l
Assuming W
0
 to be negligible in comparison to 
hc
l
i.e. 
2
max
1
mv µ
l
 Þ max
1
v µ
l
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 3 2 3 4 2 4 3 2 1 2 1 3 3 1 3 3 1 4 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.3 1 1 1 2 2 2 2 4 1 2 2 4 3 4 3 3 2 3 1
Que.41424344 4546 474849505152535455 5657 585960
Ans.1 4 4 3 3 4 3 4 4 1 4 2 2 4 2 3 2 2 4 2
Que.61626364 6566 676869707172737475 7677 787980
Ans.1 2 2 3 3 2 3 1 3 2 1 1 3 4 2 2 4 2 3 4
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.2 2 2 1 1 3 1 2 1 2 2 1 2 2 2 3 2 3 3 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.1 2 1 2 1 1 4 3 2 1 3 3 4 2 3 2 4 2 2 1
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.4 1 4 4 1 1 4 3 4 2 2 2 2 1 2 3 1 1 1 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 2 3 4 2 2 4 1 2 4 3 2 2 3 2 4 4 2 1 1
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 3 3 1 2 2 2 2 2 4 3 3 4 2 4 2 1 1 4 2
LTS/HS-2/4 0999DMD310319012
Target : Pre-Medical 2020/NEET-UG/17-11-2019
AL LEN
8.
2R 2R
2R
M
M M
22
R 22
GM 3 GM
F 3F3
(2R) 4R
éù
===
êú
ëû
9. Df  = 10pt + 8t
34
pp æö
- p+
ç÷
èø
 = 2t
12
p
p+
t = 0.5    Þ  
13
12 12
p
Df=p+ =p
10. Let the nucleus is 
z
X
A
. b
+
 decay is represented as
z
X
A
 ® 
z?1
X
A 
+ 
1
e
o 
+ n + Q
2
\     Q
2
 = [m
n
(
z
X
A
)]?m
n 
(
z-1
y
A
)?m
e
]c
2
= [m
n
(
z
X
A
) ? zm
e 
? m
n 
(
z-1
Y
A
) ? (z-1) me ? 2m
e
]c
2
=
 
[m(
z
X
A
) ? m(
z-1
Y
A
) ? 2m
e
]c
2
 = (M
x
?M
y
?2m
e
)c
2
b- decay is represented as
= 
z
X
A
 ® 
z+1
A
Y 
+ 
-1
e
0 
+ 
v
 + a
1
Q
1
 = [m
n
(
z
X
A
)] ?m
n 
(
z+1
Y
A
) ? m
e
]c
2
= [m
n
(
z
X
A
) ? zm
e 
? m
n 
(
z+1
Y
A
) ? (z+1) me]c
2
= [m(
z
X
A
) ? m(
z-1
Y
A
)]c
2
 = (M
x
?M
y
)c
2
11.
22
R 22
Gm Gm
??
F ij
(2a)
a
2
=+
æö
ç÷
èø
r
  
m
(0, 0)
m
(0, 2a)
æ
ç
è
a
2
ö
÷
ø
, 0
m
   
2
2
Gm1
??
4ij
a4
æö
=+
ç÷
èø
12.
2
23
13
1 11
RZ
nn
æö
=-
ç÷
l
èø
Þ 
2
22
1 11
R(1)
6561 2 3
éù
=-
êú
ëû
and 
2
22
1 11
R(2)
24
éù
=-
êú
l
ëû
Therefore l = 1215Å
13. 1
0
1
hc
eV = -f
l
     Þ  1
0
2
hc
3eV = -f
l
2
1
hc
3
hc
1
-f
l
=
-f
l
        Þ   
12
3hc hc
3 - f = -f
ll
12
3hc hc
2 - =f
ll
    Þ   
( )
21
12
hc
3
2
f = l -l
ll
14.
m m d
F = 
2
2
Gm
d
        
50% mass transfer
m ? 
m
2
d + 
d
2
m + 
m
2
m
2
   =
3
2
d    
3m
2
F'=
2
m 3m
G
22
3
d
2
æ öæö
ç ÷ç÷
è øèø
æö
ç÷
èø
 Þ F' = 
F
3
16. x = a sin
3
 wt
= 
3 sin3t
a sint
44
w éù
w-
êú
ëû
x = 
3a
4
 sin wt ? 
a
4
sin 3wt
So periodic but no SHM
17. Number of lines in absorption spectrum
   = (n ?1)  Þ 5 = n ? 1 Þ n = 6
\ Number of bright lines in the emission
spectrum
= 
n(n – 1) 6(6 – 1)
15
22
==
18.
235 231 231
90
92 91
X XX
-
ab
¾¾ ® ¾¾®
Therefore, one alpha and one electron are emitted.
19. K = 2
T = 
m
2
K
p
 Þ  T = 
2
2
2
p
 Þ T = 2p
20. Active fraction
No. of active nuclei
(No. of active No. of decay)nuclei +
t/Th 3
1 1 11
1 78 22
= ==
+
t = 3 Th = 3 × 1.4 × 9 yr = 4.2 × 10
9
 yr
21. By Moseley's low 
n
 = a (Z ? b)
comparing with eqn of parabola, y
2
 = 4ax
22. By using l = 
h
2mE
E = 10
?32
 J = constant for both particles.
Hence l µ 
1
m
Since m
p
 > m
e
 so l
p
 < l
e
Leader Test Series/Joint Package Course/NEET-UG/17-11-2019
AL LEN
0999DMD310319012 LTS/HS-3/4
23. Potential due to ring at its axis point
x
P
M,R
 Total potential at P
 
P
22
GM
V
Rx
=-
+
2 2 22
GM G(2M)
V
R R 4RR
--
=+
++
 
GM G(2M)
2R 5R
-
=-
 
GM12
R 25
æö
=-+
ç÷
èø
24. l
neutron
 µ
1
T
 Þ  
12
21
T
T
l
=
l
Þ  
2
(273 927) 1200
2
(273 27) 300
l+
= ==
l+
 Þ  l
2
 =
2
l
26. The electrostatic P.E. is zero when the electron
and proton are far apart from each other. Work
done in pulling electronand proton far away from
each other
W = E
?
 ? E
i
 = 0 ? E
i
 = ?
2
13.6
– eV
n
æö
ç÷
èø
Þ W = 
–19
2
13.6
1.6 10
(2)
´´ J = 3.4 × 1.6 × 10
?19
 J
27. b-particle carries one unit of negative charge and
a-particle carries 2 units of positive charge and
g-photon carries no charge, therefore electronic
energy levels of the atom changes for a and
b decay, but not for g-decay.
28. On applying constant force there is no changing
in time period force can change MP only
29. After removing one electron from helium atom it
will become hydrogen like atom.
\ E = 24.6 + (13.6) (2)
2
 = 79 eV
30. K.E acquired by the electron
K = eV = 20 × 10
3
 eV
& the energy of photon E = eV
= 0.05 × 20 × 10
3
 eV = 10
3
 eV
thus, 
hc
l
 = 10
3
 eV
33
hc 1240
10 eV 10
l==
 nm = 1.24 nm
31. Energy of the skylab in the first orbit is
GMm GMm
2(2R) 4R
- =-
Total energy required to place the skylab into the
orbit of radius 2R from the surface of earth is
GMm GMm 3GMm
4R R 4R
æö
- --=
ç÷
èø
= 
2
3gRm3
mgR
4R4
=
Energy of the skylab in the second orbit
= ?(GMm)/6R. Energy needed to shift the skylab
from the first orbit to the second orbit is
GMm GMm GMm 2 mgR
4R 6R R 24 12
- - = ´=
32. Stopping potential equals to maximum kinetic
energy. Since stopping potential is varying
linearly with the frequency. Therefore max. KE
for both the metals also vary linearly with
frequency.
33.
2 22
11
Energy KA m A
22
= =w
2
2
1g
mA
2L
ìü
ïï
=
íý
ïï
îþ
2 2g
TL L
2
g
ìü
pp
w===
ïï
íý
p
ïï
îþ
Energy = 
2
mgA
E
2L
=
(a)
()
2
mgAE
E'
22L2
==
(b)
()
2
2
mg 2A mgA
E' 4 4E
2L 2L
éù
= ==
êú
ëû
35. By conservation of momentum both particles
must have same magnetic of momentum and
Q 
D
h
P
l=
 so same l
D
36.
m
R
3m
V
at
 Surface = V
due
 to point mass + V
due
 to shell
V
surface
 = 
Gm G3m
RR
-- æ öæö
+
ç ÷ç÷
è øèø
 4Gm
R
-
=
LTS/HS-4/4 0999DMD310319012
Target : Pre-Medical 2020/NEET-UG/17-11-2019
AL LEN
37. 1 ? Df = 0.9 = e
?l(5)
....... (1)
1 ? 
19
100
 = 0.81 = e
?l(t)
ore
?lt
 = 0.81 = (0.9)
2
....... (2)
from eqn. (1) and eqn. (2)
e
?lt
 = (e
?5l
)
2
 = e
?10l
?lt = ?10l
t = 10 year
38.
3
1240 eVnm
E
nm 71 10
-
=
´
 = 17.5 keV
Q E
K
 ? E
L
 = 17.5 keV
E
L
 = E
K
 ? 17.5 keV
= 23.32 ? 17.5 keV = 5.82 keV
39.
e
50 1 2GM
VV
100 2R
==
Apply energy conservation
2
GMm 1 GMm
mV
R 2 (R h)
Þ- + =-
+
2
2GM 2GM
v
R Rh
=-
+
12GM 11
. 2GM
4 R R Rh
æö
=-
ç÷
+
èø
\
1h
4R R(R h)
=
+
 \ R + h = 4h Þ 
R
h
3
=
40. The work function has no effect on current so
long as hn > V
0
. The photoelectric current is
proportional to the intensity of light. Since there
is no change in the intensity of light, therefore
I
1
 = I
2
.
41.
2 bt/m
1
E(t) kAe
2
-
=
2 2 bt/m
111
kA kAe
222
-
´=
bt/m
1
e
2
-
=
bt/m
e2 =
bt
ln2
m
=
10
bt
2.303log 2
m
=
40
t 0.693
200
=
    Þ  t = 0.693 × 5 = 3.46s
42. Minimum l Þ series limit
Lyman Þ 
12
1 1 1 11
R1&R
4
æ ö æö
= - =-
ç ÷ ç÷
l ¥l¥
è ø èø
1
2
1
4
l
=
l
43. Q = (1.002 + 1.004 ? 1.001 ? 1.003) (931.5) MeV
    = 1.863 MeV
44. Factual theory based Q.
45. From the given figure, it is clear that slope of curve
A is greater than that of curve B. So rate of decay
is faster for A than that of B.
t
dN
dt
A
B
We know that 
dN
dt
æö
µl
ç÷
èø
, at any instant of time
hence we can say that l
A
 > l
B
. At point P shown
in the diagram the two curve intersect. Hence at
point P, rate of decay for both A and B is the same.
91. NCERT Pg. # 221, Para-13.2
92. NCERT Pg. # 234
102. NCERT Pg. # 227
103. NCERT Pg. # 231
107. NCERT Pg. # 227, Para-13.2
109. NCERT Pg. # 230
111. NCERT Pg. # 261, Para-15.1.2
114. NCERT Pg. # 232, table- 13.1
115. NCERT Pg. # 266
117. NCERT Pg. # 254, Para-14.7.1-14.7.2
119. NCERT Pg. # 255
121. NCERT Pg. # 263
125. NCERT Pg. # 243, Para-14.3
128. NCERT Pg. # 244
129. NCERT Pg. # 242, Para-14.1
130. NCERT Pg. # 243, Para-14.2
137. NCERT Pg. # 248
140. NCERT Pg. # 234
144. NCERT Pg. # 249
147. NCERT Pg. # 229
149. NCERT Pg. # 235
150. NCERT Pg. # 252
152. NCERT Pg. # 225
154. NCERT Pg. # 220
156. NCERT Pg. # 281
159. NCERT Pg. # 283
160. NCERT Pg. # 263
161. NCERT Pg. # 264
162. NCERT Pg. # 277
164. NCERT Pg. # 281
168. NCERT Pg. # 261
169. NCERT Pg. # 262
173. NCERT Pg. # 279
175. NCERT Pg. # 283
179. NCERT Pg. # 266
180. NCERT Pg. # 279
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