ALLEN TEST PAPER 7 MBBS Notes | EduRev

MBBS : ALLEN TEST PAPER 7 MBBS Notes | EduRev

 Page 1


LTS / Page 1/36
Name of the Candidate (in Capitals)
ijh{kkFkhZ dk uke (cM+s v{kjksa esa) :
Form Number : in figures
QkWeZ uEcj : vadksa esa
: in words
: 'kCnksa esa
Centre of Examination (in Capitals) :
ijh{kk dsUæ (cM+s v{kjksa esa) :
Candidate’s Signature : Invigilator’s Signature :
ijh{kkFkhZ ds gLrk{kj : fujh{kd ds gLrk{kj :
)/999DMD31/319//7)
(0999DMD310319007)
Test Pattern
egRoiw.kZ funsZ'k :
1. mÙkj i= ds i`"B-1 ,oa i`"B-2 ij /;kuiwoZd dsoy uhys@dkys ckWy
ikWbaV isu ls fooj.k HkjsaA
2. ijh{kk dh vof/k 3 ?kaVs gS ,oa ijh{kk iqfLrdk esa 180 iz'u gSaA izR;sd
iz'u 4 vad dk gSA izR;sd lgh mÙkj ds fy , ijh{kkFkhZ dks 4 vad
fn, tk,axsaA izR;sd xyr mÙkj ds fy , dqy ;ksx esa ls ,d vad
?kVk;k tk,xkA vf/kdre vad 720 gSA
3. bl i`"B ij fooj.k vafdr djus ,oa mÙkj i= ij fu'kku yxkus ds
fy, dsoy uhys@dkys ckWy ikWbaV isu dk iz;ksx djsaA
4. jQ dk;Z bl ijh{kk iqfLrdk esa fu/kkZfjr LFkku ij gh djsaA
5. ijh{kk lEiék gksus ij ] ijh{kkFkhZ d{k@gkWy NksM+us ls iwoZ
mÙkj i= fujh{kd dks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk
dsoy ijh{kk iqfLrdk dks ys tk ldrs gSaA
6. ijh{kkFkhZ lqfuf'pr djsa fd bl mÙkj i = dks eksM+k u tk , ,oa ml
ij dksbZ vU ; fu'kku u yxk,aA ijh{kkFkhZ viuk QkWeZ uEcj iz'u
iqfLrdk@mÙkj i= esa fu/kkZfjr LFkku ds vfrfjDr vU ;= u fy[ksaA
7. mÙkj i= ij fdlh izdkj ds la'kks/ku gsrq OgkbV ¶+yqbM ds iz;ksx dh
vuqefr ugha gSA
8. ;fn vki fdlh iz'u dks gy djus dk iz;kl djrs gSa rks mfpr xksys
dks uhps n'kkZ;s x;s vuqlkj xgjk dkyk djsa vU;Fkk mls [kkyh NksM+ nsaA
lgh rjhdk xyr rjhdk
Important Instructions :
1. On the Answer Sheet, fill in the particulars on Side-1
and Side-2 carefully with blue/black ball point pen only.
2. The test is of 3 hours duration and this Test Booklet
contains 180 questions. Each question carries 4 marks.
For each correct response, the candidate will get 4 marks.
For each incorrect response, one mark will be deducted
from the total scores. The maximum marks are 720.
3. Use Blue/Black Ball Point Pen only for writing
particulars on this page/marking responses.
4. Rough work is to be done on the space provided for this
purpose in the Test Booklet only.
5. On completion of the test, the candidate must
hand over the Answer Sheet to the Invigilator
before leaving the Room/Hall. The candidates are
allowed to take away this Test Booklet with them.
6. The candidates should ensure that the Answer Sheet is
not folded. Do not make any stray marks on the Answer
Sheet. Do not write your Form No. anywhere else except
in the specified space in the Test Booklet/Answer Sheet.
7. Use of white fluid for correction is not permissible on
the Answer Sheet.
8. If you want to attempt any question then circle should be
properly darkened as shown below, otherwise leave blank.
Correct Method Wrong Method
In case of any ambiguity in translation of any question, English version shall be treated as final.
iz'uksa ds vuqokn esa fdlh vLi"Vrk dh fLFkfr esa ] vaxzsth laLdj.k dks gh vafre ekuk tk,sxkA
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
Your Target is to secure Good Rank in Pre-Medical 2020
NEET(UG)
MINOR TEST # 07
29-09-2019
Read carefully the Instructions on the Back Cover of this Test Booklet.
bl ijh{kk iqfLrdk ds fiNys vkoj.k ij fn , funsZ'kksa dks /;ku ls i<+ saA
Do not open this Test Booklet until you are asked to do so.
bl ijh{kk iqfLrdk dks tc rd uk [kksysa tc rd dgk u tk,A
This Booklet contains 36 pages. bl iqfLrdk esa 36 i`"B gSaA
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
Test Type : Unit - 06
12
th
 Undergoing/Pass Students
Page 2


LTS / Page 1/36
Name of the Candidate (in Capitals)
ijh{kkFkhZ dk uke (cM+s v{kjksa esa) :
Form Number : in figures
QkWeZ uEcj : vadksa esa
: in words
: 'kCnksa esa
Centre of Examination (in Capitals) :
ijh{kk dsUæ (cM+s v{kjksa esa) :
Candidate’s Signature : Invigilator’s Signature :
ijh{kkFkhZ ds gLrk{kj : fujh{kd ds gLrk{kj :
)/999DMD31/319//7)
(0999DMD310319007)
Test Pattern
egRoiw.kZ funsZ'k :
1. mÙkj i= ds i`"B-1 ,oa i`"B-2 ij /;kuiwoZd dsoy uhys@dkys ckWy
ikWbaV isu ls fooj.k HkjsaA
2. ijh{kk dh vof/k 3 ?kaVs gS ,oa ijh{kk iqfLrdk esa 180 iz'u gSaA izR;sd
iz'u 4 vad dk gSA izR;sd lgh mÙkj ds fy , ijh{kkFkhZ dks 4 vad
fn, tk,axsaA izR;sd xyr mÙkj ds fy , dqy ;ksx esa ls ,d vad
?kVk;k tk,xkA vf/kdre vad 720 gSA
3. bl i`"B ij fooj.k vafdr djus ,oa mÙkj i= ij fu'kku yxkus ds
fy, dsoy uhys@dkys ckWy ikWbaV isu dk iz;ksx djsaA
4. jQ dk;Z bl ijh{kk iqfLrdk esa fu/kkZfjr LFkku ij gh djsaA
5. ijh{kk lEiék gksus ij ] ijh{kkFkhZ d{k@gkWy NksM+us ls iwoZ
mÙkj i= fujh{kd dks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk
dsoy ijh{kk iqfLrdk dks ys tk ldrs gSaA
6. ijh{kkFkhZ lqfuf'pr djsa fd bl mÙkj i = dks eksM+k u tk , ,oa ml
ij dksbZ vU ; fu'kku u yxk,aA ijh{kkFkhZ viuk QkWeZ uEcj iz'u
iqfLrdk@mÙkj i= esa fu/kkZfjr LFkku ds vfrfjDr vU ;= u fy[ksaA
7. mÙkj i= ij fdlh izdkj ds la'kks/ku gsrq OgkbV ¶+yqbM ds iz;ksx dh
vuqefr ugha gSA
8. ;fn vki fdlh iz'u dks gy djus dk iz;kl djrs gSa rks mfpr xksys
dks uhps n'kkZ;s x;s vuqlkj xgjk dkyk djsa vU;Fkk mls [kkyh NksM+ nsaA
lgh rjhdk xyr rjhdk
Important Instructions :
1. On the Answer Sheet, fill in the particulars on Side-1
and Side-2 carefully with blue/black ball point pen only.
2. The test is of 3 hours duration and this Test Booklet
contains 180 questions. Each question carries 4 marks.
For each correct response, the candidate will get 4 marks.
For each incorrect response, one mark will be deducted
from the total scores. The maximum marks are 720.
3. Use Blue/Black Ball Point Pen only for writing
particulars on this page/marking responses.
4. Rough work is to be done on the space provided for this
purpose in the Test Booklet only.
5. On completion of the test, the candidate must
hand over the Answer Sheet to the Invigilator
before leaving the Room/Hall. The candidates are
allowed to take away this Test Booklet with them.
6. The candidates should ensure that the Answer Sheet is
not folded. Do not make any stray marks on the Answer
Sheet. Do not write your Form No. anywhere else except
in the specified space in the Test Booklet/Answer Sheet.
7. Use of white fluid for correction is not permissible on
the Answer Sheet.
8. If you want to attempt any question then circle should be
properly darkened as shown below, otherwise leave blank.
Correct Method Wrong Method
In case of any ambiguity in translation of any question, English version shall be treated as final.
iz'uksa ds vuqokn esa fdlh vLi"Vrk dh fLFkfr esa ] vaxzsth laLdj.k dks gh vafre ekuk tk,sxkA
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
Your Target is to secure Good Rank in Pre-Medical 2020
NEET(UG)
MINOR TEST # 07
29-09-2019
Read carefully the Instructions on the Back Cover of this Test Booklet.
bl ijh{kk iqfLrdk ds fiNys vkoj.k ij fn , funsZ'kksa dks /;ku ls i<+ saA
Do not open this Test Booklet until you are asked to do so.
bl ijh{kk iqfLrdk dks tc rd uk [kksysa tc rd dgk u tk,A
This Booklet contains 36 pages. bl iqfLrdk esa 36 i`"B gSaA
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
Test Type : Unit - 06
12
th
 Undergoing/Pass Students
LTS / Page 2/36 SPACE FOR ROUGH WORK / jQ dk;Z ds fy , txg 0999DMD310319007
ALLEN
TARGET : PRE-MEDICAL 2020/NEET-UG/29-09-2019
1. In the given AC circuit  :-
I
1
X
C
I
2
R
~
I
V = V sin t
0
w
(1) Current I
2
 and V are in same phase
(2) Current I
2
 lead I
1
 by 90º
(3) Current I leads I
2
 by q < 90º
(4) Current I leads I
1
 by q < 90º
2. A ring and a disc of different masses are rotating
with the same kinetic energy. If we apply a
retarding torque t on the ring, it stops after making
n revolutions. After how many revolutions will the
disc stop if the same retarding torque on it :-
(1) n/2 (2) n
(3) 2n (4) 4n
3. In the LCR circuit shown in figure :-
X =20
C
W
~
V = 400 sin t w
R=10W
X =10
L
W
(1) Current will leads the voltage
(2) rms value of currents is 20 A
(3) Power factor of the circuit is 
1
2
(4) All of the above
TOPIC : Rotational Motion, Alternating current, Electromagnetic Waves.
1. fn;s x;s ifjiFk esa :-
I
1
X
C
I
2
R
~
I
V = V sin t
0
w
(1) /kkjk I
2
 rFkk V leku dyk esa gSA
(2) /kkjk I
2
, I
1
 ls 90º vkxs gSA
(3) /kkjk I, I
2
 ls q < 90º vkxs gSA
(4) /kkjk I, I
1
 ls q < 90º vkxs gSA
2. ,d oy; rFkk ,d pdrh leku xfrt ÅtkZ ds lkFk ?kw.kZu dj
jgh gaSA ;fn ge oy; ij voeUnd cy vk?kw.kZ t vkjksfir djrs
gS] rks ;g n pDdj yxkus ds i'pkr~ :d tkrh gS ;fn ;gh
voeUnd cy vk?kw.kZ pdrh ij yxk;k tk;s rks ;g fdrus pDdj
yxkus ds i'pkr~ :dsxh :-
(1) n/2 (2) n
(3) 2n (4) 4n
3. fp= esa iznf'kZr ifjiFk esa :-
X =20
C
W
~
V = 400 sin t w
R=10W
X =10
L
W
(1) /kkjk oksYVrk ls dyk ls vkxs gSA
(2) /kkjk dk o-ek-ew- eku 20 A gSA
(3) ifjiFk dk 'kfDr xq.kkad 
1
2
 gSA
(4) mijksDr lHkh
Page 3


LTS / Page 1/36
Name of the Candidate (in Capitals)
ijh{kkFkhZ dk uke (cM+s v{kjksa esa) :
Form Number : in figures
QkWeZ uEcj : vadksa esa
: in words
: 'kCnksa esa
Centre of Examination (in Capitals) :
ijh{kk dsUæ (cM+s v{kjksa esa) :
Candidate’s Signature : Invigilator’s Signature :
ijh{kkFkhZ ds gLrk{kj : fujh{kd ds gLrk{kj :
)/999DMD31/319//7)
(0999DMD310319007)
Test Pattern
egRoiw.kZ funsZ'k :
1. mÙkj i= ds i`"B-1 ,oa i`"B-2 ij /;kuiwoZd dsoy uhys@dkys ckWy
ikWbaV isu ls fooj.k HkjsaA
2. ijh{kk dh vof/k 3 ?kaVs gS ,oa ijh{kk iqfLrdk esa 180 iz'u gSaA izR;sd
iz'u 4 vad dk gSA izR;sd lgh mÙkj ds fy , ijh{kkFkhZ dks 4 vad
fn, tk,axsaA izR;sd xyr mÙkj ds fy , dqy ;ksx esa ls ,d vad
?kVk;k tk,xkA vf/kdre vad 720 gSA
3. bl i`"B ij fooj.k vafdr djus ,oa mÙkj i= ij fu'kku yxkus ds
fy, dsoy uhys@dkys ckWy ikWbaV isu dk iz;ksx djsaA
4. jQ dk;Z bl ijh{kk iqfLrdk esa fu/kkZfjr LFkku ij gh djsaA
5. ijh{kk lEiék gksus ij ] ijh{kkFkhZ d{k@gkWy NksM+us ls iwoZ
mÙkj i= fujh{kd dks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk
dsoy ijh{kk iqfLrdk dks ys tk ldrs gSaA
6. ijh{kkFkhZ lqfuf'pr djsa fd bl mÙkj i = dks eksM+k u tk , ,oa ml
ij dksbZ vU ; fu'kku u yxk,aA ijh{kkFkhZ viuk QkWeZ uEcj iz'u
iqfLrdk@mÙkj i= esa fu/kkZfjr LFkku ds vfrfjDr vU ;= u fy[ksaA
7. mÙkj i= ij fdlh izdkj ds la'kks/ku gsrq OgkbV ¶+yqbM ds iz;ksx dh
vuqefr ugha gSA
8. ;fn vki fdlh iz'u dks gy djus dk iz;kl djrs gSa rks mfpr xksys
dks uhps n'kkZ;s x;s vuqlkj xgjk dkyk djsa vU;Fkk mls [kkyh NksM+ nsaA
lgh rjhdk xyr rjhdk
Important Instructions :
1. On the Answer Sheet, fill in the particulars on Side-1
and Side-2 carefully with blue/black ball point pen only.
2. The test is of 3 hours duration and this Test Booklet
contains 180 questions. Each question carries 4 marks.
For each correct response, the candidate will get 4 marks.
For each incorrect response, one mark will be deducted
from the total scores. The maximum marks are 720.
3. Use Blue/Black Ball Point Pen only for writing
particulars on this page/marking responses.
4. Rough work is to be done on the space provided for this
purpose in the Test Booklet only.
5. On completion of the test, the candidate must
hand over the Answer Sheet to the Invigilator
before leaving the Room/Hall. The candidates are
allowed to take away this Test Booklet with them.
6. The candidates should ensure that the Answer Sheet is
not folded. Do not make any stray marks on the Answer
Sheet. Do not write your Form No. anywhere else except
in the specified space in the Test Booklet/Answer Sheet.
7. Use of white fluid for correction is not permissible on
the Answer Sheet.
8. If you want to attempt any question then circle should be
properly darkened as shown below, otherwise leave blank.
Correct Method Wrong Method
In case of any ambiguity in translation of any question, English version shall be treated as final.
iz'uksa ds vuqokn esa fdlh vLi"Vrk dh fLFkfr esa ] vaxzsth laLdj.k dks gh vafre ekuk tk,sxkA
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
Your Target is to secure Good Rank in Pre-Medical 2020
NEET(UG)
MINOR TEST # 07
29-09-2019
Read carefully the Instructions on the Back Cover of this Test Booklet.
bl ijh{kk iqfLrdk ds fiNys vkoj.k ij fn , funsZ'kksa dks /;ku ls i<+ saA
Do not open this Test Booklet until you are asked to do so.
bl ijh{kk iqfLrdk dks tc rd uk [kksysa tc rd dgk u tk,A
This Booklet contains 36 pages. bl iqfLrdk esa 36 i`"B gSaA
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
Test Type : Unit - 06
12
th
 Undergoing/Pass Students
LTS / Page 2/36 SPACE FOR ROUGH WORK / jQ dk;Z ds fy , txg 0999DMD310319007
ALLEN
TARGET : PRE-MEDICAL 2020/NEET-UG/29-09-2019
1. In the given AC circuit  :-
I
1
X
C
I
2
R
~
I
V = V sin t
0
w
(1) Current I
2
 and V are in same phase
(2) Current I
2
 lead I
1
 by 90º
(3) Current I leads I
2
 by q < 90º
(4) Current I leads I
1
 by q < 90º
2. A ring and a disc of different masses are rotating
with the same kinetic energy. If we apply a
retarding torque t on the ring, it stops after making
n revolutions. After how many revolutions will the
disc stop if the same retarding torque on it :-
(1) n/2 (2) n
(3) 2n (4) 4n
3. In the LCR circuit shown in figure :-
X =20
C
W
~
V = 400 sin t w
R=10W
X =10
L
W
(1) Current will leads the voltage
(2) rms value of currents is 20 A
(3) Power factor of the circuit is 
1
2
(4) All of the above
TOPIC : Rotational Motion, Alternating current, Electromagnetic Waves.
1. fn;s x;s ifjiFk esa :-
I
1
X
C
I
2
R
~
I
V = V sin t
0
w
(1) /kkjk I
2
 rFkk V leku dyk esa gSA
(2) /kkjk I
2
, I
1
 ls 90º vkxs gSA
(3) /kkjk I, I
2
 ls q < 90º vkxs gSA
(4) /kkjk I, I
1
 ls q < 90º vkxs gSA
2. ,d oy; rFkk ,d pdrh leku xfrt ÅtkZ ds lkFk ?kw.kZu dj
jgh gaSA ;fn ge oy; ij voeUnd cy vk?kw.kZ t vkjksfir djrs
gS] rks ;g n pDdj yxkus ds i'pkr~ :d tkrh gS ;fn ;gh
voeUnd cy vk?kw.kZ pdrh ij yxk;k tk;s rks ;g fdrus pDdj
yxkus ds i'pkr~ :dsxh :-
(1) n/2 (2) n
(3) 2n (4) 4n
3. fp= esa iznf'kZr ifjiFk esa :-
X =20
C
W
~
V = 400 sin t w
R=10W
X =10
L
W
(1) /kkjk oksYVrk ls dyk ls vkxs gSA
(2) /kkjk dk o-ek-ew- eku 20 A gSA
(3) ifjiFk dk 'kfDr xq.kkad 
1
2
 gSA
(4) mijksDr lHkh
0999DMD310319007 SPACE FOR ROUGH WORK / jQ dk;Z ds fy , txg LTS / Page 3/36
ALLEN
LEADER TEST SERIES/JOINT PACKAGE COURSE/NEET-UG/29-09-2019
4. Two blocks each of mass M are connected to the
ends of a light frame as shown in figure The frame
is rotated about the vertical line of symmetry. The
rod breaks if the tension in it exceeds T
0
. Find the
maximum frequency with which the frame may
be rotated without breaking the rod.
M M
L
(1) 
0
T 1
4 ML p
(2) 
0
T 1
2 ML p
(3) 
0
2T 1
6 3ML p
(4)
0
3T 1
2 2ML p
5. A 12 W resistor and a 0.21 henry inductor are
connected in series to A.C. source operating at
20 volt 50 Hz. The phase angle between the
current and source voltage is :-
(1) 30º (2) 40º (3) 80º (4) 90º
6. The electric field of an electromagnetic wave is
given by
E = (50 N/C) sin w(t–x/c)
the energy contained in a cylinder of cross-section
10cm
2
 and length 50 cm along the x-axis :-
(1) 5.5×10
–6 
J (2) 5.5 × 10
–12 
J
(3) 2.75 × 10
–6
 J (4) 2.75 × 10
–12
 J
7. A thick walled hollow sphere has outer radius R.
It rolls down an inclined plane without slipping and
its speed at the bottom is v. If the inclined plane
is frictionless and the sphere slides down without
rolling, its speed at the bottom will be 5v/4. What
is the radius of gyration of the sphere ?
(1) 
R
2
(2) 
R
2
(3) 
3R
4
(4) 
3R
4
4. fp= esa n'kkZ;s vuqlkj ,d gYdh Ýse ds fljksa ij nks CykWd tqM+s
gq, gSa] izR;sd CykWd dk nzO;eku M gSA Ýse dks m/oZ lefer
js[kk ds ifjr% ?kqek;k tk jgk gSA ;fn NM+ esa ruko T
0
 gks tk;sxk
rks NM+ VwV tk;sxhA Ýse dks ?kqekus dh vf/kdre vko`fÙk Kkr
dhft,] ftlds fy;s NM+ VwVs ughaA
M M
L
(1) 
0
T 1
4 ML p
(2) 
0
T 1
2 ML p
(3) 
0
2T 1
6 3ML p
(4)
0
3T 1
2 2ML p
5. 12 W izfrjks/k rFkk 0.21 gsujh izsjdRo 20 volt 50 Hz ds
izR;korhZ L=ksr ds lkFk Js.khØe esa tqM+s gq , gSA /kkjk ,oa oksYVrk
ds chp dyk dks.k gS :-
(1) 30º (2) 40º
(3) 80º (4) 90º
6. ,d fo|qr pqEcdh ; rjax dk fo|qr {ks= fuEukuqlkj O;Dr fd;k
tkrk gS
E = (50 N/C) sin w(t–x/c)
10cm
2
 vuqizLFk dkV rFkk x-v{k ds vuqfn'k 50 cm yEckbZ
okys csyu esa fufgr ÅtkZ gksxh :-
(1) 5.5×10
–6 
J (2) 5.5 × 10
–12 
J
(3) 2.75 × 10
–6
 J (4) 2.75 × 10
–12
 J
7. eksVh nhokj okys ,d [kks[kys xksys dh ckg~; f=T;k R gSA ;g
,d vkur ry ij fcuk fQlys uhps yq <+drk gS vkSj iSans ij bldh
pky v gSA ;fn vkur ry ?k"kZ.k jfgr gks vkSj xksyk fcuk yq <+ds
fQlys rks iSans ij bldh pky 5v/4 gksxhA xksys dh ifjHkze.k
f=T;k fdruh gS ?
(1) 
R
2
(2) 
R
2
(3) 
3R
4
(4) 
3R
4
Page 4


LTS / Page 1/36
Name of the Candidate (in Capitals)
ijh{kkFkhZ dk uke (cM+s v{kjksa esa) :
Form Number : in figures
QkWeZ uEcj : vadksa esa
: in words
: 'kCnksa esa
Centre of Examination (in Capitals) :
ijh{kk dsUæ (cM+s v{kjksa esa) :
Candidate’s Signature : Invigilator’s Signature :
ijh{kkFkhZ ds gLrk{kj : fujh{kd ds gLrk{kj :
)/999DMD31/319//7)
(0999DMD310319007)
Test Pattern
egRoiw.kZ funsZ'k :
1. mÙkj i= ds i`"B-1 ,oa i`"B-2 ij /;kuiwoZd dsoy uhys@dkys ckWy
ikWbaV isu ls fooj.k HkjsaA
2. ijh{kk dh vof/k 3 ?kaVs gS ,oa ijh{kk iqfLrdk esa 180 iz'u gSaA izR;sd
iz'u 4 vad dk gSA izR;sd lgh mÙkj ds fy , ijh{kkFkhZ dks 4 vad
fn, tk,axsaA izR;sd xyr mÙkj ds fy , dqy ;ksx esa ls ,d vad
?kVk;k tk,xkA vf/kdre vad 720 gSA
3. bl i`"B ij fooj.k vafdr djus ,oa mÙkj i= ij fu'kku yxkus ds
fy, dsoy uhys@dkys ckWy ikWbaV isu dk iz;ksx djsaA
4. jQ dk;Z bl ijh{kk iqfLrdk esa fu/kkZfjr LFkku ij gh djsaA
5. ijh{kk lEiék gksus ij ] ijh{kkFkhZ d{k@gkWy NksM+us ls iwoZ
mÙkj i= fujh{kd dks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk
dsoy ijh{kk iqfLrdk dks ys tk ldrs gSaA
6. ijh{kkFkhZ lqfuf'pr djsa fd bl mÙkj i = dks eksM+k u tk , ,oa ml
ij dksbZ vU ; fu'kku u yxk,aA ijh{kkFkhZ viuk QkWeZ uEcj iz'u
iqfLrdk@mÙkj i= esa fu/kkZfjr LFkku ds vfrfjDr vU ;= u fy[ksaA
7. mÙkj i= ij fdlh izdkj ds la'kks/ku gsrq OgkbV ¶+yqbM ds iz;ksx dh
vuqefr ugha gSA
8. ;fn vki fdlh iz'u dks gy djus dk iz;kl djrs gSa rks mfpr xksys
dks uhps n'kkZ;s x;s vuqlkj xgjk dkyk djsa vU;Fkk mls [kkyh NksM+ nsaA
lgh rjhdk xyr rjhdk
Important Instructions :
1. On the Answer Sheet, fill in the particulars on Side-1
and Side-2 carefully with blue/black ball point pen only.
2. The test is of 3 hours duration and this Test Booklet
contains 180 questions. Each question carries 4 marks.
For each correct response, the candidate will get 4 marks.
For each incorrect response, one mark will be deducted
from the total scores. The maximum marks are 720.
3. Use Blue/Black Ball Point Pen only for writing
particulars on this page/marking responses.
4. Rough work is to be done on the space provided for this
purpose in the Test Booklet only.
5. On completion of the test, the candidate must
hand over the Answer Sheet to the Invigilator
before leaving the Room/Hall. The candidates are
allowed to take away this Test Booklet with them.
6. The candidates should ensure that the Answer Sheet is
not folded. Do not make any stray marks on the Answer
Sheet. Do not write your Form No. anywhere else except
in the specified space in the Test Booklet/Answer Sheet.
7. Use of white fluid for correction is not permissible on
the Answer Sheet.
8. If you want to attempt any question then circle should be
properly darkened as shown below, otherwise leave blank.
Correct Method Wrong Method
In case of any ambiguity in translation of any question, English version shall be treated as final.
iz'uksa ds vuqokn esa fdlh vLi"Vrk dh fLFkfr esa ] vaxzsth laLdj.k dks gh vafre ekuk tk,sxkA
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
Your Target is to secure Good Rank in Pre-Medical 2020
NEET(UG)
MINOR TEST # 07
29-09-2019
Read carefully the Instructions on the Back Cover of this Test Booklet.
bl ijh{kk iqfLrdk ds fiNys vkoj.k ij fn , funsZ'kksa dks /;ku ls i<+ saA
Do not open this Test Booklet until you are asked to do so.
bl ijh{kk iqfLrdk dks tc rd uk [kksysa tc rd dgk u tk,A
This Booklet contains 36 pages. bl iqfLrdk esa 36 i`"B gSaA
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
Test Type : Unit - 06
12
th
 Undergoing/Pass Students
LTS / Page 2/36 SPACE FOR ROUGH WORK / jQ dk;Z ds fy , txg 0999DMD310319007
ALLEN
TARGET : PRE-MEDICAL 2020/NEET-UG/29-09-2019
1. In the given AC circuit  :-
I
1
X
C
I
2
R
~
I
V = V sin t
0
w
(1) Current I
2
 and V are in same phase
(2) Current I
2
 lead I
1
 by 90º
(3) Current I leads I
2
 by q < 90º
(4) Current I leads I
1
 by q < 90º
2. A ring and a disc of different masses are rotating
with the same kinetic energy. If we apply a
retarding torque t on the ring, it stops after making
n revolutions. After how many revolutions will the
disc stop if the same retarding torque on it :-
(1) n/2 (2) n
(3) 2n (4) 4n
3. In the LCR circuit shown in figure :-
X =20
C
W
~
V = 400 sin t w
R=10W
X =10
L
W
(1) Current will leads the voltage
(2) rms value of currents is 20 A
(3) Power factor of the circuit is 
1
2
(4) All of the above
TOPIC : Rotational Motion, Alternating current, Electromagnetic Waves.
1. fn;s x;s ifjiFk esa :-
I
1
X
C
I
2
R
~
I
V = V sin t
0
w
(1) /kkjk I
2
 rFkk V leku dyk esa gSA
(2) /kkjk I
2
, I
1
 ls 90º vkxs gSA
(3) /kkjk I, I
2
 ls q < 90º vkxs gSA
(4) /kkjk I, I
1
 ls q < 90º vkxs gSA
2. ,d oy; rFkk ,d pdrh leku xfrt ÅtkZ ds lkFk ?kw.kZu dj
jgh gaSA ;fn ge oy; ij voeUnd cy vk?kw.kZ t vkjksfir djrs
gS] rks ;g n pDdj yxkus ds i'pkr~ :d tkrh gS ;fn ;gh
voeUnd cy vk?kw.kZ pdrh ij yxk;k tk;s rks ;g fdrus pDdj
yxkus ds i'pkr~ :dsxh :-
(1) n/2 (2) n
(3) 2n (4) 4n
3. fp= esa iznf'kZr ifjiFk esa :-
X =20
C
W
~
V = 400 sin t w
R=10W
X =10
L
W
(1) /kkjk oksYVrk ls dyk ls vkxs gSA
(2) /kkjk dk o-ek-ew- eku 20 A gSA
(3) ifjiFk dk 'kfDr xq.kkad 
1
2
 gSA
(4) mijksDr lHkh
0999DMD310319007 SPACE FOR ROUGH WORK / jQ dk;Z ds fy , txg LTS / Page 3/36
ALLEN
LEADER TEST SERIES/JOINT PACKAGE COURSE/NEET-UG/29-09-2019
4. Two blocks each of mass M are connected to the
ends of a light frame as shown in figure The frame
is rotated about the vertical line of symmetry. The
rod breaks if the tension in it exceeds T
0
. Find the
maximum frequency with which the frame may
be rotated without breaking the rod.
M M
L
(1) 
0
T 1
4 ML p
(2) 
0
T 1
2 ML p
(3) 
0
2T 1
6 3ML p
(4)
0
3T 1
2 2ML p
5. A 12 W resistor and a 0.21 henry inductor are
connected in series to A.C. source operating at
20 volt 50 Hz. The phase angle between the
current and source voltage is :-
(1) 30º (2) 40º (3) 80º (4) 90º
6. The electric field of an electromagnetic wave is
given by
E = (50 N/C) sin w(t–x/c)
the energy contained in a cylinder of cross-section
10cm
2
 and length 50 cm along the x-axis :-
(1) 5.5×10
–6 
J (2) 5.5 × 10
–12 
J
(3) 2.75 × 10
–6
 J (4) 2.75 × 10
–12
 J
7. A thick walled hollow sphere has outer radius R.
It rolls down an inclined plane without slipping and
its speed at the bottom is v. If the inclined plane
is frictionless and the sphere slides down without
rolling, its speed at the bottom will be 5v/4. What
is the radius of gyration of the sphere ?
(1) 
R
2
(2) 
R
2
(3) 
3R
4
(4) 
3R
4
4. fp= esa n'kkZ;s vuqlkj ,d gYdh Ýse ds fljksa ij nks CykWd tqM+s
gq, gSa] izR;sd CykWd dk nzO;eku M gSA Ýse dks m/oZ lefer
js[kk ds ifjr% ?kqek;k tk jgk gSA ;fn NM+ esa ruko T
0
 gks tk;sxk
rks NM+ VwV tk;sxhA Ýse dks ?kqekus dh vf/kdre vko`fÙk Kkr
dhft,] ftlds fy;s NM+ VwVs ughaA
M M
L
(1) 
0
T 1
4 ML p
(2) 
0
T 1
2 ML p
(3) 
0
2T 1
6 3ML p
(4)
0
3T 1
2 2ML p
5. 12 W izfrjks/k rFkk 0.21 gsujh izsjdRo 20 volt 50 Hz ds
izR;korhZ L=ksr ds lkFk Js.khØe esa tqM+s gq , gSA /kkjk ,oa oksYVrk
ds chp dyk dks.k gS :-
(1) 30º (2) 40º
(3) 80º (4) 90º
6. ,d fo|qr pqEcdh ; rjax dk fo|qr {ks= fuEukuqlkj O;Dr fd;k
tkrk gS
E = (50 N/C) sin w(t–x/c)
10cm
2
 vuqizLFk dkV rFkk x-v{k ds vuqfn'k 50 cm yEckbZ
okys csyu esa fufgr ÅtkZ gksxh :-
(1) 5.5×10
–6 
J (2) 5.5 × 10
–12 
J
(3) 2.75 × 10
–6
 J (4) 2.75 × 10
–12
 J
7. eksVh nhokj okys ,d [kks[kys xksys dh ckg~; f=T;k R gSA ;g
,d vkur ry ij fcuk fQlys uhps yq <+drk gS vkSj iSans ij bldh
pky v gSA ;fn vkur ry ?k"kZ.k jfgr gks vkSj xksyk fcuk yq <+ds
fQlys rks iSans ij bldh pky 5v/4 gksxhA xksys dh ifjHkze.k
f=T;k fdruh gS ?
(1) 
R
2
(2) 
R
2
(3) 
3R
4
(4) 
3R
4
LTS / Page 4/36 SPACE FOR ROUGH WORK / jQ dk;Z ds fy , txg 0999DMD310319007
ALLEN
TARGET : PRE-MEDICAL 2020/NEET-UG/29-09-2019
8. An A.C. source of voltage V = 100 sin 100pt is
connected to a resistor of resistance 20 W. The rms
value of current through resistor is :-
(1) 10 A (2) 
10
A
2
(3) 
5
A
2
(4) None of these
9. In a plane electromagnetic wave, magnetic field
is given as follows :-
B
y
 = (2×10
–7
 T) sin (0.5 × 10
3
 x + 1.5 × 10
11
t)
Expression for electric field will be :-
(1)E
y
 = (6 N/C) sin (0.5 × 10
3
y + 1.5 × 10
11
t)
(2)E
z
 = (6 N/C) sin (0.5 × 10
3
x + 1.5 × 10
11
t)
(3)E
x
 = (60 N/C) sin (0.5 × 10
3
x + 1.5 × 10
11
t)
(4)E
z
 = (60 N/C) sin (0.5 × 10
3
x + 1.5 × 10
11
t)
10. A cyclinder of height h is placed on an inclined
plane, the angle of inclination of which is slowly
increased. It begins to slip when the angle of
inclination is 45º. What is the radius of the cylinder?
(1) h (2) 
3
h
4
(3) 
1
h
2
(4) 
1
h
4
11. The root mean square value of voltage, if an
alternating voltage is given by e = e
1
 sin wt + e
2
cos wt is :-
(1) 
22
12
ee
2
+
(2) 
22
12
ee
2
+
(3) 
22
12
ee
2
-
(4) 
22
12
ee
2
-
12. The amplitude of electric and magnetic fields in
a parallel beam of light of intensity 4.0 W/m
2
 are,
respectively :-
(1) 54.87, N/C, 1.83 × 10
–7
 T
(2) 1.83 × 10
–7
 N/C, 54.87 T
(3) 18.19 N/C, 6 T
(4) 6 N/C 18.19 T
8. V = 100 sin 100pt oksYVrk dk izR;korhZ L=ksr 20 W izfrjksèk
ds lkFk la;ksftr gSA izfrjks/k ls /kkjk dk o-ek-ew- eku gS :-
(1) 10 A (2) 
10
A
2
(3) 
5
A
2
(4) buesa ls dksbZ ugha
9. ,d lery fo|qr pqEcdh; rjax esa pqEcdh; {ks= fuEukuqlkj
O;Dr fd;k tkrk gS :-
B
y
 = (2×10
–7
 T) sin (0.5 × 10
3
 x + 1.5 × 10
11
t)
fo|qr {ks= ds fy;s O;atd gksxk :-
(1)E
y
 = (6 N/C) sin (0.5 × 10
3
y + 1.5 × 10
11
t)
(2)E
z
 = (6 N/C) sin (0.5 × 10
3
x + 1.5 × 10
11
t)
(3)E
x
 = (60 N/C) sin (0.5 × 10
3
x + 1.5 × 10
11
t)
(4)E
z
 = (60 N/C) sin (0.5 × 10
3
x + 1.5 × 10
11
t)
10. h Å¡pkbZ okyk csyu dk ur ry  ij j[kk gqvk gS] ftldk
vkufr dks.k /khjs&/khjs c<+k;k tkrk gSA tc vkufr dks.k 45º
gks tk;s ;g fQlyuk izkjEHk dj nsrk gSA csyu dh f=T;k
fdruh gS ?
(1) h (2) 
3
h
4
(3) 
1
h
2
(4) 
1
h
4
11. ;fn ,d izR;korhZ oksYVrk e = e
1
 sin wt + e
2
 cos wt ls iznf'kZr
gS] rks oksYVrk dk oxZ ek /; ewy eku gS :-
(1) 
22
12
ee
2
+
(2) 
22
12
ee
2
+
(3) 
22
12
ee
2
-
(4) 
22
12
ee
2
-
12. 4.0 W/m
2
 rhozrk okys lekukUrj izdk'k iqat esa fo|qr ,oa
pqEcdh; {ks=ksa ds vk;ke gksaxs] Øe'k% :-
(1) 54.87, N/C, 1.83 × 10
–7
 T
(2) 1.83 × 10
–7
 N/C, 54.87 T
(3) 18.19 N/C, 6 T
(4) 6 N/C 18.19 T
Page 5


LTS / Page 1/36
Name of the Candidate (in Capitals)
ijh{kkFkhZ dk uke (cM+s v{kjksa esa) :
Form Number : in figures
QkWeZ uEcj : vadksa esa
: in words
: 'kCnksa esa
Centre of Examination (in Capitals) :
ijh{kk dsUæ (cM+s v{kjksa esa) :
Candidate’s Signature : Invigilator’s Signature :
ijh{kkFkhZ ds gLrk{kj : fujh{kd ds gLrk{kj :
)/999DMD31/319//7)
(0999DMD310319007)
Test Pattern
egRoiw.kZ funsZ'k :
1. mÙkj i= ds i`"B-1 ,oa i`"B-2 ij /;kuiwoZd dsoy uhys@dkys ckWy
ikWbaV isu ls fooj.k HkjsaA
2. ijh{kk dh vof/k 3 ?kaVs gS ,oa ijh{kk iqfLrdk esa 180 iz'u gSaA izR;sd
iz'u 4 vad dk gSA izR;sd lgh mÙkj ds fy , ijh{kkFkhZ dks 4 vad
fn, tk,axsaA izR;sd xyr mÙkj ds fy , dqy ;ksx esa ls ,d vad
?kVk;k tk,xkA vf/kdre vad 720 gSA
3. bl i`"B ij fooj.k vafdr djus ,oa mÙkj i= ij fu'kku yxkus ds
fy, dsoy uhys@dkys ckWy ikWbaV isu dk iz;ksx djsaA
4. jQ dk;Z bl ijh{kk iqfLrdk esa fu/kkZfjr LFkku ij gh djsaA
5. ijh{kk lEiék gksus ij ] ijh{kkFkhZ d{k@gkWy NksM+us ls iwoZ
mÙkj i= fujh{kd dks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk
dsoy ijh{kk iqfLrdk dks ys tk ldrs gSaA
6. ijh{kkFkhZ lqfuf'pr djsa fd bl mÙkj i = dks eksM+k u tk , ,oa ml
ij dksbZ vU ; fu'kku u yxk,aA ijh{kkFkhZ viuk QkWeZ uEcj iz'u
iqfLrdk@mÙkj i= esa fu/kkZfjr LFkku ds vfrfjDr vU ;= u fy[ksaA
7. mÙkj i= ij fdlh izdkj ds la'kks/ku gsrq OgkbV ¶+yqbM ds iz;ksx dh
vuqefr ugha gSA
8. ;fn vki fdlh iz'u dks gy djus dk iz;kl djrs gSa rks mfpr xksys
dks uhps n'kkZ;s x;s vuqlkj xgjk dkyk djsa vU;Fkk mls [kkyh NksM+ nsaA
lgh rjhdk xyr rjhdk
Important Instructions :
1. On the Answer Sheet, fill in the particulars on Side-1
and Side-2 carefully with blue/black ball point pen only.
2. The test is of 3 hours duration and this Test Booklet
contains 180 questions. Each question carries 4 marks.
For each correct response, the candidate will get 4 marks.
For each incorrect response, one mark will be deducted
from the total scores. The maximum marks are 720.
3. Use Blue/Black Ball Point Pen only for writing
particulars on this page/marking responses.
4. Rough work is to be done on the space provided for this
purpose in the Test Booklet only.
5. On completion of the test, the candidate must
hand over the Answer Sheet to the Invigilator
before leaving the Room/Hall. The candidates are
allowed to take away this Test Booklet with them.
6. The candidates should ensure that the Answer Sheet is
not folded. Do not make any stray marks on the Answer
Sheet. Do not write your Form No. anywhere else except
in the specified space in the Test Booklet/Answer Sheet.
7. Use of white fluid for correction is not permissible on
the Answer Sheet.
8. If you want to attempt any question then circle should be
properly darkened as shown below, otherwise leave blank.
Correct Method Wrong Method
In case of any ambiguity in translation of any question, English version shall be treated as final.
iz'uksa ds vuqokn esa fdlh vLi"Vrk dh fLFkfr esa ] vaxzsth laLdj.k dks gh vafre ekuk tk,sxkA
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
Your Target is to secure Good Rank in Pre-Medical 2020
NEET(UG)
MINOR TEST # 07
29-09-2019
Read carefully the Instructions on the Back Cover of this Test Booklet.
bl ijh{kk iqfLrdk ds fiNys vkoj.k ij fn , funsZ'kksa dks /;ku ls i<+ saA
Do not open this Test Booklet until you are asked to do so.
bl ijh{kk iqfLrdk dks tc rd uk [kksysa tc rd dgk u tk,A
This Booklet contains 36 pages. bl iqfLrdk esa 36 i`"B gSaA
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
Test Type : Unit - 06
12
th
 Undergoing/Pass Students
LTS / Page 2/36 SPACE FOR ROUGH WORK / jQ dk;Z ds fy , txg 0999DMD310319007
ALLEN
TARGET : PRE-MEDICAL 2020/NEET-UG/29-09-2019
1. In the given AC circuit  :-
I
1
X
C
I
2
R
~
I
V = V sin t
0
w
(1) Current I
2
 and V are in same phase
(2) Current I
2
 lead I
1
 by 90º
(3) Current I leads I
2
 by q < 90º
(4) Current I leads I
1
 by q < 90º
2. A ring and a disc of different masses are rotating
with the same kinetic energy. If we apply a
retarding torque t on the ring, it stops after making
n revolutions. After how many revolutions will the
disc stop if the same retarding torque on it :-
(1) n/2 (2) n
(3) 2n (4) 4n
3. In the LCR circuit shown in figure :-
X =20
C
W
~
V = 400 sin t w
R=10W
X =10
L
W
(1) Current will leads the voltage
(2) rms value of currents is 20 A
(3) Power factor of the circuit is 
1
2
(4) All of the above
TOPIC : Rotational Motion, Alternating current, Electromagnetic Waves.
1. fn;s x;s ifjiFk esa :-
I
1
X
C
I
2
R
~
I
V = V sin t
0
w
(1) /kkjk I
2
 rFkk V leku dyk esa gSA
(2) /kkjk I
2
, I
1
 ls 90º vkxs gSA
(3) /kkjk I, I
2
 ls q < 90º vkxs gSA
(4) /kkjk I, I
1
 ls q < 90º vkxs gSA
2. ,d oy; rFkk ,d pdrh leku xfrt ÅtkZ ds lkFk ?kw.kZu dj
jgh gaSA ;fn ge oy; ij voeUnd cy vk?kw.kZ t vkjksfir djrs
gS] rks ;g n pDdj yxkus ds i'pkr~ :d tkrh gS ;fn ;gh
voeUnd cy vk?kw.kZ pdrh ij yxk;k tk;s rks ;g fdrus pDdj
yxkus ds i'pkr~ :dsxh :-
(1) n/2 (2) n
(3) 2n (4) 4n
3. fp= esa iznf'kZr ifjiFk esa :-
X =20
C
W
~
V = 400 sin t w
R=10W
X =10
L
W
(1) /kkjk oksYVrk ls dyk ls vkxs gSA
(2) /kkjk dk o-ek-ew- eku 20 A gSA
(3) ifjiFk dk 'kfDr xq.kkad 
1
2
 gSA
(4) mijksDr lHkh
0999DMD310319007 SPACE FOR ROUGH WORK / jQ dk;Z ds fy , txg LTS / Page 3/36
ALLEN
LEADER TEST SERIES/JOINT PACKAGE COURSE/NEET-UG/29-09-2019
4. Two blocks each of mass M are connected to the
ends of a light frame as shown in figure The frame
is rotated about the vertical line of symmetry. The
rod breaks if the tension in it exceeds T
0
. Find the
maximum frequency with which the frame may
be rotated without breaking the rod.
M M
L
(1) 
0
T 1
4 ML p
(2) 
0
T 1
2 ML p
(3) 
0
2T 1
6 3ML p
(4)
0
3T 1
2 2ML p
5. A 12 W resistor and a 0.21 henry inductor are
connected in series to A.C. source operating at
20 volt 50 Hz. The phase angle between the
current and source voltage is :-
(1) 30º (2) 40º (3) 80º (4) 90º
6. The electric field of an electromagnetic wave is
given by
E = (50 N/C) sin w(t–x/c)
the energy contained in a cylinder of cross-section
10cm
2
 and length 50 cm along the x-axis :-
(1) 5.5×10
–6 
J (2) 5.5 × 10
–12 
J
(3) 2.75 × 10
–6
 J (4) 2.75 × 10
–12
 J
7. A thick walled hollow sphere has outer radius R.
It rolls down an inclined plane without slipping and
its speed at the bottom is v. If the inclined plane
is frictionless and the sphere slides down without
rolling, its speed at the bottom will be 5v/4. What
is the radius of gyration of the sphere ?
(1) 
R
2
(2) 
R
2
(3) 
3R
4
(4) 
3R
4
4. fp= esa n'kkZ;s vuqlkj ,d gYdh Ýse ds fljksa ij nks CykWd tqM+s
gq, gSa] izR;sd CykWd dk nzO;eku M gSA Ýse dks m/oZ lefer
js[kk ds ifjr% ?kqek;k tk jgk gSA ;fn NM+ esa ruko T
0
 gks tk;sxk
rks NM+ VwV tk;sxhA Ýse dks ?kqekus dh vf/kdre vko`fÙk Kkr
dhft,] ftlds fy;s NM+ VwVs ughaA
M M
L
(1) 
0
T 1
4 ML p
(2) 
0
T 1
2 ML p
(3) 
0
2T 1
6 3ML p
(4)
0
3T 1
2 2ML p
5. 12 W izfrjks/k rFkk 0.21 gsujh izsjdRo 20 volt 50 Hz ds
izR;korhZ L=ksr ds lkFk Js.khØe esa tqM+s gq , gSA /kkjk ,oa oksYVrk
ds chp dyk dks.k gS :-
(1) 30º (2) 40º
(3) 80º (4) 90º
6. ,d fo|qr pqEcdh ; rjax dk fo|qr {ks= fuEukuqlkj O;Dr fd;k
tkrk gS
E = (50 N/C) sin w(t–x/c)
10cm
2
 vuqizLFk dkV rFkk x-v{k ds vuqfn'k 50 cm yEckbZ
okys csyu esa fufgr ÅtkZ gksxh :-
(1) 5.5×10
–6 
J (2) 5.5 × 10
–12 
J
(3) 2.75 × 10
–6
 J (4) 2.75 × 10
–12
 J
7. eksVh nhokj okys ,d [kks[kys xksys dh ckg~; f=T;k R gSA ;g
,d vkur ry ij fcuk fQlys uhps yq <+drk gS vkSj iSans ij bldh
pky v gSA ;fn vkur ry ?k"kZ.k jfgr gks vkSj xksyk fcuk yq <+ds
fQlys rks iSans ij bldh pky 5v/4 gksxhA xksys dh ifjHkze.k
f=T;k fdruh gS ?
(1) 
R
2
(2) 
R
2
(3) 
3R
4
(4) 
3R
4
LTS / Page 4/36 SPACE FOR ROUGH WORK / jQ dk;Z ds fy , txg 0999DMD310319007
ALLEN
TARGET : PRE-MEDICAL 2020/NEET-UG/29-09-2019
8. An A.C. source of voltage V = 100 sin 100pt is
connected to a resistor of resistance 20 W. The rms
value of current through resistor is :-
(1) 10 A (2) 
10
A
2
(3) 
5
A
2
(4) None of these
9. In a plane electromagnetic wave, magnetic field
is given as follows :-
B
y
 = (2×10
–7
 T) sin (0.5 × 10
3
 x + 1.5 × 10
11
t)
Expression for electric field will be :-
(1)E
y
 = (6 N/C) sin (0.5 × 10
3
y + 1.5 × 10
11
t)
(2)E
z
 = (6 N/C) sin (0.5 × 10
3
x + 1.5 × 10
11
t)
(3)E
x
 = (60 N/C) sin (0.5 × 10
3
x + 1.5 × 10
11
t)
(4)E
z
 = (60 N/C) sin (0.5 × 10
3
x + 1.5 × 10
11
t)
10. A cyclinder of height h is placed on an inclined
plane, the angle of inclination of which is slowly
increased. It begins to slip when the angle of
inclination is 45º. What is the radius of the cylinder?
(1) h (2) 
3
h
4
(3) 
1
h
2
(4) 
1
h
4
11. The root mean square value of voltage, if an
alternating voltage is given by e = e
1
 sin wt + e
2
cos wt is :-
(1) 
22
12
ee
2
+
(2) 
22
12
ee
2
+
(3) 
22
12
ee
2
-
(4) 
22
12
ee
2
-
12. The amplitude of electric and magnetic fields in
a parallel beam of light of intensity 4.0 W/m
2
 are,
respectively :-
(1) 54.87, N/C, 1.83 × 10
–7
 T
(2) 1.83 × 10
–7
 N/C, 54.87 T
(3) 18.19 N/C, 6 T
(4) 6 N/C 18.19 T
8. V = 100 sin 100pt oksYVrk dk izR;korhZ L=ksr 20 W izfrjksèk
ds lkFk la;ksftr gSA izfrjks/k ls /kkjk dk o-ek-ew- eku gS :-
(1) 10 A (2) 
10
A
2
(3) 
5
A
2
(4) buesa ls dksbZ ugha
9. ,d lery fo|qr pqEcdh; rjax esa pqEcdh; {ks= fuEukuqlkj
O;Dr fd;k tkrk gS :-
B
y
 = (2×10
–7
 T) sin (0.5 × 10
3
 x + 1.5 × 10
11
t)
fo|qr {ks= ds fy;s O;atd gksxk :-
(1)E
y
 = (6 N/C) sin (0.5 × 10
3
y + 1.5 × 10
11
t)
(2)E
z
 = (6 N/C) sin (0.5 × 10
3
x + 1.5 × 10
11
t)
(3)E
x
 = (60 N/C) sin (0.5 × 10
3
x + 1.5 × 10
11
t)
(4)E
z
 = (60 N/C) sin (0.5 × 10
3
x + 1.5 × 10
11
t)
10. h Å¡pkbZ okyk csyu dk ur ry  ij j[kk gqvk gS] ftldk
vkufr dks.k /khjs&/khjs c<+k;k tkrk gSA tc vkufr dks.k 45º
gks tk;s ;g fQlyuk izkjEHk dj nsrk gSA csyu dh f=T;k
fdruh gS ?
(1) h (2) 
3
h
4
(3) 
1
h
2
(4) 
1
h
4
11. ;fn ,d izR;korhZ oksYVrk e = e
1
 sin wt + e
2
 cos wt ls iznf'kZr
gS] rks oksYVrk dk oxZ ek /; ewy eku gS :-
(1) 
22
12
ee
2
+
(2) 
22
12
ee
2
+
(3) 
22
12
ee
2
-
(4) 
22
12
ee
2
-
12. 4.0 W/m
2
 rhozrk okys lekukUrj izdk'k iqat esa fo|qr ,oa
pqEcdh; {ks=ksa ds vk;ke gksaxs] Øe'k% :-
(1) 54.87, N/C, 1.83 × 10
–7
 T
(2) 1.83 × 10
–7
 N/C, 54.87 T
(3) 18.19 N/C, 6 T
(4) 6 N/C 18.19 T
0999DMD310319007 SPACE FOR ROUGH WORK / jQ dk;Z ds fy , txg LTS / Page 5/36
ALLEN
LEADER TEST SERIES/JOINT PACKAGE COURSE/NEET-UG/29-09-2019
13. A stick of length L and mass M lies on a frictionless
horizontal surface on which it is free to move in
any way. A ball of mass m moving with speed v
collides elastically with the stick as shown in the
figure. If after the collision the ball comes to rest,
then what should be the mass of the ball ?
L
(1) m = 2M (2) m = M
(3) m = M/2 (4) m = M/4
14. An AC voltage V = V
0
 sin 100 t is applied to the
circuit, the phase difference between current and
voltage is found to be 
,
4
p
 then :-
p/4
t
V, I
V I
(1) R = 100 W, C = 1 µF
(2) R = 1 kW, C = 10 µF
(3) R = 10 kW, L = 1 H
(4) R = 1 kW, L = 10 H
15. Which of the following relation is correct ?
(1) 
0 0 00
E µB e= (2) 
00 00
µ B /E e=
(3) E
0
 = 
000
µB e (4) 
00 00
µEB =e
16. A cylindrical rod of mass M, length L and radius
R has two cords wound around it whose ends are
attached to the ceiling. The rod is held horizontally
with the two cords vertical. When the rod is
released, the cords unwind and the rod rotates, the
linear acceleration of the rod as it falls, is :-
(1)g (2) g/3 (3) 2g/3 (4) g/2
13. L yEckbZ ,oa M nzO;eku okyh NM+ ,d ?k"kZ.k jfgr {ksfrt
lrg ij j[kh gq;h gS] ftl ij ;g fdlh Hkh izdkj ls xfr ds
fy;s Lora= gSA v osx ls xfr djrh gqbZ m nzO;eku dh xsan
fp= esa n'kkZ;s vuqlkj NM+ ls izR;kLFkr% Vdjkrh gSA ;fn VDdj
ds i'pkr~ xsan fojkekoLFkk esa vk tkrh gS ] rks xsan dh nzO;eku
fdruk gS ?
L
(1) m = 2M (2) m = M
(3) m = M/2 (4) m = M/4
14. ifjiFk ij ,d izR;korhZ oksYVrk V = V
0
 sin 100 t vkjksfir
dh x;h gS] /kkjk ,oa oksYVrk ds chp dykUrj 
4
p
 izkIr gksrk gS]
rks :-
p/4
t
V, I
V I
(1) R = 100 W, C = 1 µF
(2) R = 1 kW, C = 10 µF
(3) R = 10 kW, L = 1 H
(4) R = 1 kW, L = 10 H
15. fuEufyf[kr esa ls dkSulk lEcU/k lgh gS ?
(1) 
0 0 00
E µB e= (2) 
00 00
µ B /E e=
(3) E
0
 = 
000
µB e (4) 
00 00
µEB =e
16. M nzO;eku] L yEckbZ vkSj R f=T;k okyh ,d csyukdkj
NM+ ij nks Mksfj;k¡ yisVh x;h gSa ] ftuds fljs Nr ls tqM+sa gq ,
gSA Mksjh;ksa dks Å/okZ/kj j[k dj NM + dks {ksfrt j[kk x;k gSA
tc NM+ dks eqDr fd;k tkrk gS] Mksjh;k¡ [kqyrh gS vkSj NM+
tSls&tSls uhps vkrh gS ] ?kw .kZu djrh gqbZ NM + dk jSf[kd Roj.k
gS :-
(1)g (2) g/3 (3) 2g/3 (4) g/2
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