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Page 1
PAGE # 1
PART : PHYSICS
Single Choice Type (,dy fodYih; izdkj)
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. Find magnetic field at O.
O ij pqEcdh; {kS=k Kkr fdft;sA
45º
45º
O
i
i
i
A
B
D
N
M
(1)
?
?
?
?
?
?
? ?
?
?
?
2
1
R 2
i
0
(2) ? ? 1
R 2
i
0
? ?
?
?
(3)
R 2
i
0
?
(4)
?
?
?
?
?
?
? ?
?
?
2
1
R 2
i
0
Ans. (4)
Sol.
45º
45º
O
i
i
3
4 2
1
i
? ? ? ? ? ? ? ?
4 0 3 0 2 0 1 0 0
B B B B B
? ? ? ? ?
? ? ? ?
? ?
R 2
i
º 45 sin º 90 sin
R 4
i
0 0
?
? ? ?
?
?
+
R 4
i
0
?
?
(sin45º + sin90º)
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? 1
2
1
R 4
i
R 2
i
2
1
1
R 4
i
0 0 0
?
?
?
?
?
?
? ? ? ? ? ?
?
?
? 1
2
1
2
2
1
1
R 4
i
0
? ? ? ?
?
?
? 2 2
R 4
i
0
?
?
?
?
?
?
? ?
?
?
?
2
1
R 2
i
0
Page 2
PAGE # 1
PART : PHYSICS
Single Choice Type (,dy fodYih; izdkj)
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. Find magnetic field at O.
O ij pqEcdh; {kS=k Kkr fdft;sA
45º
45º
O
i
i
i
A
B
D
N
M
(1)
?
?
?
?
?
?
? ?
?
?
?
2
1
R 2
i
0
(2) ? ? 1
R 2
i
0
? ?
?
?
(3)
R 2
i
0
?
(4)
?
?
?
?
?
?
? ?
?
?
2
1
R 2
i
0
Ans. (4)
Sol.
45º
45º
O
i
i
3
4 2
1
i
? ? ? ? ? ? ? ?
4 0 3 0 2 0 1 0 0
B B B B B
? ? ? ? ?
? ? ? ?
? ?
R 2
i
º 45 sin º 90 sin
R 4
i
0 0
?
? ? ?
?
?
+
R 4
i
0
?
?
(sin45º + sin90º)
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? 1
2
1
R 4
i
R 2
i
2
1
1
R 4
i
0 0 0
?
?
?
?
?
?
? ? ? ? ? ?
?
?
? 1
2
1
2
2
1
1
R 4
i
0
? ? ? ?
?
?
? 2 2
R 4
i
0
?
?
?
?
?
?
? ?
?
?
?
2
1
R 2
i
0
PAGE # 2
2. Position of particle as a function of time is given as j
ˆ
t sin i
ˆ
t cos r ? ? ? ?
?
. Choose correct statement
about a and v , r
? ? ?
where v
?
and a
?
are velocity and acceleration of particle at time t.
(1) v
?
is perpendicular to r
?
and a
?
is towards origin
(2) a and v
? ?
are perpendicular to r
?
(3) v
?
is parallel to r
?
and a
?
parallel to r
?
.
(4) v
?
is perpendicular to r
?
and a
?
is away from origin.
fdlh d.k dh fLFkfr le; ds inksa esa j
ˆ
t sin i
ˆ
t cos r ? ? ? ?
?
ls nh tkrh gSA v , r
? ?
rFkk a
?
ds ckjs esa lgh dFku
pqfu;sA tgk v
?
rFkk a
?
, t le; ij d.k ds osx rFkk Roj.k gSaA
(1) v
?
, r
?
ds yEcor~ gS rFkk a
?
ewy fcUnq dh rjQ gSA
(2) v
?
rFkk a
?
, r
?
ds yEcor~ gSA
(3) v
?
, r
?
ds lekUrj gS rFkk a
?
, r
?
ds lekUrj gSA
(4) v
?
, r
?
yEcor~ gS rFkk a
?
dh fn’'kk ewy fcUnq ls nwj dh rjQ gSA
Ans. (1)
Sol. r
?
= j
ˆ
t sin i
ˆ
t cos ? ? ?
) j
ˆ
t cos i
ˆ
t sin (
dt
r d
v ? ? ? ? ? ? ?
?
?
) j
ˆ
t sin i
ˆ
t (cos
dt
v d
a
2
? ? ? ? ? ? ?
?
?
r a
2
? ?
? ? ? ? a
?
is antiparallel to r
?
a
?
, r
?
ds izfr lekUrj gSA
r . v
? ?
= ? (–sin ?t cos ?t + cos ?t sin ?t) = 0
so vr% r v
? ?
?
3. A Carnot engine, having an efficiency of ? = 1/10 as heat engine, is used as a refrigerator. If the work
done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is
,d dkukZsV bUtu ftldh n{krk ? = 1/10 Å"ek bUtu dh n{krk gS, jsfQztjsVj ds :i esa iz;ksx fd;k tkrk gSA ;fn
fudk; ij fd;k x;k dk;Z 10 twy gS, rc U;wu rkieku ij dq.M ls vo'kksf"kr Å"ek dh ek=kk gSA
(1) 99 J (2*) 90 J (3) 1 J (4) 100 J
Ans. 2
Sol. For Carnot engine using as refrigerator
dkuksZV btau dks jsfÝtjsVj dh Hkkfr iz;ksx djus ds fy,
1
2
2
T
W Q – 1
T
? ?
?
? ?
? ?
It is given fn;k x;k gSA ? =
1
10
? ? = 1 –
2
1
T
T
?
2
1
T 9
T 10
?
So, vr% Q 2 = 90 J (as W = 10 J)
Page 3
PAGE # 1
PART : PHYSICS
Single Choice Type (,dy fodYih; izdkj)
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. Find magnetic field at O.
O ij pqEcdh; {kS=k Kkr fdft;sA
45º
45º
O
i
i
i
A
B
D
N
M
(1)
?
?
?
?
?
?
? ?
?
?
?
2
1
R 2
i
0
(2) ? ? 1
R 2
i
0
? ?
?
?
(3)
R 2
i
0
?
(4)
?
?
?
?
?
?
? ?
?
?
2
1
R 2
i
0
Ans. (4)
Sol.
45º
45º
O
i
i
3
4 2
1
i
? ? ? ? ? ? ? ?
4 0 3 0 2 0 1 0 0
B B B B B
? ? ? ? ?
? ? ? ?
? ?
R 2
i
º 45 sin º 90 sin
R 4
i
0 0
?
? ? ?
?
?
+
R 4
i
0
?
?
(sin45º + sin90º)
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? 1
2
1
R 4
i
R 2
i
2
1
1
R 4
i
0 0 0
?
?
?
?
?
?
? ? ? ? ? ?
?
?
? 1
2
1
2
2
1
1
R 4
i
0
? ? ? ?
?
?
? 2 2
R 4
i
0
?
?
?
?
?
?
? ?
?
?
?
2
1
R 2
i
0
PAGE # 2
2. Position of particle as a function of time is given as j
ˆ
t sin i
ˆ
t cos r ? ? ? ?
?
. Choose correct statement
about a and v , r
? ? ?
where v
?
and a
?
are velocity and acceleration of particle at time t.
(1) v
?
is perpendicular to r
?
and a
?
is towards origin
(2) a and v
? ?
are perpendicular to r
?
(3) v
?
is parallel to r
?
and a
?
parallel to r
?
.
(4) v
?
is perpendicular to r
?
and a
?
is away from origin.
fdlh d.k dh fLFkfr le; ds inksa esa j
ˆ
t sin i
ˆ
t cos r ? ? ? ?
?
ls nh tkrh gSA v , r
? ?
rFkk a
?
ds ckjs esa lgh dFku
pqfu;sA tgk v
?
rFkk a
?
, t le; ij d.k ds osx rFkk Roj.k gSaA
(1) v
?
, r
?
ds yEcor~ gS rFkk a
?
ewy fcUnq dh rjQ gSA
(2) v
?
rFkk a
?
, r
?
ds yEcor~ gSA
(3) v
?
, r
?
ds lekUrj gS rFkk a
?
, r
?
ds lekUrj gSA
(4) v
?
, r
?
yEcor~ gS rFkk a
?
dh fn’'kk ewy fcUnq ls nwj dh rjQ gSA
Ans. (1)
Sol. r
?
= j
ˆ
t sin i
ˆ
t cos ? ? ?
) j
ˆ
t cos i
ˆ
t sin (
dt
r d
v ? ? ? ? ? ? ?
?
?
) j
ˆ
t sin i
ˆ
t (cos
dt
v d
a
2
? ? ? ? ? ? ?
?
?
r a
2
? ?
? ? ? ? a
?
is antiparallel to r
?
a
?
, r
?
ds izfr lekUrj gSA
r . v
? ?
= ? (–sin ?t cos ?t + cos ?t sin ?t) = 0
so vr% r v
? ?
?
3. A Carnot engine, having an efficiency of ? = 1/10 as heat engine, is used as a refrigerator. If the work
done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is
,d dkukZsV bUtu ftldh n{krk ? = 1/10 Å"ek bUtu dh n{krk gS, jsfQztjsVj ds :i esa iz;ksx fd;k tkrk gSA ;fn
fudk; ij fd;k x;k dk;Z 10 twy gS, rc U;wu rkieku ij dq.M ls vo'kksf"kr Å"ek dh ek=kk gSA
(1) 99 J (2*) 90 J (3) 1 J (4) 100 J
Ans. 2
Sol. For Carnot engine using as refrigerator
dkuksZV btau dks jsfÝtjsVj dh Hkkfr iz;ksx djus ds fy,
1
2
2
T
W Q – 1
T
? ?
?
? ?
? ?
It is given fn;k x;k gSA ? =
1
10
? ? = 1 –
2
1
T
T
?
2
1
T 9
T 10
?
So, vr% Q 2 = 90 J (as W = 10 J)
PAGE # 3
4. Two uniformly charged solid spheres are such that E 1 is electric field at surface of 1
st
sphere due to
itself. E 2 is electric field at surface of 2
nd
sphere due to itself. r 1, r 2 are radius of 1
st
and 2
nd
sphere
respectively. If
2
1
2
1
r
r
E
E
? then ratio of potential at the surface of spheres 1
st
and 2
nd
due to their self
charges is :
nks le:i vkosf'kr Bksl xksyksa dh lrg ij muds Lo;a ds vkos'’k ds dkj.k fo|qr {kS=k Êdze’'k% E1 rFkk E2 gS] ;fn
2
1
2
1
r
r
E
E
? gS] rks muds Lo;a ds vkos'kksa ds dkj.k lrgksa ij fo|qr foHkoksa dk vuqikr gksxk A
(1)
2
1
r
r
(2)
2
2
1
r
r
?
?
?
?
?
?
?
?
(3)
1
2
r
r
(4)
2
1
2
r
r
?
?
?
?
?
?
?
?
Ans. (2)
Sol.
2
1
2
1
r
r
E
E
?
2
2
1
2
1
2
1
2 2
1 1
2
1
r
r
r
r
r
r
r E
r E
V
V
?
?
?
?
?
?
?
?
? ? ? ?
5. Output at terminal Y of given logic circuit.
fn;s x;s rkfdZd ifjiFk ds fy;s] fljs Y ij fuxZr eku gSA
A=1
B=0
Y
(1) 1
(2) 0
(3) Not determine
Kkr ugh dj ldrs
(4) Oscillating between 0 and 1
0 rFkk 1 ds chp esa nksyu djsxkA
Ans. (2)
Sol. A AB Y ? ?
= A AB ?
= A AB ?
= 0 + 0
= 0
Page 4
PAGE # 1
PART : PHYSICS
Single Choice Type (,dy fodYih; izdkj)
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. Find magnetic field at O.
O ij pqEcdh; {kS=k Kkr fdft;sA
45º
45º
O
i
i
i
A
B
D
N
M
(1)
?
?
?
?
?
?
? ?
?
?
?
2
1
R 2
i
0
(2) ? ? 1
R 2
i
0
? ?
?
?
(3)
R 2
i
0
?
(4)
?
?
?
?
?
?
? ?
?
?
2
1
R 2
i
0
Ans. (4)
Sol.
45º
45º
O
i
i
3
4 2
1
i
? ? ? ? ? ? ? ?
4 0 3 0 2 0 1 0 0
B B B B B
? ? ? ? ?
? ? ? ?
? ?
R 2
i
º 45 sin º 90 sin
R 4
i
0 0
?
? ? ?
?
?
+
R 4
i
0
?
?
(sin45º + sin90º)
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? 1
2
1
R 4
i
R 2
i
2
1
1
R 4
i
0 0 0
?
?
?
?
?
?
? ? ? ? ? ?
?
?
? 1
2
1
2
2
1
1
R 4
i
0
? ? ? ?
?
?
? 2 2
R 4
i
0
?
?
?
?
?
?
? ?
?
?
?
2
1
R 2
i
0
PAGE # 2
2. Position of particle as a function of time is given as j
ˆ
t sin i
ˆ
t cos r ? ? ? ?
?
. Choose correct statement
about a and v , r
? ? ?
where v
?
and a
?
are velocity and acceleration of particle at time t.
(1) v
?
is perpendicular to r
?
and a
?
is towards origin
(2) a and v
? ?
are perpendicular to r
?
(3) v
?
is parallel to r
?
and a
?
parallel to r
?
.
(4) v
?
is perpendicular to r
?
and a
?
is away from origin.
fdlh d.k dh fLFkfr le; ds inksa esa j
ˆ
t sin i
ˆ
t cos r ? ? ? ?
?
ls nh tkrh gSA v , r
? ?
rFkk a
?
ds ckjs esa lgh dFku
pqfu;sA tgk v
?
rFkk a
?
, t le; ij d.k ds osx rFkk Roj.k gSaA
(1) v
?
, r
?
ds yEcor~ gS rFkk a
?
ewy fcUnq dh rjQ gSA
(2) v
?
rFkk a
?
, r
?
ds yEcor~ gSA
(3) v
?
, r
?
ds lekUrj gS rFkk a
?
, r
?
ds lekUrj gSA
(4) v
?
, r
?
yEcor~ gS rFkk a
?
dh fn’'kk ewy fcUnq ls nwj dh rjQ gSA
Ans. (1)
Sol. r
?
= j
ˆ
t sin i
ˆ
t cos ? ? ?
) j
ˆ
t cos i
ˆ
t sin (
dt
r d
v ? ? ? ? ? ? ?
?
?
) j
ˆ
t sin i
ˆ
t (cos
dt
v d
a
2
? ? ? ? ? ? ?
?
?
r a
2
? ?
? ? ? ? a
?
is antiparallel to r
?
a
?
, r
?
ds izfr lekUrj gSA
r . v
? ?
= ? (–sin ?t cos ?t + cos ?t sin ?t) = 0
so vr% r v
? ?
?
3. A Carnot engine, having an efficiency of ? = 1/10 as heat engine, is used as a refrigerator. If the work
done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is
,d dkukZsV bUtu ftldh n{krk ? = 1/10 Å"ek bUtu dh n{krk gS, jsfQztjsVj ds :i esa iz;ksx fd;k tkrk gSA ;fn
fudk; ij fd;k x;k dk;Z 10 twy gS, rc U;wu rkieku ij dq.M ls vo'kksf"kr Å"ek dh ek=kk gSA
(1) 99 J (2*) 90 J (3) 1 J (4) 100 J
Ans. 2
Sol. For Carnot engine using as refrigerator
dkuksZV btau dks jsfÝtjsVj dh Hkkfr iz;ksx djus ds fy,
1
2
2
T
W Q – 1
T
? ?
?
? ?
? ?
It is given fn;k x;k gSA ? =
1
10
? ? = 1 –
2
1
T
T
?
2
1
T 9
T 10
?
So, vr% Q 2 = 90 J (as W = 10 J)
PAGE # 3
4. Two uniformly charged solid spheres are such that E 1 is electric field at surface of 1
st
sphere due to
itself. E 2 is electric field at surface of 2
nd
sphere due to itself. r 1, r 2 are radius of 1
st
and 2
nd
sphere
respectively. If
2
1
2
1
r
r
E
E
? then ratio of potential at the surface of spheres 1
st
and 2
nd
due to their self
charges is :
nks le:i vkosf'kr Bksl xksyksa dh lrg ij muds Lo;a ds vkos'’k ds dkj.k fo|qr {kS=k Êdze’'k% E1 rFkk E2 gS] ;fn
2
1
2
1
r
r
E
E
? gS] rks muds Lo;a ds vkos'kksa ds dkj.k lrgksa ij fo|qr foHkoksa dk vuqikr gksxk A
(1)
2
1
r
r
(2)
2
2
1
r
r
?
?
?
?
?
?
?
?
(3)
1
2
r
r
(4)
2
1
2
r
r
?
?
?
?
?
?
?
?
Ans. (2)
Sol.
2
1
2
1
r
r
E
E
?
2
2
1
2
1
2
1
2 2
1 1
2
1
r
r
r
r
r
r
r E
r E
V
V
?
?
?
?
?
?
?
?
? ? ? ?
5. Output at terminal Y of given logic circuit.
fn;s x;s rkfdZd ifjiFk ds fy;s] fljs Y ij fuxZr eku gSA
A=1
B=0
Y
(1) 1
(2) 0
(3) Not determine
Kkr ugh dj ldrs
(4) Oscillating between 0 and 1
0 rFkk 1 ds chp esa nksyu djsxkA
Ans. (2)
Sol. A AB Y ? ?
= A AB ?
= A AB ?
= 0 + 0
= 0
PAGE # 4
6. Velocity of a wave in a wire is v when tension in it is 2.06 × 10
4
N. Find value of tension in wire when
velocity of wave become
2
v
.
fdlh rkj esa xfreku rjax ds osx dk eku v gS] tc blesa ruko 2.06 × 10
4
N gSA ;fn rjax osx dk eku
2
v
gks
tk;s rks rkj esa ruko Kkr fdft;sA
(1) 5.15 × 10
3
N (2) 8.24 × 10
4
N (3) 6 × 10
4
(4) 5.15 × 10
4
N
Ans. (1)
Sol. v ? ? T
?
2
1
2
1
T
T
v
v
? ?
T
10 06 . 2
) 2 / v (
v
4
?
?
? T = N
4
10 06 . 2
4
?
= 0.515 × 10
4
N
7. n mole of He and 2n mole of O 2 is mixed in a container. Then
mix
V
P
C
C
?
?
?
?
?
?
?
?
will be
He ds n eksy rFkk O 2 ds 2n eksy ,d ik=k esa fefJr fd;s tkrs gSa rks
mix
V
P
C
C
?
?
?
?
?
?
?
?
Kkr djksA
(1)
13
19
(2)
27
40
(3)
3
1
(4)
4
1
Ans. (1)
Sol. ?mix =
2 1
2 1
v 2 v 1
p 2 p 1
c n c n
c n c n
?
?
=
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
R
2
5
n 2 R
2
3
n
R
2
7
n 2 R
2
5
n
=
10 3
14 5
?
?
=
13
19
8. A uniform solid sphere of radius R has a cavity of radius 1m cut from it if centre of mass of the system
lies at the periphery of the cavity then
,d le:i R f=kT;k ds Bksl xksys esa fp=kkuqlkj 1m f=kT;k fd xqfgdk mifLFkr gS bl fudk; dk nzO;eku dsUnz
xqfgdk dh ifjf/k ij mifLFkr gS rks %
1
R
COM
(1) (R
2
+ R + 1) (2–R) = 1 (2) (R
2
– R – 1) (2–R) = 1
(3) (R
2
– R + 1) (2–R) = 1 (4) (R
2
+ R – 1) (2–R) = 1
Ans. (1)
Page 5
PAGE # 1
PART : PHYSICS
Single Choice Type (,dy fodYih; izdkj)
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. Find magnetic field at O.
O ij pqEcdh; {kS=k Kkr fdft;sA
45º
45º
O
i
i
i
A
B
D
N
M
(1)
?
?
?
?
?
?
? ?
?
?
?
2
1
R 2
i
0
(2) ? ? 1
R 2
i
0
? ?
?
?
(3)
R 2
i
0
?
(4)
?
?
?
?
?
?
? ?
?
?
2
1
R 2
i
0
Ans. (4)
Sol.
45º
45º
O
i
i
3
4 2
1
i
? ? ? ? ? ? ? ?
4 0 3 0 2 0 1 0 0
B B B B B
? ? ? ? ?
? ? ? ?
? ?
R 2
i
º 45 sin º 90 sin
R 4
i
0 0
?
? ? ?
?
?
+
R 4
i
0
?
?
(sin45º + sin90º)
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? 1
2
1
R 4
i
R 2
i
2
1
1
R 4
i
0 0 0
?
?
?
?
?
?
? ? ? ? ? ?
?
?
? 1
2
1
2
2
1
1
R 4
i
0
? ? ? ?
?
?
? 2 2
R 4
i
0
?
?
?
?
?
?
? ?
?
?
?
2
1
R 2
i
0
PAGE # 2
2. Position of particle as a function of time is given as j
ˆ
t sin i
ˆ
t cos r ? ? ? ?
?
. Choose correct statement
about a and v , r
? ? ?
where v
?
and a
?
are velocity and acceleration of particle at time t.
(1) v
?
is perpendicular to r
?
and a
?
is towards origin
(2) a and v
? ?
are perpendicular to r
?
(3) v
?
is parallel to r
?
and a
?
parallel to r
?
.
(4) v
?
is perpendicular to r
?
and a
?
is away from origin.
fdlh d.k dh fLFkfr le; ds inksa esa j
ˆ
t sin i
ˆ
t cos r ? ? ? ?
?
ls nh tkrh gSA v , r
? ?
rFkk a
?
ds ckjs esa lgh dFku
pqfu;sA tgk v
?
rFkk a
?
, t le; ij d.k ds osx rFkk Roj.k gSaA
(1) v
?
, r
?
ds yEcor~ gS rFkk a
?
ewy fcUnq dh rjQ gSA
(2) v
?
rFkk a
?
, r
?
ds yEcor~ gSA
(3) v
?
, r
?
ds lekUrj gS rFkk a
?
, r
?
ds lekUrj gSA
(4) v
?
, r
?
yEcor~ gS rFkk a
?
dh fn’'kk ewy fcUnq ls nwj dh rjQ gSA
Ans. (1)
Sol. r
?
= j
ˆ
t sin i
ˆ
t cos ? ? ?
) j
ˆ
t cos i
ˆ
t sin (
dt
r d
v ? ? ? ? ? ? ?
?
?
) j
ˆ
t sin i
ˆ
t (cos
dt
v d
a
2
? ? ? ? ? ? ?
?
?
r a
2
? ?
? ? ? ? a
?
is antiparallel to r
?
a
?
, r
?
ds izfr lekUrj gSA
r . v
? ?
= ? (–sin ?t cos ?t + cos ?t sin ?t) = 0
so vr% r v
? ?
?
3. A Carnot engine, having an efficiency of ? = 1/10 as heat engine, is used as a refrigerator. If the work
done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is
,d dkukZsV bUtu ftldh n{krk ? = 1/10 Å"ek bUtu dh n{krk gS, jsfQztjsVj ds :i esa iz;ksx fd;k tkrk gSA ;fn
fudk; ij fd;k x;k dk;Z 10 twy gS, rc U;wu rkieku ij dq.M ls vo'kksf"kr Å"ek dh ek=kk gSA
(1) 99 J (2*) 90 J (3) 1 J (4) 100 J
Ans. 2
Sol. For Carnot engine using as refrigerator
dkuksZV btau dks jsfÝtjsVj dh Hkkfr iz;ksx djus ds fy,
1
2
2
T
W Q – 1
T
? ?
?
? ?
? ?
It is given fn;k x;k gSA ? =
1
10
? ? = 1 –
2
1
T
T
?
2
1
T 9
T 10
?
So, vr% Q 2 = 90 J (as W = 10 J)
PAGE # 3
4. Two uniformly charged solid spheres are such that E 1 is electric field at surface of 1
st
sphere due to
itself. E 2 is electric field at surface of 2
nd
sphere due to itself. r 1, r 2 are radius of 1
st
and 2
nd
sphere
respectively. If
2
1
2
1
r
r
E
E
? then ratio of potential at the surface of spheres 1
st
and 2
nd
due to their self
charges is :
nks le:i vkosf'kr Bksl xksyksa dh lrg ij muds Lo;a ds vkos'’k ds dkj.k fo|qr {kS=k Êdze’'k% E1 rFkk E2 gS] ;fn
2
1
2
1
r
r
E
E
? gS] rks muds Lo;a ds vkos'kksa ds dkj.k lrgksa ij fo|qr foHkoksa dk vuqikr gksxk A
(1)
2
1
r
r
(2)
2
2
1
r
r
?
?
?
?
?
?
?
?
(3)
1
2
r
r
(4)
2
1
2
r
r
?
?
?
?
?
?
?
?
Ans. (2)
Sol.
2
1
2
1
r
r
E
E
?
2
2
1
2
1
2
1
2 2
1 1
2
1
r
r
r
r
r
r
r E
r E
V
V
?
?
?
?
?
?
?
?
? ? ? ?
5. Output at terminal Y of given logic circuit.
fn;s x;s rkfdZd ifjiFk ds fy;s] fljs Y ij fuxZr eku gSA
A=1
B=0
Y
(1) 1
(2) 0
(3) Not determine
Kkr ugh dj ldrs
(4) Oscillating between 0 and 1
0 rFkk 1 ds chp esa nksyu djsxkA
Ans. (2)
Sol. A AB Y ? ?
= A AB ?
= A AB ?
= 0 + 0
= 0
PAGE # 4
6. Velocity of a wave in a wire is v when tension in it is 2.06 × 10
4
N. Find value of tension in wire when
velocity of wave become
2
v
.
fdlh rkj esa xfreku rjax ds osx dk eku v gS] tc blesa ruko 2.06 × 10
4
N gSA ;fn rjax osx dk eku
2
v
gks
tk;s rks rkj esa ruko Kkr fdft;sA
(1) 5.15 × 10
3
N (2) 8.24 × 10
4
N (3) 6 × 10
4
(4) 5.15 × 10
4
N
Ans. (1)
Sol. v ? ? T
?
2
1
2
1
T
T
v
v
? ?
T
10 06 . 2
) 2 / v (
v
4
?
?
? T = N
4
10 06 . 2
4
?
= 0.515 × 10
4
N
7. n mole of He and 2n mole of O 2 is mixed in a container. Then
mix
V
P
C
C
?
?
?
?
?
?
?
?
will be
He ds n eksy rFkk O 2 ds 2n eksy ,d ik=k esa fefJr fd;s tkrs gSa rks
mix
V
P
C
C
?
?
?
?
?
?
?
?
Kkr djksA
(1)
13
19
(2)
27
40
(3)
3
1
(4)
4
1
Ans. (1)
Sol. ?mix =
2 1
2 1
v 2 v 1
p 2 p 1
c n c n
c n c n
?
?
=
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
R
2
5
n 2 R
2
3
n
R
2
7
n 2 R
2
5
n
=
10 3
14 5
?
?
=
13
19
8. A uniform solid sphere of radius R has a cavity of radius 1m cut from it if centre of mass of the system
lies at the periphery of the cavity then
,d le:i R f=kT;k ds Bksl xksys esa fp=kkuqlkj 1m f=kT;k fd xqfgdk mifLFkr gS bl fudk; dk nzO;eku dsUnz
xqfgdk dh ifjf/k ij mifLFkr gS rks %
1
R
COM
(1) (R
2
+ R + 1) (2–R) = 1 (2) (R
2
– R – 1) (2–R) = 1
(3) (R
2
– R + 1) (2–R) = 1 (4) (R
2
+ R – 1) (2–R) = 1
Ans. (1)
PAGE # 5
Sol. ? ? ?
3
1
R
3
4
M
? ? ? ? ? ? ? – 1
3
4
M
3
2
2 1
2 2 1 1
com
M M
X M X M
X
?
?
?
? ?
? ? ? ? ? ?
? ? ? ?
? ? R – 2 –
– 1
3
4
R
3
4
1 – R – 1
3
4
0 R
3
4
3 3
3 3
?
? ? ? ? ?
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
? ?
?
? ?
? ?
? ? R – 2
1 – R
1 – R
3
? (R ? 1)
? ?
? ? ? ?
R – 2
1 R R 1 – R
1 – R
2
?
? ?
(R
2
+ R + 1) (2 – R) = 1
Alternative:
Mremaining (2 – R) = Mcavity (1 – R)
? (R
3
– 1
3
) (2 – R) = 1
3
[R – 1]
? (R
2
+ R + 1) (2 – R) = 1
9. A solid sphere of mass m= 500gm is rolling without slipping on a horizontal surface. Find kinetic energy
of a sphere if velocity of centre of mass is 5 cm/sec.
m= 500gm dk ,d Bksl xksyk {ksfrt lrg ij fcuk ?k"kZ.k ds yksVuh xfr dj jgk gSA ;fn nzO;eku dsUnz dk osx 5
cm/sec gS] rks xksys dh xfrt ÅtkZ Kkr fdft;sA
(1) J 10
2
35
4 –
? (2) J 10
4
35
4 –
? (3) 35×10
–4
J (4) 35×10
–3
J
Ans. (2)
Sol.
V=5cm/sec
?=V/R
K.E. of the sphere = Translational K.E + Rotational K.E.
xksys dh xfrt ÅtkZ = LFkkukarjf.k; K.E + ?kw.kZu K.E.
=
?
?
?
?
?
?
?
?
?
2
2
2
R
K
1 mv
2
1
K = Radius of gyration ?kw.kZu f=kT;k
?
?
?
?
?
?
? ?
?
?
?
?
?
? ?
5
2
1
100
5
2
1
2
1
2
4 –
10
4
35
? J
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