PART A — PHYSICS
1.
In terms of resistance R and time T, the 1.
m
of the permeability
e
m and permittivity e is :
dimensions of ratio
2.
Öæ» A — ÖæñçÌ·¤ çߙææÙ
ÂýçÌÚUæðÏ R ¥æñÚU â×Ø T ·ð¤ ÂÎæð´ ×ð´, ¿éÕ·¤àæèÜÌæ
°ß´ çßléÌàæèÜÌæ e ·ð¤ ¥ÙéÂæÌ
(1)
[RT22]
(1)
[RT22]
(2)
[R2 T21]
(2)
[R2 T21]
(3)
[R2]
(3)
[R2]
(4)
[R2 T2]
(4)
[R2 T2]
The initial speed of a bullet fired from a 2.
rifle is 630 m/s. The rifle is fired at the
centre of a target 700 m away at the same
level as the target. How far above the
centre of the target the rifle must be aimed
m
e
m
·¤è çß×æ ãñ Ñ
°·¤ ÚUæ§È¤Ü âð Îæ»è »§ü ÕéÜðÅU ·¤è ÂýæÚUçÖ·¤ ¿æÜ
630 m/s ãñÐ ÜÿØ ·ð¤ SÌÚU ÂÚU ÜÿØ âð 700 m ÎêÚU
ÜÿØ ·ð¤ ·ð¤‹Îý ÂÚU ÚUæ§È¤Ü Îæ»è ÁæÌè ãñÐ ÜÿØ ·¤æð
Îæ»Ùð ·ð¤ çÜØð ÚUæ§È¤Ü ·¤æ çÙàææÙæ ÜÿØ ·ð¤ ·ð¤‹Îý âð
ç·¤ÌÙæ ª¤ÂÚU Ü»æÙæ ¿æçã°?
in order to hit the target ?
(1)
1.0 m
(1)
1.0 m
(2)
4.2 m
(2)
4.2 m
(3)
6.1 m
(3)
6.1 m
(4)
9.8 m
(4)
9.8 m
English : 1
Set : 11
Hindi : 1
Set : 11
3.
A body of mass 5 kg under the action of 3.
∧
∧
r
constant force F5Fx i 1 Fy j has velocity
at t50s as
(
)
∧
∧
r
v 5 6 i 22 j m/s and at
∧
r
r
t510 s as v516 j m/s. The force F is :
çSÍÚU ÕÜ
∧
∧
r
F5Fx i 1 Fy j
·¤è °·¤ ßSÌé
5 kg
)
(
∧
∧
r
v 5 6 i 22 j m/s âð
ÂÚU ßð»
∧
r
v516 j m/s âð
t50s
(1)
(23 i 14 j ) N
(2)
 3 ∧
4 ∧
i1
j N
2
5 
 5
(2)
(3)
(3 i 2 4 j ) N
 3 ∧
4 ∧
i1
j N
2
5 
 5
(3)
(3 i 2 4 j ) N
(4)
3∧
4 ∧
 i 2 j N
5 
5
(4)
3∧
4 ∧
 i 2 j N
5 
5
∧
English : 2
∧
Set : 11
∧
∧
Hindi : 2
t510 s
»çÌàæèÜ ãñÐ ÕÜ
(23 i 14 j ) N
∧
ÂÚU ßð »
»çÌàæèÜ ãñ ¥æñÚU
(1)
∧
·ð¤ ·¤æÚU‡æ ÎýÃØ×æÙ
r
F
ãñ :
∧
∧
Set : 11
4.
A small ball of mass m starts at a point A 4.
with speed v o and moves along a
frictionless track AB as shown. The
track BC has coefficient of friction m. The
ball comes to stop at C after travelling a
ÎýÃØ×æÙ m ·¤è °·¤ ÀUæðÅUè »ð´Î çՋÎé A âð ¿æÜ vo âð
ÂýæÚUÖ ·¤ÚUÌè ãñ ¥æñÚU °·¤ ƒæáü‡æãèÙ ÂÍ AB ÂÚU
»çÌàæèÜ ãñ Áñâæ ç·¤ 翘æ ×ð´ ÎàææüØæ »Øæ ãñÐ ÂÍ BC
·¤æ ƒæáü‡æ »é‡ææ¡·¤ m ãñÐ »ð´Î C ÂÚU ÎêÚUè L ¿ÜÙð ·ð¤
Âà¿æÌ÷ L¤·¤ ÁæÌè ãñ Áãæ¡ L ãñ :
distance L which is :
(1)
(2)
(3)
(4)
v2
2h
1 o
2m g
m
(1)
v2
2h
1 o
m
2m g
v2
h
1 o
m
2m g
(2)
v2
h
1 o
m
2m g
v2
h
1 o
mg
2m
(3)
v2
h
1 o
2m
mg
v2
h
1 o
2m
2m g
(4)
v2
h
1 o
2m
2m g
English : 3
Set : 11
Hindi : 3
Set : 11
5.
of the surface at a place which receives 100
ßáæü ·¤è Õê´Îæð´ ·¤æ ¥æñâÌ ÎýÃØ×æÙ 3.031025 kg ãñ
¥æñÚU ©Ù·¤æ ¥æñâÌ âè×æ‹Ì ßð» 9 m/s ãñÐ çÁâ
SÍæÙ ÂÚU °·¤ ßáü ×ð´ 100 cm ßáæü ãæðÌè ãñ ©â SÍæÙ
·ð¤ ÂýçÌ ß»ü ×èÅUÚU ÂëcÆU ÂÚU ßáæü mæÚUæ SÍæÙæ‹ÌçÚUÌ ª¤Áæü
·¤è »‡æÙæ ·¤èçÁ°Ð
cm of rain in a year.
(1)
3.53105 J
(1)
3.53105 J
(2)
4.053104 J
(2)
4.053104 J
(3)
3.03105 J
(3)
3.03105 J
(4)
9.03104 J
(4)
9.03104 J
The average mass of rain drops is 5.
3.031025 kg and their average terminal
velocity is 9 m/s. Calculate the energy
transferred by rain to each square metre
6.
is M and its mass per unit length at the
ܐÕæ§ü L ·¤è °·¤ ÂÌÜè ÀUǸ ·¤æ ÂýçÌ §·¤æ§ü ܐÕæ§ü
ÎýÃØ×æÙ l ãñ Áæð ç·¤ °·¤ çâÚðU âð ÎêÚUè ·ð¤ ¥ÙéâæÚU
ÚñUç¹·¤ÌÑ ÕɸÌæ ãñÐ ØçÎ §â·¤æ ·é¤Ü ÎýÃØ×æÙ M ãñ
¥æñÚU ãË·ð¤ çâÚðU ÂÚU ÂýçÌ §·¤æ§ü ܐÕæ§ü ÎýÃØ×æÙ lo ãñ,
ÌÕ ãË·ð¤ çâÚðU âð ÎýÃØ×æÙ ·ð¤‹Îý ·¤è ÎêÚUè ãñ Ñ
lighter end is lo, then the distance of the
centre of mass from the lighter end is :
(1)
l L2
L
2 o
2
4M
6.
A thin bar of length L has a mass per unit
length l, that increases linearly with
distance from one end. If its total mass
(1)
L
l L2
2 o
2
4M
(2)
l L2
L
1 o
3
8M
(2)
l L2
L
1 o
3
8M
(3)
l L2
L
1 o
3
4M
(3)
l L2
L
1 o
3
4M
(4)
l L2
2L
2 o
3
6M
(4)
l L2
2L
2 o
3
6M
English : 4
Set : 11
Hindi : 4
Set : 11
7.
From a sphere of mass M and radius R, a 7.
ç˜æ’Øæ
smaller sphere of radius R 2 is carved out
such that the cavity made in the original
ç˜æ’Øæ R 2 ·¤æ °·¤ ÀUæðÅUæ »æðÜæ §â Âý·¤æÚU çÙ·¤æÜ
çÜØæ ÁæÌæ ãñ ç·¤ ×êÜ »æðÜð ×ð´ ÕÙè »éãæ §â·ð¤ ·ð¤‹Îý
°ß´ ÂçÚUçÏ ·ð¤ Õè¿ ãñ (翘æ Îð¹ð´)Ð ç¿˜æ ·ð¤ çߋØæâ
·ð¤ ¥ÙéâæÚU ÁÕ ×êÜ »æðÜð ·ð¤ ·ð¤‹Îý ¥æñÚU ãÅUæØð »Øð
»æðÜð ·ð¤ ·ð¤‹Îý ·ð¤ Õè¿ ÎêÚUè 3R ãñ, ÌÕ ÎæðÙæð »æðÜæð´ ·ð¤
Õè¿ »éL¤ˆßæ·¤áü‡æ ÕÜ ãñ Ñ
sphere is between its centre and the
periphery. (See figure).
For the
configuration in the figure where the
distance between the centre of the original
R
°ß´ Îý à Ø×æÙ
M
·ð ¤ °·¤ »æð Ü ð âð ,
sphere and the removed sphere is 3R, the
gravitational force between the two
spheres is :
(1)
(1)
(2)
(3)
(4)
41 GM 2
(2)
3600 R 2
41 GM 2
(3)
450 R 2
59 GM 2
450 R
(4)
2
41 GM 2
3600 R 2
41 GM 2
450 R 2
59 GM 2
450 R 2
GM 2
225 R 2
GM 2
225 R 2
English : 5
Set : 11
Hindi : 5
Set : 11
8.
The Bulk moduli of Ethanol, Mercury and 8.
water are given as 0.9, 25 and 2.2
respectively in units of 109 Nm22. For a
given value of pressure, the fractional
9
. Which of
compression in volume is
V
the following statements about
9
for
V
°ÍÙæòÜ, ÂæÚUæ °ß´ ÂæÙè ·ð¤ ¥æØÌÙ ÂýˆØæSÍÌæ »é‡ææ¡·¤
109 Nm22 ·¤è §·¤æ§ü ×ð´ ·ý¤×àæÑ 0.9, 25 °ß´ 2.2
çÎØð ãéØð ãñ´Ð ÎæÕ ·ð¤ çÎØð ×æÙ ·ð¤ çÜ°, ¥æØÌÙ ×ð´
ç֋Ùæˆ×·¤ â´ÂèǸÙ
9
V
9
V
ãñÐ §Ù ÌèÙ Îýßæð´ ·ð¤ çÜ°
·ð¤ ÕæÚðU ×ð´ çِÙçÜç¹Ì ·¤ÍÙæð´ ×ð´ âð ·¤æñÙ âæ
âãè ãñ?
these three liquids is correct ?
9.
(1)
Ethanol > Water > Mercury
(1)
(2)
Water > Ethanol > Mercury
(2)
(3)
Mercury > Ethanol > Water
(3)
(4)
Ethanol > Mercury > Water
(4)
A tank with a small hole at the bottom has 9.
been filled with water and kerosene
(specific gravity 0.8). The height of water
is 3 m and that of kerosene 2 m. When the
hole is opened the velocity of fluid coming
out from it is nearly : (take g510 ms22
and density of water 5103 kg m23)
°ÍÙæòÜ > ÂæÙè > ÂæÚUæ
ÂæÙè > °ÍÙæòÜ > ÂæÚUæ
ÂæÚUæ > °ÍÙæòÜ > ÂæÙè
°ÍÙæòÜ > ÂæÚUæ > ÂæÙè
ÌÜè ×ð´ °·¤ ÀUæðÅðU çÀUÎý ßæÜð Åñ´U·¤ ·¤æð ÂæÙè °ß´ ç×^è ·ð¤
ÌðÜ (¥æÂðçÿæÌ ƒæÙˆß 0.8) âð ÖÚUæ »Øæ ãñÐ ÂæÙè ·¤è
ª¡¤¿æ§ü 3 m ãñ ¥æñÚU ç×^è ·ð¤ ÌðÜ ·¤è 2 mÐ ÁÕ çÀUÎý
·¤æð ¹æðÜ çÎØæ ÁæÌæ ãñ, ÌÕ çÙ·¤ÜÙð ßæÜð Îýß ·¤è
¿æÜ Ü»Ö» ãæð » è : (g510 ms 22 Üð ¥æñ Ú U
ÂæÙè ·¤æ ƒæÙˆß 5103 kg m23)
(1)
10.7 ms21
(1)
10.7 ms21
(2)
9.6 ms21
(2)
9.6 ms21
(3)
8.5 ms21
(3)
8.5 ms21
(4)
7.6 ms21
(4)
7.6 ms21
English : 6
Set : 11
Hindi : 6
Set : 11
10.
An air bubble of radius 0.1 cm is in a liquid 10.
having surface tension 0.06 N/m and
density 103 kg/m3. The pressure inside
the bubble is 1100 Nm22 greater than the
atmospheric pressure. At what depth is
ÂëcÆU ÌÙæß 0.06 N/m ¥æñÚU ƒæÙˆß 103 kg/m3
ßæÜð °·¤ Îýß ×ð´ ç˜æ’Øæ 0.1 cm ·¤æ °·¤ ßæØé ·¤æ
ÕéÜÕéÜæ ãñÐ ÕéÜÕéÜð ·ð¤ ¥‹ÎÚU ÎæÕ ßæØé×´ÇUÜèØ ÎæÕ
âð 1100 Nm22 ¥çÏ·¤ ãñÐ Îýß ·ð¤ ÂëcÆU âð ç·¤â
»ãÚUæ§ü ÂÚU ÕéÜÕéÜæ ãñ? (g59.8 ms22)
the bubble below the surface of the
liquid ? (g59.8 ms22)
11.
(1)
0.1 m
(1)
0.1 m
(2)
0.15 m
(2)
0.15 m
(3)
0.20 m
(3)
0.20 m
(4)
0.25 m
(4)
0.25 m
A hot body, obeying Newton’s law of 11.
cooling is cooling down from its peak value
808C to an ambient temperature of 308C.
It takes 5 minutes in cooling down from
808C to 408C. How much time will it take
to cool down from 628C to 328C ?
(Given ln 250.693, ln 551.609)
(1)
3.75 minutes
(2)
8.6 minutes
(3)
9.6 minutes
(4)
6.5 minutes
English : 7
Set : 11
‹ØêÅUÙ ·ð¤ àæèÌÜÙ çÙØ× ·¤æ ÂæÜÙ ·¤ÚUÌè ãé§ü °·¤ »×ü
ßSÌé ¥ÂÙð àæèáü ÌæÂ×æÙ 808C âð ÂçÚUßðàæ ÌæÂ×æÙ
308C Ì·¤ Æ´UÇUè ãæðÌè ãñÐ Øã 808C âð 408C Ì·¤
Æ´UÇUæ ãæðÙð ×ð´ 5 ç×ÙÅU ÜðÌè ãñÐ Øã 628C âð 328C
Ì·¤ Æ´UÇæ ãæðÙð ×ð´ ç·¤ÌÙæ â×Ø Üð»è?
(çÎØæ ãñ ln 250.693, ln 551.609)
(1) 3.75 ç×ÙÅU
(2) 8.6 ç×ÙÅU
(3) 9.6 ç×ÙÅU
(4) 6.5 ç×ÙÅU
Hindi : 7
Set : 11
12.
During an adiabatic compression, 830 J of 12.
work is done on 2 moles of a diatomic ideal
gas to reduce its volume by 50%. The
change in its temperature is nearly :
(R58.3 JK21 mol21)
°·¤ L¤hæððc× â´ÂèǸ٠·ð¤ ÎæñÚUæÙ, °·¤ çmÂÚU×æ‡æé·¤ ¥æÎàæü
»ñâ ·ð¤ 2 ×æðÜ ·¤æ ¥æØÌÙ 50% ·¤× ç·¤Øð ÁæÙð ×ð´
830 J ·¤æ ·¤æØü ·¤ÚUÙæ ÂǸÌæ ãñÐ §â·ð¤ ÌæÂ×æÙ ×ð´
ÂçÚUßÌüÙ ãñ ֻܻ Ñ (R58.3 JK21 mol21)
(1)
40 K
(1)
40 K
(2)
33 K
(2)
33 K
(3)
20 K
(3)
20 K
(4)
14 K
(4)
14 K
English : 8
Set : 11
Hindi : 8
Set : 11
13.
An ideal monoatomic gas is confined in a 13.
cylinder by a spring loaded piston of cross
section 8.031023 m2. Initially the gas is at
300K
and
2.4310 23
occupies
m3
a
volume
of
and the spring is in its
relaxed state as shown in figure. The gas is
heated by a small heater until the piston
moves out slowly by 0.1 m. The force
constant of the spring is 8000 N/m and the
atmospheric pressure is 1.03105 N/m2.
The cylinder and the piston are thermally
insulated. The piston and the spring are
massless and there is no friction between
the piston and the cylinder. The final
temperature of the gas will be :
°·¤ ÕðÜÙ ×ð´ ¥ÙéÂýSÍ ·¤æÅU 8.031023 m2 ·ð¤ °·¤
·¤×æÙèÎæÚU ÖæçÚUÌ çÂSÅUÙ mæÚUæ °·¤ ¥æÎàæü °·¤ÂÚU×æ‡æé·¤
»ñâ ·¤æð ÚU¹æ »Øæ ãñÐ ÂýæÚUÖ ×ð´ »ñâ 300 K ÂÚU ãñ ¥æñÚU
2.431023 m3 ¥æØÌÙ ÚU¹Ìè ãñ´ ¥æñÚU ·¤×æÙè ¥ÂÙè
çßoýæ´çÌ ¥ßSÍæ ×ð´ ãñ Áñâæ ç·¤ 翘æ ×ð´ ÎàææüØæ »Øæ ãñÐ
»ñâ ·¤æð °·¤ ÀUæðÅðU ãèÅUÚU mæÚUæ ÌÕ Ì·¤ »ÚU×U ç·¤Øæ ÁæÌæ
ãñ ÁÕ Ì·¤ ç·¤ çÂSÅUÙ ÏèÚðU âð 0.1 m ·¤è »çÌ Ù ·¤ÚU
ÜðÐ ·¤×æÙè ·¤æ ÕÜ çÙØÌæ¡·¤ 8000 N/m ãñ ¥æñÚU
ßæØé×´ÇUÜèØ ÎæÕ 1.03105 N/m2 ãñÐ ÕðÜÙ °ß´
çÂSÅUÙ ª¤c×æÚUæÏð è ãñ́Ð çÂSÅUÙ °ß´ ·¤×æÙè ÎýÃØ×æÙçßãèÙ
ãñ ¥æñÚU çÂSÅUÙ °ß´ ÕðÜÙ ·ð¤ Õè¿ ·¤æð§ü ƒæáü‡æ Ùãè´ ãñÐ
»ñâ ·¤æ ¥ç‹Ì× ÌæÂ×æÙ ãæð»æ Ñ
(ãèÅUÚU ·ð¤ ÜèÇU ÌæÚUæð´ â𠪤Áæü ·¤è ãæçÙ Ù»‡Ø ×æÙð
¥æñÚU ãèÅUÚU ·é¤‡ÇUÜè ·¤è ª¤c×æÏæçÚUÌæ Öè Ù»‡Ø ãñ) :
(Neglect the heat loss through the lead
wires of the heater. The heat capacity of
the heater coil is also negligible)
(1)
300 K
(2)
800 K
(3)
500 K
(4)
1000 K
English : 9
Set : 11
(1)
300 K
(2)
800 K
(3)
500 K
(4)
1000 K
Hindi : 9
Set : 11
14.
oscillator is given by,
°·¤ ¥ß×ç‹ÎÌ ÎæðÜ·¤ ·¤è ·¤æð‡æèØ ¥æßëçžæ §ââð Îè
ÁæÌè ãñ ,
k
r2 
v5  2
 where k is the spring
4m 2 
m
k
r2 
v5  2
 Áãæ¡ k
4m 2 
m
constant, m is the mass of the oscillator and
m ÎæðÜ·¤ ·¤æ ÎýÃØ×æÙ ãñ ¥æñÚU r ¥ß׋ÎÙ çSÍÚUæ¡·¤ ãñÐ
r is the damping constant. If the ratio
ØçÎ ¥ÙéÂæÌ
The angular frequency of the damped 14.
(1)
increases by 1%
(1)
(2)
increases by 8%
(2)
(3)
decreases by 1%
(3)
(4)
decreases by 8%
(4)
Two factories are sounding their sirens 15.
at 800 Hz. A man goes from one factory
to other at a speed of 2 m/s. The velocity
of sound is 320 m/s. The number of beats
heard by the person in one second will
ãñ, ÌÕ ¥Ùß×ç‹ÎÌ ÎæðÜ·¤
·ð¤ ×é·¤æÕÜð ¥æßÌü ·¤æÜ ×ð´ ÂçÚUßÌü٠ֻܻ ãæð»æ Ñ
r2
is 8%, the change in time period
mk
compared to the undamped oscillator is
approximately as follows :
15.
r2
5 8%
mk
·¤×æÙè çSÍÚUæ¡·¤ ãñ,
1%âð
ßëçh ãæð»è
8%âð ßëçh ãæð»è
1%âð ƒæÅðU»æ
8%âð ƒæÅðU»æ
Îæð Èñ¤€ÅUçÚUØæ¡ ¥ÂÙð âæØÚUÙ 800 Hz ÂÚU ŠßçÙÌ ·¤ÚUÌè
ãñ´Ð °·¤ ÃØç€Ì 2 m/s ·¤è ¿æÜ âð °·¤ Èñ¤€ÅUÚUè âð
Îê â ÚU è Èñ ¤ €ÅÚè Ì·¤ ÁæÌæ ãñ Ð ŠßçÙ ·¤æ ßð »
320 m/s ãñÐ °·¤ âð·´¤ÇU ×ð´ ÃØç€Ì mæÚUæ âéÙè »§ü
çßS‹Îæð´ ·¤è ⴁØæ ãñ Ñ
be :
(1)
2
(1)
2
(2)
4
(2)
4
(3)
8
(3)
8
(4)
10
(4)
10
English : 10
Set : 11
Hindi : 10
Set : 11
16.
A cone of base radius R and height h is 16.
→
located in a uniform electric field E
parallel to its base. The electric flux
¥æÏæÚU ç˜æ•Øæ R °ß´ ª¡¤¿æ§ü h ßæÜæ °·¤ àæ´·é¤ ¥æÏæÚU
→
·ð¤ â×æ‹ÌÚU °·¤â×æÙ çßléÌ ÿæð˜æ E ×ð´ çSÍÌ ãñÐ
àæ´·é¤ ×ð´ Âýßðàæ ·¤ÚUÙð ßæÜæ çßléÌ Ü€â ãñ :
entering the cone is :
(1)
1
EhR
2
(1)
1
EhR
2
(2)
EhR
(2)
EhR
(3)
2EhR
(3)
2EhR
(4)
4EhR
(4)
4EhR
English : 11
Set : 11
Hindi : 11
Set : 11
17.
A parallel plate capacitor is made of two 17.
plates of length l, width w and separated
by distance d. A dielectric slab (dielectric
constant K) that fits exactly between the
plates is held near the edge of the plates.
It is pulled into the capacitor by a force
∂U
F5 2
where U is the energy of the
∂x
capacitor when dielectric is inside the
capacitor up to distance x (See figure). If
the charge on the capacitor is Q then the
force on the dielectric when it is near the
edge is :
°·¤ â×æ‹ÌÚU ŒÜðÅU â´ÏæçÚU˜æ Îæð ŒÜðÅUæð´ âð ÕÙæ ãñ çÁÙ·¤è
ܐÕæ§ü l, ¿æñÇUæ§ü w ãñ´ ¥æñÚU °·¤ ÎéâÚðU âð d ÎêÚUè ÂÚU
ãñÐ °·¤ ÂÚUæßñléÌ Â^è (ÂÚUæßñléÌæ´·¤ K) Áæð ç·¤ ŒÜðÅUæð´
·ð¤ Õè¿ ÆUè·¤ âð â×æ ÁæÌè ãñ, ·¤æð ŒÜðÅUæð´ ·¤è çâÚðU ·ð¤
Âæâ ·¤Ç¸ ·¤ÚU ÚU¹æ ãé¥æ ãñÐ §âð â´ÏæçÚU˜æ ·ð¤ ¥‹ÎÚU
ÕÜ
F5 2
(2)
(3)
(4)
Q2 d
2 wl 2 eo
Q2 w
2 dl 2 eo
Q2 d
2 wl 2 eo
Q2 w
2 dl 2 eo
English : 12
K
(2)
( K21)
(3)
( K21)
mæÚUæ ¹è´¿æ ÁæÌæ ãñ Áãæ¡ U â´ÏæçÚU˜æ
·¤è ÌÕ ª¤Áæü ãñ ÁÕ ÂÚUæßñléÌ â´ÏæçÚU˜æ ·ð¤ ¥‹ÎÚU x ÎêÚUè
ÂÚU ãñÐ (翘æ Îð¹ð´)Ð ØçÎ â´ÏæçÚU˜æ ÂÚU ¥æßðàæ Q ãñ,
ÌÕ ÂÚUæßñléÌ ÂÚU ÕÜ, ÁÕ ßã çâÚðU ·ð¤ Âæâ ãñ, ãæð»æ Ñ
(1)
(1)
∂U
∂x
(4)
Q2 d
2 wl 2 eo
Q2 w
2 dl 2 eo
Q2 d
2 wl 2 eo
Q2 w
2 dl 2 eo
K
( K21)
( K21)
K
K
Set : 11
Hindi : 12
Set : 11
18.
In the circuit shown, current (in A) 18.
through the 50 V and 30 V batteries are,
ÎàææüØð »Øð ÂçÚUÂÍ ×ð´, 50 V °ß´
ÏæÚUæ (A ×ð´) ·ý¤×àæÑ ãñ´ Ñ
30 V
ÕñÅUçÚUØæð´ ×ð´
respectively :
19.
(1)
2.5 and 3
(2)
3.5 and 2
(3)
4.5 and 1
(4)
3 and 2.5
Three straight parallel current carrying 19.
conductors are shown in the figure. The
force experienced by the middle conductor
°ß´ 3
(2) 3.5 °ß´ 2
(3) 4.5 °ß´ 1
(4) 3 °ß´ 2.5
ÌèÙ âèÏð â×æ‹ÌÚU ÏæÚUæ ÂýßæçãÌ ¿æÜ·¤ 翘æ ×ð´ ÎàææüØð
»Øð ãñ´Ð ܐÕæ§ü 25 cm ·ð¤ Õè¿ ßæÜð ¿æÜ·¤ mæÚUæ
¥ÙéÖß ç·¤Øæ »Øæ ÕÜ ãñ :
(1)
2.5
Îæ¡Øè ¥æðÚU
631024 N Õæ¡Øè ¥æðÚU
931024 N Õæ¡Øè ¥æðÚU
àæê‹ØU
of length 25 cm is :
(1)
331024 N toward right
(1)
(2)
631024 N toward left
(2)
(3)
931024 N toward left
(3)
(4)
Zero
(4)
English : 13
Set : 11
331024 N
Hindi : 13
Set : 11
20.
Three identical bars A, B and C are made 20.
of different magnetic materials. When
kept in a uniform magnetic field, the field
lines around them look as follows :
ÌèÙ âßüâ×M¤Âè ÀUǸð A, B °ß´ C ÌèÙ çßç֋Ù
¿éÕ·¤èØ ÂÎæÍæðZ âð ÕÙè ãñ´Ð ÁÕ §‹ãð´ °·¤ °·¤â×æÙ
¿éÕ·¤èØ ÿæð˜æ ×ð´ ÚU¹æ ÁæÌæ ãñ, ÌÕ §Ù ÂÚU ÿæð˜æ ÚðU¹æ°¡
çِ٠Âý·¤æÚU âð çιÌè ãñ´ Ñ
§Ù ÀUÇæ¸ ´ð ·ð¤ ÂÎæÍæðZ ·¤æð ÂýçÌ¿éÕ·¤èØ (D), Üæðã ¿éÕ·¤èØ
(F) °ß´ ¥Ùé¿éÕ·¤èØ (P) ¥æÏæÚU ÂÚU â´»Ì ·¤ÚðU´ :
Make the correspondence of these bars
with their material being diamagnetic (D),
ferromagnetic (F) and paramagnetic (P) :
21.
(1)
A ↔ D, B ↔ P, C ↔ F
(1)
A ↔ D, B ↔ P, C ↔ F
(2)
A ↔ F, B ↔ D, C ↔ P
(2)
A ↔ F, B ↔ D, C ↔ P
(3)
A ↔ P, B ↔ F, C ↔ D
(3)
A ↔ P, B ↔ F, C ↔ D
(4)
A ↔ F, B ↔ P, C ↔ D
(4)
A ↔ F, B ↔ P, C ↔ D
0.01 s. The e.m.f. induced in the coil is :
Èð¤ÚðU °ß´ 4 cm2 Ȥܷ¤ ÿæð˜æÈ¤Ü ßæÜè °·¤
ßëžæèØ ¥ÙéÂýSÍ ·¤æÅU ·¤è ·é´¤ÇUÜè ·¤æð §â·ð¤ ¥ÿæ ·ð¤
â×æ‹ÌÚU °·¤ ¿éÕ·¤èØ ÿæð˜æ ×ð´ ÚU¹æ »Øæ ãñ Áæð ç·¤
1022 Wb m22 0.01 s ×𴠃æÅU ÁæÌæ ãñÐ ·é´¤‡ÇUÜè
×ð´ ÂýðçÚUÌ çßléÌ ßæã·¤ ÕÜ ãñ Ñ
(1)
400 mV
(1)
400 mV
(2)
200 mV
(2)
200 mV
(3)
4 mV
(3)
4 mV
(4)
0.4 mV
(4)
0.4 mV
A coil of circular cross-section having 21.
1000 turns and 4 cm2 face area is placed
with its axis parallel to a magnetic field
which decreases by 10 22 Wb m 22 in
English : 14
Set : 11
1000
Hindi : 14
Set : 11
22.
An electromagnetic wave of frequency 22.
131014 hertz is propagating along z - axis.
The amplitude of electric field is 4 V/m. If
e o58.8310 212 C 2/N-m 2, then average
energy density of electric field will
be :
23.
¥æßëçžæ 131014 ãÅüUÁ ·¤è °·¤ çßléÌ ¿éÕ·¤èØ ÌÚ´U»
z - ¥ÿæ ÂÚU â´¿ÚU‡æ ·¤ÚU ÚUãè ãñÐ çßléÌ ÿæð˜æ ·¤æ
¥æØæ× 4 V/m ãñÐ ØçÎ
eo58.8310212 C2/N-m2, ÌÕ çßléÌ ÿæð˜æ ·¤æ
¥æñâÌ ª¤Áæü ƒæÙˆß ãæð»æ :
(1)
35.2310210 J/m3
(1)
35.2310210 J/m3
(2)
35.2310211 J/m3
(2)
35.2310211 J/m3
(3)
35.2310212 J/m3
(3)
35.2310212 J/m3
(4)
35.2310213 J/m3
(4)
35.2310213 J/m3
An object is located in a fixed position 23.
in front of a screen. Sharp image is
obtained on the screen for two positions
of a thin lens separated by 10 cm. The size
of the images in two situations are in the
ratio 3 : 2. What is the distance between
°·¤ ÂÎðü ·ð¤ âæ×Ùð °·¤ çSÍÚU çSÍçÌ ×ð´ °·¤ ßSÌé çSÍÌ
ãñÐ °·¤ ÂÌÜð Üð‹â ·¤è 10 cm ÎêÚUè ÂÚU Îæð çSÍçÌØæð´
âð ÂÎðü ÂÚU SÂcÅU ÂýçÌçÕÕ ÕÙÌð ãñ´Ð ÎæðÙæð´ çSÍçÌØæð´ ×ð´
ÂýçÌçՐÕæ𴠷𤠥淤æÚU ·¤æ ¥ÙéÂæÌ 3 : 2 ãñÐ ßSÌé °ß´
ÂÎðü ·ð¤ Õè¿ ÎêÚUè €Øæ ãñ?
the screen and the object ?
(1)
124.5 cm
(1)
124.5 cm
(2)
144.5 cm
(2)
144.5 cm
(3)
65.0 cm
(3)
65.0 cm
(4)
99.0 cm
(4)
99.0 cm
English : 15
Set : 11
Hindi : 15
Set : 11
24.
Two monochromatic light beams of 24.
intensity 16 and 9 units are interfering.
The ratio of intensities of bright and dark
parts of the resultant pattern is :
25.
ÌèßýÌæ 16 °ß´ 9 §·¤æ§ü ßæÜè Îæð °·¤ß‡æèü Âý·¤æàæ Âé´Áæð´
·ð¤ Õè¿ ÃØçÌ·¤ÚU‡æ ãæð ÚUãæ ãñÐ ÂçÚU‡ææ×è ç¿˜æ ·ð¤
©ÁÜð ¥æñÚU ·¤æÜð çãSâæð´ ·¤è ÌèßýÌæ¥æð´ ·¤æ ¥ÙéÂæÌ
ãæð»æ Ñ
(1)
16
9
(1)
16
9
(2)
4
3
(2)
4
3
(3)
7
1
(3)
7
1
(4)
49
1
(4)
49
1
of the compound microscope should be :
°·¤ â´Øé€Ì âêÿ×Îàæèü ×ð´ ¥çÖÎëàØ·¤ Üð‹â ·¤è Ȥæð·¤â
ܐÕæ§ü 1.2 cm ¥æñÚU Ùðç˜æ·¤æ ·¤è Ȥæð·¤â ܐÕæ§ü
3.0 cm ãñ´Ð ÁÕ ßSÌé ·¤æð ¥çÖÎëàØ·¤ ·ð¤ âæ×Ùð
1.25 cm ·¤è ÎêÚUè ÂÚU ÚU¹æ ÁæÌæ ãñ, ÌÕ ¥ç‹Ì×
ÂýçÌçÕÕ ¥Ù‹Ì ÂÚU ÕÙÌæ ãñÐ â´Øé€Ì âêÿ×Îàæèü ·¤è
¥æßÏüÙ àæç€Ì ãæðÙè ¿æçã° Ñ
(1)
200
(1)
200
(2)
100
(2)
100
(3)
400
(3)
400
(4)
150
(4)
150
In a compound microscope the focal length 25.
of objective lens is 1.2 cm and focal length
of eye piece is 3.0 cm. When object is kept
at 1.25 cm in front of objective, final image
is formed at infinity. Magnifying power
English : 16
Set : 11
Hindi : 16
Set : 11
26.
A photon of wavelength l is scattered 26.
from an electron, which was at rest. The
wavelength shift Dl is three times of l and
the angle of scattering u is 608. The angle
at which the electron recoiled is f. The
value of tan f is : (electron speed is much
çßoýæ× ¥ßSÍæ ·ð¤ °·¤ §Üð€ÅþUæÙ âð ÌÚ´U»ÎñƒØü l ·¤æ
°·¤ ȤæðÅUæÙ Âý·¤èç‡æüÌ ãæðÌæ ãñÐ ÌÚ´U»ÎñƒØü SÍæÙæ‹ÌÚU
Dl ÌÚ´U»ÎñƒØü l ·¤æ ÌèÙ »éÙæ ãñ ¥æñÚU Âý·¤è‡æüÙ ·¤æð‡æ
u5608 ãñÐ §Üð€ÅþUæÙ f ·¤æð‡æ ÂÚU ÂýçÌçÿæŒÌ ãæðÌæ ãñÐ
tan f ·¤æ ×æÙ ãñ Ñ (§Üð€ÅþUæÙ ·¤è ¿æÜ Âý·¤æàæ ·¤è
¿æÜ âð ·¤æÈ¤è ·¤× ãñ)
smaller than the speed of light)
27.
(1)
0.16
(1)
0.16
(2)
0.22
(2)
0.22
(3)
0.25
(3)
0.25
(4)
0.28
(4)
0.28
A radioactive nuclei with decay constant 27.
100 ÙæçÖ·¤ ÂýçÌ âñç·¤‡ÇU ·¤è çSÍÚU ÎÚU âð ÿæØçSÍÚUæ¡·¤
0.5/s is being produced at a constant rate
0.5/s
of 100 nuclei/s. If at t50 there were no
nuclei, the time when there are 50 nuclei
ßæÜð ÚðUçÇUØæðâç·ý¤Ø ÙæçÖ·¤ ©ˆÂ‹Ù ãæð ÚUãð ãñ´Ð
ØçÎ t50 ÂÚU °·¤ Öè Ùæç×·¤ ©ÂçSÍÌ Ùãè´ Íæ, ÌÕ
50 ÙæçÖ·¤ ©ˆÂ‹Ù ãæðÙð ×ð´ Ü»æ â×Ø ãñ Ñ
is :
(1)
1s
(1)
1s
(2)
4
2 ln   s
3
(2)
4
2 ln   s
3
(3)
ln 2 s
(3)
ln 2 s
(4)
4
ln   s
3
(4)
4
ln   s
3
English : 17
Set : 11
Hindi : 17
Set : 11
28.
The currents I, IZ and IL are respectively
°·¤ ÁðÙÚU ÇUæØæðÇU ·¤æð °·¤ ÕñÅUÚUè °ß´ °·¤ ÜæðÇU âð
ÁæðÇ¸æ »Øæ ãñ Áñâæ ç·¤ ÂçÚUÂÍ ×ð´ ÎàææüØæ »Øæ ãñÐ ÏæÚUæØð´
I, IZ °ß´ IL ·ý¤×àæÑ ãñ´ Ñ
(1)
15 mA, 5 mA, 10 mA
(1)
15 mA, 5 mA, 10 mA
(2)
15 mA, 7.5 mA, 7.5 mA
(2)
15 mA, 7.5 mA, 7.5 mA
(3)
12.5 mA, 5 mA, 7.5 mA
(3)
12.5 mA, 5 mA, 7.5 mA
(4)
12.5 mA, 7.5 mA, 5 mA
(4)
12.5 mA, 7.5 mA, 5 mA
A Zener diode is connected to a battery 28.
and a load as shown below :
English : 18
Set : 11
Hindi : 18
Set : 11
29.
âê¿è-I (çßléÌ ¿éÕ·¤èØ çßç·¤ÚU‡æ âð âÕh ƒæÅUÙæ°¡)
·¤æð âê¿è-II (çßléÌ ¿éÕ·¤èØ SÂð€ÅþU× ·¤æ Öæ») âð
âé×ðçÜÌ ·¤èçÁ° ¥æñÚU âêç¿Øæð́ ·ð¤ Ùè¿ð çÎØð »Øð çß·¤ËÂæð́
×´ð âð âãè çß·¤Ë ¿éçÙ° Ñ
Match the List - I (Phenomenon associated 29.
with electromagnetic radiation) with
List - II (Part of electromagnetic spectrum)
and select the correct code from the
choices given below the lists :
List - I
I Doublet of sodium
ÇÏ¤Í - I
List - II
A
Visible
radiation
Wavelength
corresponding to
II temperature associated
with the isotropic
radiation filling all space
B Microwave
Wavelength emitted by
III atomic hydrogen in
interstellar space
Short
C
radiowave
Wavelength of radiation
IV arising from two close
D X - rays
energy levels in hydrogen
ÇÏ¤Í - II
I
ÇËÕ̬U½¼ œ‰Ë Ìmœ‰
A
³ÐŽ ÌÄ̜‰¿UøË
II
ǃ§ÏøËá ǼÌcªU ¼Õ™ Ǽ³ÖÌ؉
ÌÄ̜‰¿UøË œ‰Õ‰»¿UÕU ÈËÕ¾Õ ÇÕ ÇƒºÌh±
±Ë§¼Ë¾ œ‰Õ‰ Ǚ™ ± ±¿U™U ³Öç½á
B
ÇÏÏä¼ ±¿U™U III
Šü±¿U±Ë¿Uœ‰Í½ ŠËœ‰ËÅË ¼Õ™ §¿U¼ËøËÎ
Èˌ¬UãUËÕ¦¾ mË¿UË ŽíÇ̦á±
±¿U™U ³Öç½á
C
ÁìËÎ ¿UÕU̬U½ËÕ ±¿U™U Õ
IV
Èˌ¬UãUËÕ¦¾ ¼Õ™ ³ËÕ Ç¼Í§ ‰¦Ëá S±¿U˙Õ
ÇÕ Ì¾œ‰ÁÕ ÌÄ̜‰¿UøË œ‰Í ±¿U™U ³Öç½á
D X - ̜‰¿UøËՙ
(1)
(I)-(A), (II)-(B), (III)-(B), (IV)-(C)
(1)
(I)-(A), (II)-(B), (III)-(B), (IV)-(C)
(2)
(I)-(A), (II)-(B), (III)-(C), (IV)-(C)
(2)
(I)-(A), (II)-(B), (III)-(C), (IV)-(C)
(3)
(I)-(D), (II)-(C), (III)-(A), (IV)-(B)
(3)
(I)-(D), (II)-(C), (III)-(A), (IV)-(B)
(4)
(I)-(B), (II)-(A), (III)-(D), (IV)-(A)
(4)
(I)-(B), (II)-(A), (III)-(D), (IV)-(A)
English : 19
Set : 11
Hindi : 19
Set : 11
30.
In the circuit diagrams (A, B, C and D) 30.
shown below, R is a high resistance and S
is a resistance of the order of galvanometer
resistance G.
The correct circuit,
corresponding to the half deflection
ÎàææüØð ÂçÚUÂÍ ç¿˜ææð´ (A, B, C °ß´ D) ×ð´, R °·¤ ¥‹Ø
©‘¿ ÂýçÌÚUæðÏ ãñ ¥æñÚU S »ñËßñÙæð×æÂè ÂýçÌÚUæðÏ G ·¤è
·¤æðçÅU ·¤æ ÂýçÌÚUæðÏ ãñÐ »ñËßñÙæð×æÂè ·¤æ ÂýçÌÚUæðÏ °ß´
ÎÿæÌæ´·¤ çÙ·¤æÜÙð ·ð¤ ¥hü-çßÿæð‡æ çßçÏ ·ð¤ â´»Ì
âãè ÂçÚUÂÍ ç¿ç‹ãÌ ãñ §ââð Ñ
method for finding the resistance and
figure of merit of the galvanometer, is the
circuit labelled as :
(A)
(A)
(B)
(B)
(C)
(C)
English : 20
Set : 11
Hindi : 20
Set : 11
(D)
(D)
(1)
RS
Circuit A with G5 (
R 2 S)
(2)
Circuit B with G5S
(3)
Circuit C with G5S
(4)
RS
Circuit D with G5
R2 S
English : 21
Set : 11
RS
R 2 S)
(1)
G5 (
(2)
G5S ·ð¤
·ð¤ âæÍ ÂçÚUÂÍ A
(3)
âæÍ ÂçÚUÂÍ B
G5S ·ð¤ âæÍ ÂçÚUÂÍ C
(4)
G5
Hindi : 21
RS
R2 S
·ð¤ âæÍ ÂçÚUÂÍ D
Set : 11
PART B — CHEMISTRY
31.
If lo and l be the threshold wavelength 31.
and wavelength of incident light, the
velocity of photoelectron ejected from the
Öæ» B — ÚUâæØÙ çߙææÙ
ØçÎ lo ¥æñÚU l ÎãÜè$Áè ÌÚ´U»ÎñƒØü ¥æñÚU ¥æÂçÌÌ Âý·¤æàæ
·¤æ ÌÚ´U» ÎñƒØü ãæð´ Ìæð ÏæÌé SÍÜ âð çÙ·¤Üð Âý·¤æàæèØ
§Üð€ÅþUæÙæ´ð ·¤æ ßð» ãæð»æ Ñ
metal surface is :
32.
(1)
2h
( lo 2 l )
m
(1)
2h
( lo 2 l )
m
(2)
2hc
( lo 2 l )
m
(2)
2hc
( lo 2 l )
m
(3)
2 h c  lo 2 l 
m  l lo 
(3)
2 h c  lo 2 l 
m  l lo 
(4)
2h  1
1
2 

m  lo
l
(4)
2h  1
1
2 

m  lo
l
The appearance of colour in solid alkali 32.
metal halides is generally due to :
(1)
Schottky defect
(2)
Frenkel defect
(3)
Interstitial position
(4)
F-centres
English : 22
Set : 11
ÆUæðâ ÿææÚU ÏæÌé ãðÜæ§ÇUæð´ ×ð´ Ú´U» ·ð¤ Îð¹ð ÁæÙð ·¤æ ·¤æÚU‡æ
ÂýæØÑ ãæðÌæ ãñ Ñ
(1) àææòÅU·¤è Îæðá
(2) $Èýñ´¤·¤Ü Îæðá
(3) ¥‹ÌÚUæÜè SÍæÙ
(4) F-·ð¤‹Îý
Hindi : 22
Set : 11
33.
In the reaction of formation of sulphur 33.
trioxide
by
contact
process
âÂ·ü¤ çßçÏ mæÚUæ âË$ȤÚU ÅþUæ§ü¥æ€âæ§ÇU ÕÙæÙð ·¤è
¥çÖç·ý¤Øæ 2SO21O2 ì 2SO3 ×ð´ ¥çÖç·ý¤Øæ ·¤è
d [O2 ]
×æðÜ
2SO21O2 ì 2SO3 the rate of reaction was
ÎÚU ·¤æð
measured as
·ð¤ M¤Â ×ð´ ×æÂæ »ØæÐ ¥çÖç·ý¤Øæ ÎÚU [SO2]·ð¤ M¤Â ×ð´
d [O 2 ]
dt
dt
522.531024
L21 s21
×æðÜ L21 s21 ×ð´ ãæð»è Ñ
522.531024mol L21 s21. The
rate of reaction in terms of [SO2] in mol
L21s21 will be :
34.
(1)
21.25310 24
(1)
21.25310 24
(2)
22.50310 24
(2)
22.50310 24
(3)
23.75310 24
(3)
23.75310 24
(4)
25.00310 24
(4)
25.00310 24
Assuming that the degree of hydrolysis is 34.
small, the pH of 0.1 M solution of sodium
acetate (Ka=1.031025) will be :
(1) 5.0
Øã ×æÙÌð ãé° ç·¤ ãæ§ÇþUæðÜðçââ ·¤æ ·ý¤×æ´·¤ (çÇU»ýè)
‹Øê Ù ãñ , âæð ç ÇU Ø × °ð â èÅð U Å U ·ð ¤ 0.1M çßÜØÙ
(Ka=1.031025) ·¤æ pH ãæð»æ Ñ
(1)
5.0
(2)
6.0
(2)
6.0
(3)
8.0
(3)
8.0
(4)
9.0
(4)
9.0
English : 23
Set : 11
Hindi : 23
Set : 11
35.
For the reaction, 2N2O5 → 4NO21O2, the 35.
rate equation can be expressed in two
ways 2
1
36.
d[N2O5 ]
dt
5 k [ N 2 O 5 ] and
¥çÖç·ý¤Øæ 2N2O5 → 4NO21O2, ·ð¤ çÜ° ÎÚU
â×è·¤ÚU‡æ ·¤æð Îæð ÌÚUè·ð¤ âð çܹæ Áæ â·¤Ìæ ãñ
2
d [ NO 2 ]
5 k9 [ N 2 O 5 ]
dt
1
d[N2O5 ]
dt
5 k [N2O5 ]
¥æñÚU
d [ NO 2 ]
5 k9 [ N 2 O 5 ]
dt
¥æñÚU k’ ·¤æð çِ٠緤â M¤Â ×ð´ çܹæ ÁæØð»æ?
k and k’ are related as :
k
(1)
k5k’
(1)
k5k’
(2)
2k5k’
(2)
2k5k’
(3)
k52k’
(3)
k52k’
(4)
k54k’
(4)
k54k’
In some solutions, the concentration of
H3O1 remains constant even when small 36.
amounts of strong acid or strong base are
added to them. These solutions are known
as :
·é¤ÀU çßÜØÙæð´ ×ð´ ÂýÕÜ °ðçâÇU ¥Íßæ ÂýÕÜ ÿææÚU ·¤è
ÍæðǸè ×æ˜ææ ç×ÜæÙð ÂÚU Öè H3O1 ·¤æ âæ‹Îý‡æ çSÍÚU ãè
ÚUãÌæ ãñÐ §Ù çßÜØÙæð´ ·¤æð Ùæ× çÎØæ ÁæÌæ ãñ Ñ
(1)
Ideal solutions
(2)
Colloidal solutions
(1)
(3)
True solutions
(2)
(4)
Buffer solutions
(3)
(4)
English : 24
Set : 11
¥æÎàæü çßÜØÙ
·¤æðÜæØÇUè çßÜØÙ
ßæSÌçß·¤ çßÜØÙ
ÕȤÚU (Buffer) çßÜØÙ
Hindi : 24
Set : 11
37.
37.
Given
Fe31(aq)1 e2®Fe21(aq); E8=10.77 V
A l 3 1 ( a q ) 1 3 e2 ® A l ( s ) ; E 8 = 2 1 . 6 6 V
Br2(aq)12 e2®2B r2 ; E8=11.09 V
Considering the electrode potentials,
which of the following represents the
correct order of reducing power ?
38.
(1)
Fe21< Al < B r2
(2)
B r2 < Fe21< Al
(3)
Al < B r2 < Fe21
(4)
Al < Fe21< B r2
38.
The initial volume of a gas cylinder is
750.0 mL. If the pressure of gas inside the
çÎØæ »Øæ ãñ Fe31(ÁÜèØ)1 e2®Fe21(ÁÜèØ); E8 =10.77 V
Al31(ÁÜèØ)13 e2®Al(s); E8 = 21.66 V
2
Br2(ÁÜèØ)12 e ®2B r2 ; E8 = 11.09 V
§Üñ€ÅþUæðÇU çßÖßæ𴠷𤠥æÏæÚU ÂÚU çِÙæð´ ×ð´ âð ·¤æñÙ ·ý¤×
¥Â¿ØÙ àæç€ÌØæð´ ·¤æð âãè ÂýSÌéÌ ·¤ÚUÌæ ãñ?
(1)
Fe21< Al < B r2
(2)
B r2 < Fe21< Al
(3)
Al < B r2 < Fe21
(4)
Al < Fe21< B r2
°·¤ »ñ â ·ð ¤ çâç܋ÇU Ú U ·¤æ Âý æ ÚU ç Ö·¤ ¥æØÌÙ
750.0 mL ãñÐ ØçÎ çâç܋ÇUÚU ·ð¤ Õè¿ ·¤è »ñâ ·¤æ
ÎæÕ 840.0 mm Hg âð ÕÎÜ ·¤ÚU 360.0 mm Hg
ãæð ÁæÌæ ãñ Ìæð »ñâ ·¤æ ¥ç‹Ì× ¥æØÌÙ ãæð»æ Ñ
cylinder changes from 840.0 mm Hg to
360.0 mm Hg, the final volume the gas will
(1)
1.750 L
be :
(2)
3.60 L
(1)
1.750 L
(3)
4.032 L
(2)
3.60 L
(4)
7.50 L
(3)
4.032 L
(4)
7.50 L
English : 25
Set : 11
Hindi : 25
Set : 11
39.
The molar heat capacity (Cp) of CD2O is 39.
10 cals at 1000 K. The change in entropy
associated with cooling of 32 g of CD2O
vapour from 1000 K to 100 K at constant
pressure will be :
(D = deuterium, at. mass = 2 u)
40.
·¤è ×æðÜÚU ª¤c×æ ÏæçÚUÌæ (Cp) 1000 K ÂÚU
10 cals ãñÐ 32 g CD2O ßæc ·¤æð 1000 K âð
100 K Ì·¤ çSÍÚU ÎæÕ ÂÚU ÆU‡ÇUæ ·¤ÚUÙð ÂÚU âÕh
°ð‹ÅþUæÂè ÂçÚUßÌüÙ ãæð»æ Ñ
(D = çÇUØéÅUèçÚUØ×, ¥æñÚU §â·¤æ ÂÚU×æ‡æé ÎýÃØ×æÙ
= 2 ×æ˜æ·¤)
CD2O
(1)
23.03 cal deg21
(1)
23.03 cal deg21
(2)
223.03 cal deg21
(2)
223.03 cal deg21
(3)
2.303 cal deg21
(3)
2.303 cal deg21
(4)
22.303 cal deg21
(4)
22.303 cal deg21
Based on the equation :
(523 218 -
40.
1 
 1
2 2
2
 n
n 1 
2
â×è·¤ÚU‡æ Ñ
(523 218 -
the wavelength of the light that must be
absorbed to excite hydrogen electron from
level n=1 to level n=2 will be :
(h = 6.625310234 Js, C= 33108 ms21)
1 
 1
2 2
2
 n
n 1 
2
·ð¤ ¥æÏæÚU ÂÚU ãæ§ÇþUæðÁÙ ·ð¤ §Üñ€ÅþUæÙ ·¤æð SÌÚU n=1 âð
SÌÚU n=2 Ì·¤ ©žæðçÁÌ ·¤ÚUÙð ·ð¤ çÜØð Âý·¤æàæ, çÁâ
·¤æ àææðá‡æ ¥æßàØ·¤ ãæð»æ, ·¤æ ÌÚ´U» ÎñƒØü §Ù×ð´ âð €Øæ
ãæð»æ Ñ
(h = 6.625310234 Js, C = 33108 ms21)
(1)
1.325310 27 m
(1)
1.325310 27 m
(2)
1.325310210 m
(2)
1.325310210 m
(3)
2.650310 27 m
(3)
2.650310 27 m
(4)
5.300310210 m
(4)
5.300310210 m
English : 26
Set : 11
Hindi : 26
Set : 11
41.
Which of the following series correctly 41.
represents relations between the elements
çِ٠âð ·¤æñÙ âæ âèÚUè$Á Îæð ̈ßæð´ X ¥æñÚU Y ·ð¤ Õè¿
·ð¤ âÕ‹Ï ·¤æ âãè çÙM¤ÂÙ ·¤ÚUÌæ ãñ?
from X to Y ?
X ® Y
(1)
(2)
3Li
9F
X ® Y
® 19K Ionization
(3)
¥æØÙè·¤ÚU‡æ ·¤è °ð‹ÍñËÂè ÕɸÌè
ãñ
9F ® 35Br §Üñ € Åþ U æ Ù ÜæÖ ·¤è °ð ‹ Íñ Ë Âè
«¤‡ææˆ×·¤ 翋㠷ð¤ âæÍ ÕɸÌè
ãñ
6C ® 32Ge ÂÚU×æ‡æé¥æð´ ·¤è ç˜æ’Øæ°¡ ÕɸÌè ãñ´ñ
(4)
18Ar® 54Xe
enthalpy
(1)
® 35Br Electron gain enthalpy
with negative sign
(2)
increases
increases
42.
(3)
6C
(4)
18Ar ® 54Xe Noble character increases
® 32Ge Atomic radii increases
3Li
® 19K
©ˆ·ý¤C SßÖæß ÕɸÌæ ãñ
2
The correct order of bond dissociation 42.
çِ٠ÃØßSÍæ¥æð´ ×ð´ âð ç·¤â ×ð´ N2, O2,
is shown in
energy among N2, O2, O2
2
which of the following arrangements ?
¥æÕ‹Ï çßØæðÁÙ ª¤Áæü ·ð¤ âãè ·ý¤× ·¤æð çιæØæ »Øæ
ãñ ?
(1)
> O2
N2 > O2
2
(1)
N2 > O2
> O2
2
(2)
2
O 2 > O2 > N2
(2)
2
O 2 > O2 > N2
(3)
N2 > O2 > O2
2
(3)
N2 > O2 > O2
2
(4)
O2 > O2
> N2
2
(4)
O2 > O2
> N2
2
English : 27
Set : 11
Hindi : 27
O2
·¤è
Set : 11
43.
Which of the following statements about 43.
Na2O2 is not correct ?
44.
(1)
It is diamagnetic in nature.
(2)
It is a derivative of H2O2.
(3)
Na2O2 oxidises Cr31 to CrO422 in
acid medium.
(4)
It is the super oxide of sodium.
Which of the following statements about 44.
the depletion of ozone layer is correct ?
(1)
The problem of ozone depletion is less
serious at poles because NO 2
solidifies and is not available for
consuming ClO• radicals.
(2)
The problem of ozone depletion is
more serious at poles because ice
crystals in the clouds over poles act
as catalyst for photochemical
reactions
involving
decomposition of ozone by
the
Cl•
Na2O2 ·ð¤
âÕ‹Ï ×ð´ çِ٠·¤ÍÙæð´ âð ·¤æñÙ âæ ·¤ÍÙ
âãè Ùãè´ ãñ?
(1) §â ·¤è Âýßëçžæ ÂýçÌ¿éÕ·¤èØ ãñÐ
(2) Øã H2O2·¤æ °·¤ ÃØéˆÂ‹Ù ãñ
(3) ¥æÜ ×æŠØ× ×ð ´ Na 2O 2 âð Cr 31 ·¤æ
CrO422 ×𴠩¿ØÙ ãæð ÁæÌæ ãñÐ
(4) Øã âæðçÇUØ× ·¤æ ÂÚUæ-¥æ€âæ§Ç ãñÐU
¥æðÁæðÙ SÌÚU ·ð¤ ƒæÅUÙð âÕ‹Ïè çِ٠·¤ÍÙæð´ ×ð´ âð ·¤æñÙ
âæ âãè ãñ?
(1) Ïýéßè ÿæð˜ææð´ ×ð´ ¥æð$ÁæðÙ ƒæÅUÙð ·¤è â×SØæ ·¤×
×ãˆß ÚU¹Ìè ãñ €Øæð´ç·¤ NO2 Á×·¤ÚU ÆUæðâ ÕÙ
ÁæÌè ãñ ¥æñÚU ClO• ×êÜ·¤æð´ ·¤æð ãÅUæÙð ·ð¤ çÜØð
©ÂÜŽÏ Ùãè´ ãæðÌèÐ
(2) Ïýß
é è ÿæð˜ææð́ ×ð́ ¥æðÁ
$ æðÙ ·ð¤ ƒæÅUÙð ·¤è â×SØæ ¥çÏ·¤
×ãˆß ÚU¹Ìè ãñ €Øæð´ç·¤ Ïýéßæð´ ÂÚU ÕæÎÜæð´ ×ð´ ÕÈü¤
·ð¤ ç·ý¤SÅUÜæð´ ·ð¤ ãæðÙð âð Cl• ¥æñÚU ClO•
ÚðUçÇU·¤Üæð´ mæÚUæ ©ˆÂýðçÚUÌ ¥æð$ÁæðÙ çßØæðÁÙ ·¤è
Âý·¤æàæ-ÚUæâæØçÙ·¤ ¥çÖç·ý¤Øæ°¡ ãæð â·¤Ìè ãñÐ
and
ClO• radicals.
(3)
Freons, chlorofluorocarbons, are
(3)
inert chemically, they do not react
with ozone in stratosphere.
(4)
Oxides of nitrogen also do not react
with ozone in stratosphere.
English : 28
Set : 11
(4)
çÈý¤¥æÙð´ (€ÜæðÚUæðÜæðÚUæð ·¤æÕüÙ) ÚUæâæØçÙ·¤ M¤Â
×ð´ ¥ç·ý¤Ø ãæðÌè ãñ´Ð ß𠪤ÂÚUè ßæØéׇÇUÜ ×ð´
©ÂçSÍÌ ¥æð$ÁæðÙ âð ç·ý¤Øæ Ùãè´ ·¤ÚUÌè´Ð
ª¤ÂÚUè ßæØéׇÇUÜ ·¤è ¥æð$ÁæðÙ âð Ùæ§ÅUþUæðÁÙ ·ð¤
¥æ€âæ§ÇU Öè ç·ý¤Øæ Ùãè´ ·¤ÚUÌðÐ
Hindi : 28
Set : 11
45.
A gaseous compound of nitrogen and 45.
hydrogen contains 12.5%(by mass) of
hydrogen. The density of the compound
relative to hydrogen is 16. The molecular
Ùæ§ÅþUæðÁÙ ¥æñÚU ãæ§ÇþUæðÁÙ ·¤æ °·¤ »ñâèØ Øæñç»·¤
ÎýÃØ×æÙ âð 12.5% ãæ§ÇþUæðÁÙ ÚU¹Ìæ ãñÐ ãæ§ÇþUæðÁÙ
·¤è ÌéÜÙæ ×ð´ §â Øæñç»·¤ ·¤æ ƒæÙˆß 16 ãñÐ Øæñç»·¤
·¤æ ¥‡æéâê˜æ ãæð»æ Ñ
formula of the compound is :
46.
(1)
NH2
(1)
NH2
(2)
N3H
(2)
N3H
(3)
NH3
(3)
NH3
(4)
N2H4
(4)
N2H4
Shapes of certain interhalogen compounds 46.
are stated below. Which one of them is
·é¤ÀU ¥´ÌÑãñÜæðÁÙ Øæñç»·¤æ𴠷𤠥淤æÚU Ùè¿ð çܹ𠻰
ãñ´Ð §Ù×ð´ âð ·¤æñÙ âæ ·¤ÍÙ âãè Ùãè´ ãñ?
not correctly stated ?
47.
(1)
IF7 : pentagonal bipyramid
(1)
(2)
BrF5 : trigonal bipyramid
(2)
(3)
BrF3 : planar T-shaped
(3)
(4)
ICl3 : planar dimeric
(4)
Consider the following equilibrium
1
47.
´¿ÖéÁèØ çmçÂÚUæç×ÇU
BrF5 : ç˜æ·¤æð‡æèØ çmçÂÚUæç×ÇU
BrF3 : â×ÌÜèØ T-¥æ·¤æÚU ·¤æ
ICl3 : â×ÌÜèØ ÇUæ§×ðçÚU·¤ (Îæð ÃØßçSÍÌ)
IF7 :
§â âæØ
1
AgCl ¯12NH3 ì Ag ( NH 3 )2  1 Cl2
AgCl ¯12NH3 ì Ag ( NH 3 )2  1 Cl2
White precipitate of AgCl appears on
adding which of the following ?
·¤æð ŠØæÙ ÎèçÁ°Ð çِÙæð´ ×ð´ âð ç·¤âð ÇUæÜÙð ÂÚU AgCl
·¤æ àßðÌ ¥ßÿæð ÕÙð»æ?
(1)
NH3
(1)
(2)
aqueous NaCl
(2)
(3)
aqueous HNO3
(3)
(4)
aqueous NH4Cl
(4)
English : 29
Set : 11
NH3
ÁÜèØ NaCl
ÁÜèØ HNO3
ÁÜèØ NH4Cl
Hindi : 29
Set : 11
48.
Which of the following name formula 48.
çِ٠Ùæ×-âê˜æ ÁæðǸæð ×ð´ âð ·¤æñÙ âãè Ùãè´ ãñ?
combinations is not correct ?
Formula
49.
ÇÏîË
Name
(1) K2[Pt(CN)4]
Potassium
tetracyanoplatinate (II)
(1) K2[Pt(CN)4]
§ËÕªUËÌÅ˽¼ ªÖUªãUËÇ˽¾ËÕ åÁÖªU;ժU (II)
(2) [Mn(CN)5]22
Pentacyanomagnate (II) ion
(2) [Mn(CN)5]22
§ÖüªUËÇ˽¾ËÕ ¼Ö™ ˾ժU(II) ŠË½¾
(3) K[Cr(NH3)2Cl4]
Potassium diammine
tetrachlorochromate (III)
(3) K[Cr(NH3)2Cl4]
§ËÕªÖUÌÅ˽¼
¬Uˌᕼ;ªÖUªãUËþÁËÕ¿UË՜â‰ËÕ¼ÕªU‰(III)
(4) [Co(NH3)4(H2O)I]SO4
Tetraammine aquaiodo
cobalt (III) sulphate
(4) [Co(NH3)4(H2O)I]SO4
ªÖUªãUË •Õ¼Í¾ËՕþÄˊ˽ËÕ¬UË՜‰Ëպ˰ªU
(III) Ç°$$­Õ‰ªU
Consider the coordination compound, 49.
[Co(NH3)6]Cl3. In the formation of this
complex, the species which acts as the
Lewis acid is :
50.
¾Ë¼
â׋ßØè Øæñç»·¤ [Co(NH3)6]Cl3 ·¤æð ŠØæÙ ÎèçÁ°Ð
§â â´·¤ÚUU ·ð¤ ÕÙæÙð ×ð´ ÂýØé€Ì ÂÎæÍü Áæð ËØé§â °ðçâÇU
(Lewis acid) ãñ, ãæð»æ Ñ
(1)
[Co(NH3)6]31
(1)
(2)
Cl2
[Co(NH3)6]31
(2)
Cl2
(3)
Co31
(3)
Co31
(4)
NH3
(4)
NH3
Which one of the following does not have 50.
a pyramidal shape ?
çِÙæð´ ×ð´ âð 緤⠷¤æ M¤Â çÂÚUæç×ÇUèØ Ùãè´ ãñ?
(1)
(CH3)3 N
(1)
(2)
(SiH3)3 N
(CH3)3 N
(2)
(3)
P(CH3)3
(SiH3)3 N
(3)
(4)
P(SiH3)3
P(CH3)3
(4)
P(SiH3)3
English : 30
Set : 11
Hindi : 30
Set : 11
51.
51.
The following reaction
·¤æð §Ù×ð´ âð ç·¤â Ùæ× âð ÁæÙæ ÁæÌæ ãñ?
(1) ÂÚUç·¤Ù ¥çÖç·ý¤Øæ
(2) »ñÅUÚU×Ù - ·¤æò¿ $ȤÚU×èÜðàæÙ
(3) ·¤æðÜÕð ·¤è ¥çÖç·ý¤Øæ
(4) »ñÅUÚU×Ù ¥çÖç·ý¤Øæ
is known as :
52.
53.
(1)
Perkin reaction
(2)
Gattermann-Koch Formylation
(3)
Kolbe’s reaction
(4)
Gattermann reaction
¥çÖç·ý¤Øæ
The reagent needed for converting 52.
ÂçÚUßÌüÙ
is :
·ð¤ çÜØð ¥æßàØ·¤ ¥çÖ·¤æÚU·¤ ãñ Ñ
(1) ·ñ¤ÅU. ãæ§ÇþUæðçÁÙðàæÙ
(2) H2/çÜ´ÇUÜÚU ·ñ¤ÅU.
(1)
Cat. Hydrogenation
(2)
H2/Lindlar Cat.
(3)
Li/NH3
(4)
LiAlH4
Complete reduction of benzene-diazonium 53.
chloride with Zn/HCl gives :
(1)
Aniline
(2)
Phenylhydrazine
(3)
Azobenzene
(4)
Hydrazobenzene
English : 31
Set : 11
(3)
Li/NH3
(4)
LiAlH4
·ð¤ mæÚUæ Õñ‹$ÁèÙ ÇUæØæ$ÁæðçÙØ× €ÜæðÚUæ§ÇU
·¤æ ÂêÚUæ ¥Â¿ØÙ ÎðÌæ ãñ Ñ
(1) °ðÙèÜèÙ
(2) ç$ȤÙæ§Üãæ§ÇþUæ$ÁèÙ
(3) °ð$ÁæðÕñ‹$ÁèÙ
(4) ãæ§ÇþU°ð$ÁæðÕñ‹$ÁèÙ
Zn/HCl
Hindi : 31
Set : 11
54.
55.
described below :
°·¤ ¥æÚU»ñçÙ·¤ Øæñç»·¤ A, C5H8O; çِ٠Âý·¤æÚU
H2O, NH3 ¥æñÚU CH3COOH ·ð¤ âæÍ ç·ý¤Øæ ·¤ÚUÌæ
ãñ
A is :
A ãñ
(1)
(1)
(2)
(2)
(3)
(3)
(4)
(4)
An organic compound A, C5H8O; reacts 54.
with H 2 O, NH 3 and CH 3 COOH as
In
allene
(C 3 H 4 ),
the
type(s)
of 55.
hybridization of the carbon atoms is (are):
(1)
sp and sp3
(2)
sp2 and sp
(3)
only sp2
(4)
sp2 and sp3
English : 32
Set : 11
Ñ
(°ÜèÙ) ×ð´ ·¤æÕüÙ ÂÚU×æ‡æé (ÂÚU×æ‡æé¥æð´) ·ð¤
â´·¤ÚU‡æ ·¤è çßçÏ ãæðÌè ãñ Ñ
(1) sp ¥æñÚU sp3
(2) sp2 ¥æñÚU sp
(3) ·ð¤ßÜ sp2
(4) sp2 ¥æñÚU sp3
C3H4
Hindi : 32
Set : 11
56.
57.
acetaldehyde in the presence of H2SO4
·¤è ©ÂçSÍçÌ ×ð ´ €Üæð Ú Uæð Õ ñ ‹ $ Á èÙ ·¤è
ÅþU槀ÜæðÚUæð°ðçâÅU °ðçËÇUãæ§ÇU âð ¥çÖç·ý¤Øæ
The major product formed is :
×ð´ ÕÙÙð ßæÜè Âý×é¹ Øæñç»·¤ ãñ Ñ
(1)
(1)
(2)
(2)
(3)
(3)
(4)
(4)
Chlorobenzne reacts with trichloro 56.
Tischenko reaction is a modification of :
(1)
Aldol condensation
(2)
Claisen condensation
(3)
Cannizzaro reaction
(4)
Pinacol-pinacolon reaction
English : 33
Set : 11
57.
H 2 SO 4
çÅUâ¿ñ‹·¤æð ¥çÖç·ý¤Øæ °·¤ ÕÎÜæ ãé¥æ M¤Â ãñ Ñ
(1) °ðËÇUæðÜ â´ƒæÙÙ ·¤æ
(2) €Üð$Á٠ⴃæÙÙ ·¤æ
(3) ·ñ¤Ùè$ÁæÚUæð ¥çÖç·ý¤Øæ ·¤æ
(4) çÂÙæ·¤æðÜ - çÂÙæ·¤æðÜæðÙ ¥çÖç·ý¤Øæ ·¤æ
Hindi : 33
Set : 11
58.
Which one of the following is used as 58.
Antihistamine ?
59.
(1)
Omeprazole
(2)
Chloranphenicol
(3)
Diphenhydramine
(4)
Norethindrone
Which one of the following statements is 59.
çِÙæð´ ×ð´ âð 緤⠰·¤ ·¤æð °ð‹ÅUè çãSÅUæ×èÙ ·ð¤ M¤Â ×ð´
ÂýØæð» ·¤ÚUÌð ãñ´?
(1) ¥æð×èÂýæ$ÁæðÜ
(2) €ÜæðÚðUÙ $Èñ¤çÙ·¤æðÜ
(3) ÇUæ§Èñ¤Ùãæ§ÇþU°×èÙ
(4) ÙæÚU°çÍÙÇþUæðÙ
çِ٠·¤ÍÙæð´ ×ð´ âð ·¤æñÙ âæ âãè Ùãè´ ãñ?
not correct ?
(1)
(1)
°ðË·¤æðãæÜð´ ÂæÙè ·¤è ÌéÜÙæ ×ð´ ÎéÕüÜ ¥Ü ãñ´
(2)
in the following order
°ðË·¤æðãæÜæð´ ·¤è ¥Ü àæç€Ì §â ·ý¤× ×𴠃æÅUÌè
ÁæÌè ãñ -
RCH2OH > R2CHOH > R3COH
Carbon-oxygen bond length in
RCH2OH > R2CHOH > R3COH
×ñÍðÙæÜ, CH3OH ×ð´ ·¤æÕüÙ - ¥æ€âèÁÙ
Alcohols are weaker acids than
water
(2)
(3)
Acid strength of alcohols decreases
(3)
methanol, CH3OH is shorter than
that of C2O bond length in phenol.
(4)
The bond angle
in methanol
(4)
The gas liberated by the electrolysis of 60.
Dipotassium succinate solution is :
(1)
Ethane
(2)
Ethyne
(3)
Ethene
(4)
Propene
English : 34
×ðÍðÙæÜ ×ð´
¥æÕ‹Ï ·¤æð‡æ
108.98
ãæðÌæ ãñÐ
is 108.98
60.
¥æÕ‹Ï ·¤è ܐÕæ§ü ç$ȤÙæÜ ×ð´ C2O ¥æՋÏ
·¤è ܐÕæ§ü âð ÀUæðÅUè ãæðÌè ãñÐ
Set : 11
ÇUæ§ÂæðÅñUçàæØ× â€âèÙðÅU ·ð¤ çßÜØÙ ·ð¤ ßñléÌ çßÖæÁÙ
×ð´ ÂýæŒÌ »ñâ ãæðÌè ãñ Ñ
(1) §üÍðÙ
(2) §üÍæ§Ù
(3) §üÍèÙ
(4) ÂýæðÂèÙ
Hindi : 34
Set : 11
PART C — MATHEMATICS
61.
Let f be an odd function defined on the set 61.
of real numbers such that for x/0,
f(x)53 sin x14 cos x.
Then f(x) at x 5 2
62.
(1)
3
12 3
2
(2)
2
(3)
3
22 3
2
(4)
2
ÂçÚUÖæçáÌ ãñ Áãæ¡
11p
is equal to :
6
f(x) ÕÚUæÕÚU
3
12 3
2
3
22 3
2
If z1, z2 and z3, z4 are 2 pairs of complex
62.
conjugate numbers, then
z 
z 
arg  1  1 arg  2  equals :
 z4 
 z3 
(1)
0
(2)
p
2
(3)
3p
2
(4)
p
English : 35
Öæ» C — »ç‡æÌ
×æÙæ f °·¤ çßá× È¤ÜÙ ãñ Áæð ç·¤ ßæSÌçß·¤ ⴁØæ¥æð´
·ð¤ â×é‘¿Ø ÂÚU f(x)53 sin x14 cos x mæÚUæ
x/0
x5 2
ãñ Ñ
(1)
3
12 3
2
(2)
2
(3)
3
22 3
2
(4)
2
11p
6
ÂÚU
3
12 3
2
3
22 3
2
ØçÎ z1, z2 ÌÍæ z3, z4 âç×oý â´Øé‚×è ⴁØæ¥æð´ ·ð¤
Îæð Øé‚× ãñ´, Ìæðz 
z 
arg  1  1 arg  2 
 z4 
 z3 
Set : 11
ãñ, Ìæð
(1)
0
(2)
p
2
(3)
3p
2
(4)
p
Hindi : 35
ÕÚUæÕÚU ãñ Ñ
Set : 11
63.
64.
x 2 2 4 2 kx 1 2e 4 ln k2 1 5 0 for some k,
and a21b2 5 66, then a31b3 is equal to :
·ð ¤ çÜ°, â×è·¤ÚU‡æ
x 2 4 2 kx 1 2e
2 1 5 0 ·ð¤ ×êÜ ãñ´ ÌÍæ
2
2
3
a 1b 5 66, ãñ, Ìæð a 1b3 ÕÚUæÕÚU ãñ Ñ
(1)
248 2
(1)
248 2
(2)
280 2
(2)
280 2
(3)
232 2
(3)
232 2
(4)
2280 2
(4)
2280 2
If a and b are roots of the equation, 63.
64.
Let A be a 333 matrix such that
 1 2 3
A 0 2 3  5
0 1 1 
(2)
 3 2 1
3 2 0


 1 1 0 
(3)
0 1 3
0 2 3


 1 1 1 
(4)
 1 2 3
0 1 1


0 2 3 
English : 36
ÌÍæ
a
b
ç·¤âè
Set : 11
k
4 ln k
×æÙæ A °·¤ °ðâæ 333 ¥æÃØêã ãñ ç·¤
A21 ãñ
Then A21 is :
(1)
2
 1 2 3
A 0 2 3  5
0 1 1 
0 0 1 
1 0 0


0 1 0 
3 1 2 
3 0 2 


 1 0 1 
ØçÎ
0 0 1 
1 0 0

 ãñ, Ìæð
0 1 0 
Ñ
(1)
3 1 2 
3 0 2 


 1 0 1 
(2)
3 2 1
3 2 0


 1 1 0 
(3)
0 1 3
0 2 3


 1 1 1 
(4)
1 2 3
0 1 1


0 2 3 
Hindi : 36
Set : 11
65.
Let for i51, 2, 3, pi(x) be a polynomial of 65.
degree 2 in x, pi’(x) and pi’’(x) be the first
and second order derivatives of p i (x)
respectively. Let,
×æÙæ i51, 2, 3,·ð¤ çÜ° pi(x), x ×𴠃ææÌ 2 ·ð¤
ÕãéÂÎ ãñ´, pi’(x) ÌÍæ pi’’(x) ·ý¤×àæÑ ÂýÍ× ·¤æðçÅU ÌÍæ
çmÌèØ ·¤æðçÅU ·ð¤ ¥ß·¤ÜÁ ãñ´Ð
×æÙæ
 p1 ( x ) p19 ( x ) p10 ( x ) 
A( x ) 5  p2 ( x ) p29 ( x ) p20 ( x )
 p3 ( x ) p39 ( x ) p30 ( x ) 
 p1 ( x ) p19 ( x ) p10 ( x ) 
A( x ) 5  p2 ( x ) p29 ( x ) p20 ( x )
 p3 ( x ) p39 ( x ) p30 ( x ) 
and B(x) = [A(x)]T A(x). Then determinant
ÌÍæ B(x) = [A(x)] T A(x) ãñ , Ìæð B(x) ·¤æ
âæÚUç‡æ·¤ Ñ
(1) x ×𴠃ææÌ 6 ·¤æ °·¤ ÕãéÂÎ ãñÐ
(2) x ×𴠃ææÌ 3 ·¤æ °·¤ ÕãéÂÎ ãñÐ
(3) x ×𴠃ææÌ 2 ·¤æ °·¤ ÕãéÂÎ ãñÐ
(4) x ÂÚU çÙÖüÚU Ùãè´ ·¤ÚUÌæÐ
0 âð 9 Ì·¤ ·ð¤ ¥´·¤æð´ ·ð¤ ÂýØæð» âð, ¥´·¤æð´ ·¤æð ÎæðãÚUæ°
çÕÙæ, °·¤ 9 âð Öæ’Ø, ¥æÆU ¥´·¤æð´ ·¤è ⴁØæ ÕÙæÙè
ãñÐ Øã çÁÌÙð ÌÚUè·¤æð´ âð ç·¤Øæ Áæ â·¤Ìæ ãñ, ßð ãñ´ Ñ
of B(x) :
66.
(1)
is a polynomial of degree 6 in x.
(2)
is a polynomial of degree 3 in x.
(3)
is a polynomial of degree 2 in x.
(4)
does not depend on x.
An eight digit number divisible by 9 is to 66.
be formed using digits from 0 to 9 without
repeating the digits. The number of ways
in which this can be done is :
(1)
72 (7!)
(1)
72 (7!)
(2)
18 (7!)
(2)
18 (7!)
(3)
40 (7!)
(3)
40 (7!)
(4)
36 (7!)
(4)
36 (7!)
English : 37
Set : 11
Hindi : 37
Set : 11
67.
The coefficient of x 50 in the binomial 67.
(11x) 1000 1 x(11x) 999 1x 2 (11x) 998 1
expansion of
......1x1000
(11x) 1000 1 x(11x) 999 1x 2 (11x) 998 1
·ð¤ çmÂÎ ÂýâæÚU ×ð´ x50 ·¤æ »é‡ææ¡·¤ ãñ Ñ
......1x1000 is :
68.
(1)
(1000) !
(50) ! (950) !
(1)
(1000) !
(50) ! (950) !
(2)
(1000) !
(49) ! (951) !
(2)
(1000) !
(49) ! (951) !
(3)
(1001) !
(51) ! (950) !
(3)
(1001) !
(51) ! (950) !
(4)
(1001) !
(50) ! (951) !
(4)
(1001) !
(50) ! (951) !
In a geometric progression, if the ratio of 68.
the sum of first 5 terms to the sum of their
reciprocals is 49, and the sum of the first
and the third term is 35. Then the first term
°·¤ »é‡ææðžæÚU oýðɸè ×ð´ ØçÎ ÂãÜð 5 ÂÎæð´ ·ð¤ Øæð» ·¤æ
©Ù·ð¤ ÃØ鈷ý¤×æð´ ·ð¤ Øæð» âð ¥ÙéÂæ̤ 49 ãñ ÌÍæ §â·ð¤
ÂãÜð ÌÍæ ÌèâÚðU ÂÎæð´ ·¤æ Øæð»U 35 ãñ, Ìæð §â »é‡ææðžæÚU
oýðÉ¸è ·¤æ ÂýÍ× ÂÎ ãñ Ñ
of this geometric progression is :
(1)
7
(1)
7
(2)
21
(2)
21
(3)
28
(3)
28
(4)
42
(4)
42
English : 38
Set : 11
Hindi : 38
Set : 11
69.
70.
The sum of the first 20 terms common 69.
oýðç‡æØæð´U
between the series 3171111151..... and
1161111161..... ,
1161111161..... , is :
20
(1)
4000
(1)
4000
(2)
4020
(2)
4020
(3)
4200
(3)
4200
(4)
4220
(4)
4220
If
lim
70.
tan ( x 2 2 ){x 2 1 (k22)x 2 2k}
x 2 2 4x 1 4
x →2
71.
55 ,
ÂÎæð´ ·¤æ Øæð» ãñ Ñ
ØçÎ
lim
tan ( x 2 2 ){x 2 1 (k22)x 2 2k}
x 2 2 4x 1 4
x →2
then k is equal to :
ãñ, Ìæð
k ÕÚUæÕÚU
(1)
0
(1)
0
(2)
1
(2)
1
(3)
2
(3)
2
(4)
3
(4)
3
Let f(x)5x?x?, g(x) = sin x and
71.
×æÙæ
f(x)5x?x?,
h(x) 5(gof ) (x) ãñ,
(1)
h(x) is not differentiable at x50.
(1)
(2)
h(x) is differentiable at x50, but
(2)
(3)
is not continuous at x50.
h’(x) is continuous at x50 but it is
(3)
not differentiable at x50.
(4)
h’(x) is differentiable at x50.
English : 39
Set : 11
(4)
55
ãñ Ñ
h(x) 5(gof )(x). Then
h’(x)
ÌÍæU
·ð¤ Õè¿ ©ÖØçÙcÆU ÂýÍ×
3171111151.....
g(x)
=
sin
x
ÌÍæ
Ìæð
h(x), x50 ÂÚU ¥ß·¤ÜÙèØ Ùãè´ ãñÐ
h(x), x50 ÂÚU ¥ß·¤ÜÙèØ ãñ ÂÚU‹Ìé h’(x),
x50 ÂÚU âÌÌ Ùãè´ ãñÐ
h’(x), x50 ÂÚU âÌÌ ãñ, ÂÚU‹Ìé Øã x50 ÂÚU
¥ß·¤ÜÙèØ Ùãè´ ãñÐ
h’(x), x50 ÂÚU ¥ß·¤ÜÙèØ ãñÐ
Hindi : 39
Set : 11
72.
For the curve y = 3 sin u cos u, x5eu sin u, 72.
0 [ u [ p, the tangent is parallel to x-axis
when u is :
73.
(1)
3p
(2)
p
(3)
p
(4)
p
ß·ý ¤
y = 3 sin u cos u, x5e u sin u,
0 [ u [ p, ·ð¤
çÜ° SÂàæüÚðU¹æ x- ¥ÿæ ·ð¤ â×æ´ÌÚU ãñ,
ÁÕ u ÕÚUæÕÚU ãñ Ñ
(1)
3p
2
(2)
p
4
(3)
p
6
(4)
p
4
Two ships A and B are sailing straight 73.
away from a fixed point O along routes
such that ÐAOB is always 1208. At a
certain instance, OA = 8 km, OB = 6 km
and the ship A is sailing at the rate of
20 km/hr while the ship B sailing at the
rate of 30 km/hr. Then the distance
4
2
4
6
Îæð Áãæ$Á A ÌÍæ B, °·¤ çÙçà¿Ì çÕ´Îé O âð ÎêÚU âèÏð
×æ»æðZ ÂÚU §â Âý·¤æÚU Áæ ÚUãð ãñ´ ç·¤ ÐAOB âÎæ 1208
ÚU ã Ìæ ãñ Ð ç·¤âè ÿæ‡æ, OA = 8 ç·¤×è ÌÍæ
OB = 6 ç·¤×è ãñ ÌÍæ Áãæ$Á A, 20 ç·¤×è/ƒæ´ÅUæ ·¤è
¿æÜ âð ¿Ü ÚUãæ ãñ ÁÕç·¤ Áãæ$Á B, 30 ç·¤×è/ƒæ´ÅUæ
·¤è ¿æÜ âð ¿Ü ÚUãæ ãñ, Ìæð A ÌÍæ B ·ð¤ Õè¿ ·¤è ÎêÚUè
çÁâ ÎÚU (ç·¤×è/ƒæ´ÅUæ ×ð´) âð ÕÎÜ ÚUãè ãñ, ßã ãñ Ñ
between A and B is changing at the rate
(in km/hr) :
(1)
260
(2)
260 37
(3)
80
(4)
80 37
English : 40
37
37
Set : 11
(1)
260
(2)
260 37
(3)
80
(4)
80 37
Hindi : 40
37
37
Set : 11
74.
The volume of the largest possible right 74.
circular cylinder that can be inscribed in a
ç˜æ’Øæ 3 ßæÜð »æðÜð ·ð¤ ¥´Ì»üÌ, ÕǸð âð ÕǸð Ü´Õ
ßëžæèØ ÕðÜÙ ·¤æ ¥æØÌÙ ãñ Ñ
sphere of radius5 3 is :
75.
(1)
4
3 p
3
(1)
4
3 p
3
(2)
8
3 p
3
(2)
8
3 p
3
(3)
4p
(3)
4p
(4)
2p
(4)
2p
The integral
2
21  1 2 x 
x
cos

 dx (x > 0) 75.
∫
 1 1 x2 
dx , (x > 0) ÕÚUæÕÚU
ãñ Ñ
is equal to :
(1)
2 x1(11x2) tan21 x1c
(1)
2 x1(11x2) tan21 x1c
(2)
x2 (11x2) cot21 x1c
(2)
x2 (11x2) cot21 x1c
(3)
2 x1(11x2) cot21 x1c
(3)
2 x1(11x2) cot21 x1c
(4)
x2 (11x2) tan21 x1c
(4)
x2 (11x2) tan21 x1c
e
76.
2

â×æ·¤Ü ∫ x cos21  1 2 x 2 
 11x 
n
If for n/1, Pn 5 ∫ (log x ) d x , then 76.
ØçÎ
P10290P8 is equal to :
P10290P8 ÕÚUæÕÚU
(1)
29
(1)
29
(2)
10e
(2)
10e
(3)
29e
(3)
29e
(4)
10
(4)
10
n/1
·ð¤ çÜ°,
1
English : 41
Set : 11
e
Pn 5 ∫ (log x )n d x
ãñ, Ìæð
1
Hindi : 41
ãñ Ñ
Set : 11
77.
If the general solution of the differential 77.
equation
y9 5
function
F,
x
y
1 F  ,
x
y
is
for
some
given
by
ØçÎ ç·¤âè ȤÜÙ
y9 5
F
·ð¤ çÜ° ¥ß·¤Ü â×è·¤ÚU‡æ
x
y
1 F  , ·¤æ
x
y
ÃØæ·¤ ãÜ
y ln ?cx?5x,
mæÚUæ ÂýΞæ ãñ, Áãæ¡ c °·¤ Sßð‘ÀU ¥¿ÚU ãñ, Ìæð
ÕÚUæÕÚU ãñ Ñ
y ln ?cx?5x, where c is an arbitrary
F(2)
constant, then F (2) is equal to :
78.
(1)
4
(1)
4
(2)
1
4
(2)
1
4
(3)
24
(3)
24
(4)
2
(4)
2
1
4
A stair-case of length l rests against a 78.
vertical wall and a floor of a room,. Let P
be a point on the stair-case, nearer to its
end on the wall, that divides its length in
the ratio 1 : 2. If the stair-case begins to
slide on the floor, then the locus of P is :
(1)
an ellipse of eccentricity
(2)
an ellipse of eccentricity
(3)
a circle of radius
(4)
a circle of radius
English : 42
1
2
3
2
l
2
3
l
2
Set : 11
1
4
Ü´Õæ§ü l ·¤è °·¤ âèÉ¸è °·¤ ©ŠßæüÏÚU ÎèßæÚU ÌÍæ ·¤×ÚðU
·ð¤ Ȥàæü ·ð¤ âæÍ ¹Ç¸è ãñÐ ×æÙæ §â âèɸè ÂÚU °·¤
çÕ´Îé P, Áæð §â·ð¤ ÎèßæÚU ·ð¤ âæÍ Ü»Ùð ßæÜð çâÚðU ·ð¤
çÙ·¤ÅU ãñ, §â Âý·¤æÚU ãñ ç·¤ Øã âèÉ¸è ·¤è Ü´Õæ§ü ·¤æð
1 : 2 ·ð¤ ¥ÙéÂæÌ ×ð´ Õæ´ÅUÌæ ãñÐ ØçÎ âèɸè Ȥàæü ÂÚU
âÚU·¤Ùð Ü»Ìè ãñ, Ìæð P ·¤æ çÕ´Îé ÂÍ ãñ Ñ
1
ßæÜæ
2
(1)
©ˆ·ð´¤ÎýÌæ
(2)
©ˆ·ð´¤ÎýÌæ
(3)
ç˜æ’Øæ
l
ßæÜæ
2
(4)
ç˜æ’Øæ
3
l
2
Hindi : 42
°·¤ ÎèƒæüßëÌÐ
3
ßæÜæ
2
°·¤ ÎèƒæüßëÌ
°·¤ ßëÌÐ
ßæÜæ °·¤ ßëÌÐ
Set : 11
79.
The base of an equilateral triangle is along 79.
the line given by 3x14y = 9. If a vertex of
the triangle is (1, 2), then the length of a
°·¤ â×Õæãé ç˜æÖéÁ ·¤æ ¥æÏæÚU ÚðU¹æ 3x14y = 9 ·ð¤
¥ÙéçÎàæ ãñÐ ØçÎ ç˜æÖéÁ ·¤æ °·¤ àæèáü (1, 2) ãñ Ìæð
ç˜æÖéÁ ·¤è °·¤ ÖéÁæ ·¤è Ü´Õæ§ü ãñ Ñ
side of the triangle is :
80.
(1)
2 3
15
(1)
2 3
15
(2)
4 3
15
(2)
4 3
15
(3)
4 3
5
(3)
4 3
5
(4)
2 3
5
(4)
2 3
5
The set of all real values of l for which 80.
exactly two common tangents can be
drawn to the circles
x21y224x24y1650 and
x21y2210x210y1l50 is the interval :
(1)
(12, 32)
(2)
(18, 42)
(3)
(12, 24)
(4)
(18, 48)
English : 43
Set : 11
l ·ð¤
âÖè ßæSÌçß·¤ ×æÙæð´ ·¤æ â×鑿Ø, çÁÙ·ð¤ çÜ°
ßëžææð´ x21y224x24y1650 ÌÍæ
x 2 1y 2 210x210y1l50 ÂÚU ÆUè·¤ Îæð
©ÖØçÙcÆU SÂàæüÚðU¹æ°¡ ¹è´¿è Áæ â·¤Ìè ãæð´, ·¤æ Áæð
¥´ÌÚUæÜ ãñ, ßã ãñ Ñ
(1)
(12, 32)
(2)
(18, 42)
(3)
(12, 24)
(4)
(18, 48)
Hindi : 43
Set : 11
81.
Let L1 be the length of the common chord 81.
of the curves x21y259 and y258x, and
L2 be the length of the latus rectum of
y258x, then :
82.
×æÙæ L1, ß·ý ¤ æð ´ x21y259 ÌÍæ y258x, ·¤è
©ÖØçÙcÆU Áèßæ ·¤è Ü´Õæ§ü ãñ, ÌÍæ L2, y258x ·ð¤
ÙæçÖÜ´Õ ·¤è Ü´Õæ§ü ãñ, Ìæð Ñ
(1)
L1 > L 2
(1)
L1 > L2
(2)
L15L2
(2)
L15L2
(3)
L1 < L 2
(3)
L1 < L2
(4)
L1
5 2
L2
(4)
L1
5 2
L2
Let P (3 sec u, 2 tan u) and
p
, be
2
two distinct points on the hyperbola
Q (3 sec f, 2 tan f) where u 1 f 5
2
2
y
x
2
5 1 . Then the ordinate of the
9
4
point of intersection of the normals at P
82.
×æÙæ ¥çÌÂÚUßÜØ
y2
x2
2
5 1 ÂÚU
9
4
Îæð çÖóæ çÕ´Îé
P (3 sec u, 2 tan u) ÌÍæ Q (3 sec f, 2 tan f)
ãñ´, Áãæ¡
u1f 5
p
2
ãñ, Ìæð P ÌÍæ Q ÂÚU ¹è´¿ð »°
¥çÖÜ´Õæð´ ·ð¤ Âýç̑ÀðUÎÙ çÕ´Îé ·¤è ·¤æðçÅU
ãñ Ñ
(ordinate)
and Q is :
(1)
11
3
(2)
2
(3)
13
2
(4)
2
English : 44
11
3
13
2
Set : 11
(1)
11
3
(2)
2
(3)
13
2
(4)
2
Hindi : 44
11
3
13
2
Set : 11
83.
×æÙæ
A (2, 3, 5), B (21, 3, 2)
the coordinate axes, then :
ÌÍæ
C (l, 5, m) °·¤ ç˜æÖéÁ ABC ·ð¤ àæèáü ãñ´Ð ØçÎ A
âð ãæð·¤ÚU ÁæÙð ßæÜè ×æçŠØ·¤æ, çÙÎðüàææ´·¤ ¥ÿææð´ ÂÚU
â×æÙ ·¤æð‡æ ÕÙæÌè ãñ, Ìæð Ñ
(1)
5l28m50
(1)
5l28m50
(2)
8l25m50
(2)
8l25m50
(3)
10l27m50
(3)
10l27m50
(4)
7l210m50
(4)
7l210m50
Úð U ¹ æ
y 22
x 21
z 23
5
5
1
2
3
Let A (2, 3, 5), B (21, 3, 2) and 83.
C (l, 5, m) be the vertices of a DABC. If the
median through A is equally inclined to
84.
The
plane
containing
the
line
y 22
x 21
z 23
and parallel to
5
5
1
2
3
85.
84.
·¤ÚUÙð ßæÜæ ÌÍæ ÚðU¹æ
y
x
z
5 5
1
1
4
·ð¤ â×æ´ÌÚU â×ÌÜ,
y
x
z
5 5 passes through the
the line
1
1
4
point :
çÁâ çÕ´Îé âð ãæð·¤ÚU ÁæÌæ ãñ, ßã ãñ Ñ
(1)
(1,22, 5)
(1)
(1,22, 5)
(2)
(1, 0, 5)
(2)
(1, 0, 5)
(3)
(0, 3,25)
(3)
(0, 3,25)
(4)
(21,23, 0)
(4)
(21,23, 0)
→
2
If ? c ? 5 60 and
→
→
∧
∧
∧
→
c 3( i12 j15 k )50
∧
∧
, 85.
∧
ØçÎ
→
? c ?2 5 60
→
∧
then a value of c . (2 7 i 1 2 j 1 3 k ) is :
Ìæð
(1)
4 2
(1)
4 2
(2)
12
(2)
12
(3)
24
(3)
24
(4)
12 2
(4)
12 2
English : 45
Set : 11
→
ÌÍæ
∧
∧
∧
→
c 3( i12 j15 k )50 ,
∧
∧
c . (2 7 i 1 2 j 1 3 k )
Hindi : 45
·¤æ𠥋Ìçßü c ÅU
ãñ,
·¤æ °·¤ ×æÙ ãñ Ñ
Set : 11
86.
A set S contains 7 elements. A non-empty 86.
subset A of S and an element x of S are
chosen at random. Then the probability
°·¤ â×é“æØ S ×ð´ 7 ¥ßØß ãñ´Ð S ·¤æ °·¤ ¥çÚU€ˆæ
©Ââ×é“æØ A ÌÍæ S ·¤æ °·¤ ¥ßØß x , ØæÎë‘ÀUØæ
¿éÙð »° , Ìæð xÎA ·¤è ÂýæçØ·¤Ìæ ãñ Ñ
that xÎA is :
87.
(1)
1
2
(1)
1
2
(2)
64
127
(2)
64
127
(3)
63
128
(3)
63
128
(4)
31
128
(4)
31
128
If X has a binomial distribution, B(n, p) 87.
with parameters n and p such that
P(X52)5P(X53), then E(X), the mean of
ØçÎ X ·¤æ °ð â æ çmÂÎ Õ´ Å UÙ, B(n, p) ãñ ,
Áãæ¡ n, p §â·ð¤ Âýæ¿Ü ãñ´, ¥æñÚU P(X52)5P(X53)
ãñ, Ìæð ¿ÚU X ·¤æ ×æŠØ E(X) ãñ Ñ
variable X, is :
(1)
22p
(1)
22p
(2)
32p
(2)
32p
(3)
(4)
p
p
2
(3)
3
(4)
English : 46
Set : 11
p
p
Hindi : 46
2
3
Set : 11
88.
p


If 2 cos u1sin u51 u ≠ 2  ,


88.


(1)
1
2
(1)
1
2
(2)
2
(2)
2
(3)
11
2
(3)
11
2
(4)
46
5
(4)
46
5
The angle of elevation of the top of a 89.
vertical tower from a point P on the
horizontal ground was observed to be a.
After moving a distance 2 metres from P
p

2 cos u1sin u51 u ≠ 2 
7 cos u16 sin u ÕÚUæÕÚU
then 7 cos u16 sin u is equal to :
89.
ØçÎ
ãñ , Ìæð
ãñ Ñ
â×ÌÜ Öêç× ÂÚU °·¤ çÕ´Îé P âð °·¤ ª¤ŠßæüÏÚU ×èÙæÚU
·ð¤ çàæ¹ÚU ·¤æ ©‹ÙØÙ ·¤æð‡æ a ÂæØæ »ØæÐ P âð ×èÙæÚU
·ð¤ ÂæÎ ·¤è ¥æðÚU 2 ×è. ÁæÙð ÂÚU, ©‹ÙØÙ ·¤æð‡æ ÕÎÜ
·¤ÚU b ãæð ÁæÌæ ãñ, Ìæð (×è.×ð´) ×èÙæÚU ·¤è ª¡¤¿æ§ü ãñ Ñ
towards the foot of the tower, the angle of
elevation changes to b. Then the height (in
metres) of the tower is :
(1)
2 sin a sin b
sin (b2a )
(1)
2 sin a sin b
sin (b2a )
(2)
sin a sin b
cos (b2a )
(2)
sin a sin b
cos (b2a )
(3)
2 sin (b2a )
sin a sin b
(3)
2 sin (b2a )
sin a sin b
(4)
cos (b2a )
sin a sin b
(4)
cos (b2a )
sin a sin b
English : 47
Set : 11
Hindi : 47
Set : 11
90.
is logically equivalent to :
âæŠØ ~(pÚ~q)Ú~(pÚq) Ìæç·ü¤·¤ M¤Â ×ð´ çÁâ·ð¤
ÌéËØ ãñ, ßã ãñ Ñ
(1)
p
(1)
p
(2)
q
(2)
q
(3)
~p
~q
(3)
~p
~q
The
(4)
proposition
~ (pÚ ~ q)Ú ~ (pÚq)
90.
(4)
-o0o-
English : 48
-o0o-
Set : 11
Hindi : 48
Set : 11