JEE Main 2017 Question Paper 08 Apr B.Tech

PHYSICS ÷ÊÒÁÃ∑§ ÁflôÊÊŸ

1. Time (T), velocity (C) and angular 1. ÿÁŒ Œ˝√ÿ◊ÊŸ, ‹ê’Ê߸ •ÊÒ⁄U ‚◊ÿ ∑§ SÕÊŸ ¬⁄U ‚◊ÿ
momentum (h) are chosen as fundamental
(T), flª (C) ÃÕÊ ∑§ÊáÊËÿ ‚¥flª (h) ∑§Ê ◊Í‹÷ÍÃ
quantities instead of mass, length and time.
In terms of these, the dimensions of mass ⁄UÊÁ‡ÊÿÊ° ◊ÊŸ ‹¥ ÃÊ Œ˝√ÿ◊ÊŸ ∑§Ë Áfl◊Ê ∑§Ê ߟ ⁄UÊÁ‡ÊÿÊ¥
would be : ∑§ M§¬ ◊¥ ÁŸêŸ Ã⁄UË∑§ ‚ Á‹π¥ª —
(1) [ M ]=[ T−1 C−2 h ]
(2) [ M ]=[ T−1 C2 h ] (1) [ M ]=[ T−1 C−2 h ]
(3) [ M ]=[ T−1 C−2 h−1 ] (2) [ M ]=[ T−1 C2 h ]
(4) [ M ]=[ T C−2 h ] (3) [ M ]=[ T−1 C−2 h−1 ]
(4) [ M ]=[ T C−2 h ]

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2. Which graph corresponds to an object 2. ÁSÕ⁄U ´§áÊÊà◊∑§ àfl⁄UáÊ fl œŸÊà◊∑§ flª ‚ ø‹Ÿ
moving with a constant negative
flÊ‹Ë ∞∑§ flSÃÈ ∑§ Á‹ÿ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ª˝Ê$»§ ‚„Ë
acceleration and a positive velocity ?
„Ò?

(1) (1)

(2) (2)

(3)
(3)

(4)
(4)

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3. A 1 kg block attached to a spring vibrates 3. ∞∑§ ÁS¬˝¥ª ‚ ¡È«∏Ê „È•Ê 1 kg ∑§Ê ∞∑§ ªÈ≈U∑§Ê 1 Hz
with a frequency of 1 Hz on a frictionless
∑§Ë •ÊflÎÁûÊ ‚ ∞∑§ ÉÊ·¸áÊ„ËŸ ˇÊÒÁá ◊¡ ¬⁄U ŒÊ‹Ÿ
horizontal table. Two springs identical to
the original spring are attached in parallel ∑§⁄UÃÊ „Ò– ß‚Ë Ã⁄U„ ∑§Ë ŒÊ ‚◊ÊãÃ⁄U ÁS¬˝¥ªÊ¥ ‚ ∞∑§
to an 8 kg block placed on the same table. 8 kg ∑§Ê ªÈ≈U∑§Ê ¡Ê«∏∑§⁄U ©‚Ë ◊¡ ¬⁄U ŒÊ‹Ÿ ∑§⁄UÊÃ
So, the frequency of vibration of the 8 kg „Ò¥– 8 kg ∑§ ªÈ≈U∑§ ∑§Ë ŒÊ‹Ÿ •ÊflÎÁûÊ „ÊªË —
block is :

1
(1) Hz
4
1
(1) Hz
1 4
(2) Hz
2 2
1
(2) Hz
1 2 2
(3) Hz
2
1
(4) 2 Hz (3) Hz
2

(4) 2 Hz
4. An object is dropped from a height h from
the ground. Every time it hits the ground
it looses 50% of its kinetic energy. The total 4. ∞∑§ flSÃÈ ∑§Ê œ⁄UÃË ‚ h ™°§øÊ߸ ‚ ¿UÊ«∏Ê ¡ÊÃÊ „Ò–
distance covered as t→∞ is : ¡’ ÿ„ flSÃÈ ¬ÎâflË ‚ ≈U∑§⁄UÊÃË „Ò ÃÊ ¬˝àÿ∑§ ≈UÄ∑§⁄U ◊¥
(1) 2h ©‚∑§Ë 50% ªÁá ™§¡Ê¸ ˇÊÿ „ÊÃË „Ò– ÿÁŒ t →∞,
(2) ∞ flSÃÈ mÊ⁄UÊ Ãÿ ∑§Ë ªÿË ∑ȧ‹ ŒÍ⁄UË „ÊªË —
5 (1) 2h
(3) h
3 (2) ∞

8 5
(4) h (3) h
3 3

8
(4) h
3

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5. A uniform disc of radius R and mass M is 5. ∞∑§ ÁòÊíÿÊ R ÃÕÊ Œ˝√ÿ◊ÊŸ M ∑§Ë ∞∑§‚◊ÊŸ Á«US∑§
free to rotate only about its axis. A string
∑§fl‹ •¬ŸË •ˇÊ ∑§ ¬Á⁄U× ÉÊÍáʸŸ ∑§ Á‹ÿ SflÃ¥òÊ „Ò–
is wrapped over its rim and a body of mass
m is tied to the free end of the string as ÁøòÊÊŸÈ‚Ê⁄U ß‚ Á«US∑§ ∑§Ë ¬Á⁄UÁœ ¬⁄U ∞∑§ «UÊ⁄UË ‹¬≈U∑§⁄U,
shown in the figure. The body is released «UÊ⁄UË ∑§ SflÃ¥òÊ Á‚⁄U ‚ ∞∑§ Œ˝√ÿ◊ÊŸ m ∑§Ê ’Ê°œÊ ªÿÊ
from rest. Then the acceleration of the „Ò– ÿÁŒ Œ˝√ÿ◊ÊŸ ∑§Ê ÁSÕ⁄UÊflSÕÊ ‚ ¿UÊ«∏Ê ¡ÊÃÊ „Ò ÃÊ
body is : ©‚∑§Ê àfl⁄UáÊ „ÊªÊ —

2 mg 2 mg
(1) (1)
2 m +M 2m+M

2 Mg 2 Mg
(2) (2)
2 m+M 2m+M

2 mg 2 mg
(3) (3)
2 M+m 2M+m

2 Mg 2 Mg
(4) (4)
2 M +m 2M+m

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6. Moment of inertia of an equilateral 6. ÁøòÊÊŸÈ‚Ê⁄U ‚◊’Ê„È ÁòÊ÷È¡ ∑§Ë •Ê∑ΧÁà flÊ‹ ∞∑§ ¬≈U‹
triangular lamina ABC, about the axis
ABC ∑§Ê ∞∑§ •ˇÊ, ¡Ê Á’ãŒÈ O ‚ ¡ÊÃË „Ò ÃÕÊ
passing through its centre O and
perpendicular to its plane is Io as shown ¬≈U‹ ∑§ •Á÷‹ê’flØ „Ò, ∑§ ‚ʬˇÊ ¡«∏àfl •ÊÉÊÍáʸ Io
in the figure. A cavity DEF is cut out from „Ò– ß‚ ¬≈U‹ ◊¥ ‚ ∞∑§ ÁòÊ÷È¡ DEF ∑§ •Ê∑§Ê⁄U ∑§Ê
the lamina, where D, E, F are the mid ∞∑§ ¿UŒ Á∑§ÿÊ ¡ÊÃÊ „Ò– ÿ„Ê° D, E fl F ÷È¡Ê•Ê¥ ∑§
points of the sides. Moment of inertia of
the remaining part of lamina about the
◊äÿ Á’ãŒÈ „Ò¥– ß‚ ’ø „È∞ ¬≈U‹ ∑§Ê ©‚Ë •ˇÊ ∑§
same axis is : ‚ʬˇÊ ¡«∏àfl •ÊÉÊÍáʸ „ÊªÊ —

7
7 (1) Io
(1) Io 8
8
15
15 (2) Io
(2) Io 16
16
3 Io
3 Io (3)
(3) 4
4
31 I o
31 I o (4)
(4) 32
32

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7. If the Earth has no rotational motion, the 7. ÿÁŒ ¬ÎâflË ∑§Ë ÉÊÍáʸŸ ªÁà ‡ÊÍãÿ „Ò ÃÊ ∞∑§ √ÿÁÄà ∑§Ê
weight of a person on the equator is W.
÷Í◊äÿ⁄UπÊ ¬⁄U ÷Ê⁄U W „Ò– ¬ÎâflË ∑§Ë •¬ŸË •ˇÊ ∑§
Determine the speed with which the earth
would have to rotate about its axis so that ¬Á⁄U× ÉÊÍáʸŸ ∑§Ë fl„ ªÁà ôÊÊà ∑§ËÁ¡ÿ Á¡‚ ¬⁄U ©‚
the person at the equator will weigh 3
√ÿÁÄà ∑§Ê ÷Í◊äÿ⁄UπÊ ¬⁄U ÷Ê⁄U 4 W „ÊªÊ– ¬ÎâflË
3
W. Radius of the Earth is 6400 km and ∑§Ë ÁòÊíÿÊ 6400 km •ÊÒ⁄U g=10 m/s2 „Ò–
4
g=10 m/s2.
(1) 1.1×10−3 rad/s
(2) 0.83×10−3 rad/s (1) 1.1×10−3 rad/s
(3) 0.63×10−3 rad/s (2) 0.83×10−3 rad/s

(4) 0.28×10−3 rad/s (3) 0.63×10−3 rad/s
(4) 0.28×10−3 rad/s
8. In an experiment a sphere of aluminium
of mass 0.20 kg is heated upto 1508C.
Immediately, it is put into water of volume
8. ∞∑§ ¬˝ÿÊª ◊¥ 0.20 kg Œ˝√ÿ◊ÊŸ ∑§ •ÀÿÈÁ◊ÁŸÿ◊ ∑§
150 cc at 278C kept in a calorimeter of ∞∑§ ªÊ‹ ∑§Ê 1508C Ã∑§ ª◊¸ Á∑§ÿÊ ¡ÊÃÊ „Ò– ß‚∑§
water equivalent to 0.025 kg. Final ÃÈ⁄¥Uà ’ÊŒ ß‚ 278C fl 150 cc •Êÿß flÊ‹ ¬ÊŸË ‚
temperature of the system is 408C. The ÷⁄U ∞∑§ ∑Ò§‹Ê⁄UË◊Ë≈U⁄U, ¡ÊÁ∑§ 0.025 kg ¬ÊŸË ∑§ ÃÈÀÿ
specific heat of aluminium is :
(take 4.2 Joule=1 calorie) „Ò, ◊¥ «UÊ‹ ŒÃ „Ò¥– ß‚ ÁŸ∑§Êÿ ∑§Ê •ãà Ãʬ◊ÊŸ
408C „Ò– •ÀÿÈÁ◊ÁŸÿ◊ ∑§Ë ÁflÁ‡ÊC ™§c◊Ê „ÊªË —
(1) 378 J/kg – 8C
(4.2 ¡Í‹=1 ∑Ò§‹Ê⁄UË „Ò–)
(2) 315 J/kg – 8C
(1) 378 J/kg – 8C
(3) 476 J/kg – 8C
(2) 315 J/kg – 8C
(4) 434 J/kg – 8C
(3) 476 J/kg – 8C
(4) 434 J/kg – 8C

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9. A compressive force, F is applied at the 9. S≈UË‹ ∑§Ë ∞∑§ ¬Ã‹Ë ∞fl¥ ‹ê’Ë ¿U«∏ ∑§ ŒÊŸÊ¥ Á‚⁄UÊ¥ ¬⁄U
two ends of a long thin steel rod. It is
∞∑§ ‚¥¬Ë«UŸ ’‹ F ‹ªÊÿÊ ¡ÊÃÊ „Ò ÃÕÊ ‚ÊÕ „Ë ¿U«∏
heated, simultaneously, such that its
temperature increases by ∆T. The net ∑§Ê ª◊¸ ∑§⁄U∑§ ©‚∑§Ê Ãʬ◊ÊŸ ∆T ’…∏ÊÿÊ ¡ÊÃÊ „Ò–
change in its length is zero. Let l be the ß‚‚ ¿U«∏ ∑§Ë ‹ê’Ê߸ ◊¥ ∑ȧ‹ ¬Á⁄UfløŸ ‡ÊÍãÿ „Ò– ◊ÊŸÊ
length of the rod, A its area of cross-section, Á∑§ ¿U«∏ ∑§Ë ‹ê’Ê߸ l, •ŸÈ¬˝SÕ ∑§Ê≈U ∑§Ê ˇÊòÊ»§‹ A,
Y its Young’s modulus, and α its coefficient
of linear expansion. Then, F is equal to :
ÿ¥ª ¬˝àÿÊSÕÃÊ ªÈáÊÊ¥∑§ Y fl ⁄UπËÿ ¬˝‚Ê⁄U ªÈáÊÊ¥∑§ α „Ò
ÃÊ F ∑§Ê ◊ÊŸ „ÊªÊ —
(1) l2 Yα ∆T
(2) lA Yα ∆T
(1) l2 Yα ∆T
(3) A Yα ∆T
(2) lA Yα ∆T
AY
(4) (3) A Yα ∆T
α ∆T
AY
(4)
α ∆T
10. An engine operates by taking n moles of
an ideal gas through the cycle ABCDA
shown in figure. The thermal efficiency 10. ÁøòÊ ◊¥ ÁŒπÊÿ ªÿ ∞∑§ ø∑˝§Ëÿ ¬˝∑˝§◊ ABCDA ∑§
of the engine is :
•ŸÈ‚Ê⁄U n ◊Ê‹ •ÊŒ‡Ê¸ ªÒ‚ ‚ ∞∑§ ߥ¡Ÿ ø‹ÊÿÊ ¡ÊÃÊ
(Take Cv=1.5 R, where R is gas constant)
„Ò– ߥ¡Ÿ ∑§Ë ÃʬËÿ ˇÊ◊ÃÊ „ÊªË —
(ÁŒÿÊ „Ò — Cv=1.5 R, ¡„Ê° R ªÒ‚ ÁŸÿÃÊ¥∑§ „Ò–)

(1) 0.24
(2) 0.15
(1) 0.24
(3) 0.32
(2) 0.15
(4) 0.08 (3) 0.32
(4) 0.08

7 VI - PHYSICS

11. An ideal gas has molecules with 5 degrees 11. ∞∑§ •ÊŒ‡Ê¸ ªÒ‚ ∑§ •áÊÈ•Ê¥ ∑§Ë SflÊÃ¥òÊÿ ∑§ÊÁ≈U
of freedom. The ratio of specific heats at
(degrees of freedom) 5 „Ò– ß‚ ªÒ‚ ∑§Ë ÁSÕ⁄U
constant pressure (C p) and at constant
volume (Cv) is : ŒÊ’ ¬⁄U ÁflÁ‡Êc≈U ™§c◊Ê (Cp) •ÊÒ⁄U ÁSÕ⁄U •Êÿß ¬⁄U
(1) 6
ÁflÁ‡ÊC ™§c◊Ê (Cv) ∑§Ê •ŸÈ¬Êà „ÊªÊ —
(1) 6
7
(2)
2 7
(2)
2
5
(3)
2 5
(3)
2
7
(4)
5 7
(4)
5

12. The ratio of maximum acceleration to
maximum velocity in a simple harmonic 12. ∞∑§ ‚⁄U‹ •Êflø ªÁà ◊¥ •Áœ∑§Ã◊ àfl⁄UáÊ ∞fl¥
motion is 10 s −1 . At, t=0 the
displacement is 5 m. What is the
•Áœ∑§Ã◊ flª ∑§Ê •ŸÈ¬Êà 10 s−1 „Ò– ÿÁŒ t=0
maximum acceleration ? The initial phase ¬⁄U ÁflSÕʬŸ 5 m „ÒÒ ÃÊ •Áœ∑§Ã◊ àfl⁄UáÊ ∑§Ê ◊ÊŸ
π π
is . ÄÿÊ „ÊªÊ? •Ê⁄UÁê÷∑§ ∑§‹Ê ∑§Ê ◊ÊŸ 4 „Ò–
4
(1) 500 m/s2

(2) 500 2 m/s2
(1) 500 m/s2
(3) 750 m/s2
(2) 500 2 m/s2
(4) 750 2 m/s2
(3) 750 m/s2

(4) 750 2 m/s2

8 VI - PHYSICS

13. Two wires W 1 and W 2 have the same 13. ŒÊ ÃÊ⁄UÊ¥ W1 ÃÕÊ W2 ∑§Ë ‚◊ÊŸ ÁòÊíÿÊ r „Ò ÃÕÊ
radius r and respective densities ρ1 and ρ2
ÉÊŸàfl ∑˝§◊‡Ê— ρ1 •ÊÒ⁄U ρ2 ß‚ ¬˝∑§Ê⁄U „Ò¥ Á∑§ ρ2=4ρ1–
such that ρ 2 =4ρ 1 . They are joined
together at the point O, as shown in the ÁøòÊÊŸÈ‚Ê⁄U ߟ ÃÊ⁄UÊ¥ ∑§Ê Á’ãŒÈ O ¬⁄U ¡Ê«∏Ê ªÿÊ „Ò– ß‚
figure. The combination is used as a ‚¥ÿÊ¡Ÿ ∑§Ê ‚ÊŸÊ◊Ë≈U⁄U ∑§ ÃÊ⁄U ∑§ M§¬ ◊¥ ¬˝ÿÊª ∑§⁄UÃ
sonometer wire and kept under tension T. „Ò¥ •ÊÒ⁄U ß‚ ßÊfl T ¬⁄U ⁄UπÃ „Ò¥– Á’ãŒÈ O, ŒÊŸÊ¥
The point O is midway between the two
bridges. When a stationary wave is set up
‚ÃÈ•Ê¥ ∑§ ◊äÿ ◊¥ „Ò¥– ß‚ ‚¥ÿÈÄà ÃÊ⁄U ◊¥ ∞∑§ •¬˝ªÊ◊Ë
in the composite wire, the joint is found to Ã⁄¥Uª ©à¬ÛÊ ∑§Ë ¡ÊÃË „Ò¥ ÃÊ ¡Ê«∏ ¬⁄U ÁŸS¬¥Œ (node)
be a node. The ratio of the number of ’ŸÃÊ „Ò– W1 fl W2 ÃÊ⁄UÊ¥ ◊¥ ’Ÿ ¬˝S¬¥ŒÊ¥ (antinode)
antinodes formed in W1 to W2 is : ∑§Ë ‚¥ÅÿÊ ∑§Ê •ŸÈ¬Êà „ÊªÊ —

(1) 1:1
(1) 1:1
(2) 1:2
(2) 1:2
(3) 1:3
(3) 1:3
(4) 4:1
(4) 4:1

14. There is a uniform electrostatic field in a
region. The potential at various points on 14. ∞∑§ ˇÊòÊ ◊¥ ∞∑§‚◊ÊŸ ÁSÕ⁄U flÒlÈà ˇÊòÊ ©¬ÁSÕà „Ò–
a small sphere centred at P, in the region, ÿ„Ê° ∞∑§ Á’ãŒÈ P ¬⁄U ∑§ÁãŒ˝Ã ∞∑§ ªÊ‹ ∑§ ÁflÁ÷ÛÊ
is found to vary between the limits Á’ãŒÈ•Ê¥ ¬⁄U Áfl÷fl ∑§Ê ◊ÊŸ 589.0 V fl 589.8 V
589.0 V to 589.8 V. What is the potential
‚Ë◊Ê•Ê¥ ∑§ ’Ëø ¬ÊÿÊ ¡ÊÃÊ „Ò– ß‚ ªÊ‹ ∑§ ¬Îc∆U ¬⁄U
at a point on the sphere whose radius
vector makes an angle of 608 with the fl„ Á’ãŒÈ, Á¡‚∑§Ê ÁòÊíÿÊ flÄ≈U⁄U ÁfllÈà ˇÊòÊ ‚ 608
direction of the field ? ∑§Ê ∑§ÊáÊ ’ŸÊÃÊ „Ò, ¬⁄U Áfl÷fl ∑§Ê ◊ÊŸ ÄÿÊ „ÊªÊ?
(1) 589.5 V
(2) 589.2 V
(3) 589.4 V (1) 589.5 V

(4) 589.6 V (2) 589.2 V
(3) 589.4 V
(4) 589.6 V

9 VI - PHYSICS

15. The energy stored in the electric field 15. œÊÃÈ ∑§ ∞∑§ ªÊ‹ ‚ ©à¬ÛÊ ÁfllÈà ˇÊòÊ ◊¥ ‚¥Áøà ™§¡Ê¸
produced by a metal sphere is 4.5 J. If the
∑§Ê ◊ÊŸ 4.5 J „Ò– ÿÁŒ ªÊ‹ ◊¥ ÁŸÁ„à •Êfl‡Ê 4 µC
sphere contains 4 µC charge, its radius will
„Ê ÃÊ ©‚∑§Ë ÁòÊíÿÊ ∑§Ê ◊ÊŸ „ÊªÊ —
1
be : [Take : = 9 × 109 N − m 2 /C 2 ] 1
4 πo [ÁŒÿÊ „Ò — 4 π = 9 × 109 N − m 2 /C 2 ]
o
(1) 20 mm
(1) 20 mm
(2) 32 mm
(2) 32 mm
(3) 28 mm
(3) 28 mm
(4) 16 mm
(4) 16 mm

16. What is the conductivity of a
semiconductor sample having electron 16. ∞∑§ •h¸øÊ‹∑§ ◊¥ ß‹ÒÄ≈˛UÊÚŸ ÃÕÊ „Ê‹ ∑§Ê ‚¥ÅÿÊ
concentration of 5×10 18 m −3 , hole ÉÊŸàfl ∑˝§◊‡Ê— 5×1018 m−3 fl 5×1019 m−3
concentration of 5×10 19 m −3, electron
mobility of 2.0 m 2 V −1 s −1 and hole
ÃÕÊ ©Ÿ∑§Ë ªÁÇÊË‹ÃÊ∞¥ ∑˝§◊‡Ê— 2.0 m2 V−1 s−1
mobility of 0.01 m2 V−1 s−1 ? fl 0.01 m2 V−1 s−1 „Ò¥– ß‚ •h¸øÊ‹∑§ ∑§Ë
(Take charge of electron as 1.6×10−19 C) øÊ‹∑§ÃÊ ÄÿÊ „ÊªË?
(1) 1.68 (Ω - m)−1
(ÁŒÿÊ „Ò ß‹ÒÄ≈˛UÊÚŸ ¬⁄U •Êfl‡Ê=1.6×10−19 C)
(2) 1.83 (Ω - m)−1 (1) 1.68 (Ω - m)−1

(3) 0.59 (Ω - m)−1 (2) 1.83 (Ω - m)−1
(3) 0.59 (Ω - m)−1
(4) 1.20 (Ω - m)−1
(4) 1.20 (Ω - m)−1

10 VI - PHYSICS

17. 17.

A 9 V battery with internal resistance of ∞∑§ 9 V ∑§Ë ’Ò≈U⁄UË, Á¡‚∑§Ê •ÊãÃÁ⁄U∑§ ¬˝ÁÃ⁄UÊœ
0.5 Ω is connected across an infinite 0.5 Ω „Ò, ∑§Ê ÁøòÊÊŸÈ‚Ê⁄U •Ÿãà ¬Á⁄U¬Õ ◊¥ ‹ªÊÿÊ
network as shown in the figure. All
ªÿÊ „Ò– ‚÷Ë •◊Ë≈U⁄U A1, A2, A3 ÃÕÊ flÊÀ≈U◊Ë≈U⁄U
ammeters A1, A2, A3 and voltmeter V are
ideal. V •ÊŒ‡Ê¸ „Ò¥–

Choose correct statement.
‚„Ë ∑§ÕŸ øÈÁŸÿ —
(1) Reading of A1 is 2 A (1) A1 ∑§Ê ¬Ê∆˜UÿÊ¥∑§ 2 A „Ò–
(2) Reading of A1 is 18 A (2) A1 ∑§Ê ¬Ê∆˜UÿÊ¥∑§ 18 A „Ò–
(3) Reading of V is 9 V (3) V ∑§Ê ¬Ê∆˜UÿÊ¥∑§ 9 V „Ò–
(4) Reading of V is 7 V
(4) V ∑§Ê ¬Ê∆˜UÿÊ¥∑§ 7 V „Ò–

18. In a certain region static electric and
magnetic fields exist. The magnetic field 18. ∞∑§ ˇÊòÊ ◊¥ ÁSÕ⁄U ÁfllÈà ∞fl¥ øÈê’∑§Ëÿ ˇÊòÊ ©¬ÁSÕÃ
→
→
(
is given by B = B0 ∧i + 2 ∧j−4 ∧k . If a test ) (
„Ò¥– øÈê’∑§Ëÿ ˇÊòÊ B = B0 ∧i + 2 ∧j−4 ∧k „Ò– )
charge moving with a velocity ÿÁŒ ∞∑§ ¬⁄UˡÊáÊ •Êfl ‡ Ê, Á¡‚∑§Ê fl ª
→
→
( )
v =v0 3 ∧i −∧j+2 ∧k experiences no force ( )
v = v0 3 ∧i −∧j+2 ∧k , ¬⁄U ∑§Ê߸ ’‹ Ÿ„Ë¥ ‹ªÃÊ

in that region, then the electric field in the „Ò ÃÊ ß‚ ˇÊòÊ ◊¥ SI ◊ÊòÊ∑§Ê¥ ◊¥ ÁfllÈà ˇÊòÊ „ÊªÊ —
region, in SI units, is :
→
(1) (
E =−v0 B0 3 ∧i −2 ∧j−4 ∧k ) →

→
(1) (
E =−v0 B0 3 ∧i −2 ∧j−4 ∧k )
(2) (
E =−v0 B0 i + j+7 k ∧ ∧ ∧
) →

→
(2) (
E =−v0 B0 ∧i +∧j+7 ∧k )
(3) (
E = v0 B0 14 j+7 k ∧ ∧
) →

→
(3) (
E = v0 B0 14 ∧j+7 ∧k )
(4) (
E =− v0 B0 14 ∧j+7 ∧k ) →
(4) (
E =− v0 B0 14 ∧j+7 ∧k )
11 VI - PHYSICS

19. A magnetic dipole in a constant magnetic 19. ∞∑§ øÈê’∑§Ëÿ ˇÊòÊ ◊¥ ⁄Uπ „È∞ øÈê’∑§Ëÿ Ámœ˝Èfl ∑§Ë —
field has :
(1) maximum potential energy when
the torque is maximum. (1) ÁSÕÁá ™§¡Ê¸ •Áœ∑§Ã◊ „ÊªË ÿÁŒ ’‹ •ÊÉÊÍáʸ
(2) zero potential energy when the
•Áœ∑§Ã◊ „Ò–
torque is minimum. (2) ÁSÕÁá ™§¡Ê¸ ‡ÊÍãÿ „ÊªË ÿÁŒ ’‹ •ÊÉÊÍáʸ
(3) zero potential energy when the ãÿÍŸÃ◊ „Ò–
torque is maximum.
(3) ÁSÕÁá ™§¡Ê¸ ‡ÊÍãÿ „ÊªË ÿÁŒ ’‹ •ÊÉÊÍáʸ
(4) minimum potential energy when the
torque is maximum.
•Áœ∑§Ã◊ „Ò–
(4) ÁSÕÁá ™§¡Ê¸ ãÿÍŸÃ◊ „ÊªË ÿÁŒ ’‹ •ÊÉÊÍáʸ
20. A small circular loop of wire of radius a is
•Áœ∑§Ã◊ „Ò–
located at the centre of a much larger
circular wire loop of radius b. The two
loops are in the same plane. The outer loop 20. ÃÊ⁄U ‚ ’Ÿ ÁòÊíÿÊ a ∑§ ¿UÊ≈U flÎûÊÊ∑§Ê⁄U fl‹ÿ ∑§Ê ÁòÊíÿÊ
of radius b carries an alternating current b ∑§ ∞∑§ ’΄Ø flÎûÊÊ∑§Ê⁄U fl‹ÿ ∑§ ∑§ãŒ˝ ¬⁄U ⁄UπÊ ªÿÊ
I=Io cos (ωt). The emf induced in the „Ò– ŒÊŸÊ¥ fl‹ÿ ∞∑§ „Ë ‚◊Ë ◊¥ „Ò¥– ÁòÊíÿÊ b ∑§
smaller inner loop is nearly :
’Ês fl‹ÿ ◊¥ ∞∑§ ¬˝àÿÊflÃ˸ œÊ⁄UÊ I=Io cos (ωt)
πµo Io a 2 ¬˝flÊÁ„à ∑§Ë ¡ÊÃË „Ò– ÁòÊíÿÊ a flÊ‹ •ÊãÃÁ⁄U∑§ fl‹ÿ
(1) . ω sin ( ωt) ◊¥ ¬˝Á⁄Uà ÁfllÈà flÊ„∑§ ’‹ „ÊªÊ —
2 b

πµo Io a 2
(2) . ω cos (ωt) πµo Io a 2
2 b (1) . ω sin ( ωt)
2 b
a2
(3) πµo I o ω sin (ωt) πµo Io a 2
b (2) . ω cos (ωt)
2 b
πµo Io b2
(4) ω cos (ωt) a2
a (3) πµo I o ω sin (ωt)
b

πµo Io b2
(4) ω cos (ωt)
a

12 VI - PHYSICS

21. Magnetic field in a plane electromagnetic 21. ∞∑§ ‚◊Ë flÒlÈÃøÈê’∑§Ëÿ Ã⁄¥Uª ◊¥ øÈê’∑§Ëÿ ˇÊòÊ
wave is given by
→ ∧
→ ∧ B = B0 sin (k x+ω t) j T „Ò – ß‚∑ § ‚¥ ª Ã
B = B0 sin (k x+ω t) j T
ÁfllÈà ˇÊòÊ ∑§Ê ‚ÍòÊ „ÊªÊ —
Expression for corresponding electric field
will be : ÿ„Ê° c ¬˝∑§Ê‡Ê ∑§Ê flª „Ò–
Where c is speed of light.
→ ∧
(1) E = B0 c sin (k x+ω t) k V/m
→ ∧
(1) E = B0 c sin (k x+ω t) k V/m
→ B ∧
(2) E = 0 sin (k x+ωt) k V/m
→ B ∧ c
(2) E = 0 sin (k x+ωt) k V/m
c
→ ∧
(3) E =− B0 c sin (k x+ω t) k V/m
→ ∧
(3) E =− B0 c sin (k x+ω t) k V/m
→ ∧
(4) E = B0 c sin (k x−ω t) k V/m
→ ∧
(4) E = B0 c sin (k x−ω t) k V/m

22. ◊ÊŸÊ Á∑§ ∞∑§ ‚ÉÊŸ ◊Êäÿ◊ ∑§Ê ∞∑§ Áfl⁄U‹ ◊Êäÿ◊ ∑§
22. Let the refractive index of a denser medium ‚ʬˇÊ •¬fløŸÊ¥∑§ n12 „Ò ÃÕÊ ©‚∑§Ê ∑˝§ÊÁãÃ∑§ ∑§ÊáÊ
with respect to a rarer medium be n12 and
θC „Ò– ¡’ ¬˝∑§Ê‡Ê ∞∑§ •Ê¬ÃŸ ∑§ÊáÊ A ‚ ‚ÉÊŸ ‚
its critical angle be θC. At an angle of
incidence A when light is travelling from Áfl⁄U‹ ◊Êäÿ◊ ◊¥ ¡ÊÃÊ „Ò ÃÊ ©‚∑§Ê ∞∑§ ÷ʪ ¬⁄UÊflÁøÃ
denser medium to rarer medium, a part of „ÊÃÊ „Ò •ÊÒ⁄U ’øÊ „È•Ê ÷ʪ •¬flÁøà „ÊÃÊ „Ò– ¬⁄UÊflÁøÃ
the light is reflected and the rest is refracted •ÊÒ ⁄ U •¬flÁø à Á∑§⁄U á ÊÊ ¥ ∑ § ’Ëø ∑§Ê á Ê 908 „Ò–
and the angle between reflected and
refracted rays is 908. Angle A is given by : ∑§ÊáÊ A ∑§Ê ◊ÊŸ „ÊªÊ —

1
(1) −1 1
cos ( sin θC ) (1) −1
cos ( sin θC )
1
(2) −1 1
tan ( sin θC ) (2) −1
tan ( sin θC )
(3) cos−1 (sin θC)
(3) cos−1 (sin θC)
(4) tan−1 (sin θC)
(4) tan−1 (sin θC)

13 VI - PHYSICS

23. A single slit of width b is illuminated by a 23. λ Ã⁄¥UªŒÒäÿ¸ ∑§ ∑§‹Ê‚ê’h fl ∞∑§fláÊ˸ÿ ¬˝∑§Ê‡Ê ‚
coherent monochromatic light of
∞∑§ b øÊÒ«∏Ê߸ ∑§Ë Á¤Ê⁄UË ∑§Ê ¬˝ŒË# ∑§⁄UÃ „Ò¥– ÿÁŒ 1 m
wavelength λ. If the second and fourth
minima in the diffraction pattern at a ŒÍ⁄UË ¬⁄U ⁄Uπ ¬Œ¸ ¬⁄U ’Ÿ ÁflfløŸ ¬˝ÊM§¬ ◊¥ ÁmÃËÿ ∞fl¥
distance 1 m from the slit are at 3 cm and øÃÈÕ¸ ÁŸÁêŸc∆U ∑§Ë ∑§ãŒ˝Ëÿ ©ÁìÊc∆U ‚ ŒÍ⁄UË ∑˝§◊‡Ê—
6 cm respectively from the central 3 cm •ÊÒ⁄U 6 cm „Ò ÃÊ ∑§ãŒ˝Ëÿ ©Áëøc∆U ∑§Ë øÊÒ«∏Ê߸
maximum, what is the width of the central
maximum ? (i.e. distance between first
ÄÿÊ „ÊªË? (∑§ãŒ˝Ëÿ ©ÁëøD ∑§Ë øÊÒ«∏Ê߸ ©‚∑§ ŒÊŸÊ¥
minimum on either side of the central Ã⁄U»§ ∑§ ¬˝Õ◊ ÁŸÁêŸc∆U ∑§ ’Ëø ∑§Ë ŒÍ⁄UË „Ò–)
maximum)
(1) 1.5 cm
(2) 3.0 cm
(3) 4.5 cm (1) 1.5 cm

(4) 6.0 cm (2) 3.0 cm
(3) 4.5 cm

24. The maximum velocity of the (4) 6.0 cm
photoelectrons emitted from the surface
is v when light of frequency n falls on a
metal surface. If the incident frequency is 24. ¡’ •ÊflÎÁûÊ n ∑§Ê ¬˝∑§Ê‡Ê ∞∑§ œÊÃÈ ∑§ ¬Îc∆U ¬⁄U ¬«∏ÃÊ
increased to 3n, the maximum velocity of „Ò ÃÊ ©‚‚ ©à‚Á¡¸Ã »§Ê≈UÊ-ß‹Ä≈˛UÊÚŸÊ¥ ∑§Ê •Áœ∑§Ã◊
the ejected photoelectrons will be :
flª v „Ò– ÿÁŒ •Ê¬ÁÃà ¬˝∑§Ê‡Ê ∑§Ë •ÊflÎÁûÊ ’…∏Ê∑§⁄U
(1) less than 3v 3n ∑§⁄U ŒË ¡ÊÃË „Ò ÃÊ ©à‚Á¡¸Ã ß‹Ä≈˛UÊÚŸÊ¥ ∑§Ê •Áœ∑§Ã◊

(2) v
flª „ÊªÊ —

(3) more than 3v
(1) 3 v ‚ ∑§◊
(4) equal to 3v (2) v

(3) 3 v ‚ •Áœ∑§

(4) 3 v ∑§ ’⁄UÊ’⁄U

14 VI - PHYSICS

25. According to Bohr’s theory, the time 25. ’Ê⁄U (Bohr) ∑§ Á‚hÊãà ∑§ •ŸÈ‚Ê⁄U „Êß«˛UÊ¡Ÿ ¬⁄U◊ÊáÊÈ
averaged magnetic field at the centre (i.e.
∑§ ∑§ãŒ˝ (ŸÊÁ÷∑§) ¬⁄U nfl¥ ∑§ˇÊ ◊¥ ß‹Ä≈˛UÊÚŸ ∑§Ë ªÁÃ
nucleus) of a hydrogen atom due to the
motion of electrons in the n th orbit is ∑§ ∑§Ê⁄UáÊ ©à¬ãŸ ‚◊ÿ-•ÊÒ‚Ã øÈê’∑§Ëÿ ˇÊòÊ ∑§Ê
proportional to : (n=principal quantum ◊ÊŸ ÁŸêŸ ◊¥ ‚ Á∑§‚∑§ ‚◊ʟȬÊÃË „ÊªÊ — (ÿ„Ê° n
number) ◊ÈÅÿ ÄflÊã≈U◊ ‚¥ÅÿÊ „Ò–)
(1) n−4
(2) n−5 (1) n−4
(3) n−3 (2) n−5

(4) n−2 (3) n−3
(4) n−2
26. Two deuterons undergo nuclear fusion to
form a Helium nucleus. Energy released
in this process is : (given binding energy
26. ŒÊ «˜UÿÍ≈˛UÊÚŸÊ¥ ∑§ ŸÊÁ÷∑§Ëÿ ‚¥‹ÿŸ ‚ ∞∑§ „ËÁ‹ÿ◊
per nucleon for deuteron=1.1 MeV and ŸÊÁ÷∑§ ’ŸÃÊ „Ò– ß‚ ¬˝Á∑˝§ÿÊ ◊¥ ©à‚Á¡¸Ã ™§¡Ê¸ ∑§Ê
for helium=7.0 MeV) ◊ÊŸ „ÊªÊ — (ÁŒÿÊ „Ò — «˜UÿÍ≈˛UÊÚŸ ∑§Ë ¬˝ÁÃ-ãÿÍÁÄ‹•ÊÚŸ
(1) 30.2 MeV ’㜟 ™§¡Ê¸=1.1 MeV ÃÕÊ „ËÁ‹ÿ◊ ∑§Ë ¬˝ÁÃ
(2) 32.4 MeV ãÿÍÁÄ‹•ÊÚŸ ’㜟 ™§¡Ê¸=7.0 MeV)
(3) 23.6 MeV (1) 30.2 MeV

(4) 25.8 MeV (2) 32.4 MeV
(3) 23.6 MeV
(4) 25.8 MeV

15 VI - PHYSICS

27. The V-I characteristic of a diode is shown 27. ∞∑§ «UÊÿÊ«U ∑§Ê V-I •Á÷‹ˇÊÁáÊ∑§ fl∑˝§ ∑§Ê ÁøòÊ ◊¥
in the figure. The ratio of forward to
ÁŒπÊÿÊ ªÿÊ „Ò– •ª˝ÁŒÁ‡Ê∑§ ÃÕÊ ¬‡øÁŒÁ‡Ê∑§ ’Êÿ‚
reverse bias resistance is :
◊¥ ¬˝ÁÃ⁄UÊœ ∑§Ê •ŸÈ¬Êà „ÊªÊ —

(1) 10
(1) 10
(2) 10 −6
(2) 10 −6
(3) 10 6
(3) 10 6
(4) 100
(4) 100

16 VI - PHYSICS

28. A signal of frequency 20 kHz and peak 28. 1.2 MHz •ÊflÎÁûÊ ÃÕÊ 25 V Á‡Êπ⁄U flÊÀ≈UÃÊ flÊ‹Ë
voltage of 5 Volt is used to modulate a
∞∑§ flÊ„∑§ Ã⁄¥Uª ∑§Ê 20 kHz •ÊflÎÁûÊ ÃÕÊ Á‡Êπ⁄U
carrier wave of frequency 1.2 MHz and
peak voltage 25 Volts. Choose the correct flÊÀ≈UÃÊU 5 V ∑§ Á‚ÇŸ‹ ‚ ◊Ê«ÈUÁ‹Ã Á∑§ÿÊ ¡ÊÃÊ „Ò–
statement. ÁŸêŸÁ‹Áπà ◊¥ ‚ ‚„Ë ∑§ÕŸ øÈÁŸÿ–
(1) Modulation index=5, side
frequency bands are at 1400 kHz (1) ◊Ê«ÈU‹Ÿ ‚Íø∑§Ê¥∑§=5, ¬Ê‡fl¸ •ÊflÎÁûÊ ’Òá«U
and 1000 kHz
1400 kHz ÃÕÊ 1000 kHz ¬⁄U „Ò–
(2) Modulation index=5, side
frequency bands are at 21.2 kHz and
18.8 kHz (2) ◊Ê«ÈU‹Ÿ ‚Íø∑§Ê¥∑§=5, ¬Ê‡fl¸ •ÊflÎÁûÊ ’Òá«U
(3) Modulation index=0.8, side 21.2 kHz ÃÕÊ 18.8 kHz ¬⁄U „Ò–
frequency bands are at 1180 kHz
and 1220 kHz
(4) Modulation index=0.2, side
(3) ◊Ê«ÈU‹Ÿ ‚Íø∑§Ê¥∑§=0.8, ¬Ê‡fl¸ •ÊflÎÁûÊ ’Òá«U
frequency bands are at 1220 kHz 1180 kHz ÃÕÊ 1220 kHz ¬⁄U „Ò–
and 1180 kHz
(4) ◊Ê«ÈU‹Ÿ ‚Íø∑§Ê¥∑§=0.2, ¬Ê‡fl¸ •ÊflÎÁûÊ ’Òá«U
29. In a physical balance working on the 1220 kHz ÃÕÊ 1180 kHz ¬⁄U „Ò–
principle of moments, when 5 mg weight
is placed on the left pan, the beam becomes
horizontal. Both the empty pans of the
balance are of equal mass. Which of the 29. ’‹ •ÊÉÊÍáʸ ∑§ Á‚hÊãà ¬⁄U ∑§Êÿ¸ ∑§⁄UŸ flÊ‹Ë ∞∑§
following statements is correct ?
÷ÊÒÁÃ∑§ ÃÈ‹Ê ∑§ ’Ê°ÿ ¬‹«∏ ◊¥ ¡’ 5 mg ÷Ê⁄U ⁄UπÊ
(1) Left arm is longer than the right arm ¡ÊÃÊ „Ò ÃÊ ∑§◊ÊŸË ˇÊÒÁá „Ê ¡ÊÃË „Ò– ÃÈ‹Ê ∑§ ŒÊŸÊ¥
(2) Both the arms are of same length ¬‹«∏Ê¥ ∑§Ê Œ˝√ÿ◊ÊŸ ‚◊ÊŸ „Ò– ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ
(3) Left arm is shorter than the right arm ‚Ê ∑§ÕŸ ‚àÿ „Ò?
(4) Every object that is weighed using
this balance appears lighter than its (1) ’Ê°ÿË ÷È¡Ê, ŒÊ°ÿË ÷È¡Ê ‚ ‹ê’Ë „Ò–
actual weight.
(2) ŒÊŸÊ¥ ÷È¡Êÿ¥ ‚◊ÊŸ ‹ê’Ê߸ ∑§Ë „Ò¥–
(3) ’Ê°ÿË ÷È¡Ê, ŒÊ°ÿË ÷È¡Ê ‚ ¿UÊ≈UË „Ò–
(4) ¬˝àÿ∑§ flSÃÈ Á¡‚∑§Ê ß‚ ÃÈ‹Ê ¬⁄U ÃÊÒ‹Ê ¡ÊÃÊ
„Ò, ©‚∑§Ê ÷Ê⁄U •¬Ÿ flÊSÃÁfl∑§ ÷Ê⁄U ‚ ∑§◊
¬˝ÃËà „ÊÃÊ „Ò–

17 VI - PHYSICS

30. A potentiometer PQ is set up to compare 30. ∞∑§ Áfl÷fl◊Ê¬Ë PQ ∑§Ê ŒÊ ¬˝ÁÃ⁄UÊœÊ¥ ∑§Ë ÃÈ‹ŸÊ ∑§⁄UŸ
two resistances as shown in the figure. The
∑§ Á‹ÿ, ÁøòÊÊŸÈ‚Ê⁄U, ‚◊ÊÿÊÁ¡Ã Á∑§ÿÊ ¡ÊÃÊ „Ò– ÿÁŒ
ammeter A in the circuit reads 1.0 A when
two way key K3 is open. The balance point ∑È¥§¡Ë K3 ∑§Ê πÊ‹ ÁŒÿÊ ¡Êÿ ÃÊ •◊Ë≈U⁄U A ◊¥ œÊ⁄UÊ
is at a length l1 cm from P when two way 1.0 A •ÊÃË „Ò– ¡’ ÁmªÊ◊Ë ∑È¥§¡Ë K3 ∑§Ê 2 ÃÕÊ 1
key K3 is plugged in between 2 and 1, while ∑§ ’Ëø ‹ªÊÿÊ ¡ÊÃÊ „Ò ÃÊ ‚¥ÃÈ‹Ÿ Á’ãŒÈ P ‚
the balance point is at a length l2 cm from
l1 cm ŒÍ⁄UË ¬⁄U •ÊÃÊ „Ò, ¡’Á∑§ K3 ∑§Ê 3 ÃÕÊ 1 ∑§
P when key K3 is plugged in between 3
R1
’Ëø ‹ªÊŸ ¬⁄U, ‚¥ÃÈ‹Ÿ Á’ãŒÈ P ‚ l2 cm ŒÍ⁄UË ¬⁄U
and 1. The ratio of two resistances R , is R1
2 •ÊÃÊ „Ò– ŒÊŸÊ¥ ¬˝ÁÃ⁄UÊœÊ¥ ∑§ •ŸÈ¬Êà R ∑§Ê ◊ÊŸ
found to be : 2
„ÊªÊ —

l1
(1) l1
l1+l2 (1) l1+l2
l2
(2) l2
l2 −l1 (2) l2 −l1
l1
(3) l1
l1−l2 (3) l1−l2
l1
(4) l1
l2 −l1 (4) l2 −l1

18 VI - PHYSICS

CHEMISTRY ⁄U‚ÊÿŸ ‡ÊÊSòÊ

1. Among the following, correct statement
1. ÁŸêŸÁ‹Áπà ◊¥ ‚ ‚„Ë ∑§ÕŸ „Ò —
is :
(1) Brownian movement is more
pronounced for smaller particles (1) ’˝Ê©ŸË ªÁà ’«∏ ∑§áÊÊ¥ ∑§Ë ÃÈ‹ŸÊ ◊¥ ¿UÊ≈U ∑§áÊÊ¥
than for bigger–particles. ◊¥ •Áœ∑§ ÁŒπÊ߸ ŒÃË „Ò–
(2) Sols of metal sulphides are lyophilic.
(3) Hardy Schulze law states that bigger (2) œÊÃÈ ‚À»§Êß«∏Ê¥ ∑§ ‚ÊÚ‹ Œ˝fl⁄UÊªË „ÊÃ „Ò¥–
the size of the ions, the greater is its
coagulating power.
(3) „Ê«U˸-‡ÊÈÀ‚ ∑§ ÁŸÿ◊ ∑§ •ŸÈ‚Ê⁄U Á∑§‚Ë •ÊÿŸ
∑§Ê Á¡ÃŸÊ •Áœ∑§ •Ê◊ʬ „Ê ©‚∑§Ë S∑¥§ŒŸ
(4) One would expect charcoal to adsorb
chlorine more than hydrogen ‡ÊÁÄà ÷Ë ©ÃŸË „Ë •Áœ∑§ „ÊªË–
sulphide. (4) ∞‚Ë •Ê‡ÊÊ ∑§Ë ¡ÊÃË „Ò Á∑§ øÊ⁄U∑§Ê‹, „Êß«˛UÊ¡Ÿ
‚À»§Êß«U ∑§Ë ÃÈ‹ŸÊ ◊¥ Ä‹Ê⁄UËŸ ∑§Ê •Áœ∑§
2. Excess of NaOH (aq) was added to •Áœ‡ÊÊ·áÊ ∑§⁄UªÊ–
100 mL of FeCl 3 (aq) resulting into
2.14 g of Fe(OH) 3 . The molarity of
FeCl3 (aq) is : 2. 100 mL FeCl3 (¡‹Ëÿ) ◊¥ NaOH (¡‹Ëÿ)
(Given molar mass of Fe=56 g mol−1 and ∑§Ê •ÊÁœÄÿ ◊¥ «UÊ‹Ÿ ¬⁄U 2.14 g Fe(OH)3 ¬˝ÊåÃ
molar mass of Cl=35.5 g mol−1) „ÊÃÊ „Ò– FeCl3 (¡‹Ëÿ) ∑§Ë ◊Ê‹⁄UÃÊ „Ò,
(1) 0.2 M (ÁŒÿÊ ªÿÊ „Ò — Fe ∑§Ê ◊Ê‹⁄U Œ˝√ÿ◊ÊŸ=56 g mol−1
(2) 0.3 M ÃÕÊ Cl ∑§Ê ◊Ê‹⁄U Œ˝√ÿ◊ÊŸ=35.5 g mol−1)
(3) 0.6 M (1) 0.2 M
(4) 1.8 M (2) 0.3 M
(3) 0.6 M
(4) 1.8 M

1 VI - CHEMISTRY

3. Among the following, the incorrect 3. ÁŸêŸÁ‹Áπà ◊¥ ‚ •‚àÿ ∑§ÕŸ „Ò —
statement is :
(1) At low pressure, real gases show
ideal behaviour. (1) ∑§◊ ŒÊ’ ¬⁄U, flÊSÃÁfl∑§ ªÒ‚¥, •ÊŒ‡Ê¸ √ÿfl„Ê⁄U
(2) At very low temperature, real gases
Œ‡ÊʸÃË „Ò¥–
show ideal behaviour. (2) ’„Èà ÁŸêŸ Ãʬ ¬⁄U, flÊSÃÁfl∑§ ªÒ‚¥ •ÊŒ‡Ê¸
(3) At very large volume, real gases √ÿfl„Ê⁄U Œ‡ÊʸÃË „Ò¥–
show ideal behaviour.
(3) •Áœ∑§ ’«∏ •Êÿß ¬⁄U, flÊSÃÁfl∑§ ªÒ‚¥ •ÊŒ‡Ê¸
(4) At Boyle’s temperature, real gases
show ideal behaviour. √ÿfl„Ê⁄U Œ‡ÊʸÃË „Ò¥–
(4) ’ÊÚÿ‹ Ãʬ ¬⁄U, flÊSÃÁfl∑§ ªÒ‚¥ •ÊŒ‡Ê¸ √ÿfl„Ê⁄U
4. For a reaction, A(g) → A(l); ∆H=−3RT. Œ‡ÊʸÃË „Ò¥–
The correct statement for the reaction is :
(1) ∆H=∆U≠O
4. ∞∑§ •Á÷Á∑˝§ÿÊ A(g) → A(l) ∑§ Á‹∞ ∆H=−3RT.
(2) ∆H=∆U=O ß‚ •Á÷Á∑˝§ÿÊ ∑§ Á‹ÿ ‚„Ë ∑§ÕŸ „Ò —
(3) ?∆H? < ?∆U?
(1) ∆H=∆U≠O
(4) ?∆H? > ?∆U? (2) ∆H=∆U=O
(3) ?∆H? < ?∆U?
5. What is the standard reduction potential
(4) ?∆H? > ?∆U?
(E8) for Fe3+ → Fe ?
Given that :
5. Fe3+ → Fe ∑§ Á‹ÿ ◊ÊŸ∑§ •¬øÿŸ Áfl÷fl (E8)
Fe 2++2e− → Fe ; E
Fe 2+/Fe =−0.47 V ÄÿÊ „ÊªÊ?
ÁŒÿÊ ªÿÊ „Ò —
Fe 3++e− → Fe 2+ ; E
Fe 3+/Fe 2+=+0.77 V
Fe 2++2e− → Fe ; E
Fe 2+/Fe =−0.47 V
(1) −0.057 V
(2) +0.057 V
Fe 3++e− → Fe 2+ ; EFe

3+ 2+=+0.77 V
/Fe
(3) +0.30 V
(4) −0.30 V (1) −0.057 V
(2) +0.057 V
(3) +0.30 V
(4) −0.30 V

2 VI - CHEMISTRY

6. If the shortest wavelength in Lyman series 6. ÿÁŒ „Êß«˛UÊ¡Ÿ ¬⁄U◊ÊáÊÈ ∑§Ë ‹Êß◊Ÿ üÊáÊË ∑§Ë ‹ÉÊÈûÊ◊
of hydrogen atom is A, then the longest
Ã⁄¥UªŒÒÉÿ¸ A „Ê ÃÊ He+ ∑§Ë ¬Ê‡ÊŸ üÊáÊË ∑§Ë ŒËÉʸÃ◊
wavelength in Paschen series of He+ is :
Ã⁄¥UªŒÒÉÿ¸ „ÊªË —
5A
(1) 5A
9 (1)
9
9A
(2) 9A
5 (2)
5
36A
(3) 36A
5 (3)
5
36A
(4) 36A
7 (4)
7

7. 5 g of Na2SO4 was dissolved in x g of H2O.
The change in freezing point was found 7. x ª˝Ê◊ ¬ÊŸË ◊¥ 5 ª˝Ê◊ ‚ÊÁ«Uÿ◊ ‚À»§≈U ÉÊÊ‹Ê ªÿÊ–
to be 3.828C. If Na2SO4 is 81.5% ionised, ª‹ŸÊ¥∑§ ◊¥ ¬Á⁄UfløŸ 3.828C ¬ÊÿÊ ªÿÊ– ÿÁŒ
the value of x
Na2SO4 81.5% •ÊÿÁŸÃ „ÊÃÊ „Ò ÃÊ x ∑§Ê ‹ª÷ª
(K f for water=1.868C kg mol −1 ) is ◊ÊŸ „Ò — (¡‹ ∑§ Á‹∞ Kf=1.868C kg mol−1)
approximately :
(◊Ê‹⁄U Œ˝√ÿ◊ÊŸ — S=32 g mol−1 ÃÕÊ
(molar mass of S=32 g mol−1 and that of Na=23 g mol−1)
Na=23 g mol−1)
(1) 15 g
(1) 15 g
(2) 25 g
(2) 25 g
(3) 45 g
(3) 45 g
(4) 65 g
(4) 65 g

8. Addition of sodium hydroxide solution to
8. ∞∑§ ŒÈ’¸‹ •ê‹ (HA) ◊¥ ‚ÊÁ«Uÿ◊ „Êß«˛UÊÄ‚Êß«U
a weak acid (HA) results in a buffer of Áfl‹ÿŸ Á◊‹ÊŸ ‚ pH 6 ∑§Ê ’$»§⁄U ’ŸÃÊ „Ò– ÿÁŒ
pH 6. If ionisation constant of HA is 10−5, HA ∑§Ê •ÊÿŸŸ ÁSÕ⁄UÊ¥∑§ 10−5 „Ò ÃÊ, ’»§⁄U Áfl‹ÿŸ
the ratio of salt to acid concentration in ◊¥ ‹fláÊ •ÊÒ⁄U •ê‹ ∑§Ë ‚Ê¥Œ˝ÃÊ ∑§Ê •ŸÈ¬Êà „ÊªÊ —
the buffer solution will be :
(1) 4:5
(1) 4:5
(2) 1 : 10
(2) 1 : 10
(3) 10 : 1
(3) 10 : 1
(4) 5:4
(4) 5:4

3 VI - CHEMISTRY

9. The rate of a reaction A doubles on 9. ∞∑§ •Á÷Á∑˝§ÿÊ A ∑§Ë Œ⁄U, Ãʬ ∑§Ê 300 ‚ 310 K
increasing the temperature from 300 to
Ã∑§ ’…∏ÊŸ ¬⁄U ŒÈªŸË „Ê ¡ÊÃË „Ò– 300 K ‚ Ãʬ
310 K. By how much, the temperature of
reaction B should be increased from 300 K Á∑§ÃŸÊ ’…∏ÊÿÊ ¡Êÿ Á∑§ ∞∑§ ŒÍ‚⁄UË •Á÷Á∑˝§ÿÊ B ∑§Ë
so that rate doubles if activation energy Œ⁄U ÷Ë ŒÈªŸË „Ê ¡Êÿ ÿÁŒ ß‚ •Á÷Á∑˝§ÿÊ B ∑§Ë
of the reaction B is twice to that of ‚Á∑˝§ÿáÊ ™§¡Ê¸ •Á÷Á∑˝§ÿÊ A ‚ ŒÈªŸË „Ê–
reaction A.
(1) 9.84 K
(2) 4.92 K (1) 9.84 K
(3) 2.45 K (2) 4.92 K
(4) 19.67 K (3) 2.45 K
(4) 19.67 K
10. The enthalpy change on freezing of 1 mol
of water at 58C to ice at −58C is :
10. 58C ¬⁄U 1 ◊Ê‹ ¡‹ ∑§ Á„◊Ÿ ‚ −58C ¬⁄U ’»¸§
(Given ∆ fus H=6 kJ mol −1 at 08C, ’ŸÊŸ ◊¥ ∞ãÕÒÀ¬Ë ∑§Ê ¬Á⁄UfløŸ „ÊªÊ —
Cp(H2O, l)=75.3 J mol−1 K−1,
Cp(H2O, s)=36.8 J mol−1 K−1) ( ÁŒÿÊ ªÿÊ „Ò ∆ fus H=6 kJ mol−1 at 08C,
(1) 5.44 kJ mol−1 Cp(H2O, l)=75.3 J mol−1 K−1,
Cp(H2O, s)=36.8 J mol−1 K−1)
(2) 5.81 kJ mol−1
(1) 5.44 kJ mol−1
(3) 6.56 kJ mol−1
(2) 5.81 kJ mol−1
(4) 6.00 kJ mol−1
(3) 6.56 kJ mol−1
(4) 6.00 kJ mol−1
11. Which of the following is paramagnetic ?
(1) NO+
(2) CO
11. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê •ŸÈøÈê’∑§Ëÿ „Ò?
(1) NO+
2−
(3) O2 (2) CO
(4) B2 2−
(3) O2

(4) B2

4 VI - CHEMISTRY

12. The pair of compounds having metals in 12. Á¡‚ ÿÊÒ Á ª∑§ ÿÈ Ç ◊ ◊ ¥ , œÊÃÈ ∞  ¥ •¬ŸË ©ëøÃ◊
their highest oxidation state is :
•ÊÚÄ‚Ë∑§⁄UáÊ •flSÕÊ ◊¥ „Ò¥, fl„ „Ò —
(1) MnO2 and CrO2Cl2
(1) MnO2 ÃÕÊ CrO2Cl2
(2) [NiCl4]2− and [CoCl4]2−
(2) [NiCl4]2− ÃÕÊ [CoCl4]2−
(3) [Fe(CN)6]3− and [Cu(CN)4]2−
(4) [FeCl4]− and Co2O3 (3) [Fe(CN)6]3− ÃÕÊ [Cu(CN)4]2−

(4) [FeCl4]− ÃÕÊ Co2O3
13. sp3d2 hybridization is not displayed by :
(1) BrF5 13. Á¡‚∑§ mÊ⁄UÊ sp3d2 ‚¥∑§⁄UáÊ Ÿ„Ë¥ Œ‡ÊʸÿÊ ¡ÊÃÊ, fl„ „Ò —
(2) SF6
(1) BrF5
(3) [CrF6]3−
(2) SF6
(4) PF5
(3) [CrF6]3−
(4) PF5
14. Identify the pollutant gases largely
responsible for the discoloured and
lustreless nature of marble of the Taj 14. fl„ ¬˝ŒÍ·∑§ ªÒ‚¥ ¬„øÊÁŸÿ ¡Ê ÃÊ¡◊„‹ ∑§ ‚¥ª◊⁄U◊⁄U
Mahal. ∑§ ◊Á‹Ÿ fl ŒËÁåÄ˟ „ÊŸ ∑§ Á‹∞ ◊ÈÅÿ× ©ûÊ⁄UŒÊÿË
(1) O3 and CO2 „Ò–¥
(2) CO2 and NO2
(3) SO2 and NO2 (1) O3 ÃÕÊ CO2
(4) SO2 and O3 (2) CO2 ÃÕÊ NO2

(3) SO2 ÃÕÊ NO2
15. In which of the following reactions,
hydrogen peroxide acts as an oxidizing (4) SO2 ÃÕÊ O3
agent ?
(1) HOCl+H2O2 → H3O++Cl−+O2
15. ÁŸêŸÁ‹Áπà ◊¥ ‚ Á∑§‚ •Á÷Á∑˝§ÿÊ ◊¥ „Êß«˛UÊ¡Ÿ
(2) I2+H2O2+2OH−→ 2I−+2H2O+O2
¬⁄U•ÊÚÄ‚Êß«U •ÊÚÄ‚Ë∑§Ê⁄U∑§ ∑§ M§¬ ◊¥ ∑§Êÿ¸ ∑§⁄UÃË „Ò?
(3) 2MnO−
4 +3H2 O 2 → 2MnO 2 +3O 2 +
(1) HOCl+H2O2 → H3O++Cl−+O2
2H 2 O+2OH−
(2) I2+H2O2+2OH−→ 2I−+2H2O+O2
(4) PbS+4H2O2 → PbSO4+4H2O
(3) 2MnO−
4 +3H2 O 2 → 2MnO 2 +3O 2 +
2H 2 O+2OH−

(4) PbS+4H2O2 → PbSO4+4H2O

5 VI - CHEMISTRY

16. Consider the following ionization 16. ÁŸêŸÁ‹Áπà ŒÊ ÃàflÊ¥ ‘A’ ÃÕÊ ‘B’ ∑§Ë •ÊÿŸŸ
enthalpies of two elements ‘A’ and ‘B’.
∞ãÕÒÁÀ¬ÿÊ¥ ¬⁄U ÁfløÊ⁄U ∑§ËÁ¡∞–
Element Ionization enthalpy (kJ/mol)
Ãàfl •ÊÿŸŸ ∞ãÕÒÀ¬Ë (kJ/mol)
st nd rd
1 2 3
¬˝Õ◊ ÁmÃËÿ ÃÎÃËÿ
A 899 1757 14847
A 899 1757 14847
B 737 1450 7731
B 737 1450 7731
Which of the following statements is
correct ? ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ‚àÿ „Ò?
(1) Both ‘A’ and ‘B’ belong to group-1
where ‘B’ comes below ‘A’. (1) ‘A’ •ÊÒ⁄U ‘B’ ŒÊŸÊ¥ flª¸ 1 ◊¥ ©¬ÁSÕà „Ò¥ ÃÕÊ
(2) Both ‘A’ and ‘B’ belong to group-1 ‘B’, ‘A’ ∑§ ŸËø •ÊÃÊ „Ò–
where ‘A’ comes below ‘B’.
(2) ‘A’ •ÊÒ⁄U ‘B’ ŒÊŸÊ¥ flª¸ 1 ◊¥ ©¬ÁSÕà „Ò¥ ÃÕÊ
(3) Both ‘A’ and ‘B’ belong to group-2
where ‘B’ comes below ‘A’. ‘A’, ‘B’ ∑§ ŸËø •ÊÃÊ „Ò–

(4) Both ‘A’ and ‘B’ belong to group-2 (3) ‘A’ •ÊÒ⁄U ‘B’ ŒÊŸÊ¥ flª¸ 2 ◊¥ ©¬ÁSÕà „Ò¥ ÃÕÊ
where ‘A’ comes below ‘B’. ‘B’, ‘A’ ∑§ ŸËø •ÊÃÊ „Ò–

(4) ‘A’ •ÊÒ⁄U ‘B’ ŒÊŸÊ¥ flª¸ 2 ◊¥ ©¬ÁSÕà „Ò¥ ÃÕÊ
17. Consider the following standard electrode ‘A’, ‘B’ ∑§ ŸËø •ÊÃÊ „Ò–
potentials (E8 in volts) in aqueous solution :

Element M 3+ / M M+ / M 17. ÁŸêŸÁ‹Áπà ¡‹Ëÿ Áfl‹ÿŸÊ¥ ◊¥ ◊ÊŸ∑§ ß‹Ä≈˛UÊ«U
Al −1.66 + 0.55 Áfl÷flÊ¥ (E8, flÊÀ≈U ◊¥) ¬⁄U ÁfløÊ⁄U ∑§⁄¥U —
Tl +1.26 − 0.34
Ãàfl M 3+ / M M+ / M
Based on these data, which of the following
statements is correct ? Al −1.66 + 0.55
Tl +1.26 − 0.34
(1) Tl+ is more stable than Al3+
(2) Al+ is more stable than Al3+ ߟ •Ê°∑§«∏Ê¥ ∑§ •ÊœÊ⁄U ¬⁄U, ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê
(3) Tl+ is more stable than Al+ ∑§ÕŸ ‚àÿ „Ò?
(4) Tl3+ is more stable than Al3+ (1) Al3+ ‚ Tl+ •Áœ∑§ ÁSÕ⁄U „Ò–

(2) Al3+ ‚ Al+ •Áœ∑§ ÁSÕ⁄U „Ò–

(3) Al+ ‚ Tl+ •Áœ∑§ ÁSÕ⁄U „Ò–

(4) Al3+ ‚ Tl3+ •Áœ∑§ ÁSÕ⁄U „Ò–

6 VI - CHEMISTRY

18. A metal ‘M’ reacts with nitrogen gas to 18. ∞∑§ œÊÃÈ ‘M’, ŸÊß≈˛UÊ¡Ÿ ªÒ‚ ∑§ ‚ÊÕ •Á÷Á∑˝§ÿÊ
afford ‘M3N’. ‘M3N’ on heating at high
∑§⁄U∑§ M3N ©à¬ãŸ ∑§⁄UÃË „Ò– ©ëø Ãʬ ¬⁄U ª◊¸
temperature gives back ‘M’ and on
reaction with water produces a gas ‘B’. Gas ∑§⁄UŸ ¬⁄U M3N, flÊÁ¬‚ œÊÃÈ M ’ŸÊÃÊ „Ò ÃÕÊ ÿ„
‘B’ reacts with aqueous solution of CuSO4 ¡‹ ∑§ ‚ÊÕ •Á÷Á∑˝§ÿÊ ∑§⁄U∑§ ∞∑§ ªÒ‚ B ©àåÊãŸ
to form a deep blue compound. ‘M’ and ∑§⁄UÃÊ „Ò– ªÒ‚ B, ¡‹Ëÿ CuSO4 ∑§ ‚ÊÕ •Á÷Á∑˝§ÿÊ
‘B’ respectively are :
∑§⁄U∑§ ∞∑§ ª„⁄U ŸË‹ ⁄¥Uª ∑§Ê ÿÊÒÁª∑§ ©àåÊ㟠∑§⁄UÃË „Ò–
(1) Li and NH3 ‘M’ ÃÕÊ ‘B’ ∑˝§◊‡Ê— „Ò¥ —
(2) Ba and N2
(1) Li ÃÕÊ NH3
(3) Na and NH3
(2) Ba ÃÕÊ N2
(4) Al and N2
(3) Na ÃÕÊ NH3
19. The number of S=O and S−OH bonds (4) Al ÃÕÊ N2
present in peroxodisulphuric acid and
pyrosulphuric acid respectively are :
(1) (2 and 2) and (2 and 2) 19. ¬⁄U•ÊÚÄ‚Ê«UÊß‚ÀçÿÍÁ⁄U∑§ •ê‹ ÃÕÊ ¬Êÿ⁄UÊ‚ÀçÿÍÁ⁄U∑§
(2) (2 and 4) and (2 and 4) •ê‹ ◊¥ Áfll◊ÊŸ S=O ÃÕÊ S−OH •Ê’¥œÊ¥ ∑§Ë
‚¥ÅÿÊ∞¥ ∑˝§◊‡Ê— „Ò¥ —
(3) (4 and 2) and (2 and 4)
(4) (4 and 2) and (4 and 2) (1) (2 ÃÕÊ 2) ÃÕÊ (2 ÃÕÊ 2)

(2) (2 ÃÕÊ 4) ÃÕÊ (2 ÃÕÊ 4)
20. A solution containing a group-IV cation (3) (4 ÃÕÊ 2) ÃÕÊ (2 ÃÕÊ 4)
gives a precipitate on passing H 2S. A
solution of this precipitate in dil.HCl (4) (4 ÃÕÊ 2) ÃÕÊ (4 ÃÕÊ 2)
produces a white precipitate with NaOH
solution and bluish-white precipitate with
basic potassium ferrocyanide. The cation 20. ∞∑§ Áfl‹ÿŸ Á¡‚◊¥ ªÈ˝¬-IV ∑§Ê ∞∑§ œŸÊÿŸ Áfll◊ÊŸ
is : „Ò, H2S ¬˝flÊÁ„à ∑§⁄UŸ ¬⁄U ∞∑§ •flˇÊ¬ ©à¬ãŸ ∑§⁄UÃÊ
(1) Co2+ „Ò– ß‚ •flˇÊ¬ ∑§Ê ÃŸÈ HCl ◊¥ ’ŸÊ Áfl‹ÿŸ, NaOH
(2) Ni2+ ∑§ ‚ÊÕ ∞∑§ ‡flà •flˇÊ¬ ÃÕÊ ˇÊÊ⁄UËÿ ¬Ê≈UÁ‡Êÿ◊
(3) Mn2+ »§⁄UÊ‚ÊÿŸÊß«U ∑§ ‚ÊÕ ŸË‹Ê-SÊ$»§Œ •flˇÊ¬ ©à¬ãŸ
(4) Zn2+ ∑§⁄UÃÊ „Ò– ÿ„ œŸÊÿŸ „Ò —

(1) Co2+
(2) Ni2+
(3) Mn2+
(4) Zn2+

7 VI - CHEMISTRY

21. A mixture containing the following four 21. ÁŸêŸÁ‹Áπà øÊ⁄U ÿÊÒ Á ª∑§Ê ¥ ∑ § ∞∑§ Á◊üÊáÊ ∑§Ê
compounds is extracted with 1M HCl.
1M HCl ‚ ÁŸc∑§Á·¸Ã Á∑§ÿÊ ¡ÊÃÊ „Ò– fl„ ÿÊÒÁª∑§
The compound that goes to aqueous layer
is : ¡Ê ¡‹Ëÿ ¬⁄Uà ◊¥ ø‹Ê ¡ÊÃÊ „Ò, „Ò —

(I) (II) (I) (II)

(III) (IV)
(1) (I) (III) (IV)

(2) (II) (1) (I)

(3) (III) (2) (II)

(4) (IV) (3) (III)
(4) (IV)

22. The reason for “drug induced poisoning”
is : 22. •ÊÒ·œ-¬˝Á⁄Uà Áfl·ÊÄÃË∑§⁄UáÊ ∑§Ê ∑§Ê⁄UáÊ „Ò —
(1) Binding reversibly at the active site
of the enzyme
(1) ∞ã¡Êß◊ ∑§Ë ‚Á∑˝§ÿ ‚Ä ¬⁄U ©à∑˝§◊áÊËÿ
(2) Bringing conformational change in
the binding site of enzyme ‚¥ÿÊ¡Ÿ
(3) Binding irreversibly to the active site (2) ∞ã¡Êß◊ ∑§Ë ’¥œŸË ‚Ä ◊¥ ‚¥M§¬Ëÿ ¬Á⁄UfløŸ
of the enzyme
(4) Binding at the allosteric sites of the (3) ∞ã¡Êß◊ ∑§Ë ‚Á∑˝§ÿ ‚Ä ¬⁄U •ŸÈà∑˝§◊áÊËÿ
enzyme
‚¥ÿÊ¡Ÿ
(4) ∞ã¡Êß◊ ∑§Ë ∞‹ÊS≈UËÁ⁄U∑§ ‚ÄÊ¥ ¬⁄U ‚¥ÿÊ¡Ÿ

8 VI - CHEMISTRY

23. Which of the following compounds will 23. ÁãÊêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚ ÿÊÒÁª∑§ ∑§Ë ’ã$¡ËŸ ∑§ ‚ÊÕ
not undergo Friedel Craft’s reaction with
»˝§Ë«U‹ ∑˝§Êç≈˜U‚ •Á÷Á∑˝§ÿÊ Ÿ„Ë¥ „ÊªË?
benzene ?

(1)
(1)

(2)
(2)

(3)
(3)

(4)
(4)

24. Among the following, the essential amino
acid is : 24. ÁŸêŸÁ‹Áπà ◊¥ ‚ •Êfl‡ÿ∑§ ∞◊ËŸÊ •ê‹ „Ò —
(1) Alanine
(2) Valine
(1) ∞‹ÊÁŸŸ
(3) Aspartic acid (2) flÒ‹ËŸ
(4) Serine (3) ∞S¬ÊÁ≈¸U∑§ •ê‹
(4) ‚⁄UËŸ

9 VI - CHEMISTRY

25. The major product expected from the 25. ÁŸêŸÁ‹Áπà •Á÷Á∑˝§ÿÊ ∑§Ê ‚¥÷ÊÁflà ◊ÈÅÿ ©à¬ÊŒ
following reaction is :
„Ò —

(1)
(1)

(2)
(2)

(3)
(3)

(4)
(4)

10 VI - CHEMISTRY

26. The major product of the following 26. ÁŸêŸÁ‹Áπà •Á÷Á∑˝§ÿÊ ∑§Ê ◊ÈÅÿ ©à¬ÊŒ „Ò —
reaction is :

(1) CH2=CHCH2CH=CHCH3
(1) CH2=CHCH2CH=CHCH3
(2) CH2=CHCH=CHCH2CH 3
(2) CH2=CHCH=CHCH2CH3
(3) CH3CH=C=CHCH2CH3
(3) CH3CH=C=CHCH2CH3
(4) CH3CH=CH−CH=CHCH3
(4) CH3CH=CH−CH=CHCH3

27. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ÁflÃ⁄UáÊ fláʸ‹πŸ
27. Which of the following statements is not
true about partition chromatography ? ∑§ ’Ê⁄U ◊¥ ‚àÿ Ÿ„Ë¥ „Ò?
(1) Mobile phase can be a gas (1) ªÁÇÊË‹ ¬˝ÊflSÕÊ ∞∑§ ªÒ‚ „Ê ‚∑§ÃË „Ò–
(2) Stationary phase is a finely divided (2) ÁSÕ⁄U ¬˝ÊflSÕÊ ∞∑§ ’„Èà ◊„ËŸ Á¬‚Ê „È•Ê ∆UÊ‚
solid adsorbent
•Áœ‡ÊÊ·∑§ „ÊÃÊ „Ò–
(3) Separation depends upon
equilibration of solute between a (3) ¬˝ÕÄ∑§⁄UáÊ ∞∑§ ªÁÇÊË‹ ÃÕÊ ÁSÕ⁄U ¬˝ÊflSÕÊ
mobile and a stationary phase ∑§ ’Ëø Áfl‹ÿ ∑§ ‚ÊêÿŸ ¬⁄U ÁŸ÷¸⁄U ∑§⁄UÃÊ „Ò–
(4) Paper chromatography is an example
of partition chromatography
(4) ∑§Êª$¡ fláʸ‹πŸ, ÁflÃ⁄UáÊ fláʸ‹πŸ ∑§Ê ∞∑§
©ŒÊ„⁄UáÊ „Ò–
28. The IUPAC name of the following
compound is :
28. ŸËø ÁŒ∞ ÿÊÒÁª∑§ ∑§Ê IUPAC ŸÊ◊ „Ò —

(1) 1, 1-Dimethyl-2-ethylcyclohexane
(2) 2-Ethyl-1,1-dimethylcyclohexane (1) 1, 1-«UÊß◊ÁÕ‹-2-∞ÁÕ‹‚ÊßÄ‹Ê„Ä‚Ÿ
(3) 1-Ethyl-2,2-dimethylcyclohexane
(2) 2-∞ÁÕ‹-1,1-«UÊß◊ÁÕ‹‚ÊßÄ‹Ê„Ä‚Ÿ
(4) 2, 2-Dimethyl-1-ethylcyclohexane
(3) 1-∞ÁÕ‹-2,2-«UÊß◊ÁÕ‹‚ÊßÄ‹Ê„Ä‚Ÿ

(4) 2, 2-«UÊß◊ÁÕ‹-1-∞ÁÕ‹‚ÊßÄ‹Ê„Ä‚Ÿ

11 VI - CHEMISTRY

29. The major product of the following 29. ÁŸêŸÁ‹Áπà •Á÷Á∑˝§ÿÊ ∑§Ê ◊ÈÅÿ ©à¬ÊŒ „Ò —
reaction is :

(1)
(1)

(2)
(2)

(3)
(3)

(4)
(4)

12 VI - CHEMISTRY

30. The major product of the following 30. ÁŸêŸ •Á÷Á∑˝§ÿÊ ∑§Ê ◊ÈÅÿ ©à¬ÊŒ „Ò —
reaction is :

(1)
(1)

(2)
(2)

(3)
(3)

(4)
(4)

13 VI - CHEMISTRY

MATHEMATICS ªÁáÊÃ

1. Let f (x)=210⋅x+1 and g(x)=310⋅x−1. If
1. ◊ÊŸÊ f (x)=210⋅x+1 ÃÕÊ g(x)=310⋅x−1– ÿÁŒ
(fog)(x)=x, then x is equal to :
(fog)(x)=x „Ò, ÃÊ x ’⁄UÊ’⁄U „Ò —
310 − 1
(1) 310 − 1
310 − 2−10 (1)
310 − 2−10
2 10 − 1
(2) 2 10 − 1
2 10 − 3−10 (2)
2 10 − 3−10
1 − 3−10
(3) 1 − 3−10
2 10 − 3−10 (3)
2 10 − 3−10
1 − 2−10
(4) 1 − 2−10
310 − 2−10 (4)
310 − 2−10

2. Let p(x) be a quadratic polynomial such
that p(0)=1. If p(x) leaves remainder 4 2. ◊ÊŸÊ p(x) ∞‚Ê ∞∑§ ÁmÉÊÊÃË ’„ȬŒ „Ò Á¡‚∑§ Á‹ÿ
when divided by x−1 and it leaves p(0)=1 „Ò– ÿÁŒ p(x) ∑§Ê x−1 ‚ ÷ʪ ŒŸ ¬⁄U 4
remainder 6 when divided by x+1; then :
‡Ê· ⁄U„ÃÊ „Ò ÃÕÊ x+1 ‚ ÷ʪ ŒŸ ¬⁄U 6 ‡Ê· ’øÃÊ „Ò,
(1) p(2)=11 ÃÊ —
(2) p(2)=19
(1) p(2)=11
(3) p(−2)=19
(2) p(2)=19
(4) p(−2)=11
(3) p(−2)=19
(4) p(−2)=11

1 VI - MATHEMATICS

3. Let z e C, the set of complex numbers. Then 3. ◊ÊŸÊ z e C, ¡Ê ‚Áê◊üÊ ‚¥ÅÿÊ•Ê¥ ∑§Ê ‚◊ÈìÊÿ „Ò, ÃÊ
the equation, 2?z+3i?−?z−i?=0
‚◊Ë∑§⁄UáÊ 2?z+3i?−?z−i?=0 ¬˝ŒÁ‡Ê¸Ã ∑§⁄UÃÊ „Ò —
represents :

8
(1) a circle with radius .
3 (1) ∞∑§ flÎûÊ Á¡‚∑§Ë ÁòÊíÿÊ 8 „Ò–
3
10
(2) a circle with diameter .
3 (2) ∞∑§ flÎûÊ Á¡‚∑§Ê √ÿÊ‚ 10 „Ò–
3
(3) an ellipse with length of major axis
16 . (3) ∞∑§ ŒËÉʸflÎûÊ Á¡‚∑§ ŒËÉʸ •ˇÊ ∑§Ë ‹¥’Ê߸
3 16 „Ò
(4) an ellipse with length of minor axis 3

16 (4) ∞∑§ ŒËÉʸflÎûÊ Á¡‚∑§ ‹ÉÊÈ •ˇÊ ∑§Ë ‹¥’Ê߸
.
9 16
„Ò–
9
4. The number of real values of λ for which
the system of linear equations
4. λ ∑§ ©Ÿ flÊSÃÁfl∑§ ◊ÊŸÊ¥ ∑§Ë ‚¥ÅÿÊ Á¡Ÿ∑§ Á‹∞
2x+4y−λz=0 ⁄ÒUÁπ∑§ ‚◊Ë∑§⁄UáÊ ÁŸ∑§Êÿ
4x+λy+2z=0
2x+4y−λz=0
λx+2y+2z=0
4x+λy+2z=0
has infinitely many solutions, is :
λx+2y+2z=0
(1) 0
∑§ •Ÿ¥Ã „‹ „Ò¥, „Ò —
(2) 1
(1) 0
(3) 2
(2) 1
(4) 3
(3) 2
(4) 3
5. Let A be any 3×3 invertible matrix. Then
which one of the following is not always
true ? 5. ◊ÊŸÊ A ∑§Ê߸ 3×3 ∑§Ê √ÿÈà∑˝§◊áÊËÿ •Ê√ÿÍ„ „Ò ÃÊ
(1) adj (A)=?A? A−1
⋅ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ‚ŒÊ ‚àÿ Ÿ„Ë¥ „Ò?
(2) adj (adj(A))=?A?⋅A
(3) adj (adj(A))=?A?2⋅(adj(A))−1 (1) adj (A)=?A?⋅A−1
(4) adj (adj(A))=?A?⋅(adj(A))−1 (2) adj (adj(A))=?A?⋅A
(3) adj (adj(A))=?A?2⋅(adj(A))−1
(4) adj (adj(A))=?A?⋅(adj(A))−1

2 VI - MATHEMATICS

6. If all the words, with or without meaning, 6. ‡ÊéŒ QUEEN ∑§ ‚÷Ë •ˇÊ⁄UÊ¥ ∑§Ê ¬˝ÿÊª ∑§⁄U∑§ ’ŸŸ
are written using the letters of the word
flÊ‹ ‚÷Ë ‡ÊéŒ (Á¡Ÿ∑§Ê •Õ¸ „Ò •ÕflÊ fl •Õ¸„ËŸ
QUEEN and are arranged as in English
dictionary, then the position of the word „Ò¥) ∑§Ê •¥ª˝¡Ë ‡ÊéŒ∑§Ê· ∑§ •ŸÈ‚Ê⁄U ‹ªÊŸ ¬⁄U, ‡ÊéŒ
QUEEN is : QUEEN ∑§Ê SÕÊŸ „Ò —
(1) 44 th
(2) 45 th (1) 44 flʥ
(3) 46 th (2) 45 flʥ
(4) 47 th
(3) 46 flʥ

(4) 47 flʥ
7. If (27) 999 is divided by 7, then the
remainder is :
(1) 1 7. ÿÁŒ (27)999 ∑§Ê 7 ‚ ÷ʪ ÁŒÿÊ ¡Ê∞, ÃÊ ‡Ê·»§‹ „Ò —
(2) 2
(3) 3 (1) 1
(4) 6 (2) 2
(3) 3
8. If the arithmetic mean of two numbers a (4) 6
and b, a > b > 0, is five times their geometric
a+b
mean, then
a−b
is equal to : 8. ÿÁŒ ŒÊ ‚¥ÅÿÊ•Ê¥ a ÃÕÊ b, a > b > 0 ∑§Ê ‚◊Ê¥Ã⁄U
◊Êäÿ (A.M.) ©Ÿ∑§ ªÈáÊÊûÊ⁄U ◊Êäÿ (G.M.) ∑§Ê
6 a+b
(1) 5 ªÈŸÊ „Ò, ÃÊ ’⁄UÊ’⁄U „Ò —
2 a−b

3 2 6
(2) (1)
4 2

7 3 3 2
(3) (2)
12 4

5 6 7 3
(4) (3)
12 12

5 6
(4)
12

3 VI - MATHEMATICS

9. If the sum of the first n terms of the series 9. ÿÁŒ üÊáÊË 3 + 75 + 243 + 507 + .....
3 + 75 + 243 + 507 + ..... is 435 3 ,
∑§ ¬˝Õ◊ n ¬ŒÊ¥ ∑§Ê ÿÊª 435 3 „Ò, ÃÊ n ’⁄UÊ’⁄U „Ò —
then n equals :
(1) 18
(1) 18
(2) 15
(2) 15
(3) 13
(3) 13
(4) 29
(4) 29

lim 3x − 3
10. is equal to : lim 3x − 3
x →3 2x − 4 − 2 10. x →3
’⁄UÊ’⁄U „Ò —
2x − 4 − 2
(1) 3
(1) 3
1
(2) 1
2 (2)
2
3
(3) 3
2 (3)
2
1
(4) 1
2 2 (4)
2 2

11. The tangent at the point (2, −2) to the
curve, x 2y 2−2x=4(1−y) does not pass 11. fl∑˝§ x2y2−2x=4(1−y) ∑§ Á’¥ŒÈ (2, −2) ¬⁄U
through the point : πË¥øË ªß¸ S¬‡Ê¸ ⁄UπÊ ÁŸêŸ ◊¥ ‚ Á∑§‚ Á’¥ŒÈ ‚ Ÿ„Ë¥
 1 ªÈ¡⁄UÃË „Ò —
(1)  4, 
 3
 1
(1)  4, 
(2) (8, 5)  3
(3) (−4, −9) (2) (8, 5)
(4) (−2, −7) (3) (−4, −9)
(4) (−2, −7)

4 VI - MATHEMATICS

15 15 15 15
12. If y =  x+ x 2 −1  +  x − x 2 −1  , 12. ÿÁŒ y =  x+ x 2 −1  +  x − x 2 −1 
     

d2 y dy d2 y dy
then ( x 2 − 1) 2
+x is equal to : „Ò, ÃÊ ( x 2 − 1) 2
+x ’⁄UÊ’⁄U „Ò —
dx dx dx dx

(1) 125 y (1) 125 y
(2) 224 y2 (2) 224 y2
(3) 225 y2 (3) 225 y2
(4) 225 y (4) 225 y

13. If a point P has co-ordinates (0, −2) and 13. ÿÁŒ Á∑§‚Ë Á’ãŒÈ P ∑§ ÁŸŒ¸‡ÊÊ¥∑§ (0, −2) „Ò¥ ÃÕÊ ∑§Ê߸
Q is any point on the circle,
Á’ãŒÈ Q flÎûÊ x2+y2−5x−y+5=0 ¬⁄U ÁSÕà „Ò,
x2+y2−5x−y+5=0, then the maximum
value of (PQ)2 is : ÃÊ (PQ)2 ∑§Ê •Áœ∑§Ã◊ ◊ÊŸ „Ò —

25 + 6
(1)
2 25 + 6
(1)
2
(2) 14 + 5 3
(2) 14 + 5 3
47 + 10 6
(3)
2 47 + 10 6
(3)
2
(4) 8 +5 3
(4) 8 +5 3

5 VI - MATHEMATICS

14. The integral 14. ‚◊Ê∑§‹
∫ 1 + 2cot x(cosec x + cot x) dx ∫ 1 + 2cot x(cosec x + cot x) dx ,
 π  π
 0 < x <  is equal to :  0 < x <  ’⁄UÊ’⁄U „Ò —
 2  2
(where C is a constant of integration)
(¡„Ê° C ∞∑§ ‚◊Ê∑§‹Ÿ •ø⁄U „Ò)
4 log  sin  + C
x
(1) 4 log  sin  + C
x
 2 (1)
 2

2 log  sin  + C
x
(2) 2 log  sin  + C
x
 2 (2)
 2

2 log  cos  + C
x
(3) 2 log  cos  + C
x
 2 (3)
 2

4 log  cos  + C
x
(4) 4 log  cos  + C
x
 2 (4)
 2

π
8 cos 2x
∫
4
15. The integral dx π
8 cos 2x
π 3
(tan x + cot x ) 15. ‚◊Ê∑§‹ ∫ π4 dx ’⁄UÊ’⁄U „Ò —
(tan x + cot x )3
12
12
equals :

15
(1) 15
128 (1)
128
15
(2) 15
64 (2)
64
13
(3) 13
32 (3)
32
13
(4) 13
256 (4)
256

6 VI - MATHEMATICS

16. The area (in sq. units) of the smaller portion 16. fl∑˝§Ê¥ x2+y2=4 ÃÕÊ y2=3x ∑§ ’Ëø ÁÉÊ⁄U ¿UÊ≈U
enclosed between the curves, x2+y2=4
÷ʪ ∑§Ê ˇÊòÊ»§‹ (flª¸ ß∑§ÊßÿÊ¥ ◊¥) „Ò —
and y2=3x, is :

1 π
(1) + 1 π
2 3 3 (1) +
2 3 3
1 2π
(2) + 1 2π
3 3 (2) +
3 3
1 2π
(3) + 1 2π
2 3 3 (3) +
2 3 3
1 4π
(4) + 1 4π
3 3 (4) +
3 3

17. The curve satisfying the differential
equation, ydx−(x+3y 2 )dy=0 and 17. •fl∑§‹ ‚◊Ë∑§⁄U á Ê ydx−(x+3y2)dy=0 ∑§Ê
passing through the point (1, 1), also passes ‚¥ÃÈc≈U ∑§⁄UŸ flÊ‹Ë flÊ fl∑˝§, ¡Ê Á’¥ŒÈ (1, 1) ‚ „Ê∑§⁄U
through the point :
¡ÊÃË „Ò, ÁŸêŸ ◊¥ ‚ Á∑§‚ Á’¥ŒÈ ‚ ÷Ë „Ê∑§⁄U ¡ÊÃË „Ò —
(1) 1 1
 ,− 
4 2
(1) 1 1
 ,− 
 1 1 4 2
(2) − , 
 3 3
 1 1
(2) − , 
1 1  3 3
(3)  ,− 
3 3
1 1
(3)  ,− 
1 1 3 3
(4)  , 
4 2
1 1
(4)  , 
4 2

7 VI - MATHEMATICS

18. The locus of the point of intersection of the 18. ⁄UπÊ•Ê¥
straight lines, tx−2y−3t=0
tx−2y−3t=0
x−2ty+3=0 (t e R) ∑§ ¬˝ÁÃë¿UŒŸ Á’¥ŒÈ ∑§Ê Á’¥ŒÈ
x−2ty+3=0 (t e R), is : ¬Õ „Ò —
2
2 (1) ∞∑§ ŒËÉʸflÎûÊ Á¡‚∑§Ë ©à∑§ãŒ˝ÃÊ „Ò
(1) an ellipse with eccentricity 5
5
(2) an ellipse with the length of major
(2) ∞∑§ ŒËÉʸflÎûÊ Á¡‚∑§ ŒËÉʸ •ˇÊ ∑§Ë ‹¥’Ê߸ 6 „Ò
axis 6

(3) a hyperbola with eccentricity 5 (3) ∞∑§ •Áì⁄Ufl‹ÿ Á¡‚∑§Ë ©à∑§ãŒ˝ÃÊ 5 „Ò
(4) a hyperbola with the length of (4) ∞∑§ •Áì⁄U fl ‹ÿ Á¡‚∑ § ‚¥ ÿ È Ç ◊Ë •ˇÊ
conjugate axis 3 (conjugate axis) ∑§Ë ‹¥’Ê߸ 3 „Ò

19. If two parallel chords of a circle, having
diameter 4 units, lie on the opposite sides 19. ÿÁŒ ∞∑§ flÎûÊ Á¡‚∑§Ê √ÿÊ‚ 4 ß∑§Ê߸ „Ò ∑§Ë ŒÊ ‚◊Ê¥Ã⁄U
of the centre and subtend angles ¡ËflÊ∞°, ¡Ê flÎûÊ ∑§ ∑¥§Œ˝ ∑§Ë Áfl¬⁄UËà ÁŒ‡ÊÊ•Ê¥ ◊¥ „Ò¥ ÃÕÊ
cos−1   and sec −1 (7) at the centre ∑§ãŒ˝ ¬⁄U ∑˝§◊‡Ê— cos−1  1  ÃÕÊ sec−1(7) ∑§ ∑§ÊáÊ
1
7 7
respectively, then the distance between •¥ÃÁ⁄Uà ∑§⁄UÃË „Ò¥, ÃÊ ߟ ¡ËflÊ•Ê¥ ∑§ ’Ëø ∑§Ë ŒÍ⁄UË
these chords, is :
„Ò —
4
(1)
7 4
(1)
8 7
(2)
7 8
(2)
8 7
(3)
7 8
(3)
16 7
(4)
7 16
(4)
7

8 VI - MATHEMATICS

20. If the common tangents to the parabola, 20. ÿÁŒ ¬⁄Ufl‹ÿ x2=4y ÃÕÊ flÎûÊ x2+y2=4 ∑§Ë
x2=4y and the circle, x2+y2=4 intersect
©÷ÿÁŸc∆U S¬‡Ê¸ ⁄UπÊ∞° ∞∑§ Á’¥ŒÈ P ¬⁄U ¬˝ÁÃë¿UŒ ∑§⁄UÃË
at the point P, then the distance of P from
the origin, is : „Ò¥, ÃÊ P ∑§Ë ◊Í‹ Á’¥ŒÈ ‚ ŒÍ⁄UË „Ò —
(1) 2 +1
(1) 2 +1
(2) 2(3 + 2 2 )
(2) 2(3 + 2 2 )
(3) 2 ( 2 + 1)
(3) 2 ( 2 + 1)
(4) 3+2 2
(4) 3+2 2

21. Consider an ellipse, whose centre is at the
origin and its major axis is along the 21. ∞∑§ ŒËÉʸflÎûÊ, Á¡‚∑§Ê ∑¥§Œ˝ ◊Í‹ Á’¥ŒÈ ¬⁄U „Ò ÃÕÊ ŒËÉʸ
x-axis. If its eccentricity is
3
and the •ˇÊ x-•ˇÊ ∑§Ë ÁŒ‡ÊÊ ◊¥ „Ò, ¬⁄U ÁfløÊ⁄U ∑§ËÁ¡∞– ÿÁŒ
5
3
distance between its foci is 6, then the area ©‚∑§Ë ©à∑§ãŒ˝ÃÊ ÃÕÊ ŸÊÁ÷ÿÊ¥ ∑§ ’Ëø ∑§Ë ŒÍ⁄UË
5
(in sq. units) of the quadrilateral inscribed
in the ellipse, with the vertices as the 6 „Ò, ÃÊ ©‚ øÃÈ÷ȸ¡, ¡Ê ŒËÉʸflÎûÊ ∑§ •ãê¸Ã ’ŸÊ߸
vertices of the ellipse, is : ªß¸ „Ò ÃÕÊ Á¡‚∑§ ‡ÊË·¸, ŒËÉʸflÎûÊ ∑§ ‡ÊË·ÊZ ¬⁄U „Ò¥, ∑§Ê
(1) 8 ˇÊòÊ»§‹ (flª¸ ß∑§ÊßÿÊ¥ ◊¥) „Ò —
(2) 32
(3) 80 (1) 8
(4) 40 (2) 32
(3) 80
(4) 40

9 VI - MATHEMATICS

22. The coordinates of the foot of the 22. ‚◊Ë, Á¡‚◊¥ ŒÊŸÊ¥ ⁄UπÊ∞¥
perpendicular from the point (1, −2, 1)
x +1 y −1 z −3
on the plane containing the lines, = = ÃÕÊ
6 7 8
x +1 y −1 z −3
= = and
6 7 8 x −1 y −2 z −3
= = ÁSÕà „Ò¥, ¬⁄U Á’ãŒÈ
3 5 7
x −1 y −2 z −3
= = , is : (1, −2, 1) ‚ «UÊ‹ ª∞ ‹ê’ ∑§ ¬ÊŒ ∑§ ÁŸŒ¸‡ÊÊ¥∑§
3 5 7
„Ò¥ —
(1) (2, −4, 2)
(1) (2, −4, 2)
(2) (−1, 2, −1)
(2) (−1, 2, −1)
(3) (0, 0, 0)
(3) (0, 0, 0)
(4) (1, 1, 1)
(4) (1, 1, 1)

23. The line of intersection of the planes
→
( ∧ ∧
)
r . 3 i − j + k = 1 and
∧
23.
→
( ∧
‚◊ËÊ¥ r . 3 i − j + k = 1 ÃÕÊ
∧ ∧
)
r . ( i + 4 j − 2 k ) = 2, is :
→ ∧ ∧ ∧ →
( ∧ ∧ ∧
)
r . i + 4 j − 2 k = 2 ∑§Ë ¬˝ÁÃë¿UŒË ⁄UπÊ „Ò —

x − 74 y z − 75
(1) = =
−2 7 13 x − 74 y z − 75
(1) = =
−2 7 13
x − 74 y z + 75
(2) = =
2 −7 13 x − 74 y z + 75
(2) = =
2 −7 13
6
x − 13 5
y − 13 z
(3) = = 6 5
2 −7 −13 x − 13 y − 13 z
(3) = =
2 −7 −13
6
x − 13 5
y − 13 z
(4) = = 6 5
2 7 −13 x − 13 y − 13 z
(4) = =
2 7 −13

10 VI - MATHEMATICS

24. The area (in sq. units) of the parallelogram ∧ ∧
whose diagonals are along the vectors 24. ‚◊Ê¥Ã⁄U øÃÈ÷ȸ¡, Á¡‚∑§ Áfl∑§áʸ, ‚ÁŒ‡ÊÊ¥ 8 i − 6 j
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
8 i − 6 j and 3 i + 4 j − 12 k , is : ÃÕÊ 3 i + 4 j − 12 k ∑§Ë ÁŒ‡ÊÊ•Ê¥ ◊¥ „Ò,¥ ∑§Ê ˇÊòÊ»§‹
(1) 26 (flª¸ ß∑§ÊßÿÊ¥ ◊¥) „Ò —
(2) 65 (1) 26
(3) 20 (2) 65
(4) 52 (3) 20
(4) 52
25. The mean age of 25 teachers in a school is
40 years. A teacher retires at the age of
60 years and a new teacher is appointed 25. ∞∑§ ÁfllÊ‹ÿ ∑§ 25 •äÿʬ∑§Ê¥ ∑§Ë ◊Êäÿ-•ÊÿÈ
in his place. If now the mean age of the 40 fl·¸ „Ò– ∞∑§ •äÿʬ∑§ 60 fl·¸ ∑§Ë •ÊÿÈ ◊¥ ‚flÊ
teachers in this school is 39 years, then the ÁŸflÎûÊ „ÊÃÊ „Ò •ÊÒ⁄U ©‚∑§ SÕÊŸ ¬⁄U ∞∑§ Ÿÿ •äÿʬ∑§
age (in years) of the newly appointed
teacher is :
∑§Ë ÁŸÿÈÁÄà „ÊÃË „Ò– ÿÁŒ •’ ß‚ ÁfllÊ‹ÿ ∑§
•äÿʬ∑§Ê¥ ∑§Ë ◊Êäÿ-•ÊÿÈ 39 fl·¸ „Ò ÃÊ Ÿÿ •äÿʬ∑§
(1) 25
∑§Ë •ÊÿÈ (fl·ÊZ ◊¥) „Ò —
(2) 30
(3) 35 (1) 25
(4) 40 (2) 30
(3) 35
(4) 40

11 VI - MATHEMATICS

26. Three persons P, Q and R independently 26. ÃËŸ √ÿÁÄà P, Q ÃÕÊ R SflÃ¥òÊ M§¬ ‚ ∞∑§ ÁŸ‡ÊÊŸ
try to hit a target. If the probabilities of
∑§Ê ÷ŒŸ ∑§Ê ¬˝ÿÊ‚ ∑§⁄UÃ „Ò¥– ÿÁŒ ©Ÿ∑§ ÁŸ‡ÊÊŸ ∑§Ê
3 1 5
their hitting the target are , and 3 1 5
4 2 8 ÷Œ ¬ÊŸ ∑§Ë ¬˝ÊÁÿ∑§ÃÊ∞¥ ∑˝§◊‡Ê— , ÃÕÊ „Ò¥, ÃÊ
4 2 8
respectively, then the probability that the
target is hit by P or Q but not by R is : P •ÕflÊ Q ∑§ ÁŸ‡ÊÊŸÊ ÷Œ ¬ÊŸ ¬⁄UãÃÈ R ∑§ ÁŸ‡ÊÊŸÊ Ÿ
÷Œ ¬ÊŸ ∑§Ë ¬˝ÊÁÿ∑§ÃÊ „Ò —
21
(1)
64 21
(1)
64
9
(2)
64 9
(2)
64
15
(3)
64 15
(3)
64
39
(4)
64 39
(4)
64

27. An unbiased coin is tossed eight times. The
probability of obtaining at least one head 27. ∞∑§ •ŸÁ÷ŸÃ (unbiased) Á‚Ä∑§ ∑§Ê •Ê∆U ’Ê⁄U
and at least one tail is :
©¿UÊ‹Ê ¡ÊÃÊ „Ò, ÃÊ ∑§◊ ‚ ∑§◊ ∞∑§ ÁøûÊ ÃÕÊ ∑§◊ ‚
(1)
255 ∑§◊ ∞∑§ ¬≈U ¬˝Êåà ∑§⁄UŸ ∑§Ë ¬˝ÊÁÿ∑§ÃÊ „Ò —
256
255
(1)
127 256
(2)
128
127
(2)
63 128
(3)
64
63
(3)
1 64
(4)
2
1
(4)
2

12 VI - MATHEMATICS

28. If 28. ÿÁŒ
 0 cos x − sin x   0 cos x − sin x 
   
S =  x  [ 0, 2π] : sin x 0 cos x = 0  , S =  x  [ 0, 2 π] : sin x 0 cos x = 0 
 cos x sin x 0   
   cos x sin x 0 

∑ tan  3 + x  is equal to :
π
„Ò, ÃÊ ∑ tan  
then π
+ x  ’⁄UÊ’⁄U „Ò —
xȏS xȏS 3 

(1) 4 +2 3 (1) 4 +2 3
(2) −2 + 3 (2) −2 + 3
(3) −2 − 3 (3) −2 − 3
(4) −4 − 2 3 (4) −4 − 2 3

 1 + x2 + 1 − x2   1 + x2 + 1 − x2 
−1
29. The value of tan  , 29. tan−1  
 1 + x 2 − 1 − x 2   1 + x 2 − 1 − x 2 
1 1

x
< , x ≠ 0 , is equal to :
x
< , x ≠ 0 , ∑§Ê ◊ÊŸ „Ò —
2 2
π 1
(1) + cos−1 x 2 (1)
π
+
1
cos−1 x 2
4 2 4 2
π π
(2) + cos−1 x 2 (2) + cos−1 x 2
4 4
π 1
(3) − cos−1 x 2 (3)
π
−
1
cos−1 x 2
4 2 4 2
π π
(4) − cos−1 x 2 (4) − cos−1 x 2
4 4

13 VI - MATHEMATICS

30. The proposition (~p) ∨ (p ∧ ~q) is equivalent 30. ∑§ÕŸ (~p) ∨ (p ∧ ~q) ‚◊ÃÈÀÿ „Ò —
to :
(1) p ∨ ~q
(1) p ∨ ~q
(2) p → ~q
(2) p → ~q
(3) p ∧ ~q
(3) p ∧ ~q
(4) q→p
(4) q→p

-o0o-
-o0o-

14 VI - MATHEMATICS