Full jee mains 2015 online paper 10th april final

1. JEE Mains 2015 10th April (online)
Physics
Single Correct Answer Type:
1. In an ideal at temperature T, the average force that a molecule applies on the walls of a closed
container depends on 𝑇 𝑎𝑠 𝑇 𝑞
. A good estimate for q is:
(A) 2 (B)
1
2
(C) 1 (D)
1
4
Answer: (C)
Solution:
Average linear for collision to occur
𝑡 =
2𝑑
𝑢
Change in momentum in 1 collision
Δ𝑝 = 2 𝑚𝑢
∴ average force in collision
=
Δ𝑝
𝑡
𝑢 = root mean square speed
=
2 𝑚𝑢
2𝑑
× 𝑢
⇒ 𝑓 ∝ 𝑢2
∴ 𝑢2
∝ 𝑇
⇒ 𝑓 × 𝑇
⇒ 𝑞 = 1
2. In an unbiased n – p junction electrons diffuse from n-region to p-region because:
(A) Electrons travel across the junction due to potential difference
(B) Only electrons move from n to p region and not the vice – versa
(C) Electron concentration in n – region is more as compared to that in p – region
(D) Holes in p – region attract them
Answer: (C)
Solution:
In a 𝑝 − 𝑛 junction diffusion occurs due to spontaneous movement of majority charge carrier from
the region of high concentration to low concentration so option 3 in correct.
3. A 10V battery with internal resistance 1Ω 𝑎𝑛𝑑 𝑎 15𝑉 battery with internal resistance 0.6Ω are
connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to:

2. (A) 11.9 𝑉 (B) 13.1 𝑉 (C) 12.5 𝑉 (D) 24.5 𝑉
Answer: (B)
Solution:
The equivalent ems of the battery combination in given as
Equation =
𝐸1
𝑟1
+
𝐸1
𝑟2
1
𝑟1
+
1
𝑟2
=
10
1
+
15
0.6
1
1
+
1
0.6
=
10+
150
6
1+
10
6
=
105
8
= 13.1 𝑣𝑜𝑙𝑡
∴ The reading measured by voltmeter = 13.1 𝑣𝑜𝑙𝑡
4. A proton (mass m) accelerate by a potential difference V flies through a uniform transverse
magnetic field B. The field occupies a region of space by width ′𝑑′
. 𝐼𝑓 ′𝛼′ be the angle of
deviation of proton from initial direction of motion (see figure), the value of sin 𝛼 will be:
(A)
𝐵
2
√
𝑞𝑑
𝑚𝑉
(B) 𝐵𝑑√
𝑞
2𝑚𝑉
(C)
𝐵
𝑑
√
𝑞
2𝑚𝑉
(D) 𝑞 𝑉 √
𝐵𝑑
2𝑚
Answer: (B)

3. Solution:
Due to potential difference V speed acquired by proton in 𝑣0
⇒ 𝑊 = 𝑞 Δ 𝑉 = Δ𝑘
⇒ 𝑞𝑣 =
1
2
𝑚 𝑣0
2
⇒ 𝑣0 = √
2𝑞𝑣
𝑚
Radius of circular path acquired is 𝑅 =
𝑚𝑣0
𝑞𝐵
⇒ 𝑅 =
𝑚
𝑞𝐵
√
2𝑞𝑣
𝑚
= √
2𝑣𝑚
𝑞
×
1
𝐵
In ∆𝐶𝑃𝐷,sin 𝛼 =
𝑑
𝑅
= 𝑑√
𝑞
2 𝑣𝑚
𝐵 = 𝐵𝑑√
𝑞
2 𝑚𝑣
5. de – Broglie wavelength of an electron accelerated by a voltage of 50 V is close to
(|𝑒| = 1.6 × 10−19
𝐶, 𝑚 𝑒 = 9.1 × 10−31
𝑘𝑔, ℎ = 6.6 × 10−34
𝐽𝑠):
(A) 0.5 Å (B) 1.2 Å (C) 1.7 Å (D) 2.4 Å
Answer: (B)
Solution:
De broglie wavelength 𝜆 in given by
𝜆 =
ℎ
𝑝
=
ℎ
√2 𝑚𝑘
∴ 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝑘 = 𝑞 Δ𝑣
⇒ 𝜆 =
ℎ
√2𝑚𝑞∆𝑣
=
6.6 ×10−34
√2 ×9.1 × 10−3 × 1.6 ×10−19 × 50
=
6.6 ×10−34
√3.2 ×9.1 × 10−31−19 + 2
=
6.6 ×10−34
√3.2 ×9.1 × 10−48
=
6.6 ×10−34
√5.396 × 10−24
= 1.22 × 10−10
= 1.2 𝐴°
6. Suppose the drift velocity 𝑣 𝑑 in a material varied with the applied electric field E as 𝑣 𝑑 ∝ √𝐸.
Then 𝑉 − 𝐼 graph for a wire made of such a material is best given by:

4. (A)
(B)
(C)
(D)
Answer: (C)
Solution:
∴ 𝑣 𝑑 = 𝑘√𝐸 and 𝐼 = 𝑛 𝑒 𝐴 𝑣 𝑑
⇒ 𝐼 = 𝑛 𝑒𝐴 𝑘√𝐸
∴ 𝐸 =
𝑣
𝑑
⇒ 𝐼 = 𝑛𝑒𝐴𝑘 √
𝑣
𝑑
⇒ 𝐼 ∝ √ 𝑣 ⇒ 𝑣 ∝ 𝐼2
So
7. A parallel beam of electrons travelling in x – direction falls on a slit of width d (see figure). If
after passing the slit, an electron acquires momentum 𝑃𝑦 in the y – direction then for a majority
of electrons passing through the slit (h is Planck’s constant):

5. (A) |𝑃𝑦|𝑑 < ℎ (B) |𝑃𝑦|𝑑 > ℎ (C) |𝑃𝑦|𝑑 ≃ ℎ (D) |𝑃𝑦|𝑑 > > ℎ
Answer: (D)
Solution:
The electron beam will be diffractive at an angle θ
For central maxima
𝑑 sin 𝜃 = 𝜆
𝑑 sin 𝜃 =
𝑟
𝑝
Also 𝑝 sin 𝜃 = 𝑝 𝑦
⇒ 𝑑 𝑝 𝑦 = ℎ
∴ For majority of 𝑒 𝜃
′𝑠 passing through the shit lyeing in the central maxima 𝑑 𝑝 𝑦 ≈ ℎ
8. A block of mass 𝑚 = 10 𝑘𝑔 rests on a horizontal table. The coefficient of friction between the
block and the table is 0.05. When hit by a bullet of mass 50 g moving with speed v, that gets
embedded in it, the block moves and comes to stop after moving a distance of 2 m on the table.
If a freely falling object were to acquire speed
𝑣
10
after being dropped from height H, then
neglecting energy losses and taking 𝑔 = 10 𝑚𝑠−2
, the value of H is close to:
(A) 0.2 km (B) 0.5 km (C) 0.3 km (D) 0.4 km
Answer: ()
Solution:
9. When current in a coil changes from 5 A to 2 A in 0.1 s, an average voltage of 50 V is
produced. The self – inductance of the coil is:
(A) 1.67 H (B) 6 H (C) 3 H (D) 0.67 H
Answer: (A)
Solution:

6. Area of coil
𝑑 = 𝐿𝐼 ⇒
∆𝑑
∆𝑡
= 𝐿
∆𝐼
∆𝑡
∴ (𝜀𝑖𝑛𝑑) 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = |
∆𝑑
∆𝑡
| = 𝐿 |
∆𝐼
∆𝑡
|
⇒ 50 = 𝐿 ×
5−2
0.1
⇒
5
3
= 𝐿
⇒ 𝐿 = 1.674
10. 𝑥 𝑎𝑛𝑑 𝑦 displacements of a particle are given as 𝑥(𝑡) = 𝑎 sin 𝜔𝑡 𝑎𝑛𝑑 𝑦(𝑡) = 𝑎 sin 2𝜔𝑡. Its
trajectory will look like:
(A)
(B)
(C)
(D)
Answer: (C)
Solution:
∵ 𝑥 = 𝐴 sin 𝜔𝑡 ⇒ 𝑠𝑖𝑛 𝜔𝑡 =
𝑥
𝐴
Also, 𝑐𝑜𝑠 𝜔𝑡 = √1 − sin2 𝜔𝑡 = √1 −
𝑥2
𝐴2

7. ⇒ cos 𝜔𝑡 =
√𝐴2−𝑥2
𝐴
As, 𝑦 = 2𝐴 sin 𝜔𝑡 cos 𝜔𝑡
⇒ 𝑦 = 2 𝐴
𝑥
𝐴
√𝐴2 − 𝑥2
𝐴
⇒ 𝑦 =
2
𝐴
𝑥 √ 𝐴2 − 𝑥2
⇒ 𝑦 = 0 𝑎𝑡 𝑥 = 0 𝑎𝑛𝑑 𝑥 = ± 𝐴
Which in possible only in option (3)
11. Consider a thin uniform square sheet made of a rigid material. If its side is ‘a’, mass m and
moment of inertia I about one of its diagonals, then:
(A) 𝐼 =
𝑚𝑎2
24
(B)
𝑚𝑎2
24
< 𝐼 <
𝑚𝑎2
12
(C) 𝐼 >
𝑚𝑎2
12
(D) 𝐼 =
𝑚𝑎2
12
Answer: (D)
Solution:
In a uniform square plate due to symmetry moment of Inertia about all the axis passing through
centre and lying in the blank of the plate is same.
∴ 𝐼 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 = 𝐼 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑠𝑖𝑑𝑒
=
𝑚𝑎2
12
12. Diameter of a steel ball is measured using a Vernier calipers which has divisions of 0.1 cm on
its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main
scale. Three such measurements for a ball are given as:
S.No. MS (cm) VS divisions
1. 0.5 8
2. 0.5 4
3. 0.5 6
If the zero error is – 0.03 cm, then mean corrected diameter is:
(A) 0.53 cm

8. (B) 0.56 cm
(C) 0.59 cm
(D) 0.52 cm
Answer: (C)
Solution:
L.C. of Vernier calipers
=
1 𝑚𝑎𝑖𝑛 𝑠𝑐𝑎𝑙𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑛𝑠 vernier 𝑠𝑐𝑎𝑙𝑒
=
0.1
10
= 0.01 𝑐𝑚
Required of Vernier calipers
= 𝑀. 𝑆. 𝑅. +(𝐿. 𝐶) × 𝑣𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠.
∴ Measured diameter are respecting
0.52 𝑐𝑚 0.54 𝑐𝑚, 0.56 𝑐𝑚
∴ 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 =
0.58 + 0.54 + 0.56
3
=
1.68
3
= 0.56
∴ 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 0.56 − (−0.03)
= 0.56 + 0.03 = 0.59 𝑐𝑚
13. A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius
R (R < < L). A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the
galaxy and passing through its centre. If the time period of star is T and its distance from the
galaxy’s axis is r, then:
(A) 𝑇 ∝ √ 𝑟
(B) 𝑇 ∝ 𝑟
(C) 𝑇 ∝ 𝑟2
(D) 𝑇2
∝ 𝑟3
Answer: (B)
Solution:
Due to a long solid cylinder gravitational field strong can be given as:
𝑔′ =
2 𝐺 𝜆
𝑥
Where
𝜆 = 𝑙𝑖𝑛𝑒𝑎𝑟 𝑚𝑎𝑠𝑠 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑔𝑎𝑙𝑎𝑥𝑦.
𝐹𝑜𝑟 𝑡ℎ𝑒 𝑜𝑟𝑏𝑖𝑡𝑎𝑙 𝑚𝑜𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑡ℎ𝑒 𝑔𝑎𝑙𝑎𝑥𝑦.

9. 𝑓𝑔 = 𝑓𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙
⇒ 𝑚𝑔 = 𝑚 𝜔2
𝑥
⇒
2𝐺𝜆
𝑥
= 𝜔2
𝑥
⇒ 𝜔2
∝
1
𝑥2
⇒ 𝜔 ∝
1
𝑥
⇒
2𝜋
𝑇
∝
1
𝑥
⇒ 𝑇 ∝ 𝑥
So option 2 is correct
14. An electromagnetic wave travelling in the x – direction has frequency of 2 × 1014
𝐻𝑧 and
electric field amplitude of 27 𝑉𝑚−1
. From the options given below, which one describes the
magnetic field for this wave?
(A) 𝐵⃗ (𝑥, 𝑡) = (9 × 10−8
𝑇)𝑗̂ sin[1.5 × 10−6
𝑥 − 2 × 1014
𝑡]
(B) 𝐵⃗ (𝑥, 𝑡) = (9 × 10−8
𝑇)𝑖̂ sin[2𝜋(1.5 × 10−8
𝑥 − 2 × 1014
𝑡)]
(C) 𝐵⃗ (𝑥, 𝑡) = (3 × 10−8
𝑇)𝑗̂ sin[2𝜋(1.5 × 10−8
𝑥 − 2 × 1014
𝑡)]
(D) 𝐵⃗ (𝑥, 𝑡) = (9 × 10−8
𝑇)𝑘̂ sin[2𝜋 (1.5 × 10−6
𝑥 − 2 × 1014
𝑡)]
Answer: (D)
Solution:
𝑊ℎ𝑒𝑛 𝐸 = 𝐸0 𝑠𝑖𝑛 𝐶 𝑘𝑥 − 𝜔𝑡
𝑇ℎ𝑒𝑛 𝐵 = 𝐵0 𝑠𝑖𝑛 𝐶 𝑘𝑥 − 𝜔𝑡
Of light in travelling along 𝑖̂ then 𝐵⃗ in either along 𝑗 or𝑘⃗ .
∴ 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝐶 =
𝐸0
𝐵0
⇒ 𝐵0 =
𝐸0
𝐶
⇒ 𝐵0 =
27
3×108 = 9 × 10−8
𝑇
also, 𝜔 = 2𝜋 f = 2π × 2 × 1014
= 4 𝜋 × 1014
Looking into the option the correct
Answer is 𝐵⃗ = 9 × 10−8
sin2𝜋 (1.5 × 10−6
𝑥 − 2 × 1014
𝑡) 𝑘̂
15. A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If
a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the
angle formed by the image of the tower is 𝜃, then 𝜃 is close to:

10. (A) 30°
(B) 15°
(C) 1°
(D) 60°
Answer: (D)
Solution:
16. A block of mass 𝑚 = 0.1 𝑘𝑔 is connected to a spring of unknown spring constant k. It is
compressed to a distance x from its equilibrium position and released from rest. After
approaching half the distance (
𝑥
2
) from equilibrium position, it hits another block and comes
to rest momentarily, while the other block moves with a velocity 3 𝑚𝑠−1
. The total initial
energy of the spring is:
(A) 0.6 𝐽
(B) 0.8 𝐽
(C) 1.5 𝐽
(D) 0.3 𝐽
Answer: (A)
Solution: By energy conservation between compression positions 𝑥 and
𝑥
2
1
2
𝑘𝑥2
=
1
2
𝑘 (
𝑥
2
)
2
+
1
2
𝑚𝑣2
1
2
𝑘𝑥2
−
1
2
𝑘
𝑥2
4
=
1
2
𝑚𝑣2
1
2
𝑘𝑥2
(
3
4
) =
1
2
𝑚𝑣2
𝑣 = √
3𝑘𝑥2
4𝑚
= √
3𝑘
𝑚
𝑥
2
On collision with a block at rest
∵ Velocities are exchanged ⇒ elastic collision between identical masses.
∴ 𝑣 = 3 = √
3𝑘
𝑚
𝑥
2
⇒ 6 = √
3𝑘
𝑚
𝑥
⇒ 𝑥 = 6√
𝑚
3𝑘
∴ The initial energy of the spring is

11. 𝑈 =
1
2
𝑘 𝑥2
=
1
2
𝑘 × 36
𝑚
3𝑘
= 6𝑚
𝑈 = 6 × 0.1 = 0.6 𝐽
17. Shown in the figure are two point charges + Q and – Q inside the cavity of a spherical shell. The
charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If 𝜎1is
the surface charge on the inner surface and 𝑄1net charge on it and 𝜎2 the surface charge on the
other surface and 𝑄2 net charge on it then:
(A) 𝜎1 = 0, 𝑄1 = 0, 𝜎2 = 0, 𝑄2 = 0
(B) 𝜎1 ≠ 0, 𝑄1 = 0, 𝜎2 ≠ 0, 𝑄2 = 0
(C) 𝜎1 ≠ 0, 𝑄1 ≠ 0, 𝜎2 ≠ 0, 𝑄2 ≠ 0
(D) 𝜎1 ≠ 0, 𝑄1 = 0, 𝜎2 = 0, 𝑄2 = 0
Answer: (D)
Solution: By the property of electrostatic shielding in the conductors 𝜖 = 0 in the conductor.
So electric flux = 0 through a dotted Gaussian surface as shown
The net enclosed charge through Gaussian surface = 0
⇒ Net charge 𝑄1 on the inner surface = 0 but the equal and opposite induced charge on the surface
will be distributed non uniformly on the inner surface
So, 𝜎1 ≠ 0
∵ 𝑄1 = 0 on the inner surface
So, net charge 𝑄2 = 0 on the outer surface as conductor is neutral but ∵ outer surface is free from
any electric field so no charge density exists on the outer surface. So, 𝜎2 = 0.
18. You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face
and views the magnified image of the face at the closest comfortable distance of 25 cm. The
radius of curvature of the mirror would then be:
(A) 24 𝑐𝑚
(B) 30 𝑐𝑚
(C) 60 𝑐𝑚

12. (D) −24 𝑐𝑚
Answer: (C)
Solution:
If AB is the position of face of man then A ‘B’ is the position of image of face.
As image is formed at 25cm form the object.
∴ From concave mirror image is 15cm behind the mirror.
So, 𝑢 = −10 𝑐𝑚, 𝑣 = +15 𝑐𝑚
⇒
1
𝑓
=
1
𝑢
+
1
𝑣
⇒
1
𝑓
=
1
−10
+
1
15
=
−3 + 2
30
⇒ 𝑓 = −300 𝑐𝑚
So, radius of curvature = 60 𝑐𝑚
19. A thin disc of radius 𝑏 = 2𝑎 has a concentric hole of radius ‘a’ in it (see figure). It carries
uniform surface charge ′𝜎′ on it. If the electric field on its axis at height ′ℎ′
(ℎ < < 𝑎) from its
centre is given as ‘Ch’ then value of ‘C’ is:
(A)
𝜎
4 𝛼𝜖0
(B)
𝜎
𝛼𝜖0
(C)
𝜎
𝑆𝛼𝜖0
(D)
𝜎
2𝛼𝜖0
Answer: (A)
Solution: ∵ at the axial point of a uniformly charged disc electric field is given by
𝐸 =
𝜎
2𝜖0
(1 − 𝑐𝑜𝑠𝜃)

13. By superposition principle when inner disc is removed then electric field due to remaining disc is
𝐸 =
𝜎
2𝜖0
[(1 − 𝑐𝑜𝑠𝜃2) − (1 − 𝑐𝑜𝑠𝜃1)]
=
𝜎
2𝜖0
[𝑐𝑜𝑠𝜃1 − 𝑐𝑜𝑠𝜃2]
=
𝜎
2𝜖0
[
ℎ
√ℎ2 + 𝑎2
−
ℎ
√ℎ2 + 𝑏2
]
=
𝜎
2𝜖0
[
ℎ
𝑎√1 +
ℎ2
𝑎2
−
ℎ
√1 +
ℎ2
𝑏2 ]
∵ ℎ ≪ 𝑎 and b
∴ 𝐸 =
𝜎
2𝜖0
[
ℎ
𝑎
−
ℎ
𝑏
]
=
𝜎
2𝜖0
[
ℎ
𝑎
−
ℎ
2𝑎
] =
𝜎ℎ
4𝜖0 𝑎
⇒ 𝐶 =
𝜎
4𝑎𝜖0
20. An ideal gas goes through a reversible cycle 𝑎 → 𝑏 → 𝑐 → 𝑑 has the V – T diagram shown below.
Process 𝑑 → 𝑎 𝑎𝑛𝑑 𝑏 → 𝑐 are adiabatic.
The corresponding P – V diagram for the process is (all figures are schematic and not drawn to
scale) :
(A)

14. (B)
(C)
(D)
Answer: (A)
Solution: Is an adiabatic process
𝑇𝑉 𝛾−1
= 𝑐𝑜𝑛𝑠𝑡 ⇒ 𝑉𝑇
1
𝛾−1 = 𝑐𝑜𝑛𝑠𝑡
⇒ as T increase V decreases at non-uniform rate
In process 𝑎 → 𝑏 P = constant as 𝑉 ∝ 𝑇
In process 𝑐 → 𝑑 𝑃′
= constant s 𝑉 ∝ 𝑇
But since slope of V – T graph ∝
1
𝑃
since slope of ab < slope of cd
⇒ 𝑃𝑎𝑏 > 𝑃𝑐𝑑
Also in adiabatic process 𝑑 → 𝑎 as T is increasing V in decreasing
⇒ P is increasing, so P – V diagram is as below

15. 21. A uniform solid cylindrical roller of mass ‘m’ is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre. If the acceleration of the cylinder is ‘a’ and it is
rolling without slipping then the value of ‘F’ is:
(A)
3
2
𝑚𝑎
(B) 2 𝑚𝑎
(C)
5
3
𝑚𝑎
(D) 𝑚𝑎
Answer: (A)
Solution:
From free body diagram of cylinder
𝐹 − 𝑓𝑠 = 𝑚𝑎 …..(1)
∵ ∑ 𝑓𝑒𝑥𝑡 = 𝑚𝑎 𝑐𝑚
𝑎𝑙𝑠𝑜 ∑ 𝜏 𝑒𝑥𝑡 = 𝐼𝑐𝑚 ∝
⟹ 𝑓𝑠 𝑅 = 𝐼𝑐𝑚 ∝
⟹ 𝑓𝑠 𝑅 =
1
2
𝑚𝑅2
∝ ….. (2)
For rolling without slipping
𝑎 = 𝑅 ∝ …… (3)
⟹ ∝=
𝑞
𝑅
∴ 𝑓𝑠 𝑅 =
1
2
𝑚𝑅2 𝑞
𝑅
⟹ 𝑓𝑠 =
1
2
𝑚𝑎
Put in (1)
𝑓 −
1
2
𝑚𝑎 = 𝑚𝑎

16. ⟹ 𝑓 =
3
2
𝑚𝑎
22. A 25 cm long solenoid has radius 2 cm and 500 total number of turns. It carries a current of 15
A. If it is equivalent to a magnet of the same size and magnetization
𝑀⃗⃗ (𝑀𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑚𝑜𝑚𝑒𝑛𝑡 𝑉𝑜𝑙𝑢𝑚𝑒⁄ ), 𝑡ℎ𝑒𝑛 |𝑀⃗⃗ | is:
(A) 3𝜋 𝐴𝑚−1
(B) 30000 𝐴𝑚−1
(C) 30000𝜋 𝐴𝑚−1
(D) 300 𝐴𝑚−1
Answer: (B)
Solution:
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝐴𝑙
𝑀𝑎𝑔𝑛𝑒𝑡𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑀⃗⃗ =
𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑚𝑜𝑛𝑒𝑛𝑡
𝑉𝑜𝑙𝑢𝑚𝑒
=
(𝑁𝑜.𝑜𝑓 𝑡𝑢𝑟𝑛𝑠)×(𝐶𝑢𝑟𝑟𝑒𝑛𝑡)×𝐴𝑟𝑒𝑎
𝑉𝑜𝑙𝑢𝑚𝑒
=
𝑁 𝐼 𝐴
𝐴 ℓ
=
𝑁𝐼
ℓ
=
500×15×100
25
= 60 × 500
= 30 × 103
= 30000 𝐴𝑚−1
23. In the circuits (a) and (b) switches 𝑆1 𝑎𝑛𝑑 𝑆2 are closed at t = 0 and are kept closed for a long
time. The variation of currents in the two circuits for 𝑡 ≥ 0 are roughly shown by (figures are
schematic and not drawn to scale):
(A)

17. (B)
(C)
(D)
Answer: (B)
Solution:
In CR series circuit
𝑞 = 𝑞0 (1 − 𝑒
−𝑡
𝜏 )
⟹ 𝑞 = 𝐶𝐸 (1 − 𝑒
−𝑡
𝑅𝐶)
∴ 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝐼 =
𝑑𝑞
𝑑𝑡
=
𝐶𝐸
𝑅𝐶
(+𝑒
−𝑡
𝑅𝐶)
𝐼 =
𝐸
𝑅
𝑒
−𝑡
𝑅𝐶
⟹ 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑑𝑒𝑐𝑎𝑦𝑠 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙𝑙𝑦 𝑎𝑑 𝑖𝑛 𝐿𝑅 𝑠𝑒𝑟𝑖𝑒𝑠 𝑐𝑖𝑟𝑐𝑢𝑖𝑡
𝐼 = 𝐼0 (1 − 𝑒
−𝑡
𝜏 )
𝑤ℎ𝑒𝑟𝑒 𝐼0 =
𝐸
𝑅
𝑎𝑛𝑑 𝜏 =
𝐿
𝑅
𝐼 =
𝐸
𝑅
(1 − 𝑒
−𝑅𝑡
𝐿 ) ⟹ 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑔𝑟𝑜𝑤𝑠 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙𝑙𝑠
∴ 𝑓𝑜𝑟 𝐶 − 𝑅 𝑐𝑖𝑟𝑐𝑢𝑖𝑡

18. For L – R circuit
24. If two glass plates have water between them and are separated by very small distance (see
figure), it is very difficult to pull them apart. It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere. If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by:
(A)
2𝑇
𝑅
(B)
𝑇
4𝑅
(C)
4𝑇
𝑅
(D)
𝑇
2𝑅
Answer: (A)
Solution:
𝑑 = 2𝑅 𝑐𝑜𝑠𝜃
∴ 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑎 𝑑𝑜𝑢𝑏𝑙𝑒 𝑐𝑢𝑟𝑣𝑎𝑡𝑢𝑟𝑒 𝑓𝑖𝑙𝑚
∆𝑃 = 2𝑇 (
1
𝑅1
+
1
𝑅2
)
∵ 𝑅1 = 𝑅 𝑎𝑛𝑑 𝑅2 = ∞
∆𝑃 = 2𝑇 (
1
𝑅
+
1
∞
)

19. ∆𝑃 =
2𝐼
𝑅
∴ Pressure is more in the concave side hence pressure in water between the plates is lower by
2𝑇
𝑅
25. A simple harmonic oscillator of angular frequency 2 rad 𝑠−1
is acted upon by an external force
𝐹 = sin 𝑡 𝑁. If the oscillator is at rest in its equilibrium position at 𝑡 = 𝑜, its position at later
times is proportional to:
(A) sin 𝑡 +
1
2
cos 2𝑡
(B) 𝑐𝑜𝑠𝑡 −
1
2
sin2𝑡
(C) sin 𝑡 −
1
2
sin2𝑡
(D) sin 𝑡 +
1
2
sin2𝑡
Answer: (C)
Solution:
It is given that oscillator at rest at t = 0 i.e. at t = 0, v = 0
So, in option we can check by putting 𝑣 =
𝑑𝑥
𝑑𝑡
= 0
(1) 𝐼𝑓 𝑥 ∝ sin 𝑡 +
1
2
cos2𝑡
⟹ 𝑣 ∝ cos 𝑡 +
1
2
× 2 (− sin 2𝑡)
⟹ 𝑎𝑡 𝑡 = 0, 𝑣 ∝ 1 − 0 ≠ 0
(2) 𝐼𝑓 𝑥 ∝ cos 𝑡 −
1
2
sin 𝑡
⟹ 𝑣 ∝ − sin 𝑡 −
1
2
cos 𝑡
⟹ 𝑎𝑡 𝑡 = 0, 𝑣 ∝ −
1
2
≠ 0
(3) 𝐼𝑓 𝑥 ∝ sin 𝑡 −
1
2
𝑠𝑖𝑛𝜃 2𝑡
𝑡ℎ𝑒𝑛 𝜐 ∝ cos 𝑡 −
1
2
× 2 cos 2𝑡
⟹ 𝑎𝑡 𝑡 = 0, 𝑣 ∝ 1 − 1 = 0
(4) 𝐼𝑓 𝑥 ∝ sin 𝑡 +
1
2
sin2𝑡
⟹ 𝑣 ∝ cos 𝑡 +
1
2
× 2 cos2𝑡
⟹ 𝑎𝑡 𝑡 = 0, 𝑣 ∝ 1 + 1
⟹ 𝑣 ∝ 2 ≠ 0
∴ 𝑖𝑛 𝑜𝑝𝑡𝑖𝑜𝑛 (3) 𝑣 = 0 𝑎𝑡 𝑡 = 0
26. If a body moving in a circular path maintains constant speed of 10 𝑚𝑠−1
, then which of the
following correctly describes relation between acceleration and radius?
(A)

20. (B)
(C)
(D)
Answer: (D)
Solution:
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
𝑎 =
𝑣2
𝑅
⟹ 𝑎𝑅 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

21. ⟹ 𝑎 ∝
1
𝑅
27. If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter
2
√ 𝜋
𝑐𝑚 then the
Reynolds number for the flow is (density of water =103 𝑘𝑔 𝑚3⁄
𝑎𝑛𝑑 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 =
10−3
𝑃𝑎. 𝑠) close to:
(A) 5500 (B) 550 (C) 1100 (D) 11,000
Answer: (A)
Solution:
Reynolds number
𝑅 =
𝑆𝑉𝐷
𝜂
𝐷 = Diameter of litre
Also rate of flow =
𝑉𝑜𝑙𝑢𝑚𝑒
𝑡𝑖𝑚𝑒
= 𝐴 𝑉
𝑉
𝑡
=
𝜋 𝐷2
4
× 𝑉 ⇒ 𝑉 =
4𝑉
𝜋𝐷2 𝑡
∴ 𝑅 =
𝑆 𝐷
𝜂
×
4 𝑉
𝜋 𝐷2 𝑡
=
4 𝑆 𝑉
𝜋 𝜂 𝐷 𝑡
=
4 × 103
× 15 × 10−3
𝜋 × 10−3 × 2 × 5 × 60
√ 𝜋 × 102
=
10000
√ 𝜋
≈ 5500
28. If one were to apply Bohr model to a particle of mass ‘m’ and charge ‘q’ moving in a plane
under the influence of a magnetic field ‘B’, the energy of the charged particle in the 𝑛 𝑡ℎ
level
will be:
(A) 𝑛 (
ℎ𝑞𝐵
𝜋𝑚
) (B) 𝑛 (
ℎ𝑞𝐵
4𝜋𝑚
) (C) 𝑛 (
ℎ𝑞𝐵
2𝜋𝑚
) (D) 𝑛 (
ℎ𝑞𝐵
8𝜋𝑚
)
Answer: (B)
Solution:

22. For a charge q moving in a +r uniform magnetic field B
𝑓𝑚 =
𝑚𝑣2
𝑅
𝑞𝑉𝐵 =
𝑚𝑣2
𝑅
⇒ 𝑚𝑣2
= 𝑞𝑉𝐵𝑅
⇒
1
2
𝑚𝑣2
=
𝑞𝑉𝐵𝑅
2
⇒ 𝐸𝑛𝑒𝑟𝑔𝑦 =
𝑞𝑉𝐵𝑅
2
(1)
By Bohr’s quantisation condition
Angular momentum 𝐿 = 𝑛
ℎ
2𝜋
⇒ 𝑚𝑣𝑅 =
𝑛ℎ
2𝜋
⇒ 𝑣𝑅 =
𝑛ℎ
2𝜋 𝑚
(2)
Put (2) in (2)
⇒ 𝐸𝑛𝑒𝑟𝑔𝑦 =
𝑞𝐵
2
(
ℎ
2 𝜋 𝑚
)
=
𝑞𝐵 𝑛ℎ
4 𝜋 𝑚
29. If the capacitance of a nanocapacitor is measured in terms of a unit ‘u’ made by combining the
electronic charge ‘e’, Bohr radius ′𝑎0
′
, Planck’s constant ‘h’ and speed of light ‘c’ then:
(A) 𝑢 =
𝑒2 𝑎0
ℎ𝑐
(B) 𝑢 =
ℎ𝑐
𝑒2 𝑎0
(C) 𝑢 =
𝑒2 𝑐
ℎ𝑎0
(D) 𝑢 =
𝑒2ℎ
𝑐𝑎0
Answer: (A)
Solution:
∵ 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 𝐶 =
𝑄
∆𝑣
𝐴𝑙𝑠𝑜 [
ℎ𝑐
𝜆
] = [
ℎ𝑐
𝑎0
] = [𝐸𝑛𝑒𝑟𝑔𝑦]
∴ [𝐶] =
[𝑄]
[∆𝑣]
=
[𝑄] [𝑄]
[∆𝑣] [𝑄]
∵ 𝑊 = 𝑞∆𝑣 ⇒ [𝑄] [∆𝑣] = [𝐸𝑛𝑒𝑟𝑔𝑦]
∴ [𝐶] =
[𝑄2]
[𝐸𝑛𝑒𝑟𝑔𝑦]
=
[𝑄2] [𝑎0]
[ℎ𝑐]
∴ [𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 ] =
[𝑄2] [𝑎0]
[ℎ𝑐]
⇒ 𝑢 =
𝑒2 𝑎0
ℎ𝑐

23. 30. A bat moving at 10 𝑚𝑠−1
towards a wall sends a sound signal of 8000 Hz towards it. On
reflection it hears a sound of frequency𝑓. The value of 𝑓 in Hz is close to
(𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑜𝑢𝑛𝑑 = 320 𝑚𝑠−1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer: (D)
Solution:
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
𝑓 =
𝑣 + 10
𝑣 − 10
× 𝑓0
=
320 + 10
320 − 10
× 8000
=
330
310
× 8000
=
33
31
× 8000
= 8516 𝐻𝑧

24. JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1. 1.4 g of an organic compound was digested according to Kjeldahl’s method and the ammonia
evolved was absorbed in 60 mL of M/10 𝐻2 𝑆𝑂4 solution. The excess sulphuric acid required 20
mL of M/10 NaOH solution for neutralization. The percentage of nitrogen in the compound is:
(A) 24 (B)3 (C)5 (D)10
Solution: (D) 60 ×
1
10
= 6 𝑚𝑀 𝐻2 𝑆𝑂4 used
Excess 𝐻2 𝑆𝑂4 ≡ 20 ×
1
10
×
1
2
= 1 𝑚𝑀 𝐻2 𝑆𝑂4
𝐻2 𝑆𝑂4 used = 6 − 1 = 5 𝑚𝑀
2𝑁𝐻3 + 𝐻2 𝑆𝑂4 ⟶ (𝑁𝐻4)2 𝑆𝑂4
mM of 𝑁𝐻3 = 10 𝑚𝑀
Mass of 𝑁 = 10 × 10−3
× 14 (
𝑔
𝑚𝑜𝑙𝑒
) = 0.140𝑔
% 𝑁2 =
0.140
1.4
× 100 = 10%
2. The optically inactive compound from the following is:
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution: (B)
(Optically active)
(Optically inactive because of 2 − 𝐶𝐻3 groups present on same C atom)

25. (Optically active)
3. The least number of oxyacids are formed by:
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution: (B) Fluorine does not form oxyacids as it is more electronegative than oxygen.
4. Gaseous 𝑁2 𝑂4 dissociates into gaseous 𝑁𝑂2according to the reaction𝑁2 𝑂4(𝑔) ⇌ 2𝑁𝑂2(𝑔)
At 300 K and 1 atm pressure, the degree of dissociation of 𝑁2 𝑂4 is 0.2. If one mole of 𝑁2 𝑂4 gas is
contained in a vessel, then the density of the equilibrium mixture is:
(A) 3.11 g/L
(B) 1.56 g/L
(C) 4.56 g/L
(D) 6.22 g/L
Solution: (A)
𝑁2 𝑂4 ⇌ 2𝑁𝑂2
(1 − 𝛼) 2𝛼
Total moles at equilibrium = 1 − 𝛼 + 2𝛼 = 1 + 𝛼 = 1.2
M avg for equilibrium mixture =
92
𝑔
𝑚𝑜𝑙𝑒
(𝑁2 𝑂4)
1.2
𝑑 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
𝑃𝑀 𝑎𝑣𝑔
𝑅𝑇
=
1 × 76.67
0.082 × 300
=
76.67
24.6
= 3.11 𝑔𝐿−1
5. Arrange the following amines in the order of increasing basicity.
(A)

26. (B)
(C)
(D)
Solution: (C)
Most basic due to +I effect of methyl group. Methoxy group provides electron density at -
𝑁𝐻2
-𝑁𝑂2 group with draws electron density from N of -𝑁𝐻2
6.

27. A is;
(A)
(B)
(C)
(D)
Solution: (A)
7. A solution at 20 𝑜
𝐶 is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour
pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr,
respectively, then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be, respectively:
(A) 30.5 torr and 0.389
(B) 35.0 torr and 0.480
(C) 38.0 torr and 0.589
(D) 35.8 torr and 0.280
Solution: (C) 𝑋 𝐵𝑒𝑛𝑧𝑒𝑛𝑒 =
1.5
5
= 0.3
𝑋 𝑇𝑜𝑙𝑢𝑒𝑛𝑒 =
3.5
5
= 0.7
𝑃𝑡𝑜𝑡𝑎𝑙 = 0.3 × 74.7 + 0.7 × 22.3

28. = 22.41 + 15.61 = 38.02
≈ 38 𝑇𝑜𝑟𝑟
By Dalton’s law to vapour phase
𝑋 𝐵𝑒𝑛𝑧𝑒𝑛𝑒
′ (𝑣𝑎𝑝 𝑝ℎ𝑎𝑠𝑒) =
0.3 × 74.7
38
=
22.41
38
= 0.589
8. Which molecule/ion among the following cannot act as a ligand in complex compounds?
(A) 𝐶𝑁−
(B) 𝐶𝐻4
(C) 𝐶𝑂
(D) 𝐵𝑟−
Solution: (B) 𝐶𝐻4 does not have either a lone pair or 𝜋-electron pair it cannot act as ligand.
9. A compound A with molecular formula 𝐶10 𝐻13 𝐶𝑙 gives a white precipitate on adding silver
nitrate solution. A on reacting with alcoholic KOH gives compound B as the main product. B on
ozonolysis gives C and D. C gives Cannizaro reaction but not aldol condensation. D gives aldol
condensation but not Cannizaro reaction. A is:
(A)
(B)
(C)
(D)
Solution: (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 𝐴𝑔𝑁𝑂3
(Saytzeff Rule)

29. 10.
is used as:
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution: (D) Acetyl salicylic acid is analgesic.
11. An aqueous solution of a salt X turns blood red on treatment with 𝑆𝐶𝑁−
and blue on
treatment with 𝐾4[𝐹𝑒(𝐶𝑁)6], X also gives a positive chromyl chloride test. The salt X is:
(A) 𝐹𝑒𝐶𝑙3
(B) 𝐹𝑒(𝑁𝑂3)3
(C) 𝐶𝑢𝐶𝑙2
(D) 𝐶𝑢(𝑁𝑂3)2
Solution: (A)
𝐹𝑒𝐶𝐿3 + 3 𝑆𝐶𝑁𝑎𝑞
−
⇌ 𝐹𝑒(𝑆𝐶𝑁)3 + 3 𝐶𝑙−
(𝐵𝑙𝑜𝑜𝑑 𝑟𝑒𝑑)
4 𝐹𝑒𝐶𝑙3 + 3𝐾4[𝐹𝑒(𝐶𝑁)6] ⟶ 12 𝐾𝐶𝑙 + 𝐹𝑒4[𝐹𝑒(𝐶𝑁)6]3
𝑃𝑟𝑢𝑠𝑠𝑖𝑜𝑛 𝑏𝑙𝑢𝑒
2𝐹𝑒𝐶𝑙3 + 3𝐻2 𝑆𝑂4 ⟶ 𝐹𝑒2(𝑆𝑂4)3 + 6𝐻𝐶𝑙
𝐾2 𝐶𝑟2 𝑂7 + 2𝐻2 𝑆𝑂4 ⟶ 2𝐾𝐻𝑆𝑂4 + 2𝐶𝑟𝑂3 + 𝐻2 𝑂
𝐶𝑟𝑂3 + 2𝐻𝐶𝑙 ⟶ 𝐶𝑟𝑂2 𝐶𝑙2 + 𝐻2 𝑂
(𝐶ℎ𝑟𝑜𝑚𝑦𝑙𝑐ℎ𝑙𝑜𝑟𝑖𝑑𝑒)
𝐶𝑒𝑂2 𝐶𝑙2 + 4 𝑁 𝑎𝑂𝐻 ⟶ 𝑁𝑎2 𝐶𝑟𝑂4 + 2𝑁𝑎𝐶𝑙 + 2𝐻2 𝑂
(𝑦𝑒𝑙𝑙𝑜𝑤)

30. 𝑁𝑎2 𝐶𝑟𝑂4 + 𝑃𝑏(𝐶𝐻3 𝐶𝑂𝑂)2 ⟶ 𝑃𝑏𝐶𝑟𝑂4 + 2𝐶𝐻3 𝐶𝑂𝑂𝑁𝑎
(𝑦𝑒𝑙𝑙𝑜𝑤 𝑝𝑝𝑡)
12. The correct statement on the isomerism associated with the following complex ions,
(A) [𝑁𝑖(𝐻2 𝑂)5 𝑁𝐻3]2+
(B) [𝑁𝑖(𝐻2 𝑂)4(𝑁𝐻3)2]2+
and
(C) [𝑁𝑖(𝐻2 𝑂)3(𝑁𝐻3)3]2+
is
(D) (A) and (B) show only geometrical isomerism
Solution: (D) [𝑁𝑖 (𝐻2 𝑂)4(𝑁𝐻3)2]2+
Show c is & trans geometrical isomerism [𝑁𝑖 (𝐻2 𝑂)3(𝑁𝐻3)3]2+
Show facial & meridional geometrical isomerism.
13. In the presence of a small amount of phosphorous, aliphatic carboxylic acids react with 𝛼-
hydrogen has been replaced by halogen. This reaction is known as:
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution: (D) This reaction is known as HVZ reaction.
14. The reaction 2N2O5(g) → 4NO2(g) + O2(g) follows first order kinetics. The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 87.5 mm Hg in 30 min.
The pressure exerted by the gases after 60 min. Will be (Assume temperature remains
constant) :
(A) 106.25 mm Hg
(B) 125 mm Hg
(C) 116.25 mm Hg
(D) 150 mm Hg
Solution: (A)
2𝑁2 𝑂5(𝑔) ⟶ 4 𝑁𝑂2(𝑔)
(𝑝0 − 𝑥) 2𝑥
+ 𝑂2(𝑔)
𝑥
2
∑ 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝑝0 − 𝑥 + 2𝑥 +
𝑥
2
= 𝑝0 +
3𝑥
2
= 𝑝𝑡𝑜𝑡𝑎𝑙
87.5 = 50 +
3𝑥
2
3𝑥
2
= 37.5
∴ 𝑥 = 37.5 ×
2
3
= 25

31. For first order kinetics
𝑘𝑡 = ln
𝑝0
𝑝0 − 𝑥
= 𝑙𝑛
50
25
= ln 2
𝑘 =
1
𝑡
ln 2 =
1
30
ln 2
After 60 min
𝑘 =
1
𝑡′
ln
𝑝0
𝑝0 − 𝑥′
⇒
1
30
ln 2 =
1
60
ln
𝑝0
𝑝0 − 𝑥′
2 ln 2 = ln
𝑝0
𝑝0 − 𝑥′
− ln 4
𝑝0
𝑝0 − 𝑥′
= 4 ⇒ 𝑝0
= 4 𝑝0
− 4𝑥′
𝑥′
=
4𝑝0 − 𝑝0
4
=
3𝑝0
4
=
3 × 50
4
= 37.5
Σ60 𝑚𝑖𝑛 𝑇𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝑝0 +
3𝑥′
2
= 50 + 3 ×
37.5
2
= 50 + 56.25 = 106.25 𝑚𝑚
15. If the principal quantum number n = 6, the correct sequence of filling of electrons will be:
(A) ns → (n − 1) d → (n − 2) f → np
(B) ns → np → (n − 1)d → (n − 2)f
(C) ns → (n − 2)f → np → (n − 1)d
(D) ns → (n − 2)f → (n − 1)d → np
Solution: (D) As per (n + ℓ) rule when n = 6
ns subshell ⇒ 6 + 0 = 6
(n – 1) d subshell ⇒ 5 + 2 = 7
(n – 2) f subshell ⇒ 4 + 3 = 7
np subshell ⇒ 6 + 1 = 7
When n + ℓ values are same, the one have lowest n value filled first.
ns , (n − 2)f, (n − 1)d, np
(n + ℓ) values ⇒ 7 , 7 , 7
n value ⇒ 4 , 5 , 6
16. The cation that will not be precipitated by H2S in the presence of dil HCl is:
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution: (A) Co2+
precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt.)
Other are precipitated as sulphide in presence of dil HCl in group II.

32. 17. The geometry of XeOF4 by VSEPR theory is:
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution: (B) H =
1
2
(V + M − C + A)
=
1
2
(8 + 4) = 6
sp3
d2
Hybridization
4 B.P + 1 B.P (Double bonded) + 1 L.P,
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position.
18. The correct order of thermal stability of hydroxides is:
(A) Mg(OH)2 < Sr(OH)2 < Ca(OH)2 < Ba(OH)2
(B) Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2
(C) Ba(OH)2 < Sr(OH)2 < Ca(OH)2 < Mg(OH)2
(D) Ba(OH)2 < Ca(OH)2 < Sr(OH)2 < Mg(OH)2
Solution: (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases & covalent
character decreases & ionic character increases i.e. Mg(OH)2 < Ca(OH)2 < Sr(OH)2 <
Ba(OH)2
19. Photochemical smog consists of excessive amount of X, in addition to aldehydes, ketones,
peroxy acetyl nitrile (PAN), and so forth. X is:
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution: (C) Photochemical smog is the chemical reaction of sunlight, nitrogen oxides and VOCs in
the atmosphere.

33. NO2
hv
→ NO + O
O + O2 → O3
So, it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone.
20. A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is:
(atomic mass, Ba = 137 amu, Cl = 35.5 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution: (D) BaCl2 ∙ xH2O → BaCl2 + x H2O
(137 + 2 × 35.5 + 18x)
= (208 + 18x) g/mole
208 + 18 x
208
=
61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21. The following statements relate to the adsorption of gases on a solid surface. Identify the
incorrect statement among them:
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption, the residual forces on the surface are increased
Solution: (D) Adsorption is spontaneous process ∆G is –ve
During adsorption randomness of adsorbate molecules reduced ∆S is –ve
∆G = ∆H − T∆S
∆H = ∆G + T∆S
∆H is highly –ve and residual forces on surface are satisfied.
22. In the isolation of metals, calcination process usually results in:
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide

34. Solution: (A) Calcination used for decomposition of metal carbonates
M CO3
∆
→ MO + CO2 ↑
23. A variable, opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu , of potential 1.1 V. When Eext < 1.1 V and Eext > 1.1 V, respectively electrons
flow from:
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution: (B) For the Daniel cell
Ecell = 0.34 − (−0.76) = 1.10 V
When Eext < 1.10 V electron flow from anode to cathode in external circuit
When Eext > 1.10 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24. Complete hydrolysis of starch gives:
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts
(D) Glucose only
Solution: (D) On complete hydrolysis of starch, glucose is formed. Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars.
25. Match the polymers in column-A with their main uses in column-B and choose the correct
answer:
Column - A Column - B
A. Polystyrene i. Paints and lacquers
B. Glyptal ii. Rain coats
C. Polyvinyl chloride
chloride
iii. Manufacture of toys
D. Bakelite iv. Computer discs
(A) A – iii , B – i , C – ii , D – iv
(B) A – ii , B – i , C – iii , D – iv
(C) A – ii , B – iv , C – iii , D – i
(D) A – iii , B – iv , C – ii , D – i
Solution: (A) A – iii , B – i , C – ii , D – iv

35. 26. Permanent hardness in water cannot be cured by:
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgon’s methos
(D) Boiling
Solution: (D) Permanent hardness due to SO4
2−
, Cl−
of Ca2+
and Mg2+
cannot be removed by boiling.
27. In the long form of periodic table, the valence shell electronic configuration of 5s2
5p4
corresponds to the element present in:
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution: (A) 5s2
, 5p4
configuration is actually 36[Kr]5s2
, 4d10
, 5p4
i.e. 5th period and group 16 and
element Tellurium.
28. The heat of atomization of methane and ethane are 360 kJ/mol and 620 kJ/mol, respectively.
The longest wavelength of light capable of breaking the C − C bond is (Avogadro number =
6.023 × 1023
, h = 6.62 × 10−34
J s):
(A) 2.48 × 104
nm
(B) 1.49 × 104
nm
(C) 2.48 × 103
nm
(D) 1.49 × 103
nm
Solution: (D) 4 B.E (C − H) bond = 360 kJ
B.E (C − H) bond = 90 kJ/mole
In C2H6 ⇒ B. E(C−C) + 6B. E(C−H) = 620 kJ
B. E(C−C) bond = 620 − 6 × 90 = 80 kJ mole⁄
B. E(C−C) bond =
80
96.48
= 0.83 eV bond⁄
λ(Photon in Å) for rupture of
C − C bond =
12408
0.83
= 14950Å
= 1495 nm
≈ 1.49 × 103
nm
29. Which of the following is not an assumption of the kinetic theory of gases?

36. (A) Collisions of gas particles are perfectly elastic.
(B) A gas consists of many identical particles which are in continual motion.
(C) At high pressure, gas particles are difficult to compress.
(D) Gas particles have negligible volume.
Solution: (C) At high pressures gas particles difficult to compress rather they are not compressible at
all.
30. After understanding the assertion and reason, choose the correct option.
Assertion: In the bonding molecular orbital (MO) of H2 , electron density is increases between
the nuclei.
Reason: The bonding MO is ψA + ψB , which shows destructive interference of the combining
electron waves.
(A) Assertion and Reason are correct, but Reason is not the correct explanation for the Assertion.
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion.
(C) Assertion is incorrect, Reason is correct.
(D) Assertion is correct, Reason is incorrect.
Solution: (D) Electron density between nuclei increased during formation of BMO in H2.
BMO is ψA + ψB (Linear combination of Atomic orbitals) provides constructive interference.

37. JEE Mains 2015 10th April (online)
Mathematics
1. If the coefficient of the three successive terms in the binomial expansion of (1 + 𝑥) 𝑛
are in the
ratio 1 : 7 : 42, then the first of these terms in the expansion is :
1. 9 𝑡ℎ
2. 6 𝑡ℎ
3. 8 𝑡ℎ
4. 7 𝑡ℎ
Answer: (4)
Solution: Let 𝑛
𝐶𝑟 be the first term, then
𝑛 𝐶 𝑟
𝑛 𝐶 𝑟+1
=
1
7
⇒
𝑟 + 1
𝑛 − 𝑟
=
1
7
⇒ 7𝑟 + 7 = 𝑛 − 𝑟
𝑛 − 8𝑟 = 7 …..(i)
Also
𝑛 𝐶 𝑟+1
𝑛 𝐶 𝑟+2
=
7
42
=
1
6
⇒
𝑟 + 2
𝑛 − 𝑟 − 1
=
1
6
⇒ 6𝑟 + 12 = 𝑛 − 𝑟 − 1
𝑛 − 7𝑟 = 13 ……(ii)
Solving
𝑛 − 8𝑟 = 7 ….(i)
𝑛 − 7𝑟 = 13 …..(ii)
____________
−𝑟 = −6
𝑟 = 6
Hence 7 𝑡ℎ
term is the answer.
2. The least value of the product 𝑥𝑦𝑧 for which the determinant |
𝑥
1
1
1
𝑦
1
1
1
𝑧
| is non – negative, is:
1. −1
2. −16√2
3. −8
4. −2√2
Answer: (3)
Solution: |
𝑥
1
1
1
𝑦
1
1
1
𝑧
| = 𝑥𝑦𝑧 − (𝑥 + 𝑦 + 𝑧) + 2
Since 𝐴. 𝑀 ≥ 𝐺. 𝑀

38. 𝑥 + 𝑦 + 𝑧
3
≥ (𝑥𝑦𝑧)
1
3
𝑥 + 𝑦 + 𝑧 ≥ 3(𝑥𝑦𝑧)
1
3
∴ Least value of xyz will have from (when determinant non- negative terms)
𝑥𝑦𝑧 − (3)(𝑥𝑦𝑧)
1
3 + 2 ≥ 0
𝑡3
− 3𝑡 + 2 ≥ 0
(𝑡 + 2)(𝑡2
− 2𝑡 + 1)
𝑡 = −2 𝑎𝑛𝑑 𝑡 = +1
Least value of 𝑡3
= −8.
3. The contrapositive of the statement “If it is raining, then I will not come”, is:
1. If I will come, then it is not raining
2. If I will come, then it is raining
3. If I will not come, then it is raining
4. If I will not come, then it is not raining
Answer: (1)
Solution: Contrapositive of 𝑃 ⇒ 𝑞 is
~𝑞 ⇒ ~ 𝑃 So contra positive of the statement “If it is raining, then I will not come”, would be
If I will come, then it is not raining.
4. lim
𝑥→0
𝑒 𝑥2
−cos 𝑥
sin2 𝑥
is equal to:
1. 2
2.
3
2
3.
5
4
4. 3
Answer: (2)
Solution:
𝑒 𝑥2
−cos 𝑥
sin2 𝑥
=
(1 +
𝑥2
∟1 +
𝑥4
∟2 … … ) − (1 −
𝑥2
∟2 +
𝑥4
∟4 … … 𝑛)
sin2 𝑥
𝑥2 − 𝑥2
(
+3𝑥2
2
+
11 𝑥4
24
sin2 𝑥
𝑥2 ∙𝑥2
) take 𝑥2
common

39. [lim
𝑥→0
+
3
2 +
11
24 𝑥2
sin2 𝑥
𝑥2
] =
3
2
.
5. If Rolle’s theorem holds for the function 𝑓(𝑥) = 2𝑥3
+ 𝑏𝑥2
+ 𝑐𝑥, 𝑥 ∈ [−1, 1], at the point 𝑥 =
1
2
,
then 2b + c equals:
1. 2
2. 1
3. -1
4. -3
Answer: (3)
Solution: If Rolle’s theorem is satisfied in the interval [-1, 1], then
𝑓(−1) = 𝑓(1)
−2 + 𝑏 − 𝑐 = 2 + 𝑏 + 𝑐
𝑐 = −2 also 𝑓′(𝑥) = 6𝑥2
+ 2𝑏𝑥 + 𝑐
Also if 𝑓′
(
1
2
) = 0 them
6
1
4
+ 2𝑏
1
2
+ 𝑐 = 0
3
2
+ 𝑏 + 𝑐 = 0
∵ 𝑐 = −2,
𝑏 =
1
2
∴ 2𝑏 + 𝑐 = 2 (
1
2
) + (−2)
= 1 − 2
= −1.
6. If the points (1, 1, 𝜆) 𝑎𝑛𝑑 (−3, 0, 1) are equidistant from the plane, 3𝑥 + 4𝑦 − 12𝑧 + 13 = 0,
then 𝜆 satisfies the equation:
1. 3𝑥2
+ 10𝑥 + 7 = 0
2. 3𝑥2
+ 10𝑥 − 13 = 0
3. 3𝑥2
− 10𝑥 + 7 = 0
4. 3𝑥2
− 10𝑥 + 21 = 0
Answer: (3)
Solution: (1, 1, 𝜆) 𝑎𝑛𝑑 (−3, 0, 1) in equidistant from 3𝑥 + 4𝑦 − 12𝑧 + 13 = 0 then

40. |
3 + 4 − 12𝜆 + 13
√32 + 42 + 122
| = |
−9 + 0 − 12 + 13
√32 + 42 + 122
|
|20 − 12𝜆| = |−8|
|5 − 3𝜆 | = |−2|
25 − 30𝜆 + 9𝜆2
= 4
9𝜆2
− 30𝜆 + 21 = 0
3𝜆2
− 10𝜆 + 7 = 0
∴ Option 3𝑥2
− 10𝑥 + 7 = 0 Is correct
7. In a Δ𝐴𝐵𝐶,
𝑎
𝑏
= 2 + √3 𝑎𝑛𝑑 ∠𝐶 = 60 𝑜
. Then the ordered pair (∠𝐴, ∠𝐵) is equal to:
1. (105 𝑜
, 15 𝑜)
2. (15 𝑜
, 105 𝑜)
3. (45 𝑜
, 75 𝑜)
4. (75 𝑜
, 45 𝑜
)
Answer: (1)
Solution: Since
𝑎
𝑏
=
2+ √3
1
∠𝐴 > ∠𝐵.
Hence only option 1 & 4 could be correct checking for option (1)
𝑎
𝑏
=
sin105 𝑜
sin 15 𝑜
=
𝑠𝑖𝑛 (60 𝑜
+ 45 𝑜
)
sin(60 𝑜 − 45 𝑜)
=
√3 + 1
√3 − 1
𝑎
𝑏
=
2 + √3
1
Hence option (105 𝑜
, 15 𝑜) is correct.
8. A factory is operating in two shifts, day and night, with 70 and 30 workers respectively. If per
day mean wage of the day shift workers is Rs. 54 and per day mean wage of all the workers is
Rs. 60, then per day mean wage of the night shift workers (in Rs.) is :
1. 75
2. 74
3. 69
4. 66
Answer: (2)
Solution:
𝑛1 𝑥1 +𝑛2 𝑥2
𝑛1+𝑛2
= 𝑥
70 ∙ (54) + 30 (𝑥2)
70 + 30
= 60

41. = 3780 + 30 𝑥2 = 6000
∴ 𝑥2 =
6000 − 3780
30
=
2220
30
= 74.
9. The integral ∫
𝑑𝑥
(𝑥+1)
3
4 (𝑥−2)
5
4
is equal to:
1. 4 (
𝑥−2
𝑥+1
)
1
4
+ 𝐶
2. −
4
3
(
𝑥+1
𝑥−2
)
1
4
+ 𝐶
3. 4 (
𝑥+1
𝑥−2
)
1
4
+ 𝐶
4. −
4
3
(
𝑥−2
𝑥+1
)
1
4
+ 𝐶
Answer: (2)
Solution: ∫
𝑑𝑥
(𝑥+1)
3
4 (𝑥−2)
5
4
Divide & Multiply the denominator by (𝑥 + 1)
5
4.
∫
𝑑𝑥
(𝑥 + 1)2 (
𝑥 − 2
𝑥 + 1)
5
4
Put
𝑥−2
𝑥+1
= 𝑡
(
1 (𝑥 + 1) − (𝑥 − 2)(1)
(𝑥 + 1)2 ) 𝑑𝑥 = 𝑑𝑡
3
(𝑥 + 1)2
𝑑𝑥 = 𝑑𝑡
1
𝑑𝑥
(𝑥 + 1)2
= 1
𝑑𝑡
3
⇒ 1/3 ∫ 𝑡
5
4 𝑑𝑡 =
1 𝑡
1
4
3 (
−1
4
)
=
−4
3
1
𝑡
1
4
+ 𝐶
−4
3
(
𝑥+1
𝑥−2
)
1
4
+ 𝐶.
10. Let 𝑎 𝑎𝑛𝑑 𝑏⃗ be two unit vectors such that |𝑎 + 𝑏⃗ | = √3 .

42. If 𝑐 = 𝑎 + 2𝑏⃗ (𝑎 × 𝑏⃗ ), then 2|𝑐| is equal to:
1. √51
2. √37
3. √43
4. √55
Answer: (4)
Solution: As |𝑎 × 𝑏⃗ | = √3
Squaring both the sides
|𝑎|2
+ |𝑏⃗ |
2
+ 2𝑎 ∙ 𝑏⃗ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 𝜃 = 3
2𝑐𝑜𝑠𝜃 = 1
𝑐𝑜𝑠𝜃 =
1
2
𝜃 = 60
∴ Angle between 𝑎 𝑎𝑛𝑑 𝑏⃗ 𝑖𝑠 60 𝑜
Now,
|𝑐| = |𝑎 + 2𝑏 + 3(𝑎 × 𝑏)|
Squaring both the sides
|𝑐|2
= ||𝑎|2
+ 4|𝑏⃗ |
2
+ 9 (𝑎 × 𝑏)2
+ 4 𝑎 ∙ (𝑏) + 3𝑎 ∙ (𝑎 × 𝑏) + 6𝑏 ∙ (𝑎 × 𝑏)|
|𝑐|2
= |1 + 4 + 9 sin2
𝜃 + 4 𝑐𝑜𝑠𝜃 + 0 + 0 |
|𝑐|2
= |5 + 9.
3
4
+ 4.
1
2
| =
55
4
∴ 2|𝑐| = √55.
11. The area (in square units) of the region bounded by the curves 𝑦 + 2𝑥2
= 0 𝑎𝑛𝑑 𝑦 + 3𝑥2
= 1,
is equal to:
1.
3
4
2.
1
3
3.
3
5
4.
4
3
Answer: (4)
Solution:

43. Point of intersection
Put 𝑦 = −2𝑥2
𝑖𝑛 𝑦 + 3𝑥2
= 1
𝑥2
= 1
𝑥 = ± 1
The desired area would be
∫ (𝑦1 − 𝑦2) 𝑑𝑥 = ∫ ((1 − 3𝑥2) − (−2𝑥2)) 𝑑𝑥
1
−1
1
−1
∫ (1 − 𝑥2)𝑑𝑥
1
−1
(𝑥 −
𝑥3
3
)
−1
1
= ((1 −
1
3
) − (−1 +
1
3
))
2
3
− (
−2
3
)
=
4
3
.
12. If 𝑦 + 3𝑥 = 0 is the equation of a chord of the circle, 𝑥2
+ 𝑦2
− 30𝑥 = 0, then the equation of
the circle with this chord as diameter is :
1. 𝑥2
+ 𝑦2
+ 3𝑥 − 9𝑦 = 0
2. 𝑥2
+ 𝑦2
− 3𝑥 + 9𝑦 = 0
3. 𝑥2
+ 𝑦2
+ 3𝑥 + 9𝑦 = 0
4. 𝑥2
+ 𝑦2
− 3𝑥 − 9𝑦 = 0
Answer: (2)
Solution:

44. 𝑦 = −3𝑥
4𝑥2
+ 𝑦2
− 30𝑥 = 0
Point of intersection
𝑥2
+ 9𝑥2
− 30𝑥 = 0
10𝑥2
− 30𝑥 = 0
10𝑥 (𝑥 − 3) = 0
𝑥 = 0 or 𝑥 = 3
Therefore y = 0 if x = 0, and y =-9 if x = 3.
Point of intersection (0, 0) (3, -9)
Diametric form of circle,
𝑥 (𝑥 − 3) + 𝑦(𝑦 + 9) = 0
𝑥2
+ 𝑦2
− 3𝑥 + 9𝑦 = 0.
13. The value of ∑ (𝑟 + 2) (𝑟 − 3)30
𝑟=16 is equal to:
1. 7775
2. 7785
3. 7780
4. 7770
Answer: (3)
Solution: ∑ (𝑟 + 2) (𝑟 − 3)30
𝑟=16
= ∑ (𝑟2
− 𝑟 − 6) − ∑ (𝑟2
− 𝑟 − 6)15
1
30
1
Put r = 30
in (
𝑟(𝑟+1) (2𝑟+1)
6
−
𝑟(𝑟+1)
2
− 6𝑟)
30 ∙ (31)(61)
6
− 15(31) − 6(30)
9455 − 465 − 180
8810
And on putting 𝑟 = 15
We get
15∙(16) (31)
6
−
15∙16
2
− 6 ∙ (15)
= (7) ∙ (8) ∙ (31) −
15 ∙16
2
− 6 ∙ (15)

45. = 1240 − 120 − 90
= 1030
Therefore ∑ (𝑟2
− 𝑟 − 6) − ∑ (𝑟2
− 𝑟 − 6)15
1
30
1 = 8810 − 1030
= 7780.
14. Let L be the line passing through the point P(1, 2) such that its intercepted segment between
the co-ordinate axes is bisected at P. If 𝐿1 is the line perpendicular to L and passing through the
point (-2, 1), then the point of intersection of L and 𝐿1 is:
1. (
3
5
,
23
10
)
2. (
4
5
,
12
5
)
3. (
11
20
,
29
10
)
4. (
3
10
,
17
5
)
Answer: (2)
Solution:
If P is the midpoint of the segment between the axes, them point A would be (2, 0) and B would be (0,
4). The equation of the line would be
𝑥
2
+
𝑦
4
= 1
That is 2𝑥 + 𝑦 = 4 …..(i)
The line perpendicular to it would be 𝑥 − 2𝑦 = 𝑘
Since it passes through (-2, 1) −2 − 2 = 𝑘
−4 = 𝑘
∴ Line will become 𝑥 − 2𝑦 = −4 …..(ii)
Solving (i) and (ii) we get (
4
5
,
12
5
).
15. The largest value of r for which the region represented by the set {
𝜔 ∈𝐶
|𝜔−4−𝑖| ≤ 𝑟
} is contained in
the region represented by the set {
𝑧 ∈𝐶
|𝑧−1| ≤ |𝑧+𝑖|
}, is equal to :

46. 1. 2√2
2.
3
2
√2
3. √17
4.
5
2
√2
Answer: (4)
Solution:
|𝑧 − 1| ≤ |𝑧 + 𝑖|
The region in show shaded right side of the line 𝑥 + 𝑦 = 0
The largest value of r would be the length of perpendicular from A (4, 1) on the line 𝑥 + 𝑦 = 0
|
4 + 1
√2
| =
5
√2
=
5
2
√2 .
16. Let the sum of the first three terms of an A.P. be 39 and the sum of its last four terms be 178. If
the first term of this A.P. is 10, then the median of the A.P. is :
1. 26.5
2. 29.5
3. 28
4. 31
Answer: (2)
Solution: Let the A.P. be a; a + d a + 2d …………………ℓ − 3𝑑, ℓ − 2𝑑, ℓ − 𝑑, ℓ
Where a is the first term and ℓ is the last term
Sum of 1 𝑠𝑡
3 terms is 39.
3𝑎 + 3𝑑 = 39
30 + 3𝑑 = 30 as 𝑎 = 10 (Given)
𝑑 =
9
3
= 3

47. Sum of last 4 terms is 178.
4ℓ − 6𝑑 = 178
4ℓ − 18 = 178
4ℓ = 196
ℓ = 49
10, 13, 16, 19…….46, 49
Total number of the 10 + (n – 1) 3 - 49
n – 1 = 13
n = 14
So the median of the series would be mean of 7 𝑡ℎ
𝑎𝑛𝑑 8 𝑡ℎ
term
10+6∙(3)+10+7∙3
2
28 + 31
2
=
59
2
= 29.5
Alternate way
The median would be mean of 10 and 49, That is 29.5.
17. For 𝑥 > 0, let 𝑓(𝑥) = ∫
log 𝑡
1+𝑡
𝑑𝑡.
𝑥
1
Then 𝑓(𝑥) + 𝑓 (
1
𝑥
) is equal to :
1.
1
2
(log 𝑥)2
2. log 𝑥
3.
1
4
log 𝑥2
4.
1
4
(log 𝑥)2
Answer: (1)
Solution:
𝑓(𝑥) = ∫
log 𝑡
1 + 𝑡
𝑥
1
∙ 𝑑𝑡
And 𝑓 (
1
𝑥
) = ∫
log 𝑡
1+𝑡
∙ 𝑑𝑡
1
𝑥
1
Put 𝑡 =
1
𝑧
𝑑𝑡 = −
1
𝑧2
𝑑𝑡
−
1
𝑥2
𝑑𝑥 = 𝑑𝑡
𝑓(𝑥) = ∫
log 𝑧
𝑧2 (1 +
1
𝑧)
𝑧
1
∙ 𝑑𝑧

48. 𝑓(𝑥) = ∫
log 𝑧
𝑧(1 + 𝑧)
𝑑𝑧
𝑧
1
𝑓(𝑥) + 𝑓 (
1
𝑥
) = ∫ log 𝑧 [
1
1 + 𝑧
+
1
2(1 + 𝑧)
] 𝑑𝑧
𝑥
1
= ∫
1
𝑧
log 𝑧 𝑑𝑧
𝑥
1
Put log 𝑧 = 𝑃
1
𝑧
𝑑𝑧 = 𝑑𝑝
∫ 𝑃 ∙ 𝑑𝑝
𝑥
1
(
𝑃2
2
)
1
𝑥
=
1
2
(log 𝑧)1
𝑥
=
(log 𝑥)2
2
18. In a certain town, 25% of the families own a phone and 15% own a car; 65% families own
neither a phone nor a car and 2,000 families own both a car and a phone. Consider the
following three statements:
(a) 5% families own both a car and a phone.
(b) 35% families own either a car or a phone.
(c) 40, 000 families live in the town.
Then,
1. Only (b) and (c) are correct
2. Only (a) and (b) are correct
3. All (a), (b) and (c) are correct
4. Only (a) and (c) are correct
Answer: (3)
Solution: Let set A contains families which own a phone and set B contain families which own a car.
If 65% families own neither a phone nor a car, then 35% will own either a phone or a car
∴ (𝐴⋃𝐵) = 35%
Also we know that
𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵)
35 = 25 + 15 - 𝑛(𝐴 ∩ 𝐵)
𝑛(𝐴 ∩ 𝐵) = 5%
5% families own both phone and car and it is given to be 2000.
∴ 5% 𝑜𝑓 𝑥 = 2000
5
100
𝑥 = 2000

49. X = 40,000
Hence correct option is (a) (b) and (c) are correct.
19. IF 𝐴 = [
0
1
−1
0
], then which one of the following statements is not correct?
1. 𝐴3
+ 𝐼 = 𝐴(𝐴3
− 𝐼)
2. 𝐴4
− 𝐼 = 𝐴2
+ 𝐼
3. 𝐴2
+ 𝐼 = 𝐴(𝐴2
− 𝐼)
4. 𝐴3
− 𝐼 = 𝐴(𝐴 − 𝐼)
Answer: (3)
Solution: A = [
0 −1
1 0
]
𝐴2
= [
0 −1
1 0
] [
0 −1
1 0
] = [
−1 0
0 −1
]
𝐴3
= [
−1 0
0 −1
] [
0 −1
1 0
] = [
0 1
−1 0
]
𝐴4
= [
0 1
−1 0
] [
0 −1
1 0
] [
1 0
0 1
]
Option (1) 𝐴3
+ 𝐼 = 𝐴 (𝐴3
− 𝐼)
[
0
1
−1
0
] [
−1
−1
1
−1
] = [
1
−1
1
1
]
[
1
−1
1
1
] = [
1
−1
1
1
] …..Correct
Option (2) 𝐴4
− 𝐼 = 𝐴2
+ 𝐼
[
0 0
0 0
] = [
0 0
0 0
] ….Correct
Option (3) [
0 0
0 0
] = [
0 −1
1 0
] [
−2 0
0 −2
] = [
0 2
−2 0
] …..Incorrect
Option 4
𝐴3
− 𝐼 = 𝐴(𝐴 − 𝐼)
[
−1 −1
−1 −1
] = [
0 −1
1 0
] [
−1 −1
1 −1
] [
−1 1
−1 1
]
𝐴3
− 𝐼 = 𝐴4
− 𝐴
[
1 1
−1 1
] = [
1 0
0 1
] − [
0 −1
1 0
]
= [
1 1
−1 1
] ……Correct.
20. Let X be a set containing 10 elements and P(X) be its power set. If A and B are picked up at
random from P(X), with replacement, then the probability that A and B have equal number of
elements, is:
1.
(210−1)
220

50. 2.
20 𝐶10
220
3.
20 𝐶10
210
4.
(210−1)
210
Answer: (2)
Solution: The power set of x will contain 210
sets of which
10
𝐶0 will contain 0 element
10
𝐶1 will contain 1 element
10
𝐶2 will contain 2 element
⋮
⋮
10
𝐶10 will contain 10 element.
So total numbers of ways in which we can select two sets with replacement is 210
× 210
= 220
And favorable cases would be 10
𝐶0 ∙ 10
𝐶0 + 10
𝐶1
10
𝐶1 + … … 10
𝐶10
10
𝐶10 = 20
𝐶10.
Hence Probability would be =
20 𝐶10
220
Hence
20 𝐶10
220 in the correct option
21. If 2 + 3𝑖 is one of the roots of the equation 2𝑥3
− 9𝑥2
+ 𝑘𝑥 − 13 = 0, 𝑘 ∈ 𝑅, then the real
root of this equation:
1. Exists and is equal to
1
2
2. Does not exist
3. Exists and is equal to 1
4. Exists and is equal to −
1
2
Answer: (1)
Solution: If 2 + 3𝑖 in one of the roots, then 2 − 3𝑖 would be other
Since coefficients of the equation are real.
Let 𝛾 be the third root, then product of roots → 𝛼 𝛽 𝛾 =
13
2
(2 + 3𝑖) (2 − 3𝑖) ∙ 𝛾 =
13
2
(4 + 9) ∙ 𝛾 =
13
2
𝛾 =
1
2
.
The value of k will come if we
Put 𝑥 =
1
2
in the equation
2 ∙
1
8
−
9
4
+ 𝑘 ∙
1
2
− 13 = 0

51. 𝑘
2
= 15
𝑘 = 30.
∴ Equation will become
2𝑥3
− 9𝑥2
+ 30𝑥 − 13 = 0
𝛼𝛽 + 𝛽𝛾 + 𝛾𝛼 =
30
2
= 15
(2 + 3𝑖)
1
2
+ (2 − 3𝑖)
1
2
+ (2 + 3𝑖) (2 − 3𝑖) = 15
1 +
𝑖
2
+ 1 −
𝑖
2
+ 13 = 15
15 = 15
Hence option (1) is correct. ‘Exists and is equal to
1
2
‘
22. If the tangent to the conic, 𝑦 − 6 = 𝑥2
at (2, 10) touches the circle, 𝑥2
+ 𝑦2
+ 8𝑥 − 2𝑦 = 𝑘 (for
some fixed k) at a point (𝛼, 𝛽); then (𝛼, 𝛽) is :
1. (−
7
17
,
6
17
)
2. (−
8
17
,
2
17
)
3. (−
6
17
,
10
17
)
4. (−
4
17
,
1
17
)
Answer: (2)
Solution: The equation of tangent (T = 0) would be
1
2
(𝑦 + 10) − 6 = 2𝑥
4𝑥 − 𝑦 + 2 = 0
The centre of the circle is (−4, 1) and the point of touch would be the foot of perpendicular from
(−4, 1) on 4𝑥 − 𝑦 + 2 = 0
𝑥 + 4
4
=
𝑦 − 1
−1
= − (
−16 − 1 + 2
42 + 12
)
𝑥+4
4
=
15
17
and
𝑦−1
−1
=
15
17
𝑥 = −
8
17
𝑦 =
−15
17
+ 1 =
2
17

52. Hence option (−
8
17
,
2
17
) is correct.
23. The number of ways of selecting 15 teams from 15 men and 15 women, such that each team
consists of a man and a woman, is:
1. 1960
2. 1240
3. 1880
4. 1120
Answer: (2)
Solution: No. of ways of selecting 1 𝑠𝑡
team from 15 men and 15 women
15
𝐶1
15
𝐶1 = 152
2 𝑛𝑑
team- 14
𝐶1
14
𝐶1 142
and so on.
So total number of way
12
+ 22
… … … 152
=
15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct.
24. If the shortest distance between the line
𝑥−1
𝛼
=
𝑦+1
−1
=
𝑧
1
, (𝛼 ≠ −1) and 𝑥 + 𝑦 + 𝑧 + 1 = 0 =
2𝑥 − 𝑦 + 𝑧 + 3 𝑖𝑠
1
√3
, then a value of 𝛼 is :
1. −
19
16
2.
32
19
3. −
16
19
4.
19
32
Answer: (2)
Solution: Let us change the line into symmetric form.
𝑥 + 𝑦 + 𝑧 + 1 = 0 = 2𝑥 − 𝑦 + 𝑧 + 3
Put 𝑧 = 1, so we get 𝑥 + 𝑦 + 2 = 0 and 2𝑥 − 𝑦 + 4 = 0
We will get 𝑥 = −2
𝑦 = 0
∴ The point (−2, 0, 1) lies on the line and perpendicular vector will come from

53. |
𝑖 𝑗 𝑘
1 1 1
2 −1 1
| = 2𝑖 + 𝑗 − 3𝑘
So the equation line would be
𝑥 + 2
2
=
𝑦
1
=
𝑧 − 1
−3
And the other line
𝑥 − 1
𝛼
=
𝑦 + 1
−1
=
𝑧
1
Shortest distance would be
𝐷 =
[(𝑎2 − 𝑎1), 𝑏1 𝑏2]
|𝑏1 × 𝑏2|
When 𝑎1 = (−2𝑖 + 𝑜𝑗 + 1𝑘)
𝑎2 = (𝑖 − 𝑗 + 0𝑘)
𝑏1 = 2𝑖 + 𝑗 − 3𝑘
𝑏2 = 𝛼𝑖 − 𝑗 + 𝑘
|
3 −1 −1
2 1 −3
𝛼 1 −3
|
|
𝑖 𝑗 𝑘
2 1 −3
𝛼 −1 1
|
=
3(1 − 3) + 1 (2 + 3𝛼) + 1 (2 + 𝛼)
|−2𝑖 − 𝑗 (2 − 3𝛼) + 𝑘 (−2 − 𝛼)|
|
−6 + 2 + 3𝛼 + 2 + 𝛼
√4 + (2 + 3𝛼)2 + (2 + 𝛼)2
| =
1
√3
|4𝛼 − 2|
√4 + 4 + 12𝛼 + 9𝛼2 + 4 + 4𝛼 + 𝛼2
=
1
√3
|
4𝛼 − 2
√10𝛼2 + 16𝛼 + 12
| =
1
√3
(16𝛼2
− 16𝛼 + 4)3 = 10𝛼2
+ 16𝛼 + 12
48𝛼2
− 48𝛼 + 12 =
10𝛼2
+ 16𝛼 + 12
38𝛼2
− 64𝛼 = 0
𝛼(19𝛼 − 32) = 0
𝛼 =
32
19

54. 25. The distance from the origin, of the normal to the curve, 𝑥 = 2 cos 𝑡 + 2𝑡 sin 𝑡, 𝑦 =
2 sin 𝑡 − 2𝑡 cos 𝑡 𝑎𝑡 𝑡 =
𝜋
4
, is :
1. √2
2. 2√2
3. 4
4. 2
Answer: (4)
Solution: at 𝑡 =
𝜋
4
𝑥 = 2
1
√2
+ 2
𝜋
4
= (√2 +
𝜋
2√2
) = (
8 + 𝜋
2√2
)
𝑦 = 2
1
√2
− 2
𝜋
4
∙
1
√2
= (√2 −
𝜋
2√2
) − (
8 − 𝜋
2√2
)
𝑑𝑦
𝑑𝑥
= 2 cos 𝑡 − 2 [cos 𝑡 + 𝑡 (− sin 𝑡)] = 2𝑡 sin 𝑡
𝑑𝑥
𝑑𝑡
= −2 sin 𝑡 + 2 [sin 𝑡 + 𝑡 ∙ cos 𝑡] = 2𝑡 cos 𝑡
𝑑𝑦
𝑑𝑥
= tan 𝑡 𝑎𝑛𝑑 𝑡 =
𝜋
4
𝑎𝑛𝑑 tan
𝜋
4
= 1
𝑑𝑦
𝑑𝑥
= 1 Slope of tangent is 1 & therefore slope of normal would be -1.
Equation of normal 𝑦 − (
8−𝜋
2√ 2
) = −1 (𝑥 − (
8+𝜋
2√2
))
𝑥 + 𝑦 = 𝑡
(8 + 𝜋)
2√2
+ (
8 − 𝜋
2√2
)
𝑥 + 𝑦 =
16
2√2
and distance from origin
16
2√2
√2 = 4
26. An ellipse passes through the foci of the hyperbola, 9𝑥2
− 4𝑦2
= 36 and its major and minor
axes lie along the transverse and conjugate axes of the hyperbola respectively. If the product of
eccentricities of the two conics is
1
2
, then which of the following points does not lie on the
ellipse?
1. (
√39
2
, √3)
2. (
1
2
√13,
√3
2
)
3. (√
13
2
, √6)
4. (√13, 0)

55. Answer: (2)
Solution: Equation of the hyperbola
𝑥2
4
−
𝑦2
9
= 1
Focus of hyperbola (ae, 0) and (-ae, 0)
a = 2 𝑒 = √1 +
9
4
=
√13
2
∴ Focus would be (+
√13
2
, 0) 𝑎𝑛𝑑 (−
√13
2
, 0)
Product of eccentricity would be
√13
2
∙ 𝑒1 =
1
2
∴ 𝑒1 =
1
√13
.
As the major & minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be √13,
𝑒 = √1 −
𝑏2
𝑎2
𝑏2
13
=
12
13
1
√3
= √1 −
𝑏2
13
𝑏2
= 12
1
13
= 1 −
𝑏2
13
∴ Equation of the ellipse would be
𝑥2
13
+
𝑦2
12
= 1.
Option (i)
39
4 ∙(13)
+
3
12
= 1
Satisfies the equation hence it lies on the ellipse.
Option (ii)
13
4 (13)
+
3
4.12
= 1
does not lie on the ellipse.
Option (iii)
13
2(13)
+
6
12
= 1 satisfy
Option (iv)
13
13
+ 0 = 1 satisfy
So option (
1
2
√13,
√3
2
) is the answer.

56. 27. The points (0,
8
3
) , (1, 3) 𝑎𝑛𝑑 (82, 30) :
1. Form an obtuse angled triangle
2. Form an acute angled triangle
3. Lie on a straight line
4. Form a right angled triangle
Answer: (3)
Solution: The options
A B C
(0
8
2
) (1, 3) (82, 30)
Are collinear as slope f AB is equal to slope of BC
3 −
8
3
1 − 0
=
30 − 3
82 − 1
1
3
=
27
81
=
1
3
Hence option (Lie on a straight line) is correct.
28. If 𝑓(𝑥) − 2 tan−1
𝑥 + sin−1
(
2𝑥
1+𝑥2) , 𝑥 > 1, then 𝑓(5) is equal to :
1.
𝜋
2
2. tan−1
(
65
156
)
3. 𝜋
4. 4 tan−1 (5)
Answer: (3)
Solution:
2 tan−1
𝑥 + sin−1
(
2𝑥
1 + 𝑥2
) , 𝑓𝑜𝑟 𝑥 > 1.
= 2 tan−1
𝑥 + 𝜋 − 2 tan−1
𝑥 𝑎𝑠 𝑥 > 1
∴ 𝑓(5) = 𝜋
∴ Answer is 𝜋
Or 𝑓(5) = 2 tan−1 (5) + sin−1
(
10
26
)

57. = 𝜋 − tan−1
(
10
24
) + tan−1
(
10
24
)
𝜋 sin−1
(
10
26
)
29. Let the tangents drawn to the circle, 𝑥2
+ 𝑦2
= 16 from the point P(0, h) meet the 𝑥 − 𝑎𝑥𝑖𝑠 at
points A and B. If the area of Δ𝐴𝑃𝐵 is minimum, then h is equal to :
1. 4√2
2. 3√2
3. 4√3
4. 3√3
Answer: (1)
Solution:
Let the equation of the tangent be (𝑦 − ℎ) = 𝑚 (𝑥 − 0)
𝑚𝑥 − 𝑦 + ℎ = 0
|
ℓ𝑛
√𝑚2 + 1
| = 4
ℎ2
= 16𝑚2
+ 16
𝑚2
=
ℎ2
− 16
16
𝑚 =
√ℎ2 − 16
4
So co-ordinate of B would be
√
ℎ2 − 16
4
𝑥 − 𝑦 + ℎ = 0
𝑥 =
4ℎ
√ℎ2 − 16
Also of triangle
=
1
2
𝐵𝑎𝑠𝑒 𝑥 𝐻𝑒𝑖𝑔ℎ𝑡

58. Δ =
1
2
8ℎ
√ℎ2 − 16
∙ ℎ
Δ = 4
ℎ2
√ℎ2 − 16
𝑑Δ
𝑑ℎ
= 4
[
2ℎ√ℎ2 − 16 −
2ℎ ∙ ℎ2
2√ℎ2 − 16
(ℎ2 − 16)
]
= 4ℎ [
4(ℎ2
− 16) − 2ℎ2
2√ℎ2 − 16 (ℎ2 − 16)
]
=
4ℎ[2ℎ2
− 64]
2√ℎ2 − 16 (ℎ2 − 16)
For are to be minima ℎ = √32
ℎ2
= 32
ℎ = 4√2
30. If 𝑦 (𝑥) is the solution of the differential equation (𝑥 + 2)
𝑑𝑦
𝑑𝑥
= 𝑥2
+ 4𝑥 − 9, 𝑥 ≠ −2 and
𝑦(0) = 0, then 𝑦(−4) is equal to :
1. -1
2. 1
3. 0
4. 2
Answer: (3)
Solution:
(𝑥 + 2) ∙
𝑑𝑦
𝑑𝑥
= 𝑥2
+ 4𝑥 + 4 − 13
𝑑𝑦
𝑑𝑥
=
(𝑥 + 2)2
(𝑥 + 2)
−
13
(𝑥 + 2)
𝑑𝑦 = ((𝑥 + 2) −
13
𝑥𝑚
)
𝑑𝑥
𝑦 =
𝑥2
2
+ 2𝑥 − 13 log 𝑒|(𝑥 + 2)| + 𝐶
If 𝑥 = 0 then 𝑦 = 0

59. 0 = 0 + 0 − 13 𝑙𝑜𝑔|2| + 𝐶
𝑐 ∶ 13 log(2)
If 𝑥 = −4, then 𝑦
𝑦 =
16
2
− 8 − 13 log|−2| + 13 log |2|
𝑦 = 0
Hence as is option 0