Code-DD

Test Booklet Code

DATE : 06/05/2018

DD ACHLA Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.: 011-47623456 Fax : 011-47623472

Time : 3 hrs.

Answers & Solutions

Max. Marks : 720

for NEET (UG) - 2018 Important Instructions : 1.

The test is of 3 hours duration and Test Booklet contains 180 questions. Each question carries 4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response, one mark will be deducted from the total scores. The maximum marks are 720.

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1

NEET (UG) - 2018 (Code-DD) ACHLA

1.

Answer ( 1 )

A carbon resistor of (47 ± 4.7) k is to be marked with rings of different colours for its identification. The colour code sequence will be

Sol. I 

E nR  R

E R R n Dividing (ii) by (i), 10 I 

(1) Violet – Yellow – Orange – Silver (2) Yellow – Violet – Orange – Silver (3) Yellow – Green – Violet – Gold (4) Green – Orange – Violet – Gold

10 

Answer ( 2 ) S o l . (47 ± 4.7) k = 47 × 103 ± 10% 4.

A battery consists of a variable number 'n' of identical cells (having internal resistance 'r' each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n?

I

I

Sol. h 

n

I

(3)

(4)

O

n

O

n   nr r

5.

So, I is independent of n and I is constant.  I

2hm eE



t



t  m as ‘e’ is same for electron and proton.

A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27ºC two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27ºC is (1) 330 m/s

O

3.

1 eE 2 t 2 m

∵ Electron has smaller mass so it will take smaller time.

n

Answer ( 1 ) Sol. I 

(2) 5 times greater

Answer ( 1 )

O

n

An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is (3) 10 times greater (4) Equal

(2)

O

(n  1)R ⎛1 ⎞ ⎜ n  1⎟ R ⎝ ⎠

(1) Smaller

I

(1)

...(ii)

After solving the equation, n = 10

 Yellow – Violet – Orange – Silver 2.

...(i)

n A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of 'n' is

(2) 339 m/s (3) 350 m/s (4) 300 m/s Answer ( 2 ) S o l . v = 2 () [L2 – L1] = 2 × 320 [73 – 20] × 10–2

(1) 10

(2) 11

= 339.2 ms–1

(3) 20

(4) 9

= 339 m/s 2


6.

S o l . Initial de-Broglie wavelength

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s2 at a distance of 5 m from the mean position. The time period of oscillation is (1) 2 s

(2)  s

(3) 2 s

(4) 1 s

0 

E0 V0 F Acceleration of electron

Answer ( 2 ) S o l . |a| = 2y

a

 20 = 2(5)

eE0 ⎛ V  ⎜ V0  m ⎝

2 2  s  2 The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is T

So,  

(1) Independent of the distance between the plates



(2) Linearly proportional to the distance between the plates (3) Proportional to the square root of the distance between the plates



(4) Inversely proportional to the distance between the plates

h  mV

⎞ t⎟ ⎠

h eE ⎛ m ⎜ V0  0 m ⎝

h ⎡ eE0 mV0 ⎢1  ⎣ mV0 0 ⎡ eE0 ⎢1  mV ⎣ 0

⎞ t⎟ ⎠

⎤ t⎥ ⎦

⎤ t⎥ ⎦

F is Independent of the distance between plates.

When the light of frequency 20 (where 0 is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v 1 . When the frequency of the incident radiation is increased to 5 0, the maximum velocity of electrons emitted from the same plate is v2. The ratio of v1 to v2 is

An electron of mass m with an initial velocity

(1) 1 : 2



(2) 1 : 4

Answer ( 1 )

9.

S o l . For isolated capacitor Q = Constant

Fplate 

8.

eE0 m

Velocity after time ‘t’

  = 2 rad/s

7.

h mV0

Q2 2A0

V  V0 î (V 0 > 0) enters an electric field

(3) 4 : 1



(4) 2 : 1

E  –E0 î (E0 = constant > 0) at t = 0. If 0 is its

Answer ( 1 )

de-Broglie wavelength initially, then its de-Broglie wavelength at time t is (1)

S o l . E  W0 

0 ⎛ eE0 ⎜1 mV ⎝ 0

⎞ t⎟ ⎠

1 mv2 2

h(20 )  h0  h 0 

⎛ eE0 ⎞ t⎟ (2)  0 ⎜ 1  mV0 ⎠ ⎝

1 mv12 2

h(50 )  h0 

(3) 0t (4) 0

4h0 

Answer ( 1 ) 3

1 mv12 2

1 mv22 2

…(i) 1 mv22 2

…(ii)


Answer ( 4 )

Divide (i) by (ii), 1 v12  4 v22

S o l . VBE = 0 VCE = 0

v1 1  v2 2 10.

Vb = 0

20 V

For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is (1) 20

(2) 10

(3) 30

(4) 15

IC Vi

RB Ib

500 k

RC = 4 k

Vb

Answer ( 1 ) S o l . Number of nuclei remaining = 600 – 450 = 150 IC 

n

N ⎛ 1⎞  N0 ⎜⎝ 2 ⎟⎠

IC = 5 × 10–3 = 5 mA

t

Vi = VBE + IBRB

150 ⎛ 1 ⎞ t 1/2  600 ⎜⎝ 2 ⎟⎠

Vi = 0 + IBRB

t

2

20 = IB × 500 × 103

⎛ 1⎞ ⎛ 1 ⎞ t 1/2 ⎜2⎟  ⎜2⎟ ⎝ ⎠ ⎝ ⎠ t = 2t1/2 = 2 × 10

IB 

= 20 minute 11.

(20  0) 4  103



The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is (1) 1 : 1

(2) 1 : –1

(3) 2 : –1

(4) 1 : –2

13.

20  40 A 500  103

IC 25  103   125 Ib 40  106

In the combination of the following gates the output Y can be written in terms of inputs A and B as

A

Answer ( 2 )

B

S o l . KE = –(total energy)

Y

So, Kinetic energy : total energy = 1 : –1 12.

In the circuit shown in the figure, the input voltage Vi is 20 V, VBE = 0 and VCE = 0. The values of IB, IC and  are given by

(1) A  B (2) A  B  A  B

20 V

Vi

RB 500 k B

(3) A  B  A  B

RC 4 k C

(4) A  B Answer ( 2 )

E

Sol. A

B (1) IB = 40 A, IC = 10 mA,  = 250 (2) IB = 25 A, IC = 5 mA,  = 200

A B A B

(3) IB = 20 A, IC = 5 mA,  = 250 (4) IB = 40 A, IC = 5 mA,  = 125

Y  (A  B  A  B) 4

AB Y AB


14.

In a p-n junction diode, change in temperature due to heating

16.

(1) Affects only reverse resistance (2) Affects only forward resistance (3) Does not affect resistance of p-n junction (4) Affects the overall V - I characteristics of p-n junction

Three objects, A : (a solid sphere), B : (a thin circular disk) and C : (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed  about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation (1) WC > WB > WA

Answer ( 4 )

(2) WA > WB > WC

S o l . Due to heating, number of electron-hole pairs will increase, so overall resistance of diode will change.

(3) WB > WA > WC (4) WA > WC > WB Answer ( 1 )

Due to which forward biasing and reversed biasing both are changed. 15.

S o l . Work done required to bring them rest W = KE

A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to

W 

1 2 I 2

W  I for same  h

B

WA : WB : WC 

A

=

3 (1) D 2 (3)

(2) D

(4)

 WC > WB > WA

5 D 4

17.

Answer ( 4 ) Sol.

h

B A

vL

(1) 0.5

(2) 0.25

(3) 0.8

(4) 0.4

S o l . According to law of conservation of linear momentum,

T.M.EI =T.M.EF

mv  4m  0  4mv  0

1 mvL2  0 2

v 

vL2

v 4

v Relative velocity of separation 4 e  v Relative velocity of approach

2g

For completing the vertical circle, vL  5gR

h

A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

Answer ( 2 )

As track is frictionless, so total mechanical energy will remain constant

h

2 1 : :1 5 2

= 4 : 5 : 10

7 D 5

0  mgh 

2 1 MR2 : MR2 : MR2 5 2

5gR 5 5  R D 2g 2 4

e

5

1  0.25 4


18.

Which one of the following statements is incorrect?

20.

(1) Rolling friction is smaller than sliding friction. (2) Limiting value of static friction is directly proportional to normal reaction. (3) Frictional force opposes the relative motion. (4) Coefficient of sliding dimensions of length.

friction

has

Answer ( 4 )

(3) 1 m/s, 3.5 m/s

(4) 1.5 m/s, 3 m/s

A

a

–a

t=1 v = 6 ms C t=3

v=0

f N

–1

The moment of the force, F  4iˆ  5 ˆj  6kˆ at (2, 0, –3), about the point (2, –2, –2), is given by

Acceleration a 

t=2 B v=0

–a

v = –6 ms

–1

60  6 ms2 1

For t = 0 to t = 1 s,

(1) 8iˆ  4 ˆj  7kˆ

S1 

(2) 4iˆ  ˆj  8kˆ

1  6(1)2 = 3 m 2

...(i)

For t = 1 s to t = 2 s,

(3) 7iˆ  8ˆj  4kˆ

S2  6.1 

(4) 7iˆ  4 ˆj  8kˆ

1  6(1)2  3 m 2

...(ii)

For t = 2 s to t = 3 s,

Answer ( 4 ) Sol.

(2) 1 m/s, 3 m/s

Sol. t = 0

f = sN

19.

(1) 2 m/s, 4 m/s

Answer ( 2 )

S o l . Coefficient of sliding friction has no dimension.

⇒ s 

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E . Due to the force q E , its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

Y

S3  0 

A r0

Average velocity 

P

3  1 ms 1 3

Total distance travelled = 9 m

r

O   (r  r0 )  F

...(iii)

Total displacement S = S1 + S2 + S3 = 3 m

F r  r0

1  6(1)2  3 m 2

Average speed 

X

21.

...(i)

ˆ  (2iˆ  2 ˆj  2k) ˆ r  r0  (2iˆ  0ˆj  3k)  0iˆ  2 ˆj  kˆ î ˆj kˆ   0 2 1  7iˆ  4 ˆj  8kˆ 4 5 6 6

9  3 ms 1 3

A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is (1) 0.521 cm

(2) 0.525 cm

(3) 0.053 cm

(4) 0.529 cm


Answer ( 4 )

23.

S o l . Diameter of the ball = MSR + CSR × (Least count) – Zero error = 0.5 cm + 25 × 0.001 – (–0.004)

The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

= 0.5 + 0.025 + 0.004 = 0.529 cm 22.

A block of mass m is placed on a smooth inclined wedge ABC of inclination  as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and  for the block to remain stationary on the wedge is A m

(1)

2 5

(2)

2 3

(3)

1 3

(4)

2 7

Answer ( 1 ) a

S o l . Given process is isobaric

 C

dQ  n Cp dT

B

(1) a 

g cosec 

⎛5 ⎞ dQ  n ⎜ R ⎟ dT ⎝2 ⎠

(2) a 

g sin 

dW  P dV = n RdT

Required ratio 

(3) a = g cos  (4) a = g tan  Answer ( 4 )

24.

Sol.

N cos N

N sin

(2) 8 cm

 mg

(3) 12.5 cm (4) 16 cm

a



Answer ( 1 ) S o l . For closed organ pipe, third harmonic

In non-inertial frame, N sin  = ma

...(i)

N cos  = mg

...(ii)

tan  

The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is (1) 13.2 cm

 ma (pseudo)

dW nRdT 2   dQ 5 ⎛5 ⎞ n ⎜ R ⎟ dT ⎝2 ⎠



3v 4l

For open organ pipe, fundamental frequency

a g



a = g tan  7

v 2l 


Given,

27.

3v v  4l 2l 

 l  

25.

4l 2l  32 3 2  20  13.33 cm 3

(1) Reflected light is polarised with its electric vector parallel to the plane of incidence

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere?

(2) Reflected light is polarised with its electric vector perpendicular to the plane of incidence

(Given :

⎛ 1⎞ (3) i  sin1 ⎜ ⎟ ⎝⎠

Mass of oxygen molecule (m) = 2.76 × 10–26 kg Boltzmann's constant kB = 1.38 × 10–23 JK–1) (1) 2.508 × 104 K (3) 5.016 ×

104

K

(2) 8.360 × 104 K (4) 1.254 ×

104

Unpolarised light is incident from air on a plane surface of a material of refractive index ''. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

⎛ 1⎞ (4) i  tan1 ⎜ ⎟ ⎝⎠

K

Answer ( 2 )

Answer ( 2 )

S o l . Vescape = 11200 m/s

S o l . When reflected light rays and refracted rays are perpendicular, reflected light is polarised with electric field vector perpendicular to the plane of incidence.

Say at temperature T it attains Vescape

So,

3kB T  11200 m/s mO2

On solving,

T = 8.360 × 104 K 26.

i

The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is



(1) 26.8% (2) 20%

Also, tan i =  (Brewster angle)

(3) 6.25%

28.

(4) 12.5% Answer ( 1 )

T ⎞ ⎛ S o l . Efficiency of ideal heat engine,   ⎜ 1  2 ⎟ T1 ⎠ ⎝ T2 : Sink temperature T1 : Source temperature

T ⎞ ⎛ %  ⎜ 1  2 ⎟  100 T1 ⎠ ⎝

In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength  of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same  and D) the separation between the slits needs to be changed to (1) 1.8 mm (2) 1.9 mm

273 ⎞ ⎛  ⎜ 1 ⎟  100 373 ⎠ ⎝

(3) 2.1 mm (4) 1.7 mm

⎛ 100 ⎞ ⎜ ⎟  100  26.8% ⎝ 373 ⎠

Answer ( 2 ) 8


S o l . Angular width 

 d

31.

0.20 

 2 mm

…(i)

0.21 

 d

…(ii)

Dividing we get,

prism is 2 and the angle of the prism is 30°. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is

0.20 d  0.21 2 mm

(1) 60°

 d = 1.9 mm 29.

The refractive index of the material of a

(2) 45°

An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

(3) 30° (4) Zero Answer ( 2 )

(1) Small focal length and large diameter

S o l . For retracing its path, light ray should be normally incident on silvered face.

(2) Large focal length and small diameter (3) Large focal length and large diameter (4) Small focal length and small diameter Answer ( 3 )

30°

S o l . For telescope, angular magnification =

f0 fE

i

M

So, focal length of objective lens should be large.

Applying Snell's law at M,

So, objective should have large focal length (f0) and large diameter D.

sin i 2  sin30 1

An em wave is propagating in a medium with 

V  Viˆ . The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along a

velocity

 sin i  2 

sin i 

(1) –z direction (2) +z direction

32.

(3) –y direction (4) –x direction Answer ( 2 ) 



30°

 2

D should be large. Angular resolution = 1.22

30.

60°



1 2

1 2 i.e. i = 45°

An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be (1) 30 cm away from the mirror

Sol. E  B  V

(2) 36 cm away from the mirror



ˆ  (B)  Viˆ (Ej)

(3) 30 cm towards the mirror



So, B  Bkˆ

(4) 36 cm towards the mirror Answer ( 2 )

Direction of propagation is along +z direction. 9


34.

Sol.

f = 15 cm O

40 cm

1 1 1   f v1 u –

(1) 40 

(2) 25 

(3) 250 

(4) 500 

Answer ( 3 ) S o l . Current sensitivity

1 1 1  – 15 v1 40

IS 

1 1 1   v1 –15 40



Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is

Voltage sensitivity VS 

v1 = –24 cm When object is displaced by 20 cm towards mirror.

RG 

u2 = –20

35.

1 1 1   f v2 u2 1 1 1 –  –15 v2 20

1 1 1  – v2 20 15 v2 = –60 cm So, image shifts away from mirror by = 60 – 24 = 36 cm.

(2) 138.88 H

(3) 1.389 H

(4) 13.89 H

(2) 5.98 A

(3) 14.76 A

(4) 11.32 A

I

 36.

1 2 Ll 2 1  L  (60  10–3 )2 2

mg tan30 lB

0.5  9.8 0.25  3

B

(2) 0.43 W (3) 2.74 W

500 36

(4) 1.13 W Answer ( 1 )

= 13.89 H 10

 11.32 A

0°

3 n si g m 30°

s co

° 30

llB 30° llB

An inductor 20 mH, a capacitor 100 F and a resistor 50  are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is (1) 0.79 W

25  2  106  10–3 L 3600 

(1) 7.14 A

mg sin30  Il Bcos 30

S o l . Energy stored in inductor

25  10–3 

A metallic rod of mass per unit length 0.5 kg m–1 is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

S o l . For equilibrium,

Answer ( 4 )

U

IS 51 5000    250  VS 20  103 20

Answer ( 4 )

The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance (1) 0.138 H

NBA CRG

So, resistance of galvanometer

Now,

33.

NBA C


Answer ( 3 )

2

⎛V ⎞ S o l . Pav  ⎜ RMS ⎟ R Z ⎝ ⎠

S o l . We know, max T  constant (Wien's law)

2

1 ⎞ ⎛ Z  R2  ⎜ L   56  C ⎟⎠ ⎝

So, max1 T1  max2 T2

2

⎛ ⎞ 10 ⎟  50  0.79 W Pav  ⎜ ⎜ 2 56 ⎟ ⎝ ⎠ A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from



37.

⇒ 0 T 

 

⇒ T 

40.

(2) The magnetic field (3) The lattice structure of the material of the rod (4) The induced electric field due to the changing magnetic field

(3) 42.2 J (4) 84.5 J

A small sphere of radius 'r' falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to

Answer ( 2 )

(1) r3

(2) r2

41.

r5

r4

S o l . Q = U + W  54 × 4.18 = U + 1.013 × 105(167.1 × 10–6 – 0)  U = 208.7 J

Answer ( 3 ) 2 S o l . Power = 6 rVT iVT  6 rVT

VT  r 2 39.

A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 105 Nm–2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

(2) 208.7 J

S o l . Energy of current source will be converted into potential energy of the rod.

(4)

Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by l on applying a force F, how much force is needed to stretch the socond wire by the same amount?

⇒ Power  r 5

(1) 9 F

The power radiated by a black body is P and it radiates maximum energy at wavelength, 0. If the temperature of the black body is now changed so that it radiates maximum energy 3 at wavelength  0 , the power radiated by it 4 becomes nP. The value of n is

(2) 6 F

3 (1) 4 (3)

256 81

(3) 4 F (4) F Answer ( 1 ) S o l . Wire 1 : A, 3l

4 (2) 3 (4)

4

P2 ⎛ T  ⎞ 256 ⎛4⎞ ⎜ ⎟ ⎜ ⎟  P1 ⎝ T ⎠ 81 ⎝3⎠

(1) 104.3 J

Answer ( 1 )

(3)

4 T 3 4

So,

(1) The current source

38.

3 0 T 4

F

Wire 2 :

81 256

3A, l

11

F


Answer ( 2 )

For wire 1,

⎛ F ⎞ l  ⎜ ⎟ 3l ⎝ AY ⎠

B

Sol. …(i)

perihelion A

For wire 2,

Point A is perihelion and C is aphelion. So, VA > VB > VC

⎛ F ⎞  l  ⎜ ⎟l ⎝ 3AY ⎠

So, KA > KB > KC …(ii)

44.

From equation (i) & (ii),

⎛ F ⎞ ⎛ F ⎞ l  ⎜ ⎟ 3l  ⎜ 3AY ⎟ l AY ⎝ ⎠ ⎝ ⎠

42.

C aphelion

S VA

F l Y 3A l



VC

A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (K t ) as well as rotational kinetic energy (K r ) simultaneously. The ratio Kt : (Kt + Kr) for the sphere is (1) 7 : 10 (2) 5 : 7

F  9F

(3) 10 : 7

A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?

(4) 2 : 5 Answer ( 2 ) S o l . Kt 

1 mv 2 2

(1) Angular velocity

Kt  Kr 

(2) Moment of inertia (3) Rotational kinetic energy

1 1 1 1⎛ 2 ⎞⎛ v ⎞ mv2  I2  mv2  ⎜ mr 2 ⎟⎜ ⎟ 2 2 2 2⎝5 ⎠⎝ r ⎠

(4) Angular momentum



Answer ( 4 ) S o l . ex = 0 So,

dL So, 0 dt i.e. L = constant

45.

So angular momentum remains constant. 43.

The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then

S

7 mv2 10

Kt 5  Kt  Kr 7

If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct? (1) Raindrops will fall faster (2) Walking on the ground would become more difficult (3) Time period of a simple pendulum on the Earth would decrease

B A

2

(4) ‘g’ on the Earth will not change Answer ( 4 )

C

S o l . If Universal Gravitational constant becomes ten times, then G = 10 G

(1) KA < KB < KC

(2) KA > KB > KC

So, acceleration due to gravity increases.

(3) KB < KA < KC

(4) KB > KA > KC

i.e. (4) is wrong option. 12


46.

Answer ( 3 )

Which of the following is an amino acid derived hormone? (1) Epinephrine

(2) Ecdysone

(3) Estradiol

(4) Estriol

S o l . Estrogen promotes the activity of osteoblast and inhibits osteoclast. In an ageing female osteoporosis occurs due to deficiency of estrogen. Parathormone promotes mobilisation of calcium from bone into blood. Excessive activity of parathormone causes demineralisation leading to osteoporosis.

Answer ( 1 ) S o l . Epinephrine is derived from tyrosine amino acid by the removal of carboxyl group. It is a catecholamine. 47.

50.

Which of the following structures or regions is incorrectly paired with its functions?

Ciliates differ from all other protozoans in (1) using flagella for locomotion

(1) Medulla oblongata : controls respiration and cardiovascular reflexes.

(2) having a contractile vacuole for removing excess water

(2) Limbic system

(4) having two types of nuclei

(3) Hypothalamus

(4) Corpus callosum

(3) using pseudopodia for capturing prey

: consists of fibre tracts that interconnect different regions of brain; controls movement.

Answer ( 4 ) S o l . Ciliates differs from other protozoans in having two types of nuclei. eg. Paramoecium have two types of nuclei i.e. macronucleus & micronucleus.

: production of releasing hormones and regulation of temperature, hunger and thirst.

51.

Identify the vertebrate group of animals characterized by crop and gizzard in its digestive system (1) Amphibia

: band of fibers connecting left and right cerebral hemispheres.

(2) Reptilia (3) Aves (4) Osteichthyes

Answer ( 2 )

Answer ( 3 )

S o l . Limbic system is emotional brain. It controls all emotions in our body but not movements.

S o l . The digestive tract of Aves has additional chambers in their digestive system as crop and Gizzard.

48.

The transparent lens in the human eye is held in its place by

Crop is concerned with storage of food grains.

(1) ligaments attached to the ciliary body

Gizzard is a masticatory organ in birds used to crush food grain.

(2) ligaments attached to the iris (3) smooth muscles attached to the iris

52.

(4) smooth muscles attached to the ciliary body Answer ( 1 )

Which of the following organisms are known as chief producers in the oceans? (1) Dinoflagellates

(2) Diatoms

(3) Cyanobacteria

(4) Euglenoids

S o l . Lens in the human eye is held in its place by suspensory ligaments attached to the ciliary body.

Answer ( 2 )

49.

53.

S o l . Diatoms are chief producers of the ocean.

Which of the following hormones can play a significant role in osteoporosis?

Which one of these animals is not a homeotherm?

(1) Aldosterone and Prolactin

(1) Macropus

(2) Progesterone and Aldosterone

(2) Chelone

(3) Estrogen and Parathyroid hormone

(3) Camelus

(4) Parathyroid hormone and Prolactin

(4) Psittacula 13


Answer ( 2 )

57.

S o l . Homeotherm are animals that maintain constant body temperature, irrespective of surrounding temperature.

(1) Wildlife safari parks (2) Sacred groves

Birds and mammals are homeotherm.

(3) Botanical gardens

Chelone (Turtle) belongs to class reptilia which is Poikilotherm or cold blood. 54.

(4) Seed banks Answer ( 2 )

Which of the following animals does not undergo metamorphosis? (1) Earthworm

(2) Tunicate

(3) Moth

(4) Starfish

Sol.   58.

Answer ( 1 ) S o l . Metamorphosis refers to transformation of larva into adult.

In earthworm development is direct which means no larval stage and hence no metamorphosis. Which of the following features is used to identify a male cockroach from a female cockroach?

Represent pristine forest patch as protected by Tribal groups.

Match the items given in Column I with those in Column II and select the correct option given below : Column-II

a. Eutrophication

i.

UV-B radiation

b. Sanitary landfill

ii. Deforestation

c. Snow blindness

iii. Nutrient enrichment

d. Jhum cultivation iv. Waste disposal a

b

c

d

(1) Presence of a boat shaped sternum on the 9th abdominal segment

(1) ii

i

iii

iv

(2) i

iii

iv

ii

(2) Presence of caudal styles

(3) iii

iv

i

ii

(3) Forewings with darker tegmina

(4) i

ii

iv

iii

(4) Presence of anal cerci

Answer ( 3 )

Answer ( 2 )

S o l . a. Eutrophication

S o l . Males bear a pair of short, thread like anal styles which are absent in females. Anal/caudal styles arise from 9th abdominal segment in male cockroach. 56.

Sacred groves – in-situ conservation.

Column-I

Animal that perform metamorphosis are said to have indirect development.

55.

All of the following are included in ‘ex-situ conservation’ except

59.

b. Sanitary landfill

iv.

Waste disposal

c. Snow blindness

i.

UV-B radiation Deforestation

In a growing population of a country, (1) pre-reproductive individuals are more than the reproductive individuals.

(1) Commensalism (2) Mutualism

(2) reproductive individuals are less than the post-reproductive individuals.

(3) Parasitism

(3) reproductive and pre-reproductive individuals are equal in number.

(4) Amensalism Answer ( 4 )

(4) pre-reproductive individuals are less than the reproductive individuals.

S o l . Amensalism/Antibiosis (0, –)



Nutrient enrichment

d. Jhum cultivation ii.

Which one of the following population interactions is widely used in medical science for the production of antibiotics?



iii.

Antibiotics are chemicals secreted by one microbial group (eg : Penicillium) which harm other microbes (eg : Staphylococcus)

Answer ( 1 ) S o l . Whenever the pre-reproductive individuals or the younger population size is larger than the reproductive group, the population will be an increasing population.

It has no effect on Penicillium or the organism which produces it. 14


60.

Answer ( 1 )

Which part of poppy plant is used to obtain the drug “Smack”? (1) Flowers

(2) Latex

(3) Roots

(4) Leaves

S o l . The extraembryonic or foetal membranes are amnion, chorion, allantois and Yolk sac.

Answer ( 2 )

Amnion is formed from mesoderm on outer side and ectoderm on inner side.

S o l . ‘Smack’ also called as brown sugar/Heroin is formed by acetylation of morphine. It is obtained from the latex of unripe capsule of Poppy plant.

Chorion is formed from trophoectoderm and mesoderm whereas allantois and Yolk sac membrane have mesoderm on outerside and endoderm in inner side.

61.

64.

Hormones secreted by the placenta to maintain pregnancy are

(1) In spermiogenesis spermatids are formed, while in spermiation spermatozoa are formed.

(1) hCG, hPL, progestogens, prolactin (2) hCG, hPL, estrogens, relaxin, oxytocin (3) hCG, hPL, progestogens, estrogens (4) hCG, progestogens, glucocorticoids

(2) In spermiogenesis spermatozoa are formed, while in spermiation spermatids are formed.

estrogens,

Answer ( 3 )

(3) In spermiogenesis spermatozoa from sertoli cells are released into the cavity of seminiferous tubules, while in spermiation spermatozoa are formed.

S o l . Placenta releases human chorionic gonadotropic hormone (hCG) which stimulates the Corpus luteum during pregnancy to release estrogen and progesterone and also rescues corpus luteum from regression. Human placental lactogen (hPL) is involved in growth of body of mother and breast. Progesterone maintains pregnancy, keeps the uterus silent by increasing uterine threshold to contractile stimuli. 62.

(4) In spermiogenesis spermatozoa are formed, while in spermiation spermatozoa are released from sertoli cells into the cavity of seminiferous tubules. Answer ( 4 ) S o l . Spermiogenesis is transformation of spermatids into spermatozoa whereas spermiation is the release of the sperms from sertoli cells into the lumen of seminiferous tubule.

The contraceptive ‘SAHELI’ (1) blocks estrogen receptors in the uterus, preventing eggs from getting implanted.

65.

(2) increases the concentration of estrogen and prevents ovulation in females. (3) is an IUD.

Which of the following options correctly represents the lung conditions in asthma and emphysema, respectively? (1) Inflammation of bronchioles; Decreased respiratory surface

(4) is a post-coital contraceptive.

(2) Increased number of bronchioles; Increased respiratory surface

Answer ( 1 ) S o l . Saheli is the first non-steroidal, once a week pill. It contains centchroman and its functioning is based upon selective Estrogen Receptor modulation. 63.

The difference between spermiogenesis and spermiation is

The amnion of mammalian embryo is derived from

(3) Increased respiratory Inflammation of bronchioles

surface;

(4) Decreased respiratory Inflammation of bronchioles

surface;

Answer ( 1 )

(1) ectoderm and mesoderm

S o l . Asthma is a difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles. Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased.

(2) endoderm and mesoderm (3) mesoderm and trophoblast (4) ectoderm and endoderm 15


66.

S o l . Tidal volume is volume of air inspired or expired during normal respiration. It is approximately 500 mL. Inspiratory reserve volume is additional volume of air a person can inspire by a forceful inspiration. It is around 2500 – 3000 mL. Expiratory reserve volume is additional volume of air a person can be expired by a forceful expiration. This averages 1000 – 1100 mL.

Match the items given in Column I with those in Column II and select the correct option given below : Column I

Column II

a. Tricuspid valve

i.

b. Bicuspid valve

ii. Between right ventricle and pulmonary artery

c. Semilunar valve

a

b

c

(1) iii

i

ii

(2)

i

iii

ii

(3) i

ii

iii

(4) ii

i

iii

Between left atrium and left ventricle

Residual volume is volume of air remaining in lungs even after forceful expiration. This averages 1100 – 1200 mL.

iii. Between right atrium and right ventricle

68.


Column II

a. Glycosuria

i.

b. Gout

ii. Mass of crystallised salts within the kidney

c. Renal calculi

iii. Inflammation in glomeruli

d. Glomerular nephritis

iv. Presence of in glucose urine

Answer ( 1 ) S o l . Tricuspid valves are AV valve present between right atrium and right ventricle. Bicuspid valves are AV valve present between left atrium and left ventricle. Semilunar valves are present at the openings of aortic and pulmonary aorta. 67.

Match the items given in Column I with those in Column II and select the correct option given below: Column I

Column II

a. Tidal volume

i. 2500 – 3000 mL

b. Inspiratory Reserve

ii. 1100 – 1200 mL

a

b

c

d

(1)

iii

ii

iv

i

(2)

i

ii

iii

iv

(3)

ii

iii

i

iv

(4)

iv

i

ii

iii

Answer ( 4 )

volume c. Expiratory Reserve

S o l . Glycosuria denotes presence of glucose in the urine. This is observed when blood glucose level rises above 180 mg/100 ml of blood, this is called renal threshold value for glucose. Gout is due to deposition of uric acid crystals in the joint.

iii. 500 – 550 mL

volume d. Residual volume a

Accumulation of uric acid in joints

iv. 1000 – 1100 mL

b

c

d

(1) iii

ii

i

iv

(2) iii

i

iv

ii

(3) i

iv

ii

iii

(4) iv

iii

ii

i

Renal calculi are precipitates of calcium phosphate produced in the pelvis of the kidney. Glomerular nephritis is the inflammatory condition of glomerulus characterised by proteinuria and haematuria.

Answer ( 2 ) 16


69.

Match the items given in Column I with those in Column II and select the correct option given below: Column I

Column II

(Function)

(Part of Excretory system)

a. Ultrafiltration

i.

b. Concentration

ii. Ureter

71.

(1) Enzymes of TCA cycle are present in mitochondrial matrix (2) Glycolysis occurs in cytosol (3) Glycolysis operates as long as it is supplied with NAD that can pick up hydrogen atoms

Henle's loop

(4) Oxidative phosphorylation takes place in outer mitochondrial membrane

of urine c. Transport of

iii. Urinary bladder

Answer ( 4 )

iv. Malpighian

S o l . Oxidative phosphorylation takes place in inner mitochondrial membrane.

urine d. Storage of urine

72.

corpuscle v.

Nissl bodies are mainly composed of

Proximal

(1) Proteins and lipids

convoluted tubule

(2) DNA and RNA

a

b

c

d

(3) Nucleic acids and SER

(1) iv

v

ii

iii

(4) Free ribosomes and RER

(2) iv

i

ii

iii

(3) v

iv

i

ii

(4) v

iv

i

iii

Answer ( 4 ) S o l . Nissl granules are present in the cyton and even extend into the dendrite but absent in axon and rest of the neuron.

Answer ( 2 )

Nissl granules are in fact composed of free ribosomes and RER. They are responsible for protein synthesis.

S o l . Ultrafiltration refers to filtration of very fine particles having molecular weight less than 68,000 daltons through malpighian corpuscle. Concentration of urine refers to water absorption from glomerular filtrate as a result of hyperosmolarity in the medulla created by counter-current mechanism in Henle's loop.

73.

Which of the following terms describe human dentition? (1) Thecodont, Diphyodont, Homodont (2) Thecodont, Diphyodont, Heterodont

Urine is carried from kidney to bladder through ureter.

(3) Pleurodont, Monophyodont, Homodont

Urinary bladder is concerned with storage of urine. 70.

Which of these statements is incorrect?

(4) Pleurodont, Diphyodont, Heterodont Answer ( 2 )

Which of the following events does not occur in rough endoplasmic reticulum?

S o l . In humans, dentition is

(1) Protein folding (2) Protein glycosylation (3) Cleavage of signal peptide



Thecodont : Teeth are present in the sockets of the jaw bone called alveoli.



Diphyodont : Teeth erupts twice, temporary milk or deciduous teeth are replaced by a set of permanent or adult teeth.



Heterodont dentition : Dentition consists of different types of teeth namely incisors, canine, premolars and molars.

(4) Phospholipid synthesis Answer ( 4 ) S o l . Phospholipid synthesis does not take place in RER. Smooth endoplasmic reticulum are involved in lipid synthesis. 17


74.

Select the incorrect match : (1) Lampbrush

S o l . During proliferative phase, the follicles start developing, hence, called follicular phase.

– Diplotene bivalents

Secretory phase is also called as luteal phase mainly controlled by progesterone secreted by corpus luteum. Estrogen further thickens the endometrium maintained by progesterone.

chromosomes (2) Allosomes

– Sex chromosomes

(3) Submetacentric – L-shaped chromosomes chromosomes (4) Polytene chromosomes

– Oocytes of amphibians

Menstruation occurs due to decline in progesterone level and involves breakdown of overgrown endometrial lining.

Answer ( 4 ) S o l . Polytene chromosomes are found in salivary glands of insects of order Diptera. 75.

78.

Many ribosomes may associate with a single mRNA to form multiple copies of a polypeptide simultaneously. Such strings of ribosomes are termed as

(1) Multiple step mutations (2) Saltation (3) Phenotypic variations

(1) Polysome

(4) Minor mutations

(2) Polyhedral bodies (3) Plastidome

Answer ( 2 )

(4) Nucleosome

S o l . As per mutation theory given by Hugo de Vries, the evolution is a discontinuous phenomenon or saltatory phenomenon/ saltation.

Answer ( 1 ) S o l . The phenomenon of association of many ribosomes with single m-RNA leads to formation of polyribosomes or polysomes or ergasomes. 76.

79.

All of the following are part of an operon except (1) an operator

(2) structural genes

(3) an enhancer

(4) a promoter

Sol. • •

A woman has an X-linked condition on one of her X chromosomes. This chromosome can be inherited by (1) Only daughters (2) Only sons (3) Only grandchildren

Answer ( 3 )

77.

According to Hugo de Vries, the mechanism of evolution is

(4) Both sons and daughters

Enhancer sequences are present in eukaryotes.

Answer ( 4 )

Operon concept is for prokaryotes.

Sol. •


Column II

80.

a. Proliferative Phase i. Breakdown of endometrial lining

Woman is a carrier

•

Both son X–chromosome

•

Although only son be the diseased

ii. Follicular Phase

(1) AGGUAUCGCAU

c. Menstruation

iii. Luteal Phase

(2) UGGTUTCGCAT

b

c

(1) iii

ii

i

(2) i

iii

ii

(3) ii

iii

i

(4) iii

i

ii

daughter inherit

AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA?

b. Secretory Phase a

&

(3) ACCUAUGCGAU (4) UCCAUAGCGUA Answer ( 1 ) S o l . Coding strand and mRNA has same nucleotide sequence except, ‘T’ – Thymine is replaced by ‘U’–Uracil in mRNA.

Answer ( 3 ) 18


81.

Which of the following gastric cells indirectly help in erythropoiesis?

S o l . Silicosis is due to excess inhalation of silica dust in the workers involved grinding or stone breaking industries.

(1) Chief cells

Long exposure can give rise to inflammation leading to fibrosis and thus causing serious lung damage.

(2) Mucous cells (3) Goblet cells (4) Parietal cells

Anthrax is a serious infectious disease caused by Bacillus anthracis. It commonly affects domestic and wild animals. Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased.

Answer ( 4 ) S o l . Parietal or oxyntic cell is a source of HCl and intrinsic factor. HCl converts iron present in diet from ferric to ferrous form so that it can be absorbed easily and used during erythropoiesis.

Botulism is a form of food poisoning caused by Clostridium botulinum.

Intrinsic factor is essential for the absorption of vitamin B 12 and its deficiency causes pernicious anaemia. 82.

84.


(i) Osmotic balance

b. Globulin

(ii) Blood clotting

c. Albumin

(iii) Defence mechanism

a

(1) Binds to troponin to remove the masking of active sites on actin for myosin. (2) Activates the myosin ATPase by binding to it.

Column II

a. Fibrinogen

b

c

(1) (iii)

(ii)

(i)

(2) (i)

(ii)

(iii)

(3) (i)

(iii)

(ii)

(4) (ii)

(iii)

(i)

Calcium is important in skeletal muscle contraction because it

(3) Detaches the myosin head from the actin filament. (4) Prevents the formation of bonds between the myosin cross bridges and the actin filament. Answer ( 1 ) Sol. 

Signal for contraction increase Ca++ level many folds in the sarcoplasm.



Ca++ now binds with sub-unit of troponin (troponin "C") which is masking the active site on actin filament and displaces the sub-unit of troponin.



Once the active site is exposed, head of the myosin attaches and initiate contraction by sliding the actin over myosin.

Answer ( 4 ) S o l . Fibrinogen forms fibrin strands during coagulation. These strands forms a network and the meshes of which are occupied by blood cells, this structure finally forms a clot. Antibodies are derived from -Globulin fraction of plasma proteins which means globulins are involved in defence mechanisms. Albumin is a plasma responsible for BCOP. 83.

protein

85.

mainly

(1) Psoriasis

Which of the following is an occupational respiratory disorder? (1) Anthracis

(2) Silicosis

(3) Botulism

(4) Emphysema

Which of the following is not an autoimmune disease?

(2) Rheumatoid arthritis (3) Alzheimer's disease (4) Vitiligo Answer ( 3 )

Answer ( 2 ) 19


S o l . Rheumatoid arthritis is an autoimmune disorder in which antibodies are produced against the synovial membrane and cartilage.

86.

89.

Vitiligo causes white patches on skin also characterised as autoimmune disorder.

(1) Homology

Psoriasis is a skin disease that causes itchy or sore patches of thick red skin and is also autoimmune whereas Alzheimer's disease is due to deficiency of neurotransmitter acetylcholine.

(3) Convergent evolution

(2) Analogy (4) Adaptive radiation Answer ( 1 )

Among the following sets of examples for divergent evolution, select the incorrect option :

S o l . In different vertebrates, bones of forelimbs are similar but their forelimbs are adapted in different way as per their adaptation, hence example of homology.

(1) Forelimbs of man, bat and cheetah

90.

(2) Heart of bat, man and cheetah (3) Brain of bat, man and cheetah

Which of the following characteristics represent ‘Inheritance of blood groups’ in humans? a. Dominance

(4) Eye of octopus, bat and man

b. Co-dominance

Answer ( 4 )

c. Multiple allele

S o l . Divergent evolution occurs in the same structure, example - forelimbs, heart, brain of vertebrates which have developed along different directions due to adaptation to different needs whereas eye of octopus, bat and man are examples of analogous organs showing convergent evolution. 87.

The similarity of bone structure in the forelimbs of many vertebrates is an example of

d. Incomplete dominance e. Polygenic inheritance (1) b, c and e (2) a, b and c (3) b, d and e

Conversion of milk to curd improves its nutritional value by increasing the amount of

(4) a, c and e Answer ( 2 )

(1) Vitamin D

Sol. 

(2) Vitamin A

IAIO, IBIO

-

Dominant–recessive relationship

(3) Vitamin B12



IAIB

-

Codominance

(4) Vitamin E



IA, IB & IO

-

3-different allelic forms of a gene (multiple allelism)

Answer ( 3 ) Sol.   88.

Curd is more nourishing than milk. 91.

It has enriched presence of vitamins specially Vit-B12.

Which of the following flowers only once in its life-time? (1) Bamboo species

In which disease does mosquito transmitted pathogen cause chronic inflammation of lymphatic vessels?

(2) Jackfruit (3) Mango

(1) Elephantiasis

(4) Papaya

(2) Ascariasis

Answer ( 1 )

(3) Ringworm disease

S o l . Bamboo species are monocarpic i.e., flower generally only once in its life-time after 50100 years.

(4) Amoebiasis Answer ( 1 )

Jackfruit, papaya and mango are polycarpic i.e., produce flowers and fruits many times in their life-time.

S o l . Elephantiasis is caused by roundworm, Wuchereria bancrofti and it is transmitted by Culex mosquito. 20


92.

S o l . Offset is a vegetative part of a plant, formed by mitosis.

Which of the following pairs is wrongly matched? (1) Starch synthesis in pea : Multiple alleles

–

Meiotic divisions do not occur in somatic cells.

(2) ABO blood grouping

: Co-dominance

(3) XO type sex

: Grasshopper

–

Parthenogenesis is the formation of embryo from ovum or egg without fertilisation.

: Linkage

–

Parthenocarpy is the fruit formed without fertilisation, (generally seedless)

determination (4) T.H. Morgan Answer ( 1 ) 96.

S o l . Starch synthesis in pea is controlled by pleiotropic gene.

93.

Which of the following has proved helpful in preserving pollen as fossils?

Other options (2, 3 & 4) are correctly matched.

(1) Pollenkitt

Select the correct statement

(3) Oil content

(1) Franklin Stahl coined the term ‘‘linkage’’

(4) Sporopollenin

(2) Cellulosic intine

(2) Punnett square was developed by a British scientist

Answer ( 4 ) S o l . Sporopollenin cannot be degraded by enzyme; strong acids and alkali, therefore it is helpful in preserving pollen as fossil.

(3) Spliceosomes take part in translation (4) Transduction was discovered by S. Altman Answer ( 2 )

Pollenkitt – Help in insect pollination.

S o l . Punnett square was developed by a British geneticist, Reginald C. Punnett.

Cellulosic Intine – Inner sporoderm layer of pollen grain known as intine made up cellulose & pectin.

–

94.

Franklin Stahl proved semi-conservative mode of replication.

–

Transduction was discovered by Zinder and Laderberg.

–

Spliceosome formation is part of posttranscriptional change in Eukaryotes

Oil content – No role is pollen preservation. 97.

Select the correct match (1) Alec Jeffreys

- Streptococcus pneumoniae

(2) Alfred Hershey and

The experimental proof for semiconservative replication of DNA was first shown in a

- TMV

Martha Chase (3) Matthew Meselson

(1) Fungus

- Pisum sativum

and F. Stahl

(2) Bacterium

(4) Francois Jacob and - Lac operon

(3) Plant

Jacques Monod

(4) Virus Answer ( 2 )

Answer ( 4 )

S o l . Semi-conservative DNA replication was first shown in Bacterium Escherichia coli by Matthew Meselson and Franklin Stahl.

S o l . Francois Jacob and Jacque Monod proposed model of gene regulation known as operon model/lac operon.

95.

Offsets are produced by

–

Alec Jeffreys – DNA fingerprinting technique.

–

Matthew Meselson and F. Stahl – Semiconservative DNA replication in E. coli.

–

Alfred Hershey and Martha Chase – Proved DNA as genetic material not protein

(1) Meiotic divisions (2) Mitotic divisions (3) Parthenocarpy (4) Parthenogenesis Answer ( 2 ) 21


98.

101. A ‘new’ variety of rice was patented by a foreign company, though such varieties have been present in India for a long time. This is related to

In India, the organisation responsible for assessing the safety of introducing genetically modified organisms for public use is (1) Indian Council of Medical Research (ICMR)

(1) Co-667

(2) Sharbati Sonora

(2) Council for Scientific and Industrial Research (CSIR)

(3) Lerma Rojo

(4) Basmati

(3) Research Committee Manipulation (RCGM)

on

Answer ( 4 )

Genetic

S o l . In 1997, an American company got patent rights on Basmati rice through the US patent and trademark office that was actually been derived from Indian farmer’s varieties.

(4) Genetic Engineering Appraisal Committee (GEAC) Answer ( 4 )

The diversity of rice in India is one of the richest in the world, 27 documented varieties of Basmati are grown in India.

S o l . Indian Government has setup organisation such as GEAC (Genetic Engineering Appraisal Committee) which will make decisions regarding the validity of GM research and safety of introducing GM-organism for public services. (Direct from NCERT). 99.

Indian basmati was crossed with semi-dwarf varieties and claimed as an invention or a novelty. Sharbati Sonora and Lerma Rojo are varieties of wheat.

Which of the following is commonly used as a vector for introducing a DNA fragment in human lymphocytes?

102. Select the correct match (1) Ribozyme

(1) Retrovirus

- Nucleic acid

(2) F2 × Recessive parent - Dihybrid cross

(2) Ti plasmid (3)  phage (4) pBR 322

(3) T.H. Morgan

- Transduction

(4) G. Mendel

- Transformation

Answer ( 1 )

Answer ( 1 )

S o l . Ribozyme is a catalytic RNA, which is nucleic acid.

S o l . Retrovirus is commonly used as vector for introducing a DNA fragment in human lymphocyte.

103. Use of bioresources by multinational companies and organisations without authorisation from the concerned country and its people is called

Gene therapy : Lymphocyte from blood of patient are grown in culture outside the body, a functional gene is introduced by using a retroviral vector, into these lymphocyte.

(1) Bio-infringement (2) Biopiracy (3) Biodegradation

100. The correct order of steps in Polymerase Chain Reaction (PCR) is

(4) Bioexploitation

Answer ( 2 ) S o l . Biopiracy is term used for or refer to the use of bioresources by multinational companies and other organisation without proper authorisation from the countries and people concerned with compensatory payment (definition of biopiracy given in NCERT).

(1) Extension, Denaturation, Annealing (2) Annealing, Extension, Denaturation (3) Denaturation, Extension, Annealing (4) Denaturation, Annealing, Extension Answer ( 4 )

104. Natality refers to

S o l . This technique is used for making multiple copies of gene (or DNA) of interest in vitro.

(1) Death rate

Each cycle has three steps

(2) Birth rate

(1) Denaturation

(3) Number of individuals leaving the habitat

(2) Primer annealing

(4) Number of individuals entering a habitat Answer ( 2 )

(3) Extension of primer 22


S o l . Natality refers to birth rate. •

Death rate

– Mortality

•

Number of individual entering a habitat is

– Immigration

•

Number of individual leaving the habital

– Emigration

S o l . UV rays act on CFCs, releasing Cl atoms, chlorine reacts with ozone in sequential method converting into oxygen Carbon, oxygen and Fe are not related to ozone layer depletion 108. World Ozone Day is celebrated on (1) 5th June (2) 21st April

105. Niche is

(3) 16th September

(1) all the biological factors in the organism's environment

(4) 22nd April Answer ( 3 )

(2) the physical space where an organism lives

S o l . World Ozone day is celebrated on 16 th September.

(3) the range of temperature that the organism needs to live

5th June - World Environment Day 21st April - National Yellow Bat Day

(4) the functional role played by the organism where it lives

22nd April - National Earth Day

Answer ( 4 )

109. Which of the following is a secondary pollutant?

S o l . Ecological niche was termed by J. Grinnel. It refers the functional role played by the organism where it lives. 106. What type of ecological pyramid would be obtained with the following data?

(1) CO

(2) CO2

(3) SO2

(4) O3

Answer ( 4 ) S o l . O3 (ozone) is a secondary pollutant. These are formed by the reaction of primary pollutant.

Secondary consumer : 120 g Primary consumer : 60 g

CO – Quantitative pollutant

Primary producer : 10 g

CO2 – Primary pollutant

(1) Inverted pyramid of biomass

SO2 – Primary pollutant

(2) Pyramid of energy

110. The two functional groups characteristic of sugars are

(3) Upright pyramid of numbers

(1) Hydroxyl and methyl

(4) Upright pyramid of biomass

(2) Carbonyl and methyl

Answer ( 1 ) Sol. •

(3) Carbonyl and phosphate

The given data depicts the inverted pyramid of biomass, usually found in aquatic ecosystem.

•

Pyramid of energy is always upright

•

Upright pyramid of biomass and numbers are not possible, as the data depicts primary producer is less than primary consumer and this is less than secondary consumers.

(4) Carbonyl and hydroxyl Answer ( 4 ) S o l . Sugar is a common term used to denote carbohydrate. Carbohydrates are polyhydroxy aldehyde, ketone or their derivatives, which means they have carbonyl and hydroxyl groups. 111. Which among the following is not a prokaryote?

107. In stratosphere, which of the following elements acts as a catalyst in degradation of ozone and release of molecular oxygen?

(1) Saccharomyces (2) Mycobacterium

(1) Carbon

(2) Cl

(3) Nostoc

(3) Fe

(4) Oxygen

(4) Oscillatoria Answer ( 1 )

Answer ( 2 ) 23


S o l . Saccharomyces i.e. yeast is an eukaryote (unicellular fungi)

117. Stomata in grass leaf are (1) Dumb-bell shaped

Mycobacterium – a bacterium

(2) Kidney shaped

Oscillatoria and Nostoc are cyanobacteria.

(3) Rectangular

112. Stomatal movement is not affected by

(4) Barrel shaped

(1) Temperature (2) Light

Answer ( 1 )

(3) O2 concentration

S o l . Grass being a monocot, has Dumb-bell shaped stomata in their leaves.

(4) CO2 concentration Answer ( 3 )

118. Casparian strips occur in

S o l . Light, temperature and concentration of CO2 affect opening and closing of stomata while they are not affected by O2 concentration.

(2) NADH

(3) NADPH

(4) Oxygen

(2) Pericycle

(3) Cortex

(4) Endodermis

Answer ( 4 )

113. Which of the following is not a product of light reaction of photosynthesis? (1) ATP

(1) Epidermis

Sol. • •

Answer ( 2 )

Endodermis have casparian strip on radial and inner tangential wall. It is suberin rich.

119. Plants having little or no secondary growth are

S o l . ATP, NADPH and oxygen are products of light reaction, while NADH is a product of respiration process.

(1) Grasses (2) Deciduous angiosperms

114. The Golgi complex participates in (1) Fatty acid breakdown

(3) Conifers

(2) Formation of secretory vesicles

(4) Cycads

(3) Respiration in bacteria

Answer (1)

(4) Activation of amino acid

S o l . Grasses are monocots and monocots usually do not have secondary growth.

Answer ( 2 ) S o l . Golgi complex, after processing releases secretory vesicles from their trans-face.

Palm like monocots secondary growth.

115. Which of the following is true for nucleolus?

(1) Halophytes

(2) It is a membrane-bound structure

(2) Free-floating hydrophytes

(3) It takes part in spindle formation

(3) Carnivorous plants

(4) It is a site for active ribosomal RNA synthesis

(4) Submerged hydrophytes

Answer ( 4 )

Answer (1)

S o l . Nucleolus is a non membranous structure and is a site of r-RNA synthesis.

Sol. 

116. The stage during which separation of the paired homologous chromosomes begins is



(2) Diplotene

(3) Diakinesis

(4) Zygotene

anomalous

120. Pneumatophores occur in

(1) Larger nucleoli are present in dividing cells

(1) Pachytene

have

Halophytes like pneumatophores.

mangrooves

have

Apogeotropic (–vely geotropic) roots having lenticels called pneumathodes to uptake O2.

121. Sweet potato is a modified

Answer ( 2 )

(1) Stem

(2) Adventitious root

S o l . Synaptonemal complex disintegrates. Terminalisation begins at diplotene stage i.e. chiasmata start to shift towards end.

(3) Tap root

(4) Rhizome

Answer ( 2 ) 24


•

Rhizomes are underground modified stem

S o l . In Pinus, winged pollen grains are present. It is extended outer exine on two lateral sides to form the wings of pollen. It is the characteristic feature, only in Pinus.

•

Tap root is primary root directly elongated from the redicle

Pollen grains of Mustard, Cycas & Mango are not winged shaped.

122. Secondary xylem and phloem in dicot stem are produced by

126. After karyogamy followed by meiosis, spores are produced exogenously in

S o l . Sweet potato is a modified adventitious root for storage of food

(1) Apical meristems (2) Vascular cambium (3) Phellogen

(4) Axillary meristems

• •

(2) Alternaria

(3) Agaricus

(4) Saccharomyces

Answer ( 3 )


(1) Neurospora

Sol. 

In Agaricus (a genus of basidiomycetes), basidiospores or meiospores are produced exogenously.



Neurospora (a genus of ascomycetes) produces ascospores as meiospores but endogenously inside the ascus.)



Alternaria (a genus of deuteromycetes) does not produce sexual spores.



Saccharomyces (Unicellular ascomycetes) produces ascospores, endogenously.

Vascular cambium is partially secondary Form secondary xylem towards its inside and secondary phloem towards outsides. 4 – 10 times more secondary xylem is produced than secondary phloem.

123. Which of the following statements is correct? (1) Ovules are not enclosed by ovary wall in gymnosperms (2) Selaginella is heterosporous, while Salvinia is homosporous

127. Match the items given in Column I with those in Column II and select the correct option given below:

(3) Horsetails are gymnosperms

Column I

(4) Stems are usually unbranched in both Cycas and Cedrus

Column II

a. Herbarium

(i) It is a place having a collection of preserved plants and animals

b. Key

(ii) A list that enumerates methodically all the species found in an area with brief description aiding identification

c. Museum

(iii) Is a place where dried and pressed plant specimens mounted on sheets are kept

d. Catalogue

(iv) A booklet containing a list of characters and their alternates which are helpful in identification of various taxa.

Answer ( 1 ) Sol. • •

Gymnosperms have naked ovule. Called phanerogams without womb/ovary

124. Select the wrong statement : (1) Cell wall is present in members of Fungi and Plantae (2) Mushrooms belong to Basidiomycetes (3) Pseudopodia are locomotory and feeding structures in Sporozoans (4) Mitochondria are the powerhouse of the cell in all kingdoms except Monera Answer ( 3 ) S o l . Pseudopodia are locomotory structures in sarcodines (Amoeboid) 125. Winged pollen grains are present in

a

b

c

d

(1)

(i)

(iv)

(iii)

(ii)

(2)

(iii)

(ii)

(i)

(iv)

(1) Mustard

(2) Cycas

(3)

(ii)

(iv)

(iii)

(i)

(3) Mango

(4) Pinus

(4)

(iii)

(iv)

(i)

(ii)

Answer ( 4 )

Answer ( 4 ) 25


Sol. •

Herbarium

–

Dried and pressed plant specimen

•

Key

–

Identification of various taxa

•

Museum

–

Plant and animal specimen are preserved

•

Catalogue

–

132. Double fertilization is (1) Fusion of two male gametes of a pollen tube with two different eggs (2) Fusion of one male gamete with two polar nuclei (3) Fusion of two male gametes with one egg

Alphabetical listing of species

(4) Syngamy and triple fusion Answer ( 4 )

128. Which one is wrongly matched?

S o l . Double fertilization is a unique phenomenon that occur in angiosperms only.

(1) Uniflagellate gametes – Polysiphonia (2) Biflagellate zoospores – Brown algae (3) Gemma cups

Syngamy + Triple fusion = Double fertilization

– Marchantia

133. Oxygen is not produced during photosynthesis by

(4) Unicellular organism – Chlorella Answer ( 1 )

(1) Green sulphur bacteria

Sol. •

(2) Nostoc

•

Polysiphonia is a genus of red algae, where asexual spores and gametes are non-motile or non-flagellated.

(3) Cycas (4) Chara

Other options (2, 3 & 4) are correctly matched

Answer ( 1 ) S o l . Green sulphur bacteria do not use H 2O as source of proton, therefore they do not evolve O2.

129. Which one of the following plants shows a very close relationship with a species of moth, where none of the two can complete its life cycle without the other? (1) Hydrilla

(2) Yucca

(3) Banana

(4) Viola

134. What is the role of NAD + in cellular respiration? (1) It functions as an enzyme.

Answer ( 2 )

(2) It functions as an electron carrier.

S o l . Yucca have an obligate mutualism with a species of moth i.e. Pronuba.

(3) It is a nucleotide source for ATP synthesis. (4) It is the final electron acceptor for anaerobic respiration.

130. Pollen grains can be stored for several years in liquid nitrogen having a temperature of (1) –120°C

(2) –80°C

(3) –196°C

(4) –160°C

Answer ( 2 ) S o l . In cellular respiration, NAD+ act as an electron carrier. 135. In which of the following forms is iron absorbed by plants?

Answer ( 3 ) S o l . Pollen grains can be stored for several years in liquid nitrogen at –196°C (Cryopreservation)

(1) Ferric (2) Ferrous

131. Which of the following elements is responsible for maintaining turgor in cells? (1) Magnesium

(2) Sodium

(3) Potassium

(4) Calcium

(3) Free element (4) Both ferric and ferrous Answer ( 1 * ) S o l . Iron is absorbed by plants in the form of ferric ions. (According to NCERT)

Answer ( 3 ) S o l . Potassium helps in maintaining turgidity of cells.

*Plants absorb iron in both form i.e. Fe++ and Fe+++. (Preferably Fe++) 26


Answer ( 4 )

136. Which of the following statements is not true for halogens?

Sol.

(1) All form monobasic oxyacids Elements Atomic radii (pm)

(2) All are oxidizing agents (3) All but fluorine show positive oxidation states

Ga 135

Al 143

In 167

Tl 170

140. The correct order of N-compounds in its decreasing order of oxidation states is

(4) Chlorine has the highest electron-gain enthalpy

(1) HNO3, NO, N2, NH4Cl (2) HNO3, NO, NH4Cl, N2

Answer ( 3 )

(3) HNO3, NH4Cl, NO, N2

S o l . Due to high electronegativity and small size, F forms only one oxoacid, HOF known as Fluoric (I) acid. Oxidation number of F is +1 in HOF.

(4) NH4Cl, N2, NO, HNO3 Answer ( 1 ) 5

(1) Fe

(2) Zn

(3) Mg

(4) Cu

(4) Three

 

(1) C2H5OH, C2H6, C2H5Cl

 

 

(4) In

142. The compound A on treatment with Na gives B, and with PCl 5 gives C. B and C react together to give diethyl ether. A, B and C are in the order

(2) C2H5OH, C2H5Cl, C2H5ONa

 

(3) C2H5Cl, C2H6, C2H5OH

Cl

F

(3) B

Hence, the correct option is (3).

S o l . The structure of ClF3 is

 

(2) Al

S o l . ∵ 'B' has no vacant d-orbitals in its valence shell, so it can't extend its covalency beyond 4. i.e. 'B' cannot form the ion like MF6 3(–) i.e. BF63(–).

Answer ( 2 )

F

(1) Ga

Answer ( 3 )

138. In the structure of ClF3, the number of lone pair of electrons on central atom ‘Cl’ is

(3) Four

–3

141. Which one of the following elements is unable to form MF63– ion?

S o l . The metal which is more reactive than 'Al' can reduce alumina i.e. 'Mg' should be the correct option.

(2) Two

0

Hence, the correct option is (1).

Answer ( 3 )

(1) One

2

S o l . H N O , N O, N2 , NH Cl 3 4

137. Considering Ellingham diagram, which of the following metals can be used to reduce alumina?

F

 

(4) C2H5OH, C2H5ONa, C2H5Cl  

 

B 85

Answer ( 4 )

 

S o l . C2H5OH (A)

The number of lone pair of electrons on central Cl is 2.

Na

C2H5O Na+ (B)

PCl5

139. The correct order of atomic radii in group 13 elements is

C2H5Cl (C)

(1) B < Al < In < Ga < Tl

C2H5O Na+ + C2H5Cl (B) (C)

(2) B < Al < Ga < In < Tl (3) B < Ga < Al < Tl < In

SN2

C2H5OC2 H5

So the correct option is (4) (4) B < Ga < Al < In < Tl 27


146. Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations :

143. Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is (1) CH  CH

(2) CH2  CH2

(3) CH3 – CH3

(4) CH4

Answer ( 4 ) Br2/h

S o l . CH4 (A)

CH3Br Na/dry ether Wurtz reaction CH3 — CH3

M M HCl + 40 mL NaOH 10 10

b. 55 mL


c. 75 mL


d. 100 mL

Hence the correct option is (4)

Br /Fe

Zn/HCl

2 2 C7H8   A   B  C

(1) b

(2) a

(3) d

(4) c

Answer ( 4 )

The product 'C' is Sol. •

(1) m-bromotoluene (2) o-bromotoluene (3) 3-bromo-2,4,6-trichlorotoluene (4) p-bromotoluene Answer ( 1 )

CH3

CCl3

(C7H8)

1 Meq of HCl = 75   1 = 15 5

•

1 Meq of NaOH = 25   1 = 5 5

•

Meq of HCl in resulting solution = 10

•

Molarity of [H+] in resulting mixture

CCl3 =

3Cl 2 

Sol.

Br2 Fe

(A)


pH of which one of them will be equal to 1?

144. The compound C7H8 undergoes the following reactions: 3Cl / 

a. 60 mL

(B)

10 1  100 10

 1 pH = –log[H+] =  log   = 1.0  10 

Br

Zn HCl

147. On which of the following properties does the coagulating power of an ion depend?

CH3

(1) The magnitude of the charge on the ion alone

(C)

Br

(2) Size of the ion alone (3) The sign of charge on the ion alone

So, the correct option is (1)

(4) Both magnitude and sign of the charge on the ion

145. Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity?


(2) NO2

Coagulation of colloidal solution by using an electrolyte depends on the charge present (positive or negative) on colloidal

(3) N2O

particles as well as on its size.

(1) N2O5

(4) NO

•

Answer ( 1 ) S o l . Fact 28

Coagulating power of an electrolyte depends on the magnitude of charge present on effective ion of electrolyte.


148. The solubility of BaSO4 in water is 2.42 × 10–3 gL–1 at 298 K. The value of its solubility product (Ksp) will be (Given molar mass of BaSO4 = 233 g mol–1) (1) 1.08 × 10–10 mol2L–2 (2) 1.08 × 10–12 mol2L–2

a

b

c

d

(1)

iv

v

ii

i

(2)

i

ii

iii

iv

(3)

iv

i

ii

iii

(4)

iii

v

i

ii

Answer ( 1 )

(3) 1.08 × 10–14 mol2L–2

S o l . Co3+ = [Ar] 3d6, Unpaired e–(n) = 4

(4) 1.08 × 10–8 mol2L–2

4(4  2)  24 BM

Answer ( 1 )

Spin magnetic moment =

2.42  103 S o l . Solubility of BaSO4, s = (mol L–1) 233

Cr3+ = [Ar] 3d3, Unpaired e–(n) = 3

3(3  2)  15 BM


= 1.04 × 10–5 (mol L–1)

Fe3+ = [Ar] 3d5, Unpaired e–(n) = 5

Ba2  (aq)  SO 24(aq) BaSO 4 (s)


Ksp = [Ba2+] [SO42–]= s2

Ni2+ = [Ar] 3d8, Unpaired e–(n) = 2

s

s

= (1.04 × 10–5)2

151. Iron carbonyl, Fe(CO)5 is

149. Given van der Waals constant for NH3, H2, O2 and CO2 are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied?

(1) Tetranuclear

(2) Mononuclear

(3) Trinuclear

(4) Dinuclear

Answer ( 2 )

(1) NH3

S o l . • Based on the number of metal atoms present in a complex, they are classified into mononuclear, dinuclear, trinuclear and so on.

(2) H2 (3) O2 (4) CO2

eg: Fe(CO)5 : mononuclear

Answer ( 1 )

•

2(2  2)  8 BM


= 1.08 × 10–10 mol2 L–2

Sol. •

5(5  2)  35 BM

Co2(CO)8 : dinuclear

van der waal constant ‘a’, signifies intermolecular forces of attraction.

Fe3(CO)12: trinuclear

Higher is the value of ‘a’, easier will be the liquefaction of gas.

Hence, option (2) should be the right answer.

150. Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code :

152. The geometry and magnetic behaviour of the complex [Ni(CO)4] are

Column I

(1) Square planar geometry and diamagnetic

Column II

(2) Tetrahedral geometry and diamagnetic a. Co3+

i.

8 BM

b. Cr3+

ii.

35 BM

c. Fe3+

iii.

3 BM

Answer ( 2 )

d. Ni2+

iv.

24 BM

S o l . Ni(28) : [Ar]3d8 4s2

v.

15 BM

(3) Square planar paramagnetic

geometry

and

(4) Tetrahedral geometry and paramagnetic

∵ CO is a strong field ligand Configuration would be : 29


• As per given option, type of isomerism is geometrical isomerism.

3

sp -hybridisation ××

×× ×× ××

CO

CO CO CO

155. Identify the major products P, Q and R in the following sequence of reactions:

For, four ‘CO’-ligands hybridisation would be sp 3 and thus the complex would be diamagnetic and of tetrahedral geometry.

Anhydrous AlCl3

+ CH3CH2CH2Cl

CO

(i) O2

P P

Ni OC

,

(1)

(2)

(3) MnO4–

(2)

,

OH , CH3CH(OH)CH3

,

(3)

S o l . CrO42–  Cr6+ = [Ar]

OH

CH(CH3)2

Unpaired electron (n) = 0; Diamagnetic 

COOH

,

CH(CH3)2

(4) MnO42–

Cr6+

CHO

Cr2O72–

Answer ( 4 )

Cr2O72–

CH3CH2 – OH

,

CH2CH2CH3

153. Which one of the following ions exhibits d-d transition and paramagnetism as well?

R

CHO

(1)

CO

CrO42–

Q

CH 2CH 2CH3

CO

Q+R

(ii) H3O+/

,

(4)

,

CH3 – CO – CH3

= [Ar] Answer ( 4 )

Unpaired electron (n) = 0; Diamagnetic

Cl

MnO42– = Mn6+ = [Ar] 3d1 S o l . CH CH CH – Cl + 3 2 2

Unpaired electron (n) = 1; Paramagnetic

Al Cl

Cl

MnO4– = Mn7+ = [Ar] Unpaired electron (n) = 0; Diamagnetic

+

CH3 – CH – CH3

154. The type of isomerism shown by the complex [CoCl2(en)2] is

1, 2–H Shift

+

CH3CH2CH2

Cl

(Incipient carbocation)

–

AlCl3

Cl

(1) Geometrical isomerism

–

AlCl3

(2) Coordination isomerism

Now,

(3) Ionization isomerism

CH3

(4) Linkage isomerism

CH – CH3

Answer ( 1 )

O2

CH3 – CH – CH3

S o l . In [CoCl2(en)2], Coordination number of Co is 6 and this compound has octahedral geometry.

(P) CH3 O

+

H /H2O

CH3 – C – CH3 + (R) 30

HC –C – O– O –H 3

OH

(Q)

Hydroperoxide Rearrangement


159. Which of the following is correct with respect to – I effect of the substituents? (R = alkyl) (1) – NH2 < – OR < – F (2) – NR2 < – OR < – F (3) – NH2 > – OR > – F (4) – NR2 > – OR > – F Answer ( 1 * ) S o l . –I effect increases on increasing electronegativity of atom. So, correct order of –I effect is –NH2 < – OR < – F. *Most appropriate Answer is option (1), however option (2) may also be correct answer.

156. Which of the following compounds can form a zwitterion? (1) Aniline

(2) Acetanilide

(3) Benzoic acid

(4) Glycine

Answer ( 4 ) Sol.





H3N – CH2 – COOH pKa = 9.60

H3N – CH2 – COO

–

(Zwitterion form)

pKa = 2.34

H2N – CH2 –

160. Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s2 2s2 2p3, the simplest formula for this compound is

COO–

157. Which of the following molecules represents the order of hybridisation sp2, sp2, sp, sp from left to right atoms? (1) HC  C – C  CH

(1) Mg2X3

(2) MgX2

(3) Mg2X

(4) Mg3X2

Answer ( 4 )

(2) CH2 = CH – C  CH

S o l . Element (X) electronic configuration

(3) CH2 = CH – CH = CH2

1s2 2s2 2p3

(4) CH3 – CH = CH – CH3

So, valency of X will be 3.

Answer ( 2 ) sp2

Valency of Mg is 2.

sp2

sp

Formula of compound formed by Mg and X will be Mg3X2.

sp

S o l . CH2  CH – C  CH

161. Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is

Number of orbital require in hybridization = Number of -bonds around each carbon atom.

158. Which of the following carbocations is expected to be most stable? (1)

NO2

NO2 

(1)

 Y

H

(2)

4 2

4 3 3 2

(4)

1 2

H S o l . For BCC lattice : Z = 2, a 

NO2 H

Y

3 3

(2)

Answer ( 3 )

Y

NO2

(3) H

(3)

3 2

4r 3

For FCC lattice : Z = 4, a = 2 2 r



(4) Y 



Answer ( 3 ) S o l . –NO2 group exhibit –I effect and it decreases with increase in distance. In option (4) positive charge present on C-atom at maximum distance so –I effect reaching to it is minimum and stability is maximum.

d25C d900C



 ZM   N a3   A  BCC  ZM   N a3   A  FCC 3

3 3 2  2 2 r     4r  4 2  4    3 

31


162. Which one is a wrong statement?

164. In the reaction

(1) Total orbital angular momentum of electron in 's' orbital is equal to zero (2) An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers 1

2s2

2px

1

2py

CHO

+ CHCl3 + NaOH

(3) The electronic configuration of N atom is 1s2

O–Na+

OH

The electrophile involved is

1

2pz





(1) Dichloromethyl cation CHCl2



(4) The value of m for dz2 is zero



(2) Formyl cation CHO

Answer ( 3 )





S o l . According to Hund's Rule of maximum multiplicity, the correct electronic configuration of N-atom is





(3) Dichloromethyl anion CHCl2



(4) Dichlorocarbene : CCl2  Answer ( 4 )

2

1s

2s

2

2p

3

S o l . It is Reimer-Tiemann reaction. The electrophile formed is :CCl2 (Dichlorocarbene) according to the following reaction

OR

.–. CCl3  H2 O CHCl3  OH– 2

2

3

1s 2s 2p  Option (3) violates Hund's Rule.

.–.

CCl3   : CCl2  Cl– Electrophile

163. Consider the following species : CN+, CN–, NO and CN

165. Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their

Which one of these will have the highest bond order? (1) NO (3)

CN+

(2) CN–

(1) Formation of intramolecular H-bonding

(4) CN

(2) Formation of carboxylate ion

Answer ( 2 ) S o l . NO : (1s)2, (1s)2, (2s)2,(2s)2,(2pz)2, (2px)2 = (2py)2,(2px)1 = (2py)0

(3) More extensive association of carboxylic acid via van der Waals force of attraction

10  5  2.5 2 CN– : (1s)2, (1s)2, (2s)2,(2s)2, (2px)2 = (2py)2,(2pz)2

(4) Formation of intermolecular H-bonding

BO =

Answer ( 4 ) S o l . Due to formation of intermolecular H-bonding in carboxylic acid, association occurs. Hence boiling point increases and become more than the boiling point of aldehydes, ketones and alcohols of comparable molecular masses.

10  4 3 2 CN : (1s)2, (1s)2, (2s)2,(2s)2, (2px)2 = (2py)2,(2pz)1

BO =

9 4  2.5 2 CN+ : (1s)2, (1s)2, (2s)2,(2s)2, (2px)2 = (2py)2

BO =

166. Compound A, C8H10O, is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with characteristic smell.

8 4 2 2 Hence, option(2) should be the right answer. BO =

A and Y are respectively 32


(1) H3C



CH2 – OH and I2

(2)

CH2 – CH2 – OH and I2

(3)

CH – CH3 and I2

For second order reaction, t 1/2 

1 , k[A0 ]

which depends on initial concentration of reactant. 168. Among CaH2, BeH2, BaH2, the order of ionic character is (1) BeH2 < CaH2 < BaH2 (2) CaH2 < BeH2 < BaH2

OH CH3

(3) BeH2 < BaH2 < CaH2 (4) BaH2 < BeH2 < CaH2

(4) CH3

Answer ( 1 )

OH and I2

S o l . For 2nd group hydrides, on moving down the group metallic character of metals increases so ionic character of metal hydride increases.

Answer ( 3 ) S o l . Option (3) is secondary alcohol which on oxidation gives phenylmethyl ketone (Acetophenone). This on reaction with I2 and NaOH form iodoform and sodium benzoate.

Hence the option (1) should be correct option. 169. Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :

2NaOH  I2  NaOI  NaI  H2 O CH – CH3 OH (A)

NaOI

– BrO4

C – CH3 O

– Br

Acetophenone

Sodium benzoate

Iodoform (Yellow PPt)

1.0652 V

1.5 V

Br2

NaOH

167. The correct difference between first and second order reactions is that

undergoing

(1) BrO3

(2) BrO4

(3) Br2

(4) HBrO

1

0

o S o l . HBrO   Br2 , EHBrO/Br  1.595 V 2

1

5

(2) The half-life of a first-order reaction does not depend on [A] 0 ; the half-life of a second-order reaction does depend on [A]0

HBrO   BrO3 , Eo

(3) A first-order reaction can catalyzed; a second-order reaction cannot be catalyzed

o o Ecell  EHBrO/Br  Eo

BrO3 /HBrO

 1.5 V

o for the disproportionation of HBrO, Ecell 2

BrO3 /HBrO

= 1.595 – 1.5 = 0.095 V = + ve Hence, option (4) is correct answer.

(4) The rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations

170. In which case is number of molecules of water maximum?

Answer ( 2 )

(1) 18 mL of water

For first order reaction, t 1/2  which is independent concentration of reactant.

1.595 V

Answer ( 4 )

(1) The rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations

Sol. 

HBrO

Then the species disproportionation is

I2

COONa + CHI3

– BrO3

1.82 V

of

(2) 0.18 g of water

0.693 , k

(3) 0.00224 L of water vapours at 1 atm and 273 K

initial

(4) 10–3 mol of water 33


Answer ( 1 )

173. Which of the following oxides is most acidic in nature? (1) MgO (2) BeO (3) BaO (4) CaO Answer ( 2 ) S o l . BeO < MgO < CaO < BaO  Basic character increases. So, the most acidic should be BeO. In fact, BeO is amphoteric oxide while other given oxides are basic. 174. The difference between amylose and amylopectin is (1) Amylopectin have 1  4 -linkage and 1 6 -linkage (2) Amylose have 1  4 -linkage and 1  6 -linkage (3) Amylopectin have 1  4 -linkage and 1  6 -linkage (4) Amylose is made up of glucose and galactose Answer ( 1 ) S o l . Amylose and Amylopectin are polymers of D-glucose, so -link is not possible. Amylose is linear with 1  4 -linkage whereas Amylopectin is branched and has both 1  4 and 1  6 -linkages. So option (1) should be the correct option. 175. A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be (1) 1.4 (2) 3.0 (3) 2.8 (4) 4.4 Answer ( 3 )

S o l . (1) Mass of water = 18 × 1 = 18 g Molecules of water = mole × NA =

18 NA 18

= NA (2) Molecules of water = mole × NA =

0.18 NA 18

= 10–2 NA 0.00224 = 10–4 22.4 Molecules of water = mole × NA = 10–4 NA

(3) Moles of water =

(4) Molecules of water = mole × NA = 10–3 NA 171. Regarding cross-linked or network polymers, which of the following statements is incorrect? (1) They contain covalent bonds between various linear polymer chains. (2) They are formed from bi- and tri-functional monomers. (3) Examples are bakelite and melamine. (4) They contain strong covalents bonds in their polymer chains. Answer ( 4 ) S o l . Cross linked or network polymers are formed from bi-functional and tri-functional monomers and contain strong covalent bonds between various linear polymer chains, e.g. bakelite, melamine etc. Option (4) is not related to cross-linking. So option (4) should be the correct option. 172. Nitration of aniline in strong acidic medium also gives m-nitroaniline because (1) Inspite of substituents nitro group always goes to only m-position. (2) In electrophilic substitution reactions amino group is meta directive.

Conc.H2 SO4 S o l . HCOOH   CO(g)  H2 O(l) 1  1  mol 2.3 g or  mol  20  20 

(3) In absence of substituents nitro group always goes to m-position.

COOH

(4) In acidic (strong) medium aniline is present as anilinium ion.

COOH

CO(g) + CO2 (g) + H2O(l) 1 mol 20

1 mol 20

 1  4.5 g or  mol   20 

Answer ( 4 )

NH2 Sol.

Conc.H2SO4

NH3

Gaseous mixture formed is CO and CO2 when it is passed through KOH, only CO 2 is absorbed. So the remaining gas is CO. So, weight of remaining gaseous product CO is

H Anilinium ion

–NH3 is m-directing, hence besides para

2  28  2.8 g 20 So, the correct option is (3)

(51%) and ortho (2%), meta product (47%) is also formed in significant yield. 34


Answer ( 1 )

176. For the redox reaction

X2 (g); H  x kJ S o l . A2 (g)  B2 (g)

MnO4  C2 O24  H   Mn2   CO2  H2 O

On increasing pressure equilibrium shifts in a direction where pressure decreases i.e. forward direction.

The correct coefficients of the reactants for the balanced equation are

MnO4

C2 O24

H+

(1) 16

5

2

(2) 2

5

16

(3) 2

16

5

(4) 5

16

2

On decreasing temperature, equilibrium shifts in exothermic direction i.e., forward direction. So, high pressure and low temperature favours maximum formation of product. 179. The bond dissociation energies of X2, Y2 and XY are in the ratio of 1 : 0.5 : 1. H for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X2 will be

Answer ( 2 )

Reduction +3

+7

S o l . MnO4– + C2O42– + H+

2+

+4

(1) 200 kJ mol–1

Mn + CO2 + H2O

(2) 100 kJ mol–1

Oxidation  n-factor of MnO4  5

(3) 800 kJ mol–1

n-factor of C2 O24  2

(4) 400 kJ mol–1 Answer ( 3 )

 Ratio of n-factors of MnO4 and C2 O24 is 5 : 2 So, molar ratio in balanced reaction is 2 : 5

S o l . The reaction for fH°(XY) 1 1  XY(g) X2 (g)  Y2 (g)  2 2

 The balanced equation is

2MnO4  5C2 O24  16H  2Mn2   10CO2  8H2 O

Bond energies of X2, Y2 and XY are X,

177. The correction factor ‘a’ to the ideal gas equation corresponds to

respectively

(1) Density of the gas molecules

X X H      X  200 2 4

(2) Volume of the gas molecules



(3) Electric field present between the gas molecules

On solving, we get

(4) Forces of attraction between the gas molecules



Answer ( 4 )

X , X 2

X X   200 2 4

 X = 800 kJ/mole

2   S o l . In real gas equation,  P  an  (V  nb)  nRT  V2   van der Waal’s constant, ‘a’ signifies intermolecular forces of attraction.

180. When initial concentration of the reactant is doubled, the half-life period of a zero order reaction

178. Which one of the following conditions will favour maximum formation of the product in the reaction,

(1) Is halved

(2) Is doubled

(3) Is tripled

(4) Remains unchanged

Answer ( 2 ) S o l . Half life of zero order

X2 (g) r H   X kJ? A2 (g)  B2 (g)

t 1/2 

(1) Low temperature and high pressure (2) Low temperature and low pressure

[A0 ] 2K

t 1/2 will be doubled on doubling the initial concentration.

(3) High temperature and high pressure (4) High temperature and low pressure

35