AIEEE – 2012 PAPER & SOLUTIONS PAPER – 1 : PHYSICS, CHEMISTRY& MATHEMATICS Do not open this Test Booklet until you are asked to do so. Read carefully the Instructions on the Back Cover of this Test Booklet

Important Instructions: 1. Immediately fill in the particular on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited 2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particular carefully. 3. The Test Booklet is of 3 hours duration 4. The Test Booklet consists of 90 questions. The maximum marks are 360. 5.

There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (Four) marks for each correct response.

6.

Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each question. ¼ (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in 6the answer sheet.

7.

There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.

8.

Use Blue/Black Point Pen only for writing particularly/marking response on Side–1 and Side – 2 of the Answer sheet. Use of Pencil is strictly prohibited.

9.

No candidate is allowed o carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc., except the Admit card inside the examination hall/room.

10. Rough work is to be done on the same provided for this purpose in the Test Booklet only. This space is given at the bottom of each page and in 3 pages (Page 21 – 23) at the end of the booklet. 11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take this Test Booklet with them. 12. The CODE for this booklet is A. Make sure that CODE Printed on Side–2 of the Answer Sheet is the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the Invigilator for replacement of both the Teat Booklet and the answer sheet. 13. Do not fold or make any stray marks the Answer Sheet.

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PART – 1 – PHYSICS 1.

1.

2.

2.

Two electric bulbs marked 25W-220V and 100W-220V are connected in series to a 440V supply. Which of the bulbs will fuse? (1) 100W (2) 25W (3) Neither (4) both (2) The 25 W bulb has a resistance 4 times that of 100 W bulb. When they are connected in series the voltage that will appear across 2 bulbs is 4 V25   440  352 5 1 V   440  88 100 5 Since 25 W bulb cannot with stand more than 220 volts and hence will fuse. A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be: (1) 10 m (2) 10 2 m (3) 20 m (4) 20 2 m (3) Let u be the speed with which the boy can throw the ball. Then for maximum height  = 90° u2 2g Where as for maximum horizontal range  = 45°



And 3.

h max 

R max 

u2 g

=

2h

max

 20 m

Truth table for system of four NAND gates as shown in figure is: A

Y

B

(1)

(3) 2

A 0

B 0

Y 0

A 0

B 0

Y 1

0

1

0

0

1

1

1

0

1

1

0

0

1

1

1

1

1

0

A 0

B 0

Y 1

A 0

B 0

Y 0

0

1

0

0

1

1

1

0

0

1

0

1

1

1

1

1

1

0

(2)

(4)

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3.

Option (4) is correct Y  (A. A. B) . (B. A. B)  (A. A. B)  (B. A. B)

A . (A  B)  B . (A  B)  A . A  A . B  B . A  B . B Y  0  A . B  B . A  0

4.

This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements. Statement 1: Davisson – Germer experiment established the wave nature of electrons. Statement 2: If electrons have wave nature, they can interfere and show diffraction. (1) (2)

4.

5.

Statement 1 is true, Statement 2 is false Statement 1is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1 (3) Statement 1is true, Statement 2 is true, Statement 2 is not correct explanation for Statement 1 (4) Statement 1is false, Statement 2 is true Davisson – Germer experiment established the wave nature of electrons using diffraction of electrons by crystals. In Young’s double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If Im be the maximum intensity, the resultant intensity I when they interfere at phase difference  is given by

Im  Im  2  2  (2) 1  2cos  1  4cos  3  2 5  2 Im  Im 2  (3) (4) (4  5 cos ) 1  8cos  9  2 9 If A and 2A are the amplitude of the two beams then amplitude at the maximum is 3A therefore I or A 2  m I max  9A 2 9 Intensity at any phase difference  is given by 5I 4I m  m cos  I = A 2  4A 2  2A . 2 A cos  = 5A 2  4A 2 cos  = 9 9 Im Im  = [ cos   2 cos  ( / 2)  1] [5  4 cos ] = [1  8cos 2 ] 9 9 2 (1)

5.

6.

6.

If a simple pendulum has significant amplitude (up to a factor of 1/e of original) only in the period between t = 0s to t =  s, then  may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with ‘b’ as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds: 1 (1) b (2) b 2 0.693 (3) (4) b b (3) Bansal Tutorials Pvt. Ltd. 65 Kalu Sarai, Sarvpriya Vihar, Hauz Khas, New Delhi – 16, Ph: 41828700

3

m

d2x dt

m

  kx  b

2

d2 x

 b

2

dx dt

dx  kx  0 here b is demping coefficient dt

dt This has solution of type x  et substituting this m 2  b  k  0 b 2  4mk 2m On solving for x, we get  

b 



b

x  e 2m

t

a cos (1t  )

1 

02   2 where 0 

  

b 2

So, average life, 7.

k m

2 b

This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements. If two springs S1 and S2 of force constants k1 and k 2 , respectively, are stretched by the same force, it is found that more work is done on spring S1 than on spring S2 . Statement 1: If stretched by the same amount, work done on S1 , will be more than that on S 2 . Statement 2: k1  k 2 . (1) (2) (3) (4)

7.

Statement 1 is true, Statement 2 is false Statement 1is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1 Statement 1is true, Statement 2 is true, Statement 2 is not correct explanation for Statement 1 Statement 1is false, Statement 2 is true

(4) Work done W when stretching a spring is given by 1 2 1 F2 kx  2 2 k As per the information given if the force applied is same then work in stretching S is more than that 1 for S implies k  K 2 1 2 Also we know that W 

1 2 kx if stretched by the same amount then work done on S is less than 1 2

that on S 2 .  Statement 1 is wrong and 2 is right Option (4) is correct. 4

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8.

8.

An object 2.4m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film? (1) 2.4 m (2) 3.2 m (3) 5.6 m (4) 7.2 m Re al thickness Apparent thickness =  And hence apparent shift = Real thickness – apparent thickness  1 = Re al thickness 1     

1 cm 3 1 35 Now the new position of the image = 12   cm 3 3 Since the focal length of lens is constant we get 3 1 1 1 =    35 u 12 240  u = –5.6 m 1   1 1   1 .5  

=

9.

=

In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as a function of distance from the centre. The graph which would correspond to the above will be E

(1)

E

r

R

(2)

E

(3) 9.

R

r

R

r

E

R

r

(4)

The electric field inside the sphere is given by

r which is linearly proportional to r. Whereas 30

outside the sphere it obey the inverse square law. Option (2) is correct.

10.

10.

A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating, it is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to (1) induction of electrical charge on the plate (2) shielding of magnetic lines of force as aluminium is a paramagnetic material. (3) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping. (4) development of air current when the plate is placed. The EMI uses Eddy current and hence causes a loss of Energy and hence it stops faster. Thus option (3) is correct. Bansal Tutorials Pvt. Ltd. 65 Kalu Sarai, Sarvpriya Vihar, Hauz Khas, New Delhi – 16, Ph: 41828700

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11.

11.

12.

A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading: 58.5 degree Vernier scale reading : 09 divisions Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data (1) 58.77 degree (2) 58.65 degree (3) 59 degree (4) 58.59 degree 0.5 1 Least count =  deg ree 30 60 9 The angle of the prism = 58.5  60 = 58.65 degree A diatomic molecule is made of two masses m1 and m 2 which are separated by a distance r. If we calculate its rotational energy by applying Bohr’s rule of angular momentum quantization, its energy will be given by (n is an integer) n 2h 2

(1)

2(m1  m 2 )r 2 (m1  m 2 )n 2 h 2

(3) 12.

(2) (4)

2m1m 2 r 2

2n 2 h 2 (m1  m 2 )r 2 (m1  m 2 )2 n 2 h 2

m1r1  m 2 r2 r1  r2  r m2 r  r1  m1  m 2 m1r r2  m1  m 2 m1r 1  E  = (m1r12  m2 r22 ) . 2 m1  m2 2

2m12 m 22 r 2

….(1)

h  I 2 nh   2I mvr 

 E 

1 n 2h 2 n 2h 2 1 I.  2 2 2 2 2 4 I 8 (m1r1  m 2 r22 )

n2h2 8 2

1 m1

m 22 r02

 m2

m12 r 2 (m1  m 2 )2

(m1  m 2 )2 (m1  m 2 )2 (m1  m 2 )n 2 h 2 n2h2   82 r 2 m1m 2 (m1  m2 ) 82 r 2 m1m 2

Note: Here choices should have  (=

h ) instead of h and therefore a factor of 42 is missing 2

in the denominator. 6

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Potential difference V in volts

25 20 15 10 5 0

50

100

13.

150

200

250

300

Time in T seconds

The figure shows an experimental plot for discharging of a capacitor in an R-C- circuit. The time constant  of this circuit lies between (1) 0 and 50 sec (2) 50 sec and 100 sec (3) 100 sec and 150 sec (4) 150 sec and 200 sec 13.

During discharging of a capacitor, voltage V across the capacitor is given by –

t

V  Vo e  t

– V Or e  Vo V t  ln o  V 

t V ln o V Here at t = 200s, V = 5V and Vo = 25V 



or  

200 200  125s 2.303  ln 5 2.303  [1 – 0.3010]



lies between 100 sec and 150 sec

Hence option (3) is correct. 14.

A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F(t)  F0 e  bt in the x direction. Its speed v(t) is depicted by which of the following curves F0 b m

F0 mb

v(t)

v(t)

(1)

(3)

t

(2)

t

F0 mb

F0 mb

v(t)

v(t) t

(4)

t

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7

14.

F dv  0  e  bt dt m F0 bt dv  e dt m  bt F 0  e V =  C m b F = – 0 e bt  C mb At t = 0 V = 0 F 0 = – 0  C mb F 0 C = mb F 0 (1  e  bt ) V = mb  Choice (2) is correct.





15.

Two cars of masses m1 and m 2 are moving in circles of radii r1 and r2 , respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is (1) (2) r1 : r2 m1 : m 2 (3) 1:1 (4) m1r1 : m 2 r2

15.

Centripetal ace = r 2 Since their time periods are equal, their  are also equal a1 r   1 a2 r2 Hence (2) is correct.

16.

16.

A radar has a power of 1 kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth = 6.4  106 m ) is (1) 16 km (2) 40 km (3) 64 km (4) 80 km (4) Let the height of the radar from the center of the earth is r  r=R+h x  (R  h) 2 – R 2  h 2  2hR

Since h