8/10/2019 Aiot Jee Adv Sol Eng 05-05-2013

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SOLAIOT050513 - 1

PAPER-1

PART-I (Physics)

1. (A)Sol:

Let the elongation of the spring be x.From the N.L.M. of the block

mg - kx = m (g - t )kx = mt

or k(l - l0) = mt

Differentiating w.r.t. time

mdt

dlk !"

#

$%&

'

( ) m'VVk !*

V = 'Vk

m+ = ( )dttg

k

m t

0, ++

=2

tgt

k

m 2*+

2. (C)

Sol. From the graph it can be seen that the max. value - is at060!. . - is max. when the rod Q is tangent on the circle

on which the ring attached to P moves.

from the fig. d = 1 sec 60= 2 m

3. (E)

4. (B)

Sol. w = dzfdyfdxf

f

c

z

e

b

y

d

a

x ,,, **

No w dxf

d

a

x, is positive since fx> 0 and d > a

dyf

e

b

y, is negative since fy> 0 ; e < b

dzf

f

c

z

,is negative f

z< 0; f > c

/ w = + A1+ - A2- A3

HINTS & SOLUTIONS

ALL INDIA OPEN TEST (AIOT)JEE ADVANCED

5. (B)

Sol = +

Mass of sphere if its density was 0 would be 1 kg.

Ycm

=7

3x12x6 *= m

7

15

6. (B)

Sol: g/cmcm/Ag/A aaa *!

= a.2

R*1 = 23

456

7*222

3

4

555

6

7

m

F

I

2

R3xF

2

R

=m

F

I4

FR3 2

*

7. (D)Sol. From energy conservation w.r.t.

an observer in the trolley we haveFrom A to BW

mg+ W

)= 8K.E.

mg2R + maRsin.= 0

a =.sin

g2

8. (C)Sol. F = shear strength x area on which shear stress acts

=46 10x4x10x345 +

= 138000 N= 13800 Kg

9. (B)Sol From the conservation of the energy we have ,

Initial internal energy= Dissociation energy + final internal en-ergy

500242

5

999 = 2000 + 'T2223

999 + 'T2325

999

T=21

8000= 100 T ; T = 4

10. (A)

Sol: , **9!)1,2,4(

)1,3,3(

E dz3ydy2xdx2W = 3

11. (B)

Sol. Reading V2= 2

C

2

R VV * =2

C

2 XRI *

22 44I28 *! = 24I/ I = 2 Amp

DATE : 05-05-2013

COURSE : JP, JF, JR & JCC TARGET : JEE (ADVANCED)-2013

8/10/2019 Aiot Jee Adv Sol Eng 05-05-2013

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SOLAIOT050513 - 2

12. (D)

Sol. V1= ( )2CL

2 XXRI +*

= ( )22 4742 +* = 2 x 5 = 10 volt

13. (D)Sol. Just after closing the switch, the

inductor will not allow currentHence the equivalent circuitwill be as shown

R2i

:!

14. (A)Sol long after closing the switch the p.d.

across the inductor will zero

"#

$%&

'

:!

3

R5i

15. (B)

16. (A)17. (4)

Sol t =pE

I

4

2

4

T ;! ; I = pE

t42

2

;

18. (4)

Sol. Both the slits