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 !"
#
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'
( ) 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
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'
:!
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