Sec: ISR.

IIT_*COSC(MODEL-A) WAT-42 Date: 21-04-24

Time: 3 Hrs 2020_P1 Max. Marks: 180
KEY SHEET
MATHEMATICS
1 B 2 D 3 C 4 C 5 B
6 D 7 AB 8 ABCD 9 BC 10 ABC
11 ABD 12 AC 13 4 14 8 15 3
16 3 17 3 18 0

PHYSICS
19 B 20 A 21 A 22 A 23 D

24 C 25 AC 26 AC 27 BD 28 ABC

29 D 30 AB 31 12 32 20 33 8
20.70
34 22.5 35 TO 36 2
20.90

CHEMISTRY
37 C 38 A 39 A 40 D 41 A

42 B 43 BD 44 ABD 45 BD 46 AD

47 ABCD 48 ABC 49 5 50 1 51 4

52 4 53 4 54 7
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SOLUTIONS
MATHEMATICS
1.

2.

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3.

4.

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5.

6.

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7.

8.

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9.

10.

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11.

12.

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13.

14.

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15.

16.

17.

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18.

PHYSICS
19. Conceptual

20. R0    3 
A
When wire is stretched to twice its original length, then
R '  n 2 R0
 R '   2   3  12
2

We have to find resistance between A and B


AOB  600 
3
 12    
 R1   a 
 2 a   3 
 R1  2

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 12   5 
And R2   a 
 2 a   3 
 R2  10
1 1 1 6
   
Req 2 10 10
5
 Req  
3
Hence, the correct answer is (A)
22. The resistance R of fuse wire of length  , cross sectional area A and resistivity  is

R
A
Here   22  10 cm  22  108 m
6

A  0.2 mm 2  0.2 106 m 2  2  107 m2

Since, Q  I 2 Rt (in joule)
  
Q  I 2  t ......(1)
 A
Again if m is mass of wire, d is its density, c its specific heat and T , the rise of temperature, then
Q  mcT
 Q  Volume  density   c   327  20 
 Q   Adc  307  ....(2)
From (1) and (2), we get
 
 Ad  c  307  I 2  t
 AJ 

t 
 A cd   307  J
2

....(3)
I 2
Here density d  11.34 gcm 3  11.34  103 kgm 3 ,
Current I  20 A,   22  106 cm  22  108 m
And specific heat c  0.032 calg 1  0 C 
1

 c   0.032  4200  Jkg 1  0 C  ,

1

A  0.2 mm 2  2  10 7 m 2
Substituting given values in (3), we get

t
 2 10  7 2
 4.2  103  11.34 103  0.032  307
 20 
2
 22  108
 t  0.095 s  95 103 s  95ms
23. Conceptual
24. Conceptual
25. Consider a cylindrical element of radius r, thickness dr. If dR is the resistance of this element then
 r
dR 
2 r dr
Total resistance of the cylinder is given by
2 3
a
1 1
Rtotal  dR   0
  r dr

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1 2  a  4
  
Rtotal    4 

2 
 Rtotal 
 a4
Rtotal 2
 
  a 2 
2

2
R
A2
V
Since E  (in magnitude)

R 
 E  I  total 
  
2 I
E
A2
Hence, (A) and (C) are correct.
26. Consider the ‘n’ sided polygon, shown below for the points that face each other like (A, D), (B, E)
and (C, F), we have
RMAX  RAD  RBE  RCF
 R  n  R
Resistance for the upper half between A and D is     and for the lower half between A
 n  2  2
R
and D is also .
2

These both are in parallel across A and D, hence

 R  R 
   R
2 2
RMAX  RAD  RBE  RCF     
R R 4

2 2
We shall get the minimum value of resistance between the adjacent points of the polygon i.e., AB,
BC, CD, DE, EF or FA. Between two adjacent points we have two resistors connected in parallel one
R R
having resistance and other having resistance  n  1 .
n n
R R
   n  1  
n
So, RMIN  RADJACENT     n    n 1  R
R R  2 
POINTS
  n  1  n 
n n
Hence, (A) and (C) are correct.
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27. Let R1 and R2 be the resistance of the two heaters. Let, initially, the heat produced be H, then
V 2  V 2 
H    t A    tB ......(1)
 R1   R2 
When used in series,
 V2 
H  T ......(2)
 R1  R2 
When used in parallel
V 2 V 2 
H   t .......(3)
 R1 R2 
Using (1), (2) and (3) we get the desired results
Hence, (B) and (D) are correct.
500 1
28. Speed of car, C  60 kmh 1  ms
3

At a point S, between P and Q

   C cos      C cos  
vM'  vM   and vN  vN 
'

     
  cos  
 v   v N  v M   1  C
  
Similarly, between Q and R, we have
  cos  
v   vN  vM  1  C
  
d  v    d 
    vN  vM  C sin   
dt   dt 
At P and R, the car is very far from M and N, So   00 . Hence, slope of graph is zero.
d  d   
At Q, we have   900 , so sin  is maximum. Also value of is maximum as    
dt  dt Q rmin 10
constant
Hence, slope is maximum at Q.
  
At P, vP  v   vN  vM  1  C 
  
  00 

  
At R, vR  v   vN  vM  1  C 
  
  00 

At Q, vQ  v   N   M   900 
From these equations, we get
vP  vR  2vQ
Hence, (A), (B) and (C) are correct.

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
29. Minimum resonance length is 1 
4
   41
   f    244  41 
Where 1   0.350  0.005  m
So,  lies between 336.7 ms 1 to 346.5 ms 1
 RT
Now,   , here M is molar mass of gas in gram
M  103
10
   100 RT 
M
For monoatomic gas i.e., for neon and argon,   1.67
10
   640 
M
For diatomic gas i.e., for oxygen and nitrogen,   1.4
10
   590 
M
7 17
 Ne  640   448 ms 1 ,  Ar  640   340 ms 1
10 32
9 3
O2  590   331.8 ms 1 ,  N 2  590   354 ms 1
16 5
Only possible answer is Argon.
Hence, the correct answer is (D)
30. Conceptual
 30 Rx 
 30  R   30 Rx 
31. V1  V30    x
V   V
 30 Rx  20   50 Rx  600 
 30  R 
 x 

Now, power generated in Rx is

V12 900 RxV 2
P 
Rx  50 Rx  600 2
For P to be constant, we have
dP
0
dRx


 50 Rx  600 
2
 900V   1800  50   R V   50 R
2
x
2
x  600 
0
 50 Rx  600 
4

 50 Rx  600  100 Rx  0
 Rx  12 
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32. Conceptual
33.
34. Conceptual
35. Conceptual
36. Conceptual
CHEMISTRY
37.
O O
OH

H O CH3 OH
LiA/H4

O OH
OH ONa

O
Br .
Na .

38.
OH OH OH

NO2

HNO3/H2SO4 +

(B)
(A)
NO 2
39.
.
.
. O
. O
t-BuOK
.
Cl
+
H
.
.

O
O
.
.
Intra

+
40.
41. n-butane(iv)<Methoxy ethane(v)<Propanal(i)<Acetone(iii)<Propan-1-o1
+
H
CH3-CH=CH . OH

42.
Br
.
Br CH 3CH 2-CH.
.
CH 3-CH2-CH OH OH

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43. B (Friedal-Crafts) and
D (Reimer-Tieman)
44. A,B,D
KCN
CH 3CH 2Cl CH 3CH 3CN
OH
O
KCN
H3C CH3
H3 C CH3
CN
HO
O

H3 C H CN
KCN

H2 C
45. B,C
OH

+
18 H 18
O. . +
H2 O OH

46. AD
47. A)
O O

A) DIBAL-H
+C2H5OH

H3C OC 2H5 H3C H

O

O3
B) CH3-CH=CH-CH3
Zn/H2O H3 C2 H
O O
O

+
C) +
(CH3COO)2Ca+(HCOO)2Ca H H H 3C CH3
H3C H
O

Stephen
D) CH3- C N
H3C H
48.
O .
O

.
1) (CH3)2CuLi
.
A) +
2) H3 O

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O .

O
CH3
.
AlCl3
B) Cl
+ H C
3

HO
. O
O .
PCC
C) C 6H5MgBr + Ph
+ . Ph .
. H H3O
O OH
.
+ PCC
D) H3O No Reaction
(CH3)2CH MgBr + Ph CH3 Ph .
.
49.
O
OH

CH3
O HO
OH
P
Q)
OH

OH
5
P=2, ; Q=3
50. P-nitro benzaldehyde,
51. Formaldehyde, cyclopropanone
Ninhydrin, hexathroacetone
52. 4
53.
OH
OH ;
;
; .
Ph OH

OH
; OH
. .

54.
Cn H 2 n O  12n  2n  16  100
14n  84
n6
O
.
.
.
. . . .
.

O
O OH

. .
.

.
.

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