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SOLUTIONSTO
TEST- II
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(Solutions)PHYSICS
SECTION ISINGLE CORRECT CHOICE TYPE
1. VOLUME OF BALL
=m
V
ACCELERATION OF BALL INSIDE THE LIQUID
mm
Fa
weightupthrustnet ==
g
v=0
a
H
h
ORm
mggm
a
=))(3(
= 2G(UPWARDS)
VELOCITY OF BALL WHILE REACHING AT SURFACE
ghahv 42 ==
THE BALL WILL JUMP TO A HEIGHT
hg
gh
g
vH 2
2
4
2
2
===
(C)
2. Maximum heat is released when block reaches till pointA . glv = 220 , glv = 20 (b)
3. Loss in potential energy = gain in kinetic energy
22
22
123
= MLQL ;
ML
Q212=
( ) QLjkQLkji
Ep 3
330
00
===rrr
unit vector along the direction of
is
2
jk
( )jkML
Q 26 =r
(b)
4. BY CONSERVATION OF ENERGY
FOR 0 0 ga =
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2
max
2
2
mg
E
g
vt ==
TIME PERIOD IS 21 2tT + (C)
5. gRMMRTdt 2=
( )2
22
=MR
RdtT R
g
2=
(b)
6. mumu
dtN +=2
, mvmvdtN = ' ,
ghm
2
2
3 = )'( vvm , ghvv 23.0'=
(c)
SECTION II
One or more than one type
7. LET AT ANY INSTANT, THE VELOCITY IS vr
.
jvivv yx +=
rAND ( )BvqF
rrr
=
ixvjxva yx =
rALSO 2
22vvv yx =+
xvdy
dvv
dt
dva x
y
y
y
y ===
yyxx dvvdvv = (I)
xvdy
dvv xyy = (II)
xvdx
dvv
yx
x= (III)
FROM (I), (II) & (III)2
2x
vy = AND4
422 x
vvx =
ALSO,2
vvx =
2
3va = AND B= 0
WHEN VX= 0, X= (2V)
1/2
(B, C)
8. K.E. 2222
8
3
422
1mv
mRmR =
+=
ANGULAR MOMENTUM =
+= 2
22
4
3
42mR
mRmR
(A, C)9. (A, B, C, D)
10. NET = 0
CROSSING = E2RL
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R
lE
++= sin2
cossin2
32
1 AA
AEcrossing
++= sin
2cos2sin
2
22R
RlR
E
Er
A2
A1
A3
( )+= cos2sin2 RlREcrossing
(A3/2)sin
A2
A1
A2cos
(A1/2)sin
A3
(A, B, C)SECTION III
Reasoning type
11. (D)12. (A)13. (A)
14. (A)
SECTION IV
Comprehension Type
PASSAGE I
AA
Ndt
dN)3( = ;30
t
A eNN=
AB
BNN
dt
dN +=1
12
2
11 =A
B
N
N
ALSO,
AB
BNN
dt
dN+= 2
2
2 22 =A
B
N
N
15. (B)
16. (C)17. (D)
PASSAGE-II
18. As the process in cylinderA is adiabatic
PV
= constant
Initial pressure inA, P0
= 105
N/m2
Initial volume inA , V0 = 10-2
m2
Final volume inA = V0/4 = 25 104
Let final pressure isA = P1
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5.1
01
5.1
004
=
VPVP ;
018PP =
For equilibrium of pistons (let pressure in cylinderB finally = P2)
P1SP2(2S)
212PP =
024PP =
For gas in cylinderA, 000 RTM
mVP = (i)
For gas in cylinderB, 002'
2
5RT
M
mVP = (ii)
kg110' == mm (d)
19.15.1
48
1
020
01122
=
=VP
VP
r
VPVPu
==00
2 VPu 2000 J
(a)
20. N800080
== SPF (C)
PASSAGE-III
APPLYING KIRCHOFFS LAW IN A LOOP Qxyppp'
1243 1 =++ rIRI (1)
APPLYING KIRCHOFFS LAW IN A LOOP QQxypQ'
04
)( 11 =+R
IIrI (2)
SOLVING (1) AND (2) WE GET 1244
311 =+ rIRI
SINCE Rr>> 1241
=rI ; VrI 31
=
V
Q
P
yx
Q
IP
II1I1
I1
I
SIMILARLY IN THE SECOND CASE WHEN VOLTMETER IS CONNECTED AS IN FIGURE (2)READING OF VOLTMETER IS 9V.
21. (c)
22. (b)
23. (b)
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(Solutions)CHEMISTRY
SECTION ISINGLE CORRECT CHOICE TYPE
24. (B) INITIALLY THE PH WILL INCREASE SLOWLY DUE TO NEUTRALISATION OF WEAK ACID
BY NAOH. AFTER THE END POINT IS REACHED, PH WILL INCREASE RAPIDLY DUE TOADDITION OF NAOH.
25. (b)
26. (c)
27. (c) AH3
12
Ka;AHH + + AH2 2
2 Ka;HAH + + 2HA 3
3 Ka;AH + +
At second equivalence point, the only species present in appreciable concentration is2HA .
102
128
2
pKpKpH 32
aa=
+=
+=
1010H
+ =
and[ ]
[ ][ ] 7
321
3
3
310
KaKaKa
H
A
AH +
=
=
28. (b)1
1d
P3U =
( )12RMS PPd
3U =
( )20030075.03 =
2001004 == m/s
29. (c) (A) is Lunar caustic i.e. AgNO3
2AgNO3(A)
2NO2(B) + O2(C) + 2Ag(D)
AgNO3(A) + KCN AgCN + KNO3(white ppt.)
AgCN + KCN K[Ag(CN)2](Soluble complex
Potassium argentocyanide)
SECTION II
More than one type
30. (a, c, d)
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O
S
O
HO OS
O
OH,
OO
O
S
O
HO OS
O
OH,
O
O
S
O
HO O OH, CrO
OO O
O
31. (a, c, d)
C + O2 CO + CO2t = 0 0.1
t = t 0 x (0.1 x)
5CO + I2O5 5CO2 + I2t = 0 x
t = t 0 x x/5
Moles of I2 liberated =2
1moles of hypo consumed =
2
1 120 103 0.1 = 60 104
So,x = 5 60 104 = 0.03
Percentage of C converted to CO is 30% .
Moles of CO2 left = [0.1 x]1st reaction + [x]IInd reaction
= [0.1 0.03] + [0.03] = 0.1
CO2 + H2O H2CO30.1 0.1
H2CO3 + Ca(OH)2 CaCO3 + 2H2O0.1 0.1
32. (a, b, c)
Since, product is an alcohol, so compound (A) is a 10 aliphatic amine.
33. (c, d)
Le Chateliers principle is not quantitative. If both the stresses are caused in the same direction, the shift is
determinable. If both the stresses are caused in the opposite direction, the shift is undeterminable.
SECTION III
Reasoning type
34. (c)
35. (a) The conductivity due to free electrons is more than due to charged ions.
36. (b)
37. (a)0
11
1
10 epn + (-particle). -particle emission occurs when n/p ratio is higher than 1.6.
SECTION IV
Comprehension Type
PASSAGE I
THE REQUIRED REACTIONS OCCURING AT THE ELECTRODES ARE
CATHODE: 2H+
+ 2E
H2
CU2+
+ 2E
CU
ANODE: 2OH
O2 + H2O + 2E
38. (B) THE MOLAR VOLUME OF A GAS AT 1 ATM AND 273 K IS 22.4 L/MOLE. SINCE, THE
VOLUME OF H2 COLLECTED AT 0.1 ATM AND 273 K IS 2.024 CM3, SO THE VOLUME OF H2
COLLECTED AT 1 ATM AND 273 K (STP CONDITIONS) WOULD HAVE BEEN 0.2024 CM3.
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45. (c)
( )[ ] ( )pptYellowcolourBlue
NH4K2CNCuKCN2NHCu 322
43+++ ++
( ) ( )pptWhite
CNCuCN2CNCu2 2decomposes
2 +
( )[ ]Colourless
K3CNCuKCN3CuCN 34+ ++
1Cu + has a d10 configuration, so it is tetrahedral in shape with sp3 hybridization.
46. (d)
Gravimetric estimation of Cu 2+ would be possible only when the given procedure gives an insoluble compound
on treatment with a given reagent.
( )pptBlack
K2SCNCuKSCN2Cu 22 ++ ++
( ) ( )pptWhite
SCNCuSCN2SCNCu2 2decomposes
2 +
With other reagents, either there is no reaction or even if it reacts, then the product is a soluble compound.
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(Solutions)MATHEMATICS
SECTION ISINGLE CORRECT CHOICE TYPE
47. (C)
48. (B)
49. (A)
50. (A) LET THREE NUMBERS ARE 2,2 AND 2
( )5
37
3
.......321=
++++++
n
n
( )
( )35
37
2
1
+
=++ nnn
(NMUST BE IN THE FORM OF ( )35 +N ) AND ++ MUST BE NATURAL NUMBER LESSTHAN 33 n THUS POSSIBLE VALUES OF NARE 8 AND 13.
51. (D) ( ) dxkkxx
xkkkI
++=
0
22
ln LET ktx =
( )
++=
0
21
lndt
tt
ktkkI
( ) ( ) ( )ktt
dt
IkkI ln11 20
++=
( )22
0
2
3
2
1
ln
+
+
=
t
dtk
( )6
2
3
2ln
3
=
k
23ek=
52. (A)1
2. =n
n nS NOW FOR ANY VALUE OF N,
nS WILL END IN TWO ZEROS IF THE PRODUCT.
( )( )times1.....2.2.22. 1 = nnn n CONTAINS EXACTLY TWO 5 AS FACTORS.
THIS WILL HAPPEN FOR N= 25, 50, 75, 100
So required probability = 2511004 = SECTION II
One or more than one type
53. (B), (C)
Z= 0 IS ONE OF THE PLANE PERPENDICULAR TO 1,0,0 ===+ xyxyx
NOW THE TRIANGLE FORMED IS TAKE ONE LET ON Z-AXIS I.E., ( )k,0,0 , THEN2222 211 =++ k 2=k
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NOW THE DIRECTION RATIO OF PLANE HAVING POINT ( ( )0,1,1,2,0,0 AND ( )0,1,1 IS( )1,0,2 AND THE DIRECTION RATIO OF PLANE HAVING POINT ( ) ( )0,1,1,2,0,0 AND ( )0,1,1
IS ( )1,0,2 .
(1, 1, 0)
(1, 1, 0)
(0, 0, 0)
B
A
C
54. (A), (B), (C)
( ) ( ) ( ) ( ) ( )( )yxyxyyxfyxyxfyx +=++ 2 LET vyxuyx =+= ;
( ) ( ) ( )uvuvuvfvuf = 2 ( ) ( )uv
u
uf
v
vf=
( ) ( )
=
u
u
ufv
v
vf=CONSTANT
LET( )
= xx
xf
( ) ( 2xxxf += ( ) 21 =f
21 =+ 1= ( ) xxxf += 2
55. (B) LET X [0, 2]. SINCE ( )xf SATISFIES ALL THE CONDITIONS OF L.M.V. THEOREM ON [0, 2].
IT ALSO SATISFIES ON [0, X] [0, 2]
( ) ( )
( )10
0xf
x
fxf =
WHERE 20 1
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SECTION III
Reasoning type
57. (A) STATEMENT-1: ( ) ( ) ( ) abcbacacbcba
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WHEN GAME ENDS WITH TWO HEADS THE PROBABILITY IS .......9
4
9
4
9
432
+
+
+
THEN FOR MINIMUM NUMBER OF THROWS, PROBABILITY IS
( )
9
5
941
1
9
4
94
=
63. (D) GAME END WITH EITHER 2 HEADS OR 2 TAILS THEN IT IS 1.
PASSAGE-II
E D
C
B
A
O
SOL.:Q bOEaODbOCbaOBaOArrrrrr
===+== ,,,,
bABr
= AND aBCr
=
|||| BCAB = |||| barr
=
NOW, OCAB = AND CBOA =
AND CBOAOCAB ===
OABCIS A RHOMBUS
AND ODOA, AND OCEO, ARE COLLINEAR
OAOD =[ AND ]OCOE =
53== ABCAOC AND 52== OCBOAB
I.E., ANGLE BETWEEN ar
AND br
IS5
3
NOW bbaABDCrrr
==
( ) ( ) ( )2222 .2 bbabarrrrr
=+ 5
2cos2
=
64. (B)( )
5
2cos211
1 +==
=
a
a
BC
ADr
r
65. (A)
66. (D)
PASSAGE-III
67. (A)68. (A)
69. (C)
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