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NEET/JEE Class 11 Practice Test_PCM11T005 - 2022-23

Time : 180 Min Phy : Units, Dimensions and Errors in Measurement Marks : 300
Hints and Solutions

01) Ans: C) 8 %
1 08) Ans: C) C
Sol: Kinetic energy, E  mv2 Sol: C quantity has maximum power. So it brings
2
maximum error in P.
E m v
 1 0 0   1 0 0 2  100
E m v
09) Ans: D) L2MT 2 
 2  2  3  8%  
Sol: Torque = force  distance = [ML2 T 2 ]
02) Ans: A) ML2 T 2 
 
Sol: Couple = Force  Arm length 10) Ans: A) dyne  cm4
Couple  [MLT2 ][L]  [ML2 T2 ] Sol: Units of a and PV2 are same and = dyne × cm4

3 1 11) Ans: C) energy.

03) Ans: D)  , ,1
2 2 12) Ans: B) Joule-s
Sol: By adding the dimension of each quantity,
we get T  [ML1T2 ]a [L3M]b [MT2 ]c 13) Ans: A)  16%
By solving, we get a = - 3/2, b = 1/2 and c = 1 Sol: Given, H  I2Rt
H  2I R t 
04) Ans: A) angle   100      100
Sol: As we have dimensions as
H  I R t 
[E]  [ML2 T2 ], [m]  [M], [l]  [ML2T 1] and Error in the measurement of
H  (2  3  4  6)%  16%
[G]  [M1L3 T 2 ]
Then, by putting the dimension of above quantities
14) Ans: B) M0L3 T 1
in the given formula: 4
 3 6 4   Pr ML1T 2L4
El2 [ML2 T 2 ][ML2 T 1 ]2 M L T  Sol: V    M0L3 T 1
   [M0L0 T0 ] 8 nl ML1T 1L
m5 G2 [M5 ][M1L3 T 2 ]2 M3L6 T 4 
 
15) Ans: D) W m2 K 4
05) Ans: C) 0.1
E
F L dyne 105 N Sol: Stefan's law is E  (T4 )   
Sol: Y  .    0.1N / m2 T4
A L cm2 104 m2
Energy Watt
where, E  
06) Ans: D) k1/2a 3/2 / T
Area  Time m2
Watt-m2
Sol: Suppose, n  ka a b Tc  4
 Watt  m2K 4 = W m2 K 4 .
K
where []  [ML3 ], [a]  [L] and [T]  [MT 2 ]
By comparing both the sides, we get 16) Ans: B) 7%
1/2 3/2
1 3 1 k a  R  V I
a ,b and c   Sol:   100    100   100
2 2 2 T  R max V I
5 0.2
07) Ans: D) x  1, y  1, z  1   100   100
100 10
Sol: By putting the dimension of given quantities   (5  2)% = 7%
[ML1T2 ]x [MT3 ]y [LT1 ]z  [MLT]0
By comparing the power of M, L, T in both sides 17) Ans: C) M0L0 T 
x  y  0 .....(i)  
L
x  z  0 .....(ii) Sol:  Time constant
2x  3y  z  0 ....…(iii) R
Only values of x, y, z satisfy the equation (i), (ii)
and (iii) corresponds to (4). 18) Ans: C) N / m2

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Sol: N / m2

19) Ans: B) 0.35 %

Sol: Percentage error in volume is
0.01 0.01 0.01
 100   100   100  0.35 %
15.12 10.15 5.28

20) Ans: C) the error in the measurement of l is

twice that due to the error in the measurement of d
0.5
Sol: d  l  mm  0.005 mm
100
4MLg  Y   l   d 
Y 2
      2 
ld  Y max  l   d 
 l  0.5 /100 2d
 l   0.25  0.02 and d
 
 2 0.5 /100  l d
  0.02 or  2.
0.5 l d

21) Ans: C) 0.94  0.02 cm

Sol: 0.94  0.02 cm

22) Ans: A) solar constant.

Sol: Energy received per unit area per unit time
[ML2 T 2 ]
means the Solar constant i.e.  [M1T 3 ]
[L2 ][T]

23) Ans: C) ( b1  c1  d1  e1 )%

Sol: Given, a  b c /d e
Hence, the maximum error in a is given by
 a  b c
 a  100   .
b
 100  .
c
 100
 max
d e
.  100  .  100
d e
maximum error   b1  c1  d1  e1  %

24) Ans: D) velocity of sound (332 m /sec) .

Velocity of object
Sol: Mach number 
Velocity of sound

25) Ans: B) 3%
Sol: Volume of cylinder V  r2l
Percentage error in volume
V 2r l
 100   100   100
V r l
 0.01 0.1 
 2   100   100 
 2.0 5.0 
Percentage error  (1  2)% = 3%

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NEET/JEE Class 11 Practice Test_PCM11T005 2022-23
Time : 180 Min Chem : Some Basic Concepts in Chemistry Marks : 300
Hints and Solutions

26) Ans: A) Law of multiple proportion 35) Ans: C) 48

27) Ans: C) 1500 mg 36) Ans: B) is not changed

Sol: 1kg  1000g The total nitrogen present in 1 Sol: After a chemical reaction, the total mass of
10 7 products and reactants is not changed.
kg bread  1000   g
100 100 37) Ans: A) Atomic weight
So, N converted to urea everyday
10 7 10 38) Ans: D) 44
 1000    g  0.7 g 28 g N yield
100 100 100 Sol: Here, MW = 2 x V. D = 2 x 22 = 44.
60g urea
60 39) Ans: B) Atomic number
0.7 g N yields  0.7g urea
28
40) Ans: B) 0.05 M
Sol: 0.1 M AgNO3 will react with 0.1 M NaCl to
28) Ans: D) 106 g form 0.1 M NaNO3. But when the volume doubled,
Sol: 1  106 g. 0.1
concentration of NO3   0.05 M .
2
29) Ans: D) M
2 2.5 41) Ans: A) 500
Sol: 2Na 2S2O3  I2  Na 2S4O6  2NaI
Sol: N1  0.5N  10 mg per mL
M M
n  2  0.5  1 E   M 10  103 g
n  factor 1 N2   1000  0.25 N V1  500 ml, V2  ?
40  1
30) Ans: B) Exa N1V1  N2V2 ; 0.5  500  0.25  V2  V2  1000 mL
So, final volume water added
31) Ans: A) N/40  1000  500  500 mL.
Sol: By using, NV  N1V1  N2V2  N3 V3
1 1 42) Ans: A) 49
N  1000  1  5   20   30  5  10  10  25 Sol: The equivalent weight of
2 3
molecular weight
N H3PO4 
N  0.025  2
40
As, mole wt of H3PO4 = 3 + 31 + 64 = 98
32) Ans: A) Gay-Lussac's law of gaseous volumes. 98
 Equivalent weight =  49
Sol: Gay- Lussac's law: The volumes of the reacting 2
gases and those of the gaseous products bear the
simple ratio (also known as the law of gaseous 43) Ans: B) 6.02  1023
volumes).
Sol: Because this is a fact.
33) Ans: C) 40
44) Ans: A) Nal
Sol: Since, 40 gm NaOH contains 16 gm of oxygen
16 45) Ans: C) 0.001 M
 100 gm of NaOH contains  100 = 40%
40 W  gm   1000
oxygen. Sol: Molarity 
molecular wt.  V  ml 
34) Ans: D) energy per unit volume 2.65  1000
  0.1M
Sol: We know, 106  250
Force [MLT 2 ] 10 ml of this solution is diluted to
Pressure    [ML1T 2 ]
Area 2
[L ] 1000 ml M1V1  M2V2  1 0  0 . 1 1 0 0 
0 x

[ML2 T 2 ] 0.1  10
Energy per unit volume   [ML1T 2 ] x  0.001M.
1000
[L3 ]

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46) Ans: B) 24.3

Sol: Since the given sulphate is isomorphous with
ZnSO4.7H2O, its formula would be MSO4.7H2O. Let
'm' is the atomic weight of M, then molecular
weight of MSO4.7H2O = m + 32 + 64 + 64 + 126 =
m + 222
m
 % of M   100  9.87 (given)
m  222
 100 m = 9.87 m + 222 x 9.87
222  9.87
 90.13 m = 222 x 9.87  m   24.3
90.13

47) Ans: B) 32
Sol: SO2  2 H2O  S 2 H2O2
4 0

M 64
 EW    16
4 4
Twice = 16  2  32

48) Ans: A) 31.6 g

M 158
Sol: Here, E    31.6 gm .
5 5

49) Ans: B) 45  6.02  1023 atoms/mole

Sol: 1 mole of sucrose contains 6.023  1023
molecules.
Since 1 molecule of sucrose has 45 atoms,
 6.023  1023 molecule of sucrose has
45  6.023  1023 atoms/mole

50) Ans: C) MnO2

2 4
Sol: In this case, Mn SO4  Mn O2
Here, change of valency  4  2  2
M
 Equivalent weight 
2

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NEET/JEE Class 11 Practice Test_PCM11T005 2022-23
Time : 180 Min Maths : Set Theory, Relations and Functions Marks : 300
Hints and Solutions

51) Ans: B) 3  (b,a)  R and (b,a)  S , [ R and S are

Sol: n(A  B)  n(A)  n(B)  n(A  B)  12  9  4  17 symmetric]
Now, n((A  B)C )  n(U)  n(A  B)  20  17  3  (b,a)  R  S  ( a , b) R S
52) Ans: B) 121  (b,a)  R  S for all (a,b)  R  S .
Sol: f(x)  (1  x)2 and g(x)  x 2  1 So, R  S is symmetric on A.
Transitivity: Suppose a,b,c  A such as
and fog(3)  f[g(3)]  f[9  1]  f[10]
(a,b)  R  S and (b,c)  R  S .
 fog(3)  f[10]  [11]2  121
Then, (a,b)  R  S and (b,c)  R  S
53) Ans: C) 2|x|  {((a,b)  R and (a,b)  S )} and
Sol: f(2x)  2(2x)|2x| 4x  2|x| , {((b,c)  R and (b,c)  S}
f(x)  2x|x| 2x|x|,  {(a,b)  R,(b,c)  R} and {(a,b)  S,(b,c)  S}
f(x)  2x |x|  f(2x)  f(x)  f(x)  (a,c)  R and (a,c)  S
f(2x)  4x  2|x| |x| 2x  2x |x|  2|x| As R and S are transitive,
54) Ans: A) {(3, 3), (3, 1), (5, 2)}  (a, b)  R and (b,c)  R  (a,c)  R
55) Ans: D)  (a, b)  Sand (b,c)  S  (a,c)  S

Sol: x2  16  x  4 and 2x  6  x  3  (a,c)  R  S

There is no value of x which satisfies both the Thus, (a,b)  R  S and
above equations. Hence, A   . (b,c)  R  S  (a,c)  R  S .
 R  S is transitive on A.
Hence, R is an equivalence relation on A.

58) Ans: D) (A  B) - (A  B)
Sol: (A - B)  (B - A) = (A  B) - (A  B)

59) Ans: C) no (a, b), a  b, R

Sol: No (a, b), a  b, R

60) Ans: D)  0,1

Sol: We have, f  x   1  b2  x2  2bx  1
2
b2
 


 1  b2  x 
b 
2 
1 b 
1
1  b2
1
m (b) = minimum value of f  x   is positive
1  b2
57) Ans: C) R  S is an equivalence relation on A and m (b) varies from 1 to 0, so range = (0, 1]
Sol: Given, R and S are relations on set A.
 R  A  A and S  A  A  R  C  A  A 61) Ans: A) {7p  3 : p  Z}
 R  S is also a relation on A. Sol: x  3 ( mod 7)  x  3  7p, (p  z)
Reflexivity: Suppose a be an arbitrary element of A.  x  7p  3, p  z means {7p  3 : p  z)
Then, a  A  (a, a)  R and (a, a)  S
[ R and S are reflexive] 62) Ans: B) 7
 (a, a)  R  S Sol: R is reflexive if it has (1, 1), (2, 2), (3, 3)
As (1, 2)  R, (2, 3)  R ,
 (a,a)  R  S for all a  A .
 R is symmetric if (2, 1), (3, 2)  R
So, R  S is a reflexive relation on A.
Now R  {(1,1), (2, 2), (3, 3), (2,1), (3, 2), (2, 3), (1, 2)}
Symmetry: Suppose a,b  A such as
R will be transitive if (3, 1); (1, 3)  R.
(a,b)  R  S . So R becomes an equivalence relation by adding (1,
Then, (a,b)  R  S  (a,b)  R and (a,b)  S 1) (2, 2) (3, 3) (2, 1) (3,2) (1, 3) (3, 1).
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Hence, the total no. of ordered pairs is 7. 1  xy 1

 a     a x  y  a   
 xy  xy
  a
2   2  
63) Ans: D) A and complement of B are always
non-disjoint 1 1 ax ay 
 a x a y  x y  y  x 
Sol: Since A is not a subset of B, some point of A 2  a a a a 
will not be a point of B so that point will being to
1 x y 1 1 1 y 
Bc . Thus A and complement of B are always  a  a  y   x  y  a  
2  a  a  a 
non-disjoint.
1 x 1  1   a x  a x   a y  a y
 a  x  ay  y   2   
2 
64) Ans: A) Reflexive and transitive only 
a  a   2  2 
Sol: Here (3, 3), (6, 6), (9, 9), (12, 12) are [Reflexive]
and (3, 6), (6, 12), (3, 12) are [Transitive]. Thus,  2f1  x  .f1  y 
reflexive and transitive only.
70) Ans: A) 2f  x  .f  y 
x 1
65) Ans: D) Sol: We have f  x  y   f  x  y 
3x  2
2x  1 1  xy
Sol: Let y  f(x)  y   y  3xy  2x  1  a  a x y  a x y  a x  y 
1  3x 2 

 x
y 1
3y  2
 f 1(y) 
y 1
3y  2
1
  
 a x a y  a  y  a x a y  a  y 
2  
 f 1(x) 
x 1 
1 x
2
  
a  a x a y  a  y  2f  x  f  y 
3x  2

66) Ans: C) Y 71) Ans: C) 1,7 / 3

Sol: As 4n  3n  1  (3  1)n  3n  1 x2  x  2
n n n1 n n 2 n n Sol: Consider y  f  x   ,x R
 3  C1 3  C2 3  .....  Cn13  Cn  3n  1 x2  x  1
x2  x  2
n 2 n 3 n n y
 C2 3  C3 .3  ...  Cn 3 , x2  x  1
(n C0  nCn , n
C1 = n
Cn-1 etc) 1
y 1 2 i.e.y  1 ...  i 
 9[ C2  C3 (3)  ..... n Cn 3n1 ]
n n x  x 1
 yx 2  yx  y  x 2  x  2
 4n  3n  1 is a multiple of 9 for n  2 .
For n  1, 4n  3n  1 = 4  3  1  0 ,  x 2  y  1  x y  1  y  2  0 , x R

For n  2, 4n  3n  1 = 16  6  1  9 As, x is real, D  0

  y  1  4  y  1 y  2  0
2
 4n  3n  1 is a multiple of 9 for all n  N
Here X contains elements which are not multiples   y  1  y  1
  4 y  2  0
of 9 and clearly Y contains all multiples of 9.
 X  Y i.e. X  Y  Y 7
  y  1 3y  7   0 1  y  . .. i i
3
67) Detail solution given at the end
 7
So, from Eqs. (i) and (ii), Range  1, 
68) Ans: D) NNP  I  0  3
Sol: N NP   N  NP    NP  N 
 1, 2, .. .  1  I
.. . 2, 0

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73) Ans: B) [6]  [13]

1
Sol: 8x  6  14P, (x  Z)  x  [14P  6] , (x  Z)
8
1
x= (7P  3)  x = 6, 13, 20, 27, 34, 41, 48, .….
4
 Solution set ={6, 20, 34, 48,...}  {13, 27, 41, ....}
= [6]  [13], where [6], [13] are equivalence classes
of 6 and 13 respectively.

75) Ans: D) 6, 3
Sol: As 2m  2n  56  8  7  23  7
 2n (2mn 1)  23  7 ,  n  3
And 2mn  8  23
 m n  3  m3  3  m  6;
m  6, n  3

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