RT-22

SOLUTIONS
MATHEMATICS
1. No. of words formed when word starts from B= 5!
No. of words formed when word starts from C = 5!
No. of words formed when word starts from L = 5!
No. of words formed when word starts from PB = 4!
No. of words formed when word starts from PC = 4!
No. of words formed when word starts from PI = 4!
No. of words formed when word starts from PL = 4!
No. of words formed when word starts from PUBC = 2!
No. of words formed when word starts from PUBI = 2!
No. of words formed when word starts from PUBLC = 1
Now the word comes is PUBLIC = 1
 Serial number of word PUBLIC= 4x 5!+ 4x 4! + 2 x 2! + 1+1 = 582
2. Case I : All the three digits are same i.e., 111,333,555,888. All the numbers are
divisible by 3.
Case II : Two digits are same
We can see that when 2 digits are same then numbers having digits 5,5,8 or 8,8,5 will
be divisible by 3 as their sum is divisible by 3.
Now for each number, We have 3!/2!=3 arrangements.
 In this case we have 6 numbers divisible by 3.
Case III : All digits are distinct.
i.e., (1,3,5), (1,3,8)
So, number formed = 2 x 3! =12
Hence, total 3-digit number formed by digit 1,3,5,8 and divisible by 3 is 4+6+12 =22
3. Given, y+z = 5 …….(i)
1 1 5
And   ,yz …….(ii)
y z 6
Sloving (i) and (ii), we get y=3, z=2
 n  2 x.33.52  (2.2.2....)(3.3.3)(5.5)
 Number of odd divisiors = (3+1)(2+1) = 4 x 3 =12
4. Here, f(k) = multiple of 3 (3,6,9,12,15,18)
For k= 4,8,12,16,20, there are 6x5x4x3x2=6! Ways For rest numbers, there are 15!
Ways  Total ways = 6!(15!)
5. Required number of ways = 7! – 6! X 2 = 6! (7-2) = 5 x 6!
6. Since, nC4 , nC5 and nC6 are in A.P.  2  nC5  nC4  nC6
n n n 2 1 1
 2     
5 n5 4 n4 6 n6 5(n  5) (n  4)( n  5) 6  5
2 (n  4)(n  5)
 (n  4)  1   12( n  4)  30  n 2  9n  20
5 65
 21n  n  98  0  n 2  21n  98  0
2
 (n  14)( n  7)  0  n  7,14
But only n=14 lies in the options.
7. x1  x2  x3  x4  10
The number of positive integral solution is 6  4  1C41  9C3
It is the same as the number of ways of choosing any 3 places drom 9 different
places.
8. There are three ways to make four letter words from the letters of the word
BARRACK.
(i). When two letters are same i.e., A,A,R,R
2C2 .4!
So, number of words  6
2!2!
(ii) When one letter is same and other is different
2C1  4C2  4!
so number of words   144
2!
(iii). When all letters are different
So, number of words  5C4  4!  120
 Total number of words= 6+144+120=270
9. We can do case work on number of ladies and men to be invited.
X, Y can satisfy the condition in 4 ways
(i) X invites 3 ladies and Y invites 3 men.
(ii) X invites 2 ladies, 1 man and Y invites 1 lady 2 men.
(iii) X invites 1 lady, 2 men and Y invites 2 ladies, 1 man.
(iv) X invites 3 men and Y invites 3 ladies.
The number of ways= 4C3.4C3  4C2 .3C1.3C1.4C2  4C1.3C2 .3C2 .4C1  3C3.3C3
= 16+324+144+1=485
10. Since, n digit numbers are formed using 2,5 and 7 digits.
The number of ways in which it can be done = 3n
Now, 36  900 and 37  900  n  7
11. We know that the sum of all n- digit numbers formed by using n- digits from the
Digits 1,2,3,4,5,6,7,8,9 is
10n  1 
= ( sum of the digits) (n-1) !=  
 10  1 
105  1  (100000  1)
Hence, required sum= (1+2+3+4+5)  4!     360
 10  1  9
 40  999999  3999960.
12. We observe that first five places can be filled in ways
I II III IV V
The sum of the digits in first five places can be of the following forms:
5m or 5m+1 or 5m+2 or 5m+3 or 5m+4
The possible values taken by the last digit corresponding to each of the possible
values are shown below:
Sum of the digits last digit
5m 0 or 5
 5m +1 4 or 9
5m+2 3 or 8
5m+3 2 or 7
5m+4 1 or 6
Thus, corresponding to each way of filling first five places there are two ways of
filling the last digit.
Hence, required number of numbers= 9  104  2  180000
13. Each question can be omitted or one of the two parts can be attempted i.e., it can be
taken in 3 ways.
8
So the 8 questions can be attempted in 3  1  6550 ways.
14. We have, 9 letters 3 a’s , 2b’s and 4 c’s . These 9 letters can be arranged in
9!
 1260 ways.
3!2! 4!
15. The hall can be illuminated by switching on at least one of the 10 bulbs.
Required number of ways is 210  1  1023
16. Let n=2m+1, Then the common differences of the A.P can be 1,2,3……m. The
number of A.P.s with 1,2,3…m common differences are
(2m-1),(2m-3)….3,1respectively
2
2  n 1
So,total number of A.Ps=(2m-1)+…..+1= m  
 2 
17. Four letters can be selected in the following ways
(i) all different i.e., C,O,R,G
(ii) 2 like and 2 different i.e., two O,1 R and 1 G
(iii) 3 like and 1 different i.e., three O and 1 from R, G and C.
The number of ways in (i) is 4C4  1
The number of ways in (ii) is 3C2 .2C2  3
The number of ways in (iii) is 3C3  2C1  3
So the required number of ways = 1+3+3 = 7
18. There are 98 sets of 3 consecutive integers and 97 sets of 4 consecutive integers
Number of permutations of b1, b2 , b3 , b4 number of permutations when b1, b2 , b3
Are consecutive + number of permutations when b2 , b3 , b4 are consecutive –
number of permutations when b1, b2 , b3 , b4 are consecutive.
= 97 x 98 + 97 -97 [ bi  b j , i  j ]
= 97 x 195 =18915
19. We have the following cases:
(1,1,8),(1,2,7),(1,3,6),(1,4,5),(2,2,6),(2,3,5),(2,4,4,),(3,3,4) and
(1,9,0),(2,8,0),(3,7,0),(4,6,0),(5,5,0)
Required number of 3- digits numbers
4  3!
 4  3!  4  2! 2  2!  24  12  16  2  54
2!
 a. 10C2  4C2  1  45  6  1  40
b. 1 10C2  45
20.
c. 2  6C2  30
d . 6C2  4  60
y

(5,3)
(1,2)


A (a,o) x
21.
3 2

5a 1 a
a  13
5
22. Solving both equations centre C  1, 1
Also  r 2  154  r  7
Equation of a circle  x  1   y  1  49
2 2

23. Total number of such matrices is

9c5  9c4  9c6  3c2  9c7  414
24. x  x  6  0  x  R   2,3
2
------ 1
The equation becomes x  x  6  x  2  x  2, 4 ------- 2
2

1, 2  x  2, 4
Let x 2  x  6  0  x   2,3 ---------- 3
The equation becomes x 2  x  6  x  2  0  x  2, 2 --------- 4
3, 4  x  2, 2
 The roots are 2, 2, 4 sum  4

sin x.tan x sin x.tan x

25. When x  0  1 but 1
x2 x2
 sin x.tan x 
 Lt    1
x 0
 x2
 PHYSICS
26). a x V W A  x
2 2

1 1 1
27). Springs are in series  
kq k1 k2
Springs are in parallel keq  k1  k 2
28). Heat energy lost by hot body is equal o heat energy gained by cold body
29). Hr=constant
30). K .E  1 / 2 kx 2 V  w A2  x 2 p.E  1 kx 2
2
31).

T 
Time taken by pendulum in going from A to B where T  2
4 g
T
Time taken by pendulum in going from B to C 
12
 T T  2T 2  2
 Time period of pendulum  2     .  sec
 4 12  3 3 5 15
Altier :

240 2
T .T  T
360 3
fs=kx
32).
(Where fs is frictional force on 20 kg block xis instantaneous elongation or compression
in spring)
Fs=k (A cos wt) | Fs | kA | cos t |
33).

ma   mg sin 
a   g sin  or a   g tan  …….(1)
(as  is small)
dy x
Now, X  4ay
2
 
dx 2a
x gx g
a  g  2 x   
2a 2a 2a
 1
y  4 cos 2   sin 1000t   2 1  cos t  sin 1000t 
34).  2
y  2sin 1000t   2 cos t sin1000t  2sin 1000t   2sin  999t   sin(1001t )  it is the super of three
k 2 2 1 2
35). w    ,w  ,T   2 2
M m  3  1 4 2 w
36). Use ansulan S.H.M equation
  mL  2
2
L  L  L   mL  2 k
  ky1  L ky2  I    k     LkL   w  L2    k    w 
2 2 2   3  4   3 
1 1 1
37). springs are in series  
keq k1 k2
1 1 1
Springs are in parallel  
keq k1 k2
38). Write the torque about contact point
2 
R  kn   I , R  k  R    mR 2  mR 2  
5 
7 5k 5k
kR 2    mR 2   w2   w
5 7m 7m
39). Motion of a body along the tunnel of the earth. Potential energy of the spring is +ve
Total displacement
40). V  a   w2 x
Total timetaken
1
41). x  A sin  wt    A  A sin  wt   
2
 sin  0      6
2
42). A  A0 e bt /2 m
2l
43). Time taken to collide on left wall and get back to the mass attached with spring is t1  ;
v
T 2 m m
Time to get the compressed once and to come back is t2   
2 2 k k
2l m

v k l  m
T  t1  t2   
2 v 2 k
44). P.E=- M .B
1
45). 7  xL  xC  WL 
wc
46). Let the frictional force be in the forward direction, then

MR 2
f  M 0 And f R  0
2
f
0  and  0  R 0  
m
f 2f Ma
  a f 
m m 3
 For pure rolling, f  fL   Mg
ma
  Mg  a  3 g  amax  9m / s 2
3
47). The given circuit can be redrawn as follows

24  8
Resistance between A and B  32  6
48
Current between A and B=current between X and Y  i  6  8 A
Resistance between X and Y   3  10  6  1  20
 Potential difference between X and Y  8  20  160V
   
      

48). Result ant force on the charged particle is, F net  qE  q V  B   2q i  3q j   q  V  B 
   
The particle is moving in x-y plane while the magnetic field is in y-z plane, so that, the
work done by magnetic field
will be equal to zero. So, the required work done is given by,
    
  
W F net . S   2q i  3q j  .  i  j 
  
49). The given circuit can be simplified as follows

All three points B have same potential and all three points C have same potential.

50). Use the gauss law

 CHEMISTRY
51). Distilled water is almost deionized water, so least conductance, so least conductance.
52). Species having higher SRP will be the strongest oxidizing agent.
53). 1G o  nFE ocell.n  2
 2  96500  2  384000, J  384 KJ
54). The given graph is for weak electrolyte and HCOOH is the weak electrolyte
55). On product side it should be KI instead of NaBr
56). Salt of divalent cation  A2  and anion  B2  is AB. It dissociates in water
AB  A2   B 2 
As per Kohlrausch law
 
 o m  AB    o m A2  m
o
B 2  
 o m  AB   60  70  130
57). In electrochemical series Li is present above Hydrogen, so it has max tendency to
liberate H 2
Other three metal  Cu, Ag , Au  are below hydrogen in electrochemical series
0.059
58). E 2   Eo 2  log  mg 2  
Mg / mg mg / mg 2  
Y C + m x
0.059
So, y intercept © = E o and slope (m) is 
Mg 2  / mg 2
59). During the electrolysis of dilute Sulphur acid above given three reaction occurs each of
which represent Particular reaction either oxidation half cell reaction or reduction
half cell reaction. At anode reaction with lower values
o
Of E cell will undergo faster oxidation. Hence oxidation of water occurs preferentially at
anode and H  Undergoes reduction at cathode
60). At Anode Cu s   Cu 2q  2e
 

At cathode Cu 2aq  2e   Cu
  s
61). Conceptual
1
62). At cathode H  e  H
 aq  2 2 g 
1
At anode CIaq   CI 2  e  Ecell
o
 1.36V
2
2 H 2O  O2  4H   4e  Eo  1.23V
g  aq   aq 
o
Ecell for this reaction has lower value but formation of oxygen at anode required over
potential
63). Change in stoichiometric coefficients will not change SRP. For metal electrodes doubling
of metal ion Concentration doubles metal electrode potential
64). When 2 faradays of electricity is passed 2 equivalent of oxygen is released 1.
2  5.6 L  11.2 L. Molecular weight of O2 is 32
65). As SRP value increases oxidizing power increases
66). Over reaction
Fe  Fe2  2e  G1o
s
1
2 H   2e  O2  H 2O l G2
2
1
Fe s   2 H   O2  Fe2  H 2OG3
2
G1o  G2o  G3o
G3o    2 F  0.44    2 F 1.33  322310 J
 322KJ
67). o
Ecell  E o  Eo
 X /X  M / M    0.44  0.33  0.11
o
Ecell   ve, G o  ve, Spontaneous reaction
2
 X   Y 
68). o
Ecell  Ecell 
0.059
log
P  0.29 
0.059
log   2
n R 2
PX 2 . Y  
2
 

102  102 
2
0.059
 0.29  log  0.35V
1 10 
2 2
2

69). k  3.06 106 mhocm 1

k  1000 3.06  106  1000
 eq  1.53mhocm 2eq 1,  eq  ,1.53 
N N

 
2
N  2  103 = solubility(s), K sp  BaSO4   S 2  2  10 3  4  106
o o
G o nFEcell 2  96500  Ecell
70). Efficiency   0.84   , E o  1.24V .
H o H o 285  1000
71). Structures of (b) , (c) and ( e) have asymmetric . c-atoms
1 T1 7  105 273
72).   CRT ,  T  ;  2  51.8  104 Nm2
2 T2 2 283
73). N 2 , C2 , O2 , O2 H 2 , have one unpaired e .
74). Pm is radioactive amongst lanthanides all 14 actinides are radioactive
75). a) Mohr’s salt FeSO4.  NH 4 2 SO4 .6 H 2 O
c) Best reducing agent in aqueous solution amongst elements –Lithium
Mol.Wt
d) Equivalent weight of KMnO4 in acidic medium is .
5