(Batches: e-SANKALP-2325 S1 & T1)

I I T – JEE, 2 3 2 5 Paper Code

100805
(CLASS XI)
Time: 2 Hours Maximum Marks: 126
INSTRUCTIONS
A. General
1. Write your Name, Enrolment number in the space provided on this booklet as soon as you get the paper.
2. Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers, and electronic
gadgets of any kind are NOT allowed in the examination hall.
3. Use a ball point pen do darken the bubbles on OMR sheet as your answer besides Name, Enrolment
number, Phase, Paper sequence, Venue, Date along with your signature on OMR sheet.

B. Question Paper Format

The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of
three sections.
4. Section–1 (01 – 06) contains (06) Multiple Choice Questions which have Only One Correct answer. Each
question will be evaluated according to the following marking scheme.
Full Marks : +3 If only (all) the correct option(s) is (are) chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : –1 In all other cases
5 Section–2 (07 – 09) contains (03) Multiple Choice Questions which have one or more than one
correct answer. Each question will be evaluated according to the following marking scheme.
Full Marks : +4 If only (all) the correct option(s) is (are) chosen;
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen;
Partial Marks : +2 If three or more options are correct but ONLY two options are chosen and
both of which are correct;
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a
correct option;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : –2 In all other cases.
6. Section–3 (10 – 12) contains (03) Numerical Value Questions, the answer to each question is a
Numerical Value. For each question, enter the correct numerical value corresponding to the answer
and each question carries +4 marks for correct answer. There is no negative marking.

Enrolment No. :

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e-Sankalp2325 S1 & T1-XI-PCM-(100805)-2

PART I : PHYSICS
SECTION – 1 : (Only One Option Correct Type)

This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

1. A surface consisting of curve inclined and A D

two semicircular part in vertical plane is as
shown in diagram. There is no friction R/16
anywhere and ball smoothly crosses from
one semicircular part to other at C. C
[Radius of ball is negligible. What is the
minimum value of h so that ball leaves the h
R
surface from point D?

B
67 5
(A) R (B) R
32 2
17 19
(C) R (D) R
8 8

2. A block lying on a smooth surface with spring connected to it is K = 100 N/m

pulled by an external force as shown. Initially the velocity of ends A B A
and B of the spring are 4 m/s and 2 m/s respectively. If the energy
of the spring is increasing at the rate of 20 J/sec, then the stretch in
the spring is
(A) 1.0 cm (B) 10 cm
(C) 2.0 cm (D) 20 cm

Space for Rough work

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 e-Sankalp2325 S1 & T1-XI-PCM-(100805)-3

3. A particle of mass m is projected with velocity u at an angle  with horizontal. During the period
when the particle descends from highest point to the position where its velocity vector makes an

angle with horizontal. Work done by gravity force is
2
1 1 
(A) mu2 tan2  (B) mu2 tan2
2 2 2
1  1 
(C) mu2 cos2  tan2 (D) mu2 cos2 sin2 
2 2 2 2

4. Consider an elliptically shaped rail PQ in the vertical plane with OP = 3 m and Q

OQ = 4 m. A block of mass 1 kg is pulled along the rail from P to Q with a force
of 18 N, which is always parallel to line PQ (see the figure given). Assuming no
frictional losses, the kinetic energy of the block when it reaches Q is (n×10) 4m
Joules. The value of n is (take acceleration due to gravity = 10 ms –2) 90°
O 3m P
(A) 3 (B) 5
(C) 7 (D) 9

5. A constant power P is applied to a particle of mass m. The distance travelled by the particle when
its velocity increases from v1 to v2 is (neglect friction)
(A)
m

3P 2

v 2  v12 (B)
m
3P
 v 2  v1 
(C)
3P

m 3

v 2  v13 (D)
3P

m 2

v 2  v12

6. Two blocks of masses m1 = 1 kg and m2 = 2kg are m2 m1

connected by non-deformed light spring. They are lying F
on a rough horizontal surface. The coefficient of friction
between the blocks and the surface is 0.4. What
minimum constant force F has to be applied in
horizontal direction to the block of mass m1 in order to
shift the other block? (Take g = 10 m/s2)
(A) 8 N (B) 15 N
(C) 10 N (D) 25 N

Space for Rough work

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e-Sankalp2325 S1 & T1-XI-PCM-(100805)-4

SECTION – 2 : (One or More Than One Options Correct Type)

This section contains 3 multiple choice type questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.

7. A small rubber eraser is placed at one edge of a quarter-circle-shaped R

45
track of radius R that lies in a vertical plane and has its axis of A
symmetry vertical (see figure); it is then released. The coefficient of
B
friction between the eraser and the surface of the track is μ = 0.6.
0.6mgR
(A) If the particle slides from A to B work done by frictional force will be .
2
0.6mgR
(B) If the particle slides from A to B work done by frictional force will be greater than .
4 2
(C) The particle will never be able to go from A to B.
(D) If  is 2 the particle will not begin to slide.

8. A particle is constraint to move along x-axis acted by a force whose potential energy is given by
U = 2x3 15x2 + 36 x + 100
Now mark the correct option about particle
(A) If it is slightly displaced from x = 2 it will oscillate.
(B) If it is slightly displaced from x = 3 it will oscillate.
(C) The minimum K.E. required at origin to reach at x = 3 will be 27 J
(D) The minimum K.E. required at origin to reach at x = 3 will be 28 J.

9. A massless spring of spring constant k has two ends P and Q as k

A B
shown in the figure. End Q is being pulled by person B with constant
speed u towards right. End P is being pushed by person A with P Q
constant acceleration a towards right starting from rest. Both starts
a u
simultaneously. Choose the correct option at the instant when B
appears to be at rest with respect to A
(A) work done by B = ku4/3a2
(B) work done by A = 5ku4/24a2
(C) work done by A = +5ku4/24a2
(D) elastic potential energy stored in spring = ku4/8a2

Space for Rough work

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 e-Sankalp2325 S1 & T1-XI-PCM-(100805)-5

SECTION – 3: (Numerical Answer Type)

This section contains THREE questions. The answer to each question is a NUMERICAL VALUE. For
each question, enter the correct numerical value (in decimal notation, truncated/rounded‐off to the
second decimal place; e.g. xxxxx.xx).

10. A light rod of length L, is hanging from the vertical smooth wall of a vehicle moving with
acceleration 3 g having a small mass attached at it’s one end is free to rotate about an axis
passing through the other end. The minimum velocity given to the mass at it’s equilibrium position
so that it can complete vertical circular motion is KgL . Find the value of K.

11. In the setup shown, an almost inertia-less bar is

suspended horizontally with the help of two identical
springs each of stiffness k, two light inextensible cords
that pass over fixed ideal pulleys and two k k
counterweights each of mass m. A small disc of mass 1.01m
0.01m is placed at the midpoint of the bar. A block of
mass 1.99m is suspended from the disc with the help m 1.99m m
of a light cord that passes through a hole in the bar. If
this cord is cut, the maximum height will the disc jump
1000Pmg
is . Find value of P.
k

12. A horizontal frictionless thin rod wearing a sleeve of mass m 

is being rotated at a constant angular velocity  about a
stationary vertical axis through one of its ends. The sleeve is
held stationary with respect to the rod at a distance r = r1 from r
the axis with the help of a light cord as shown in the figure. At
some instant of time, the cord is cut. The work done by the
external agency in maintaining the angular velocity of the rod
constant until the sleeve slides to a distance r = 3r1 is
40km2r 2 . Find value of k

Space for Rough work

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e-Sankalp2325 S1 & T1-XI-PCM-(100805)-6

PART II : CHEMISTRY
SECTION – 1 : (Only One Option Correct Type)

This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

1. Select the incorrect statement about the following chemical species:

O2 , O2 , N2 , NO, H2 and H2
(A) Magnetic moment of NO is greater than that of O2
(B) The bond length of O2 is shorter than that of O2
(C) The ionization energy of N2 is greater than that of N
(D) H2 is more stable than H2 although both have the same bond order

2. Which of the following order is correct, against the indicated properties?

(A) AgF  AgCl  AgI : Covalent character
(B) NaHCO3  KHCO3  RbHCO3 : Solubility in water
(C) NaF  AlF3  MgF2 : Melting point
(D) MgC2O 4  CaC2O4  BeC2 O4 : Solubility in water

3. The number of nodal planes in a pxorbital is ….?

(A) One (B) Two
(C) Three (D) Four
Space for Rough work

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 e-Sankalp2325 S1 & T1-XI-PCM-(100805)-7

4. The dipole moment of “O–H” bond & “O–D” bond is ‘x’ & ‘y’ respectively. If net of H2O = ‘Z’
Debyes, then net of D2O will be [Assume bond angle of H 2O = D2O =  & cos  = – 0.24]…..?
(A) z + 1.52 (y - x) (B) z - 1.52 (y - x )
(C) z - 1.52 (y + x) (D) z + 1.52 (y + x)

5. The geometrical shapes of XeF5 , XeF6 and XeO 64  respectively are….?

(A) Trigonal bipyramidal, distorted octahedral and square planar
(B) Square pyramidal, distorted octahedral and octahedral
(C) Planar pentagonal, distorted octahedral and square antiprismatic
(D) Square pyramidal, distorted octahedral and square antiprismatic.

6. F
Cl

C a b
C H
1 H 2
H
H
H
H
Which of the following is correct about the above?
(A) a > b (B) a = b
(C) 2 > 1 (D) 1 > 2

Space for Rough work

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e-Sankalp2325 S1 & T1-XI-PCM-(100805)-8

SECTION – 2 : (One or More Than One Options Correct Type)

This section contains 3 multiple choice type questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.

7. The correct order among the following is/are….?

(A) CO  CO 2  CO32  (C – O bond length)
(B) O 22   O2  O 2  O2 (Bond order)
(C) O2 < O3 < H2O2 (O – O bond length)
(D) NH4  NH3  NH2 (Bond angle)

8. If equatorial plane in PCl5 molecule is the XY - plane, the orbitals hybridizing to produce axial
bonds will be….?
(A) pz (B) dx2  y 2
(C) py (D) dz2

9. Which of the following is/are correct statement(s)?

(A) B22 and C2 both are paramagnetic with bond order = 2.
(B) Bond order for CO and N2 are 3.0 and 2.5 respectively.
(C) During the formation of N2 from N2 , bond length increases.
(D) Solubility order of hydroxides of Na, Mg and Al in water are : Al  OH3  Mg  OH2  NaOH .

Space for Rough work

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 e-Sankalp2325 S1 & T1-XI-PCM-(100805)-9

SECTION – 3: (Numerical Answer Type)

This section contains THREE questions. The answer to each question is a NUMERICAL VALUE. For
each question, enter the correct numerical value (in decimal notation, truncated/rounded‐off to the
second decimal place; e.g. xxxxx.xx).

10. Find the number of chemical species which are planar and any of the dxy , dyz or dxz orbirtal of
central atom, (also) participates in the hybridization.
XeF5 , XeF5 , XeF6 , XeF4 , SF4 , ICl4

11. What is the total number of bonds in the molecules/ions which are non-linear from the following?
ICl2 , AsF2 , NOF, N3 , CO 2 , COS, H2 O, OF2 , SCl2 , SO 2

12. Total number of lone pairs present on all the atoms of each of the molecule of XeF 6, SF6, SF4 are.
Space for Rough work

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e-Sankalp2325 S1 & T1-XI-PCM-(100805)-10

PART III : MATHEMATICS

SECTION – 1: (Only One Options Correct Type)

This section contains 6 multiple choice type questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE is correct.

1. The circles which can be drawn to pass through (3, 0) and (5, 0) and to touch the y-axis, intersect
at an angle , then cos is equal to
7 1
(A) (B)
8 2
(C) 0 (D) none of these

2. The number of rational point(s) (a point (a, b) is called rational, if a and b both are rational
numbers) on the circumference of a circle having center (, e) is
(A) at most one (B) at least two
(C) exactly two (D) infinite

3. The equation of the circle passing through the point of intersection of the curves (2x+3y+19)
(9x+6y-17) = 0 and xy = 0 is
(A) x2+ y2+137x+63y-303 = 0 (B) 4x2+4y2+137x+63y-323 = 0
(C) 18x +18y +137x+63y –323 = 0
2 2 (D) None of these

4.  
For a, b  R the maximum  a  1b  1  1  1  a2 1  1  b2 
 

(A) 2 2 (B) 2 2
(C) 3  2 (D) 3  2 2

Space for rough work

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 e-Sankalp2325 S1 & T1-XI-PCM-(100805)-11

5. A circle (x – 2)2 + (y – 3)2 = 16 is given for which two lines L1: 2x + 3y = 7 and L2: = 3x + 2y = 11
act as chords. If infinite more chords are drawn parallel to L1, then to L2. A curve is drawn by
joining midpoints of chords parallel to L1 and another curve is drawn by joining mid points of
chords parallel to L2, then the acute angle between the curves at the point of intersection is
  5 
(A) (B) tan1  
2  12 
 7  
(C) tan1   (D)
 12  6

6. Three parallel chords of a circle have lengths 2, 3, 4 and subtend angles ,,  +  at the centre
respectively (given  +  < ), then cos  is equal to
15 17
(A) (B)
31 35
17
(C) (D) none of these
32

SECTION – 2: (Multi Correct Choice Type)

This section contains 3 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE OR MORE may be correct.

7. Let AC and BD be two chords of a circle with centre O such that they intersect at right angle
inside the circle at point M. Suppose K and L are the mid points of the chords AB and CD
respectively, then
(A) OK = LM (B) OKLM is a parallelogram
(C) OKML is a trapezium (isosceles) (D) none of these

8. AC is a diameter of a circle. AB is a tangent BC meets the circle again at D. If AC = 1, AB = a,

CD = b, then
(A) ab > 1 (B) ab < 1
b 1 b 1
(C)  (D) 
a 2 1 a 2 1
a  a 
2 2

Space for rough work

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e-Sankalp2325 S1 & T1-XI-PCM-(100805)-12

9. Let A(–2, 0) & B(4, 0) & let L be any line passing through (6, 2) having slope m. The values of m
for which there are two points on the line L at which AB subtends a right angle can be
(A) 0 (B) – 1/7
(C) 6/11 (D) 4/5

SECTION – 3: (Numerical Answer Type)

This section contains THREE questions. The answer to each question is a NUMERICAL VALUE. For
each question, enter the correct numerical value (in decimal notation, truncated/rounded‐off to the
second decimal place; e.g. xxxxx.xx).

10. A ray of light is incident at the point (–2, – 1) gets reflected from the tangent at (0, – 1) to the
circle x2 + y2 = 1. The reflected ray touches the circle. If the equation of the incident ray is
ax + by + c = 0, then find the value of c – a – b. (b & c are prime numbers)

11. A circle is drawn to pass through the vertex ‘A’ and to touch the sides BC and CD at P and Q
respectively of a rectangle ABCD. If the length of the perpendicular drawn from A upon PQ is
13 units, if area of the rectangle ABCD is A, then A – 7 is ________

 1 1 8 8
12. If a chord of the circle x2 + y2 – 4x – 2y – c = 0 is trisected at the points  ,  and  ,  then c
 3 3  3 3
equals to _____

Space for rough work

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 e-Sankalp2325 S1 & T1-XI-PCM-(100805)-13

(Batches: e-SANKALP-2325 S1 & T1)

I I T – JEE, 2 3 2 5 Paper Code

100805
(CLASS XI)
ANSWERS
PHYSICS (PART-I)
1. B 2. B 3. C 4. B

5. C 6. A 7. C, D 8. B, D

9. A, B, D 10. 8.00 11. 0.1 12. 0.2

CHEMISTRY (PART-II)

1. A 2. B 3. A 4. A
5. B 6. D 7. A, B, C 8. A, D
9. C 10. 1 11. 17.00 12. 50.00

MATHEMATICS (PART-III)

1. A 2. A 3. C 4. D
5. B 6. C 7. A, B 8. B, C
9. A, B, C, D 10. 4 11. 6 12. 20

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e-Sankalp2325 S1 & T1-XI-PCM-(100805)-14

HINTS AND SOLUTION

PHYSICS (PART-I)

1. Considering point C vC  gR
gR
Considering point D: vD 
16
1
 mgh > mv 2C  2mg2R
2
1  R
mgh > mv D2  mg  2R  
2  8
 5 
 h  5R/2  h  R  2  
 32 
answer is greater of two

2. Let xA and xB the position of ends A and B at time t from the block, then stretched length of the
spring will be
f = x A  x B
and So the stretch
 = f  i = (xA  xB)  i (i Natural length of the spring.)
1 1
So, U = k 2 = k  x A  xB    i 
2

2 2
du 1  dx dx 
P= = k  2  X A  XB   i   A  B 
dt 2  dt dt 
P = F(vA  vB)
P
F=
v A  vB
F P 20
 =  
k  v A  v B  k  4  2   100
 = 0.1 m = 10 cm

4. Using work energy theorem

Wmg + WF = KE
– mgh + Fd = KE
–1×10×4 + 18(5) = KE
KE = 50
n=5

7. Underestimate the work done against friction, and compare it with the initial gravitational potential
energy of the eraser.
First of all, we investigate whether the rubber eraser will start moving at all. It will do so provided
that mg sin α>μmg cos α, i.e. μ< tan α = tan 45◦ = 1. This is clearly the case, since μ = 0.6. So,
the eraser will start moving. The trouble is that the determination of how the normal force acting
on the eraser varies with position is difficult. A calculation of the work done against friction can be
carried out, to any given degree of accuracy, only by using a computer. The trouble is that the
determination of how the normal force acting on the eraser varies with position is difficult.
However, it is certain that the frictional force is always larger than its initial value of μmg cos α =
μmg cos 45◦ This is because, after the initial release, the angle with the horizontal made by the
slope on which the eraser moves decreases, and, in addition, the track has to provide a
centripetal force for the moving eraser. The path to reach the lowest point of the track would be

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 e-Sankalp2325 S1 & T1-XI-PCM-(100805)-15

one-eighth of a circle, with a length of Rπ/4. The work done against friction can now be
underestimated by taking the normal force as if it always had its initial value:
 mg  R
W f  Wund   0.333mgR
2 4
The gravitational potential energy difference between the initial position and the bottom of the
track is ΔEp = mgR(1 − cos α) = mgR(1 − cos 45◦) ≈ 0.293mgR.
It can be seen that |Wf| >ΔEp, i.e. the work to be done against friction is clearly larger than what
can be provided by the gravitational potential energy. So, the rubber eraser cannot reach the very
lowest part of the track.

8. P.E. of the situation is

U in Joule
128

127
100

x=2 x=3 x

u/a  1  ku4
9. Work done by B = 0 k  ut  at 2  udt  2
 2  3a
u/a  1  5ku4
Work done by A=   k  ut  at 2  atdt  
0
 2  24a2
2
1  u2  ku4
Energy stored in spring =  k    2
2  2a  8a

1    3
10. Conservation of energy mv 2  mg  mg  m 3  2 g
2 2 2  2 

v  8g 60

CHEMISTRY (PART-II)

2. Correct order is AlF3  MgF2  NaF .

3. pxorbital has one nodal plane in the yz plane.

4. Resultant net = net (O–H) + net (lone pairs)

= x 2 + x 2 + 2xx cos q + a
=Z= 2x (1+ cos q) + a
= Z = x 1.52 + a  (1)

For D2O
Resultant net (D2O) = net (O–D) + net (lone pairs)
Z = y 2 + y 2 + 2yy cos q + a
Z ¢= y 1.52 + a  (2)
From equation (1) & (2), eliminating ‘a’, we get
Z ¢= Z + (y - x) 1.52
7. The correct bond angle order is

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e-Sankalp2325 S1 & T1-XI-PCM-(100805)-16

NH2  NH3  NH4

9. (A) B22 and C2 both are diamagnetic with bond order = 2.

(D) Correct solubility order is NaOH  Mg  OH2  Al  OH3 .

10. In XeF5 , central atom is sp3 d3 hybridized with two lone pairs at axial positions of pentagonal
bipyramidal geometry of the species.

11. Except N3 , CO 2 and COS all are non-linear.

Non-linear species ICl AsF NOF H2O OF2 SCl2 SO2
2 2
No. of bonds 2 2 3 2 2 2 4

12. F
F F
F F
F F F
Xe S S
F F F F F
F F
1  6  3  19 F
6  3  18 1  4  3  13
Total lone pairs = 50

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 e-Sankalp2325 S1 & T1-XI-PCM-(100805)-17

MATHEMATICS (PART-III)

1. A
Let A  (3, 0), B  (5, 0) and C1, C2 be the centres of
circles passing through A, B and touching the y-axis
at P1 and P2.
Radius of both the circles will be same.
Let C1A = C2A = r C1
P1
Now C1   4, 15  , C2   4,  15 
7 7
If C1AC2 =   cos = –  cos = .
8 8 O A B

P2
C2

2. A
If there are more than one rational points on the circumference of the circle x 2+y2-2x – 2ey + c =
0, (as (, e) is the centre), then e will be a rational multiple of , which is not possible. Thus the
number of rational points on the circumference of the circle is at most one.

3. C
Any curve passing through the point of intersection of the given curves is
(2x+3y+19) (9x+6y-17) +xy = 0
For it to be circle, coefficient of x2 = coefficient of y2 (already satisfied), and coefficient of xy = 0
 (12 + 27 +  ) = 0
  = -39

4. D
Let 1  a2  y1  a2 + y12  1

Let 1  b2  y 2  a2 + y 22  1
Hence (a, y1) and (b, y2) can be chosen on circle x2 + y2 = 1
so, a = cos, y1 = sin
b = cos, y2 = sin
 L = (cos  1) (cos  1) + (1  sin) (1  sin)
L = (2  (cos + cos + sin + sin)) + (coscos + sinsin)
      
L  2  2  sin      sin        cos     
  4  4  
5
L is maximum if  =  =
4
 Lmax = 2 + 1 + 2 2 = 3  2 2

5. B
For the circle (x – 2)2 + (y – 3)2 = 16
Then chords through midpoint (h, k) are
(h – 2)x + (k – 3)y + (2h + 3k – h2 – k2) = 0
when chords are parallel to 2x + 3y = 7 locus is

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e-Sankalp2325 S1 & T1-XI-PCM-(100805)-18

h2 2
     3x – 2y = 0
k 3 3
when chords are parallel to 3x + 2y = 11 locus is
h2 3
     2x – 3y + 5 = 0.
 k  3  2
Clearly both the locus meet at (2, 3) and being straight lines acute angle is
 5 
  tan1   .
 12 

6. C
Using the property that equal chords subtends equal angles at
centre of circle, then problem can be converted to the diagram in
adjoining figure. O
AB = 4, AC = 2, BC = 3
ABC = /2 4
A B
9  16  4 7 3
cos(  / 2)   cos  = 2cos2(/2)–1 2
23 4 8
49 98  64 34 17 C
 2 1  cos   
64 64 64 32

7. A, B
Choose M as origin.
AC as x axis and BD is y-axis.
Equation of circle x2 + y2 + 2gx + 2fy + c = 0.

  
Co-ordinate of A is g  g2  c, 0 and C is g  g2  c, 0 . 
  
Co-ordinate of B and D are 0,  f  f 2  c are 0,  f  f 2  c . 
 g  g2  c f  f 2  c   g  g2  c  f  f 2  c 
Thus K is  ,  , L is  , .
 2 2   2 2 
   
 g f 
Mid point of KL is   ,   .
 2 2
 g f
Mid point of OM is   ,   .
 2 2

8. B, C
AD AB AD a B
tanC =     AD = ab and AD < AC
DC AC b 1 D
 ab < 1
Now (AC)2 = (AD)2 + (CD)2 = a2 b2 + b2 = 1
1 b2 1 A C
 b2  2  2  4
a 1 a a  a2
b2 1
 2
 2
a  2 1
 a 
 2 

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 e-Sankalp2325 S1 & T1-XI-PCM-(100805)-19

9. A, B, C, D
The line y – 2 = m(n – 6) should cut the
circle whose diameter is AB at 2 points
m  0  2  6m (6, 2)
i.e. 3
m2  1 A B
43 5 43 5 
 m , 
 8 8
 
10. 4
Any line through (–2, –1) is y + 1 = m(x + 2) …(1)
2m  1 4
It touches the circle if  1  m = 0 or
1 m 2 3
4
Equation of reflected ray is y + 1 = (x + 2)
3
 4x – 3y + 5 = 0 …(2)
a point on equation (2) is (–5, –5).
Its image by the line y = – 1 is (–5, 3).
3 1 
Hence equation of incident ray is y – 3 = x  5   4x + 3y + 11 = 0.
5  2
11. 6
Equation of circle is x2 + y2– 2ax – 2ay + a2 = 0
Satisfying by A(, ) A(, )
D
 2 + 2 – 2a – 2a + a2 = 0 …(1)
Also equation of PQ = x + y – a = 0 (a, a)
 a Q
 13 
2
 ( +  – a)2 = 26. C
If we subtract (1), then P B
2 = 26   = 13 = area of the rectangle.

12. 20
 1 1
Let B   , 
3 3
8 8
C  , 
3 3
Q
Centre of circle Q (2, 1) radius of circle = 5c
P is mid point of AD
3 3
P  ,  A D
2 2 B P C
BC 7
AP = 3  
2 2
1
QP =
2
AQ2 = AP2 + QP2
49 1
5+c=   c = 20
2 2

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