02-08-14 sr.iplco jee main (2013) ptm-1 final key & solutions

8/9/2019 02-08-14 Sr.iplcO Jee Main (2013) PTM-1 Final Key & Solutions

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Sri Chaitanya IIT Academy., India.A right Choice for the Real Aspirant

Central Office , Madhapur – HyderabadSec: Sr. IPLCO JEE MAIN – 2!"

#t: 2$%$!&'i(e : " )o*rs P'M$! Ma+Mar,s : "-

KEY SHEET

E/ S)EE'

P)/SICS C)EMIS'R/ MA')S

0.NO ANS1ER 0.NO ANS1ER 0.NO ANS1ER

! &

"! !

-! 22 ! "2 " -2 !

" 2 "" 2 -" && ! "& ! -& "

" " " - 2

- ! "- ! -- !3 " "3 " -3 !

% 2 "% ! -% &

4 " "4 " -4 !! 2 & ! 3 "

!! ! &! ! 3! 2!2 " &2 ! 32 !

!" & &" ! 3" "!& 2 && " 3& !

! " & " 3 2

!- ! &- 2 3- !!3 2 &3 ! 33 "

!% " &% & 3% 2

!4 ! &4 " 34 !2 & " % !

2! ! ! " %! !

22 " 2 & %2 "

2" 2 " 2 %" &2& ! & & %& "

2 " " % 2

2- 2 - 2 %- &23 ! 3 & %3 "

2% & % & %% 2

24 2 4 & %4 &" " - 2 4 &



Sri Chaitan5a II' Aca6e(5 2$%$!&7Sr.IPLCO7JEEMAIN782!"97P'M$!

)IN'S

P)/SICS 1. Conceptual

2. Conceptual

3. Conceptual4. Conceptual

5. Let the particles meet after a time t1 after the projection of the second particle. Then for their motion

upto their meeting, we have

first particle second particle

!nitial velocit" u u

acceleration #g g#

time of motion t$t1 t1

final velocit" ( )1u g t t − + 1u tg −

The particle will meet, when first particle is comparing down and second particle is going up. Their

final velocit" must %e e&ual in magnitude, so( )1 1

u tg u g t t − = − − +

12

u t t

g

∴ = − ÷

!n this time, height

2

1 1

1

2h ut gt = −

2

2 2 2

u t g u t u

g g

= − − − ÷

2 2

2 '

u gt or h

g

= −

Thus 12

u t t t t

g

+ = + − ÷

2

u t

g

= +

Sr.IPLCO_JEE-MAIN_Solutions Page2




(. Consider motion of stones with respect to the %alloon. )t the instant of release of stones, the initial

velocit" of %oth stones w.r.t. the %alloon is *ero. The acceleration of stone w.r.t. the %allon

[ ] ( ) stone balloona g a g a= − − = +

+here a is the acceleration of %alloon which is 15.- ms 2

/ow [ ] 21 110

2 stone balloon s a t = +

( )1 4 ( 10t s= + =

∴ ( ) 2

1

110

2 s g a X = +

( ) 21.' 15.- 10

2 X m= +

and ( ) 2

2 2

10

2 s g a t = + +

+here2

(t s=

∴ ( ) 2

2

1.' 15.- (

2 s X = +

The distance %etween 1 2 s and s

1 2 s s s= −

( ) 2 21.' 15.- 10 (

2 = + −

'1( m

-. The displacement, 2 cos s R θ = and acceleration, cosa g θ =





∴ 21

2 s at =

or ( ) 212 cos cos

2 R g t θ θ =

∴ 2 Rt g

=

'. if u is the velocit" of projection, then

0 3.5u a X = −or 3.5 3.5 -.5u a X = = 2(.25ms

/ow 21

1 12

y u X a X = −

21

2(.25 -.5 22.52

X X l m= − =

)lso 21

2 22

y h u X X a X + = −

212(.25 2 -.5 22

X X X = −∴ 15h m=

. Conceptual

10. Conceptual

11. 2.5dv

vdt

= −

⇒ 2.5dv

dt v

= −

⇒ 0

12

(.25 02.5

t

v dv dt − = −∫ ∫ ⇒ [ ]

012

0 (.252.5 2

t t v − =

⇒ 2t s=

12. 21

2h gt =

v gt = − and after the collision, v gt = straight line

Collison is perfectl" elastic then %all reaches to same height again and again with same velocit".





13. ( )2 0sin2 (0

2 2

u R

g =

23

4

u

g =

2 2 0 2sin (0 3

2 '

u u H

g g = =

2 22 21

2 '

R u AB H

g

∴ = + = ÷

14. rom the graph,

22.522.5 .

150a s or = −

(0

0 0

22.5. 22.5

150

v

v dv X s ds = − ÷ ∫ ∫

∴ ( )

22 (022.522.5 (0

2 150 2

v X X = −

4(.4- v m s∴ =15. 6elative velocit" of 7 with respect to 8 should %e along 78 or a%solute velocit" components

perpendicular to 78 should %e same.

02sin sin 30

3

uuθ ∴ =

∴ 1 3

sin4

θ − = ÷

1(. .dv

f v a bxdx

= = −

or ( )0 0

v v

vdv a bx dx= −∫ ∫ ∴ 22v ax bx= −)t other station, v0

⇒ 2a

xb

=

urther acceleration will change its direction when,

f0 or a#%90 ora

xb

=

)t this 9, velocit" is ma9imum.

:sing ;&. i2

ma9 2 a a a

v a bb b b

= − = ÷ ÷

1-. Conceptual





25. Conceptual

2(. rom the graph

2

25 ( 5 (

da d d s

dt dt dt

= = ÷

!ntegrate the a%ove

2

2

5

(

d s

t dt =

!ntegrate again25

12

ds t

dt =

!ntegrate again35

3(

t s =

=istance travelled from 0 to (s 3

1

5( 30

3( s m= × =

<elocit" at25

( 15 12

t s v t m s= = =

rom ( to 12 s, 215 5 u m s a m s= =2

2

11'0

2 s ut at m= + =

Total distance travelled 30 1'0 210 m+ =

2-. 6esistance force2 2dv

v m kvdt

µ ⇒ = − 2dv k

vdt m

= −

dv

k mv

= −,

0 0

v h

v

dv k dx

v m= −∫ ∫ ,

0log v kh

v m=

0

log

vm

k h v= )lso 2

dv k

dt v m= −0

2

0

v t

v

k

v dv dt m

− −

⇒ =∫ ∫ ( )0

00 ln

h v vt

vvv

v

−

=

.

2'. 2 2 1 1cos cos x u t u t θ θ + =

1 1 2 2cos cos

xt

u uθ θ ⇒ =

− also 1 1 2 2sin sinu uθ θ =

)fter solving we get ( )2

1 2 1

sin

sin

x

t u

θ

θ θ = −

2. Conceptual

30.

30, 5.0 25 .e M f cm and D cm= = = +e >now

0 0 0

25

1 30 1 5.0ee

D

M m X m m or m f

= = + = + ÷ ÷

+hich gives 0 5.0m =





C)EMIS'R/ 31. ! 0.5 mole 3 324O g O=

ii 0.5g atom of o9"gen 'g

iii23

223

3.011 1032 1(

(.022 10

X X g O

X =

iv 2 2

5.(44 11

22.4 X g CO g CO=

32. /o.of moles of C?2 in 100 g mi9ture

((

1.544

= =

/o.of moles of @2 in 100g mi9ture

34

1-2

= =

100

5.401'.5

average M = =

5.4

. . 2.-2

V D = =

33. Let 1 mole of mi9ture has 2 4 x mole N O

2 9 2-.( ( ) ( )2 1 4( 0.2 x x x+ − =34. Conceptual

35. 1 mole 2 4 N H loses ' mole e− 1 mole / loses 4 mole of e

−

∴ /ew o9idation state of / is 2 4 2− + ⇒3(.

5 3

2 2 - 3

n Cr O X X O Cr + + − ++ → +

( )3 3( 10 ( 5 10 1 X X n X X n− −= − ⇒ =

3-. 3 2 22 12 10 ( ( BrO H Br Br H O− + −+ + → +

10 mole e− re&uired for formation of ( moles of Ar 2

∴ 2

10 5

( 3

n factor of Br − = =

e&. wt.. . 3

5 3 5

mol !t m M

n= = =

3'. Aalancing the e&uation, we have

( )2 2 - 2 4 4 2 4 2 4 233 Cr O H "O "O "O Cr "O H O+ + → + +

3. 1 1 2 2 N V N V =0.5 20 0.4 25 X X =

40. 8ince, phenolphthalein indicates onl" conversion of 2 3 Na CO into 3 NaHCO hence, x m# , of @Cl

will %e further re&uired to convert 3 NaHCO to 2 3 H CO . 8o, total volume of @Cl re&uired to convert

2 3 Na CO into 2 3 2 H CO x x x m#= + = 2 3 H CO .





41.1 (.'

. . 3.42 2

n $ % $ eV = = − = −

2

13.(3.4

n $

n

−⇒ = −Q

n2 or first e9cited state42. Conceptual

43.10 13

10 10 m g kg − −= ⇒

( 2 14 10.000110 10 10 sec

100V X X m− − − −∆ = =

.4

h x &

π = ∆ ∆ =

4 4 .

h h x

& m vπ π ⇒ ∆ = =

∆ ∆34

13 14

(.(2 10

4 3.14 10 10

X x X X X

−

− −∆ =34

'

2-

(.(2 10 5.2 10

12.5( 10 x X x X m

−−

−∆ = ∆ =

44.2 2

2 2 2 2

1 2

1 1 1 1 13

3 R' RX

n nλ

= − = − ∞ 1

3 3

R R

λ ⇒ =

45. Conceptual

4(. 2l =

num%er of degenerate or%itals 2 1 l + =

ma9imum total spins 1

2

X

ma9imum multiplicit" 2 1" +

2 1 102

X = + =

minimum total spins 0

minimum multiplicit" 2 0 1 1´ + =

4-. Conceptual

4'. ?r%ital angular momentum ( )1 12

h

l l l π = + = for p#or%ital.

4. Conceptual

50. Conceptual

51. Conceptual

52. Conceptual

53. Conceptual

54. Conceptual

55. Conceptual

5(. Conceptual

5-. Conceptual

5'. Conceptual5. Conceptual





(0. Bass of acetic acid adsor%ed %" 2g charcoal 100 10#3 0.(#0.5 (0

0.(0.( 0.3

2

x

m⇒ = ⇒

MA')EMA'ICS

(1. The e&uation of the line L %e ( )2 ' , 0 y m x m− = − <

coordinates of 7 and D are2

' , 0 % m

− ÷

and ( )0, 2 '( m− .

8o( )

( )( )

( )2 2 2

' 2 ' 10 ' 10 2 ' 1'O% O( m m X mm m m

+ = − + − = + + − ≥ + − ≥− −

8o, a%solute minimum value of 1'O% O(+ =

(2. 8olving given e&uations we get5

3 4 x

m=

+ x is an integer, if 3 4 1, 5m+ = ± ±

∴ 2 4 2 '

, , ,4 4 4 4

m = − − −

8o m has two integral values.

(3. !f 1 2 & and & %e the distance %etween parallel sides and E %e the angle %etween adjacent sides, then

6e&uired area 1 2 cos & & ecθ

+here( )

12

11

&m

= + ,( )

22

11

&n

= +

distance %etween F F lines

and tan1

m n

mnθ

−=

+6e )u*red area∴

Sr.IPLCO_JEE-MAIN_Solutions




(4. !mage of the point 7 a%out the line ( )1, 4 y x *s (=∴ Transformation through a 2 units along the positive direction of x ax*s− , then new point

( )1 2, 4 R + ie, 63, 4

G 5OR OR∴ = =and tan 4 3θ =

4 3sin cos

5 5

and θ θ ∴ = =

Then , final position of the point is ( ) ( )( )G cos 4 , Gsin 4OR ORπ θ π θ + +

1 1 1 15 cos sin , 5 cos sin

2 2 2 2θ θ θ θ

⇒ − + ÷ ÷ 1 -

,2 2

⇒ − ÷

(5. ;&uation of L is 1 x y

a b+ = and let the a9is %e rotated through an angle θ and let ( ), X + %e the new

coordinates of an" point ( ), % x y in the plane, thencos sin , x X + θ θ = − sin cos , y X + θ θ = − the e&uation of the line with reference to original

coordinates is 1 x y

a b+ =

ie,cos sin sin cos

1 X + X +

a b

θ θ θ θ − ++ = HHHHHHi

and with reference to new coordinates is

1 X +

& )+ = HHHHHHHHii

Comparing ;&s. i and ii, we get

cos sin 1

a b &

θ θ + = HHHHHHHH.iii

sin cos 1

a b )

θ θ −+ = HHHHHHHH.iv

8&uaring and adding ;&s. iii and iv, we get

2 2 2 2

1 1 1 1

a b & )+ = +

((. Let the coordinates of ) %e a, 0. Then the slope of the reflected ra" is

( )3 0

tan5

saya

θ −

=−

HHHHHHH..i





Then the slope of the incident ra"

( )2 0

tan1 a

π θ −

= = −−

HHHHHHHHii

rom ;&s. i and ii,

( )tan tanθ π θ + −⇒ 3 2

05 1a a

+ =− −

⇒ 133 3 10 2 0

5a a a− + − = =

Thus, the coordinate of ) is13

, 05

÷

(-. 2 2 1 2

1II 3I-I 5I5I I

m

n+ + =Q

1 2 10I 2 10I 10I 2

10I 1II 3I-I 5I5I I

m X X

n

+ + =

{ }10 10 10

1 3 5

1 22 2 2

10I I

m

C C C n

⇒ + + =

{ }10 10 10 10 10

1 3 5 -

1 2

10I I

m

C C C C C n

+ + + + =

( )10 11 2

210I I

m

n

−⇒ =

∴ 10m and n

= =@ence, 9#"$10 and 3 0 x y+ + = are perpendicular to each other, then orthocenter is the point of

intersection which is #2, #1∴ #22m#2n

and #1m#n

∴ 7oint is ( )2 2 ,m n m n− −

('. +e have ( ) ( )2cos cos 2 cos 1 y x x x= + − +

( ) ( ){ }212 cos cos 2 2 cos 1

2 y x x x= + − +

( ) ( ){ }

1

cos 2 2 cos 2 1 cos 2 22 x x= + + − − +

( )1

cos 2 12

= −

( )211 2 sin 1 1

2= − −

2sin 1= −

+hich is a straight line passing through2, sin 1

2

π − ÷

and parallel to the x ax*s−





(. 1 , 22 2

% α α + + ÷

Q lies %etween the parallel lines 2 1 x y+ = and 2 4 15 x y+ = , then

1 2 2 12 2

02 1 4 2 15

2 2

α α

α α

+ + + − ÷ ÷

< + + + − ÷ ÷

34

2 0(

52

α

α

+⇒ <

− +

4 2

30

5 2

(

α

α

+ ÷

⇒ <

− ÷

4 2 5 2

3 (α ∴ − < <

-0. ( ),a a∴ fall %etween the lines 2 x y+ = , then

20

2

a a

a a

+ −<

+ +1

0 1 11

aor a

a

−⇒ < − < <

+1a∴ <

-1. 8ince =istance %etween parallel lines are same∴ 7arallelogram is a rhom%us

!n rhom%us diagonals are perpendicular to each other.

@ence, angle %etween diagonals 2π =-2. +hen x is rotational

let x)

ρ =

Then, ( )I 1.2.3....... . int &

n x n an eger )

= =

( ) ( )( )2cos I 1 cos 1

nn x nπ π ∴ = = −Q

( ) ( )2lim lim cos I 1 1m

m nn xπ

∞

→∞ →∞∴ = =

+hen x is irrational

Then, In x is not an integer

( )20 cos I 1n xπ ∴ ≤ <

( )2lim lim cos I 0m

m nn xπ

→∞ →∞∴ =

Thus, points are ( ) ( ) ( )2, 1 , 2, 1 1, 0and −

∴ 6e&uired area

2 1

2 11 1F F 4 2 .

1 02 2

2 1

s) un*t

−

= = − =

−





;&uation of ?7 is y x=Then, 1 1 1 2O% %% = =

+e have, ( ) ( ) ( )2 2 2

1 1n n n nOM O% % M − −= +

( ) { }2

2 nO% y x= =Q

( )22 n sayα =

)lso, ( ) ( ) ( )2 2 2

1 1 1 1n n n nO% OM % M − − − −= +

2 2 2

1 1

12

2n n nα α α − −= +

2 2

1

12

2 n nα α −⇒ =

1

1

2n nα α −⇒ =

1

1

2n n nO% α α −∴ = =

2 32 3

1 1

2 2n nα α − −= =

HHHHHHHHHHHHH.

HHHHHHHHHHHHH.

HHHHHHHHHHHHH.

11

1

2n

α −=

1

1 1 1

2 2 2n n− = = ÷

-(. rom figure

( )tan ........2

R" R" *

%R r α = =

( )tan ........

2 2

%( %(**

%R r

π α

− = = ÷

Bultipl"ing ;&s. i and ii,





( ) 2

.tan cot

2

R" %(

r α α =

∴ ( ) ( )2r R" %(=

--. Let ( ),C h k %e the centre of the circle passing through the end points of the rod )A and 7D oflengths a and % respectivel", CL and CB %e perpendiculars from C on )A and 7D respectivel". Then

C)C7 radii of the same circle

2 22 2 2 2

4 4

a bk h A# a and M% b⇒ + = + ∴ = =

( )2 2 2 24 h k a b⇒ − = −

-'. Let AO# θ =( )cos , sin A a aθ θ ∴ =

( )cos , sin M a a aθ θ ∴ = +cos x a a θ = +

( ) ( )cos .......... x a a *θ ⇒ − =

( )sin ..........and y a **θ =rom ;&s. i and ii,

( ) 2 2 2 x a y a− + =

2 2 2 0 x y ax⇒ + − =

2 2 2 x y ax⇒ + =





-. Let ( ),h k %e an" point in the set, then e&uation of circle is ( ) ( )2 2

x h y k − + − =

Aut ( ) 2 2, 25h k l*e on x y+ =

then 2 2 25h k + =2∴ ≤ =istance %etween the centres of two circles

≤ '

( )2 22 'h k ⇒ ≤ + ≤2 24 (4h k ≤ + ≤

( ) 2 2, 4 (4locus of h k *s x y∴ ≤ + ≤

'0. Let coordinates of ) and A are ( )α β and ( ),γ δ respectivel".22 ,a bα γ αγ + = − = −

and2

2 , & )β δ βδ + = − = −;&uation of circle with )A as diameter.

( ) ( ) ( ) ( ) 0 x x y yα γ β δ − − + − − =

( ) ( )2 2 0 x y x yα γ β δ αγ βδ ⇒ + − + − + + + =2 2 2 22 2 0 x y ax &y b )⇒ + + + − − =

( )2 2 2 2 Rad*us a & b )∴ = + + +

( )2 2 2 2a b & )= + + +

'1. Centre and radius of the circle2 2 2 2 0 x y gx fy+ + + = are ( ) ( )2 2

1 1,C g f and r g f − − = + and

centre and radius of the circle ( )2 2 2 2

1 1 2 1 12 2 0 x y g x f y are C g f + + + = +

( )2 2

2 1 1and r g f = + Circles touch each other.

∴ 1 2 1 2C C r r = ±

( ) ( ) ( ) ( )2 2 2 2 2 2

1 1 1 1 g g f f g f g f ⇒ − + − = + ± +

s&uaring %oth sides then

( ) ( )2 2 2 2 2 2 2 2 2 2 2 2

1 1 1 1 1 1 1 12 2 2 g f g f gg ff g f g f g f g f + + + − − = + + + ± + +

( ) ( ) ( )2 2 2 2 2

1 1 1 1 gg ff g f g f ⇒ + = + +2 2 2 2 2 2 2 2 2 2 2 2

1 1 1 1 1 1 1 12 g g f f gg ff g g g f f g f f ⇒ + + = + + +2 2 2 2

1 1 1 12 0 g f f g gg ff ⇒ + − =

( )1 1 0 gf fg ⇒ − =

1 1 gf fg ⇒ ='2. ;&uation of chord can %e written as x y a± = ±

8o, length of perpendicular from the centre 0, 0 of the circle to the chordJradius

ie, ' 42

aa

±< ⇒ <





'3.

8uppose chord %isect at ( ), 0 M λ , then

other end point of chord is ( ),h )−

where2

& hλ

+=

+hich lie on2 2 x y &x )y+ = +

or2 2 2h ) &h )+ = −

2 2

2 0h &h )⇒ − + =for two chords,

2 4 0 B AC − >or

2 24.1.2 0 & )− <or

2 2' & )>

'4. Let B/ %e the diameter of the circle whose e&uation is

4 - y x= + HHHHHHHi

and coordinates of ) and A are #3, 4 and 5, 4 respectivel".

;&uation of ⊥ %isector of )A is ( )1, 4 # =

( )1

4 10

y x− = − − ( )0 slo&e of AB =Q

∴ 1 x = HHHHHHHii

8olving ;&s. i and ii we get the coordinates of the centre of the circle as 1, 2

( ) ( )2 2

1 1 4 2 2O#∴ = − + − =

2 4 BC O# un*ts∴ = =

' AB un*ts=

tan 4 ' Area of rec gle ABCD X ∴ =

32 s&. unit.

'5. Let the circle %e ( ) ( )2 2 2 x h y k r − + − =

then( )2 2

1lh mk r l m+ +=

+

( ) ( )2 2 2 2 2 2 2 2 2 1 0or l h r m k r lmhk lh mk − + − + + + + =





also 2 24 5 ( 1 0l m l − + + =

Comparing the coefficients of similar terms2 2 2 24, 5, 0h r k r hk − = − = − =

2 (, 2 0h k = =

20, 3 4k h and r ∴ = = − =5r ∴ =

@ence, centre ( )3, 0 and radius 5 .

'(. ?C is the diameter of the circle

0 0,

2 2

g f C*rcumcentre

− − ∴ = ÷

,2 2

g f = − − ÷

'-. +e have,2 22 3 2 0 x xy y− − =

( ) ( ), 2 2 0,*e x y x y− + = which represents a pair of 7erpendicular lines passing through the origin.

( ) ( )2 2

2 16e

4 4)u*red area

π π ∴ = −

3

4 s) un*t π =

''. Let line is 1 x y

a b+ =

( ) ( ), 0 0, A a and B b∴;&uation of circle since )A is diameters is

( ) ( ) ( ) ( )0 0 0 x a x y y b− − + − − =2 2 0 x y ax by⇒ + − − =

Tangent at 0, 0 is

0ax by+ =2

2 2

a AM m

a b∴ = =

+





and

2

2 2

b BN n

a b= =

+

( )2 2m n a b d*ameter of c*rcle⇒ + = + =

'. Q The point [ ] [ ]( )1 , & &+ lies inside the circle2 2 2 15 0 x y x+ − − = , then

[ ] [ ] [ ]2 2

1 2 1 15 0 % % % + + − + − < page no. '15 &.no#1

[ ]( ) [ ] [ ]( )2 2

1 2 1 15 0 % % % ⇒ + + − + − <

[ ] [ ]2 2

2 1( 0 ' % % ⇒ − < ⇒ < HHHHHH.i

∴ Circles are concentric.

∴ 7oint [ ] [ ]( )1 , % % + out side the circle

2 22 - 0 x y x+ − − =

∴ [ ] [ ] [ ]( )2 21 2 1 - 0 % % % + + − + − >

[ ]( ) [ ] [ ]( )2 2

1 2 1 - 0 % % % ⇒ + + − + − >

[ ] 2

2 ' 0 % ⇒ − >

[ ] ( )2

4 ........... % **∴ >rom ;&s. i and ii, we get

[ ] 2

4 ' % < < which is impossi%le

∴ or no value of K7the point will %e within the region.

0. The image of the circle has same radius %ut centre different. !f centre is ( ),α β , then

( )2 2

2 3 2 13 2

1 1 1 1

α β − + −− −= =

+3 2 14α β ⇒ − = − =1-, 1(α β ∴ = =

∴ 6e&uired circle is ( ) ( )2 2

1- 1( 1 x y− + − =

02-08-14 sr.iplco jee main (2013) ptm-1 final key & solutions

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