b.mat Fst IV Main Solns

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1

∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 1

1. (1) When the rod is in horizontalposition, its total mechanicalenergy is +K.E. P.E Here =K.E. 0 but

= L

P.E. Mg2

relative to the lowest

position of its centre of mass,when the rod is in verticalposition

= = ω21P.E. 0 and K.E. I

2

where I is the moment of inertia.By law of conservation of energy

ω21 1MgL = I

2 2

= ω

2 21 1mL

2 3

∴ω =3gL

velocity of the centre of mass is

=cmL 3g

v 2 L

13gL

2=

The velocity of the lowest positionon the rod has a velocity

=cm2v 3gL

Now =3 x 10 x L 6 (given)

∴ =3 x10 L 36

L 1.2 m=∵

2. (2)

θ

When the rod makes an angle θ with the vertical, the forcesacting on the rod are

BRILLIANT’S

FULL SYLLABUS TEST 4FOR OUR STUDENTS

TOWARDS

JOINT ENTRANCE EXAMINATION, 2013

JEE 2013B.MAT 4 (MAIN) SOLNS

PHYSICS MATHEMATICS CHEMISTRY

SOLUTIONS

PART A : PHYSICS



2


(1) weight mg

(2) normal reaction R and

(3) tangential friction force F

Taking moments about O we have

τ = θ = α0L

mg sin I2

… (i)

where 0I is the moment of inertia

about O and α the angular

acceleration.

Now

22

0mL L

I m12 2

= + by parallel axis

theorem

22mL Lm

12 4= +

2mL3

=

Nowω ω θ ω

α = = = ωθ θ

d d d d.

dt d dt d

ω∴ θ = ω θ

2L mL dmg sin x ,2 3 d

on using (1)

Nowω θ

ω ω = θ θ∫ ∫0 0

3 gd sin d

2 L

( )ω∴ = − θ

2 3g1 cos

2 2L

( )∴ω = − θ2 3g1 cos

L

3 x 10 1 x 95 2

3= =

−∴ω = 13 rads

3. (3) The angular acceleration

β = − +2120t 48t 16

ω∴ = − +2d

120t 48t 16dt

( )ω

∴ ω = − +∫ ∫t

2

0 0d 120t 48t 16 dt

∴ω = − + +3 240t 24t 16t c

Since =t 0,

ω = 0, =c 0

∴ω = − +3 240t 24t 16t

Then linear velocity

= ω v r

( )∴ = − +3 2 v 1x 40t 24t 16t

[ ]=∵ r 1m

Tangential acceleration

( )3 2t

dv da 40t 24t 16t

dt dt= = − +

= − +2120t 48t 16

=∴ = − +

t 1

dv 120 48 16dt

288ms −=

4. (4)

θ

θ

m g s i n θ

mgcos θ

2a 3 ms −=

The forces acting on the block are asshown in figure.

(1) weight mg(2) pseudo force ma

(3) friction µ N, where N is normalreaction.



3


Normal reaction

N mg cos ma sin= θ + θ

∴ friction force

( )= µ θ + θmgcos masin

For equation of block, we have

θ + µ = θma cos N mg sin

∴µ = θ − θN mg sin ma cos

i.e, ( )µ θ + θmgcos mas in

= θ − θmgsin macos

i.e, ( )µ θ + θg cos a sin

( )= θ − θg sin a cos

i.e,

µ + 4 3

10 x 3 x 5 5

= −

3 410 x 3 x

5 5

∴ µ =49 18

∴µ = 1849

5. (1) For the particle to move with

constant velocity, we should

have

i.e, V x A mg− −

=

i.e, × θ = V A sin mg

mg V

A sin∴ =

θ

For V to be a minimum ( )θ =sin 1 max

mg V

A ∴ =

6. (2) Let A be the weight of the cable.

Then mass per unit length of the

cable is A L

. Consider a length

dx of the cable. The mass of the

cable is A

dx.L

Let this length

dx of the cable be raised through

a distance x measured from A.

Then work done against gravity

A dw dx x g

L

= × ×

Then total work done in winding

up the entire cable is

w L

o o

A dw dx x g

L= × ×∫ ∫

∴ = ×

L2

o

A x w g

L 2

2 A Lg

L 2= × ×

A g L2

× ×=

3310 9.81 20

98.1 10 J2

× ×= = ×

98.1kJ=



4


7. (3)

1 0

α

Let the pipe make an angle α tothe horizon. Let the cart movewith a velocity 1 v along x–axis

Then

c 1 v v i=

Let the rain fall vertically down.Then

r 2 v v j= −

When the rain goes along theaxis of the pipe making an angleα to the horizon, then

= − α − α rc v v cos i v sin j (1) where v

is the magnitude of the velocityof the rain

But rc r c v v v = −

2 1 v j v i= − − ...(2)

From (1) and (2) we get

2 1 v cos i v sin j v j v i− α − α = − −

6j 2i= − −

v cos 2∴− α = −

v cos 2∴ α = …(3)

v sin 6− α = −

i.e, v sin 6α = ….(4)

(4)gives tan 3

(3)∴ α =

From the3

ABC, sin10

∆ α =

v sin 6∴ α =

i.e, 6 6 v 10 2 10

sin 3= = =

α

−= 140 ms

8. (4) Let u be the initial velocity and θ ,the angle of projection. Thedirection of the projectilebecomes horizontal after 3seconds. Here the projectile is at

its greatest height.usin

3,g

θ∴ = where θ is the

angle of projection

u sin 30∴ θ = …(1)

Let v be the velocity of theprojectile after 2 seconds. Then

= θ v cos 30 u cos …(2)

(horizontal component)

Also v sin u sin gtθ = θ −

30 10 2= − ×

10=

10 v 20 m / s

sin 30∴ = =

Putting the value of v in (2), weget

× = θ3

20 u cos2

∴ θ =u cos 10 3 …(3)

From (1) and (3) we gettan 3θ =

60∴θ =



5


9. (1)

The given figure can be dividedinto two rectangles namely,OABC and CFED and one square,namely GHIJ.

Let 1 2 3C ,C , and C be the centres

of mass of the respective areasmentioned. The area of the

rectangle OABC is 24 1 4 cm× =

The area of the rectangle CFED is23 1 3 cm× =

The area of the square GHIJ is21 1 1cm× =

The mass of any section will beproportional to its area. The

coordinates of 1C are 1

,22

The coordinates of 2C are

3 9

,2 2

The coordinates of 3C are

3 5,

2 2

The x coordinates of the centreof mass are

× + × + ×=

+ +

1 3 34 3 1

2 2 2 x 4 3 1

= ≈8

cm 1cm8

The y coordinates of the centreof mass are

× + × + ×=

+ +

9 54 2 3 1

2 2 y 4 3 1

+ + =

27 58

2 28

16 27 5 483 cm

2 8 16

+ += = =

×

10. (2) Considering angular momentumabout the centre of mass of thebar, we get

+ = ω1 2m u y m u y I …(1)

Here2

2MlI 2 my

12

= +

220.16 (1.5)

2 0.08 (0.5)12×

= + × ×

1.44 0.1648 4

= +

1.44 1.92 3.3648 48+= =

20.07 kg m=

∴ × × + × ×0.08 10 0.5 0.08 6 0.5

= ω0.07

i.e, + = ω0.4 0.24 0.07

∴ω = 0.640.07

−= 19.1rad s



6


11. (2)

d θ

dx

θ

Consider an element dx on thering. Let it subtend an angle θd

at the centre of curvature C.Then

= θdx R d

The mass of dx isM

dx Rπ

= θπM

R dR

= θπM

d

The force on m due to dx is

dx

dx

dF cos

dF sin

dF sin

= θπM

dF G m d along the radius

towards dx. Consider a similarelement dx locatedsymmetrically below P. The sin θ

components on resolution get

cancelled, but cos θ components get reinforced.Hence the interaction force on mdue to M is

π

−π= θ θ∫

π

/ 2

2 / 2

GMmcos d

R

π= θ θ∫

π

/ 2

20

GMm2 cos d

R

= ⋅

π 2GMm

2 1R

Hence the interaction force on m

due to M is 2 in units ofπ 2

GMm

R.

12. (3) When the glass ball is placed invacuum the change in pressure is

5 2P 1 10 Nm∆ = × . Corresponding

to this change in pressure there

will be a fractional change V

V ∆

in the volume of the ball. Thepercentage change is

V 100.

V ∆

× we have ∆

=∆

PB

V V

where P∆ is the stress and V

V ∆

the strain.

Now 5 2P 1 10 Nm (given)∆ = ×

510 1 10

4.0 10 V

V

×∴ × =

∆

55

10 V 1 10

0.25 10 V 4 .0 10

−∆ ×∴ = = ×

×

i.e, 5 V 100 0.25 10 100

V −∆

× = × ×

525 10 −= ×

42.5 10 −= ×



7


13. (4)

From continuity equation wehave

=1 1 2 2 A V A V

Now ( )= π × =22 2 22 A V V Qr

where Q is the volume rate offlow

Hence =22

Q V

A

( )

4

26.43 10

0.04

−×=

π ×

0.128 m / s=

By virtue of equation of continuity

= 2 2

1 1

A V V

A

=

22

1 221

r V V

r

20.040.128

0.1

= ×

40.128

25

= ×

0.02m/s=

By Bernoulli’s equation,

+ ρ = + ρ2 21 21 2

1 1P P V V

2 2

( ) ( )ρ

∴ − = × −2 2

1 2 12P P V V

2

But ( )1 2P P− is the work done per

unit volume.

( ) ( )3

2 2 W 1.25 100.128 0.02

V 2×

∴ = −

= 310.0 J/ m

14. (1) The work done in forming thebubble under isothermalconditions is stored as surfaceenergy of the drop. The surfaceenergy as per unit area is T,where T is the surface tension. Thesurface area of the bubble is

22 4 R= × π

28 R= π

The surface energy is 28 R T= π

∴ work done ∝ 2R (square of the

radius)

15. (2) The situation is shown in figurebelow.

A

P

C

B

echo

Let v be the velocity of themotorist. He travels a distance ofv metre in one second. The firstecho is heard by the motoristafter one second, that is, afterthe sound has travelled adistance of 330 m through theleast path. Hence the path takenby sound is AB+BC as shown inthe figure.



8


It is clear that, = v

AP2

AB BC 330 m+ =

and = +2 2 2 AB AP PB

165PB

2= , where 165 m is the

width of the road.

∴ + =

2 2 v 1652 330

2 2

( )

+ =

2 22 v 165

1652 2

( ) ( )∴ + =22

2165 v 165

4 4

( )∴ = × 22 v 3 165

∴ = × v 3 165

= v 286 m / s

16. (3) We have 1 1 1u v t

+ =

i.e, ( )− + = =∵1 1 1 r 2 f10 v 10.5

∴ = +1 1 1 v 10.5 10

×∴ = =

10.5 10 v 5.12 m

20.5

For magnification( )

−= =

− v v

mu u

5.120.51

10= =

The radius of the circular image is0.51

The circumference of the circularimage is

2 0.51π ×

Time taken = 2s

∴ Speed v2 0.51

2π ×

=

0.51 1.6 m / sπ × =

Since the image is erect, theparticle’s sense of rotation asseen from A, is clockwise in theimage.

17. (4) As per the problem, the conditionfor nth order fringe of longerwavelength Lλ to coincide with

the (n + 1)th order of the shorterwavelength λ s is (n + 1) λ = λs Ln

( )+∴ = =

n 1 6000 4n 4500 3

i.e, 3n 3 4n+ =

n 3∴ =

18. (3)

Let A be the amplitude of theresultant simple harmonic motion.If ω is the angular frequency ofthe particle, then

ω = A 1910

ω =2 A 599736

∴ω = 599736

1910

314 rad / s=

Amplitude1910

A 6.1cm314

= =



9


Now ( ) ( )

= −∑ ∑ 22

x y A A A

Now x A 4 3 cos= + θ

= θ y A 3 sin

( )∴ 26.08

( )= + θ + θ2 24 3 cos 3 sin

= + θ + θ2 237 16 9 cos 9 sin

+ θ24 cos

16 9 24 cos∴= + + θ

θ = =12 1

cos24 2

603π

∴θ = =

19. (1) For an ideal gas PV RT=

The given law is 2oP P V = − α

Substituting forRT

V P

= in the

given law

2

oRTP PP = − α

2 2

o 2R T

PP

= − α

22

o2R

T P PP

∴α = −

2 2 2 3oR T P P P∴α = − …(1)

Differentiating both sides w.r.t., Pwe get

2 2o

dTR T 2 P P 3 PdPα = −

For dTdP

to be maximum

dT0

dP=

2o2 P P 3 P 0∴ − =

o2 P 3 P∴ =

i.e, o2

P P3

= …(2)

substituting (2) in (1) we get

2 32 2 o o4P 8P

R T P3 27

α = −

= −3 3

o o4P 8P

9 27

−= =

3 3 3o o o12P 8P 4P

27 27

32 o

24P

T27 R

∴ =α

12o o2P P

T3R 3

∴ =

α

20. (2) Work done W in an adiabaticchange is

−= γ =γ −

p2 2 1 1 v

CP V P V W where1 C

For an adiabatic process

γ γ =1 1 2 2P V P V

( )γ

γ ∴ = =

1

2 1 12

V P P P 3

V

Further = +P v C C R

3R 5RR

2 2= + =

∴ γ = =5R 523R 32



10


γ × −∴ = γ −

1 1 1P 3 P V W 1

( )γ −=

γ −

5110 3 V

1

i.e,( )5

110 6.2 V 957

23

× −=

153 957

6.2 V 2 10

∴ × = −

∴ = − × 53 9576.22 10

1 V

6.2 0.014355= −

= − =1V 6.2 0.01 .6 19

But 1

2

V 3

V =

∴ = = ≈26.19

V 2.06 2 litres3

21. (3) We have toN N e −λ= (law of

Radioactivity)

Rate of disintegration R is

todNR N edt

−λ= − = λ

When t 0,= Rate = − = λo odN

R Ndt

At = = − X x dN

t 2hr, for R , Rdt

− λ= λ 2oN e

Since Too

NN e

2−λ=

1T Te 2 or e 2λ λ= =

i.e,−

−λ =1

Te e

( )

2

T x oR N 2 as T 1hr

−

∴ = − λ = 2R2−=

−

= − λ2T

y oR N 2

( )−= =1R2 as T 2 hr

2 x

1 y

R 2 2 1R 4 22

−

−∴ = = =

22. (4) ( )= +N MSD N 1 VSD

∴ =+N

1VSD MSDN 1

LC 1MSD 1VSD∴ = −

Na a

N 1

= −

+

( )( )

a N 1 NaN 1+ −

=+

( )+ −

=+

a N a NaN 1

( )a

L.C.N 1

∴ =+

23. (1) The resistance of each coil afterdivision is

1 2 nR 6

R R ... Rn n

= = = = =

When the coils are connected inparallel effective resistance effR

becomes 21 R Rn n n

=

Now steady state current is

= = 2

e ff

E 12I . n

R R

i.e, 2128A . n A

R=

2 212n 2 n

6= =

2n 4 i.e, n 2∴ = =



11


24. (3)

Let v be the maximum velocity ofthe bob in the equilibrium

position. In the extreme position,its kinetic energy transforms intopotential energy

∴ =21mv mgh

2

∴ = v 2gh …(1)

From the figure,

( )h CA 1 cos= = − θ

2h 2 sin / 2= θ …(2)


= θ 2 v 4g sin / 2

∴ = θ v 2 sin / 2 g

The emf induced at any instant is= = θ e Bv B 2 sin / 2 g

θ∴ = θ × ×

d d 12 B g cos / 2

dt dt 2

( )= ω θ B g cos / 2, where ω is

angular velocity

Now max d e when 0dt = θ =

Thusmax

doccurs at 0

dt

θ =

25. (4) The equation for the saw–toothwave is

oo o

2V 2t V t V V 1

T T

= − = −

∫

= = −∫ ∫

T2 T

20av oT

02

0

Vdt2 2t

V V 1 dtT T

dt

= − =

oav o

2 T T V V V

T 4 2 2

26. (4) When the two equal capacitorsare connected in series across asource emf E,

then p.d. across each isE2

…(1)

When one of the capacitors isfilled with a dielectric thecapacitance of that capacitorbecomes EC. Then the effectivecapacitance of the seriescombination is given by

( )ε ε= =

+ ε + εeffC C C

CC C 1

The charge on any one of the

capacitors is ε= =+ εeff E CQ C E

1

Then potential of the capacitorwith the dielectric is

ε= =

ε + ε εQ E C 1

.C 1 C

=+ εE

1 …(2)

∴ Decrease in potential of thecapacitor with the dielectric

= −+ ε

E E2 1

( ) + ε −=

+ ε 1 2E2 1

( )( )ε −

=+ ε

1E

2 1



12


27. (2) Let 1 2R and R be the resistance

of the two coils. Then the highestresistance in series is

1 2R R 16+ = …(1)

The least resistance is obtainedby connecting 1 2R and R inparallel.

Then =+

1 2

1 2

R R3

R R

1 2R R 3 16 48∴ = × = …(2)

21

48R

R∴ = …(3)


+ =11

48R 16R

∴ − + =21 1R 16 R 48 0

21 11 12 R 4 R 48 0R∴ − − + =

( ) ( )∴ − − − =1 1 1R R 12 4 R 12 0

( ) ( )∴ − − =1 1R 12 R 4 0

Hence =1R 4 or 12

Hence the resistance values are4 and 12 Ω

28. (3) The current ABCi in part ABC of

the outer coil is equal to thecurrent ADCi which is equal to2.5 A. The current in the portionGPHi of the inner coil is equal to

the current in the portion GQHi ofthe inner coil. Since the currentsin the two halves of the coils arein opposite direction, the netmagnetic induction at O byeach coil is zero. Hencemagnetic induction at O due to

the long wires is µ

= π o

o

i2

4 r

whereor is the radius of the outer

coil7

52

4 10 2.52 10 T

4 5 10

−−

−

π ×= × × = π ×

29. (4) ∝ ∝n n2

1E and J n

n

∴ ∝n 2n

1E

J

n 2n

1E k

J= where k is a constant

Let nE y = and

2n

1 x. then

J=

y kx = which is a straight line.

30. (1) Photon energy = ν =λ

hcE h where

c is the velocity of light

( ) ( )34 8

7

6.6 10 3 10

4.8 10

−

−

× × ×=

×

194.125 10 J−= ×

The rate of emission of photonfrom the source

1819

1.02.425 10 sec

4.125 10 −= = ××

Number of photons striking per

square metre per second on theplate is

18

22.425 10

n cos 604 3.14 2

×= ×

× ×

182.425 10n

32 3.14×

=×

16n 2.41 10= ×

In multiple of 1610 , n 2.41=



13


31. (3) If , δγ are the roots, then

( )sin 2, 1 sin+ δ = α − γδ = −γ + α

sum of squares of the roots,

( )22 2S 2= γ + δ = γ + δ − γδ

( ) ( )2sin 2 2 1 sin= α − + + α

( )22sin 2 sin 6 sin 1 5= α − α + = α − +

S is least sin 12

π⇒ α = ⇒ α =

32. (2) Let ( ) ( ) ( )1 2 3P t , Q t and R t

Equation of normal to

2 3y 4ax is y xt 2at at= + = +

This passes through (h, k)3k th 2at at⇒ + = +

( )3at 2a h t k 0⇒ + − − =

1 2 3t , t , t are the roots of this

equation.

1 2 3t t t 0∴ + + =

Now, centroid PQR∆ is

G ( ) ( )2 2 21 2 3 1 2 3

a 2at t t , t t t

3 3

+ + + +

We have

( )1 2 3 1 2 32at t t 0 t t t 03

+ + = ⇒ + + =

⇒ G lies on y = 0

33. (1) The given equation is linear in yand can be written as

2 2dy x ax

ydx 1 x 1 x

+ =− −

Its integrating factor is

( )22

x dx 1/ 2 log 1 x1 xe e

⌠

⌡ − −− =

2

22

1if 1 x 1

1 x1

if x 1x 1

− < <−=

>−

( )3 / 22 2

1 ax dxy

1 x 1 x

⌠

⌡

=− −

( )3 / 22

1 2xa. dx

2 1 x

⌠

⌡

−= −

−

2 2

1 ay c

1 x 1 x⇒ = +

− −

2y a c 1 x⇒ = + −

( ) ( )2 2 2y a c 1 x⇒ − = −

( )2 2 2 2

y a c x c⇒ − + = , whichrepresents an ellipse for 1 x 1− < < .

If 2x 1,> then the solution is of the

form ( )2 2 2 2y a c x c− − − = , which

represents a hyperbola.

34. (1) 2x iy 1 2t i 4t 2t 2+ = − + + +

2x 1 2 t and y 4t 2t 2∴ = − = + +

Eliminating t,

22 3 7

y x 2 4

− − =

which is a hyperbola.

PART B: MATHEMATICS





15


41. (3) If the last two digits are equal to

zero, then the first digit may beany digit from 1 to 9.

If the last two digits are equal toany number other than 0, thenthe first digit may be selected in8 ways.

∴ The required number

= 9 9 8 81+ × =

42. (1) Required number of

arrangements

= 1 1 1 15!2! 3! 4! 5!

− + −

60 20 5 1= − + − = 44

43. (3) + + +2 30 1 2 3a a x a x a x

( )n2n 2

2n.... a x 1 x x+ + = + +

Differentiating w.r.t. x,

21 2 3a 2a x 3a x ....+ + +

2n 12n2na x −+

( ) ( )n 12n 1 x x 1 2x

−= + + +

x = 1

( )⇒ + + + +1 2 3 2na 2a 3a ... 2n a

n 13 n . 3 −=

nn . 3=

44. (1) Equations of planes in twosystems are

x y z x y z1and 1

a b c P q r+ + = + + =

Since the origin is the same, the

distance of origin from the planeis also the same.

2 2 2 2 2 2

1 11 1 1 1 1 1

a b c p q r

∴ =+ + + +

⇒ (1) is the correct choice.

45. (2) c 3a 4b, 2c a 3b= + = −

⇒ = −

5c 13b 13

c b5

⇒ = −

c and b have opposite

direction and13

c b 2 b5

= >

46. (1)1 1 1

a b c1 i

a b c+ + = +

2 2 2

2 2 21 1 1

a b c

a b c⇒ + +

1 1 1 1 1 1

ab bc ca2i 2 a b b c c a

= − + +

1 1 1

1 1 1

c b aabc2i 2

a b c c b a

= − + +

2i 0 2i= − =

47. (3) 183! has 10 as a factor.

Now, ( )91183 23 3 3=

( ) ( )913 10 1 3 10 k 1= − = − 10t 7= +

∴ The required digit at the unit

place is 7.



16


48. (4) ( ) ( )x 0

f 0 0, Lt f x→

=

( )22

x 0Lt x x→

= −

( ) ( )− →→

= − = −x 0x 0

Lt f x Lt 0 1 1

( ) ( )+ →→

= − =x 0x 0

Lt f x Lt 0 0 0

( )f x⇒ is discontinuous at x = 0

and also discontinuous on z.

Now,

( ) ( ) ( )→ → = = − = 22x 1 x 1

f 1 0, Lt f x Lt x x 0

( )f x∴ is discontinuous at all

integers except at 1.

49. (2) Let⌠

⌡

π+

=+

/ 2

0

3 sec x 5 cos ec xI dxsec x cos ec x

⌠

⌡

π+

=+

/ 2

0

3 cosec x 5 s ec xdx

sec x cos ec x

( )

/ 2/ 2

00

f x 2x f x dx2

⌠ ⌠ ⌡

⌡

π π π = −

∵

/ 2

0

8 sec x 8 cos ec x2 dx

sec x cos ec x

⌠

⌡

π+

∴ Ι =+

8. 42π

= = π

2∴ Ι = π

50. (3) ( )5002003 4 32 2 . 2=

( ) ( )500 50020032 8. 16 8 17 1⇒ = = −

20032 8.⇒ = ( ) ( )−1

500 499C17 500 17

( ) + − +499C... 500 17 1

20032 88 ,

17 17⇒ = Ι + where

( ) ( )Ι = −1

499 498C17 500 17

+ −500499... C

such that I is an integer

20032 817 17

∴ =

51. (2) ( ) ( ) 21 1

f x x f x 1x x

′= + ⇒ = −

( ) 21

f x 0 1 0x

′ = ⇒ − =

x 1⇒ = ±

( )3

2f x

x′′ =

( )f 1 2 0′′ = >

( )f x⇒ has minimum at x = 1.

52. (2)

( )2B z′

As ABC is an isosceles righttriangle, right angled at B,



17


BA = BC and ABC 90º∠ =

1 2 3 2z z z z⇒ − = − and

arg 3 2

1 2

z zz z 2

− π= ±

−

3 2 3 2

1 2 1 2

z z z zz z z z

− −⇒ =

− −

cos i sin i2 2

+π +π + = ±

( ) ( )2 2

3 2 1 2z z z z⇒

− = − −

( ) ( )2 21 2 2 3z z z z 0⇒ − + − =

53. (3) y x= α + β is a tangent to

2 22 2 2 2

2 2x y

1 a ba b

− = ⇒ β = α −

∴ Locus of ( ),α β is

2 2 2 2y a x b= −

( )2 2 22

1x y ba

⇒ = + , which is a

parabola.

54. (1) ( )f x is continuous in a,b and

differentiable in (a, b).

Also ( )′ = +f x 2 x m

∴ By Lagrange’s Mean value

Theorem, there exists ( )0,1θ∈

such that ( ) ( ) ( )−

′ + θ = −f b f a

f a h b a

where h = b – a.

( )2. a h m⇒ + θ +

( )2 2b mb n a ma n

b a

+ + − + +=

−

( ) ( )2 a h m b a m⇒ + θ + = + +

a ba h2+

⇒ + θ =

( )b a b ah b a

2 2− −

⇒ θ = ⇒ θ − =

1

2⇒ θ =

55. (4) The given equation can be

written as 2 2x dy y dx

dxx y

−= −

+

2 2

2

x dy y dx 1dx

x y1x

−⇒ × = −

+

2

2

1 d ydx

dx xy1x

⇒ = −

+

Integrating, we get

1 ytan x cx

− = − +

( )y x tan c x⇒ = −

56. (4) Let a = cis A, b = cis B, c = cis C

Then

a + b+ c = cis A + cis B + cis C

( )cos A cos B cos C= + +

( )i sin A sin B sin C+ + +

= 0



18


3 3 3a b c 3 abc⇒ + + =

cos 3A cos 3B cos 3C⇒ + +

( )3 cos A B C= + +

57. (3) 21 1 3 1 1 3

A 5 2 6 5 2 62 1 3 2 1 3

= − − − − − −

0 0 03 3 91 1 3

=− − −

30 0 0 1 1 3

A 3 3 9 5 2 61 1 3 2 1 3

=

− − − − − −

0 0 00 0 00 0 0

=

⇒ A is a nilpotent matrix ofindex 3.

58. (2) AP = PQ = QB

The coordinates of P are (a, 0),and of Q are (2a, 0). Equations ofthe circles on AP, PQ and QB asdiameters are respectively

( ) ( ) 2x 0 x a y 0,− − + =

( ) ( ) 2x a x 2a y 0− − + = and

( ) ( ) 2x 2a x 3a y 0− − + =

So if (h, k) be any point on thelocus, then

( ) ( )( )2h h a k h a h 2a− + + − − +

( ) ( )2 2 2k h 2a h 3a k b+ − − + =

( )2 2 2 23 h k 9ah 8a b⇒ + − + =

∴ The required locus of (h, k) is

( )2 2 2 23 x y 9ax 8a b 0+ − + − =

59. (2) ( )A 7=

07,16,25, 34, 43, 52, 61, 70=

( )B 0=

00, 01, ... 09,10, 20,....90=

( ) ( ) A 7 B 0 07, 70= ∩ = =

( )P A 7/B 0∴ = =

( ) ( ) ( )

P A 7 B 0 2P B 0 19

= ∩ == =

=

60. (3) ( ) ( )2 2log sin x log cos x−

( )22log 1 tan x 1− − = −

2 2tanx

log 11 tan x

⇒ = −

−

12

tan x 12

21 tan x−⇒ = =

−

22tan x 11 tan x⇒ =

−

tan 2x 1⇒ =



19


61. (2)

(2) Mass of 1 mole of sulphur

( )8S 32 8 256g= × =

(1) Mass of 1 gram atom ofsodium = 23 g

(3) 22.4 L of 2N at STP = 1 molar

mass of 2N 28g=

(4) 1 Avogadro number of

2 2CO 1mole of CO 44 g= =

62. (1)

2 2 7 2 4K Cr O 4H SO+ →

2 23SO 3 O 3H O+ + →

+ + 24H O 3 O

2 43H SO

+ + →2 2 7 2 4 2K Cr O H SO 3SO

+ 2H O

( )+2 4 2 4 3K SO Cr SO

( )+2 4 2 4 3K SO Cr SO

63. (3)h

x p4

∆ ∆ =π

hx

4 p

∆ =π ∆

−

− −

×=

× × ×

34

5 16.626 10 Js

4 3.14 1 10 kg ms

305.27 10 m−= ×

64. (1)

65. (3) 2 2H O Bond order 1, largest bond

length

3O Bond order 1.5, medium

bond length

2O Bond order 2.0, shortest bond

length.

Correct order of – O – O- bond

length

2 2 3 2H O O O> >

66. (1)

67. (2)

68. (4) + 2 2 3 1N 3H 2NH ; K …I

Reversing the above equation

+3 2 21

12 NH N 3H ;

K …II

2 2 2N O 2NO ; K+ …III

2 2 31H O H O ; K2

+ …IV

Multiply equation IV by 3

32 2 4 33H 3 / 2 O 3H O K K+ → = = …V

3 2 2 55

2NH O 2NO 3H O; K2

+ +

× ×= = =

3 32 3 2 3

5 51 1

1 K K K KK K

K K

69. (4) PMd ,RT

= Pressure = 82 atmos,

Molar mass of =2H 2, T = 27 + 273 =

300K. R = 0.082 Latmos 1 1deg mol− −

82 2d 6.667 g/L0.082 300

×= =

×

PART C : CHEMISTRY



20

∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/PMC(MAIN)/Solns - 20

70. (3)

71. (4) Unit of rate constant and rate of

reaction are same 1 1mol L s− − for

zero order reactions.

72. (2) ∆ =f fT i K m

∆ =fTP 1.1º

=f 2K H O 1.86

m 0.2=

∆= = =

×f

f

T 1.1i 2.95K m 1.86 0.2

( )= + − αi 1 n 1

n = number of ions formed by the

dissociation of 1 mole of

electrolyte

+ −−= = + =2 4 4n K SO 2K SO 3

α - degree of dissociation of

electrolyte in solution.

( )= + − α2.95 1 3 1

α =2 1.95

α = 0.975 or 97.5%

73. (1)

(1) Salt formed by the neutralization

of weak base with strong acid will

be acidic

( ) ( )+ −+ 4 s 4 aqNH C NH C

( )+ ++ +4 2 4 aqNH H O NH OH H

( )( ) ( ) ( )

( ) ( )

+ −−

+

+

+

4 aq4 2 4 24 aq

2

3 4 3 4aq

2 NH C O 2NH C ONeutral.

3 CH COONH CH COO NH

(4)( ) ( )

+ −+ aq aq

KC K C

74. (4) ( ) ( ) ( )

+ −+2H O

S aq aqNaOH Na OH

complete ionization.

( )1

NaOH0.4 1000

M 0.1 1040 1000

−×= = =

×

1OH − in solution 110 M−

−+ −

−− = = =

1413

1kW 10

H 1010OH

+ − = − = − = 13pH log H log 10 13

75. (3) ( ) ( ) ( )g g gC CΙ → Ι +

Bond energy =

∑Bond energy of reactants

Bond energy of product− ∑

( ) −= + − 1107 121 18 J mol

= 1210 J mol −

76. (4) Negatively charged particles are

precipitated by oppositely

charged + ions only

77. (4)



21


78. (1) 3 2CH CH CHO HCN+

→ − − − ≡|

3 2|

H

CH CH C C N

OH

→ − −4LiA H3 2 2 2CH CH CHOHCH NH

°1 amine

79. (2) meso–2, 3 – butane diol

≡

MeOH

HO H

H

Me

HO

HO H

H

Me

MeMe

HH

OHOH

Me

≡

A C

80. (4)

81. (1)

82. (2) More the electron withdrawingnitrogroups by resonance and

–I effects at ortho and para

positions with respect to Br, make

it to ionise easily facilitatingnucleophilic substitution. 2NO

group at meta position can actas –I group only.

83. (4) + → 2 5 2RCH OH PC RCH C

3 2POC H O+ +

5RCOOH PC RCOC+ →

3 2POC H O+ +

→

3 5 2 3RCOCH 2 PC RCC CH+ →

3 22POC 2H O+ +

− − − + →3 2 3 5CH CH O CH PC

3 3CH CHC O CH− −

+ +3 2POC H O

84. (2) It is a linear complex with Ag in sp

hybridised state

85. (2)

86. (2)

87. (4) Pure copper is obtained by

electrolysis

88. (2) 4 2CaSO .2H O is used in cement

to delay the setting of cement

89. (2) In metals valency electrons form

of pool of electrons responsible

for specific properties of metals

called metallic bond.

90. (1)

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b.mat Fst IV Main Solns