Download - b.mat Fst IV Main Solns
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 1/21
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
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 2/21
2
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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.
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 3/21
3
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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=
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 4/21
4
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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 =
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 5/21
5
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 6/21
6
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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 −= ×
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 7/21
7
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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.
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 8/21
8
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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
= =
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 9/21
9
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 10/21
10
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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∴ = =
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 11/21
11
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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)
substituting (2) in (1) we get
= 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
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 12/21
12
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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)
substituting (3) in (1) we get
+ =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=
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 13/21
13
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 15/21
15
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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.
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 16/21
16
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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,
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 17/21
17
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 18/21
18
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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⇒ =
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 19/21
19
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 20/21
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)
8/12/2019 b.mat Fst IV Main Solns
http://slidepdf.com/reader/full/bmat-fst-iv-main-solns 21/21
21
∆ Brilliant Tutorials Pvt. Ltd. IIT/BMAT4/ PMC(MAIN) /Solns - 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)