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[1]TAITS / JM / PT-4 -A/ 2016IITians TAPASYA...IITians creating IITians

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JEE MAIN PART TEST - 4 SOLUTIONS PART - I PHYSICSQ.1 Ans.(B)Sol. Conservation of angular momentum of door + Bullet system about Hinge

2 2l l MImu m w2 2 3

Q.2 Ans. (B)Sol. Surface Tension decreases on increasing temperature.

Q.3 Ans.(C)

Sol. s s s b b 2b b b s s

l F l A Y 3 1 1aI F l A Y 2 b c

Q.4 Ans.(A)

Sol. For velocity v total energy of sphere is

22 2 2

21 1 2 v 5K mv mR mv2 2 3 R 6

2 2

2

5 6ghmv mgh v6 5

v sin2R Put h 1.8mg

Q.5 Ans.(C)

Sol. 2v r I

v r l2v r l

4l F 2.5 10l AY

Q.6 Ans.(C)

Sol. If wax is coated then Adhesive forces decreases. This increases .

Since 2Tcosh if hgr

Q.7 Ans.(A)

Sol. L m r v

drvdt



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Q.8 Ans.(D)

Sol. cc

F(I)l 1mmAY

s s cF(0.5) F(0.5) 1l mmAY A 2Y 4

l 1 0.25 1.25 mm

Q.9. Ans.(C)

Sol. Since dt5

then

C = 1 and a = 2Now compare dimensions on Both sides to find b.

Q.10 Ans.(A)

Sol.

21 1 1

22 2 2

Kx w l x ...(i)

Kx w l x ...(ii)

Now divide and put values.

Q.11 Ans.(B)

Sol.2mM TT g,m M A

Q.12 Ans.(C)

Sol.

Fv

Fv

2(mg B)

10cm/s 10cm/s

(mg B) (mg B)

Q.13 Ans.(C)

Sol. 2 2i1K MR w2

Apply cons. of Angular momentum

2 2 2 1MR w 2mR MR w

2 2 12f 1K MR 2mR w2

Q.14 Ans.(C)Sol. If v1 is velocity after collision



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1mVv

m M

if system moves for distance x

211 m M v m M gx2

and V 2gH10

Q.15 Ans.(C)

Sol.

Fa

Taking Torque about bottom point.

23 aFR mR2 R

3F ma2

.

Q.16 Ans.(C)

Sol.2maI

12

Q.17 Ans.(D)

Sol.T 2TP

radius R

Q.18 Ans.(B)Sol. In each collision velocity is interchanged. So it can assumed as if a single particle is moving on

the wire with velocity v time taken to cover L distance L 2nr tV

distance is reduced by the

factor = diameter x No. of Beads in each collision with rod momentum transferred = 2 mv timeinterval between two successive time interval between two successive collision = 2t

Average force = 22mV mv mv

2t t L 2nr

.

Q.19 Ans.(C)

Sol. 2 2L 2 0.6 12 0.6 0.8 14.4 .

Q.20 Ans. (A)

Sol.3.2

6

10

10 e 210e

3.22g



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Q.21 Ans.(D)

Sol.usin37

eucos37

usin37

just before collision just after collision

ucos37u

Net velocity upwards after collision is

y(eucos37)cos37 (usin37)sin37 v

2yvH

2g

Q.22 Ans.(A)

Sol. A has less deformation after elastic limit.

Q.23 Ans.(A)

Sol.2w 2mgTension3 3

Stress = 2mg3 A

u = 2stress1

2 y

Q.24 Ans.(A)

Sol.R R

R/2R/2

8M/7 M/7

mass density 33

M4 RR3 8

222 8M M RI R

5 7 7 2



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Q.25 Ans.(C)

Sol.

mg

A

N1

N2

N1

N2

1 2

2 1

A

1 1

N N .....(i)N N mg .....(ii)

0lN lsin N lcos mg cos ...(iii)2

Q.26 Ans.(C)

Sol. Change in length for both the strings are same.

Since left string has less length then strain and stress is more.

l1 l212

mg

C

2T T

B DA

c

1 21 1

3T mg ...(i)0

l l 122Tl T l 12 mg ....(ii)2

Q.27 Ans.(C)

Sol.

a

mgsin37

f = mgcos37

2

a g(sin37 cos37)fR I

mRmgcos37R2

for pure rolling after sometime

v = wR 0at R w t

Q.28 Ans.(C)

Sol. 2 2 r Ri l R kv v j

L m(r v)



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Q.29 Ans.(C)

Sol. 2 3

2 1/3

44 R d r3

r (3R d)2T / rratio4T /R

Q.30 Ans.(B)

Sol.2

43

dv 7 10F A 1 (100 10 ) 0.7Ndy 10



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SOLUTIONS PART - II CHEMISTRY

Q.31 Ans. (A)Soln First order reaction.

o2.303 PK lott P

Q.32 Ans. (A)Soln Rate a b0 0[A] [B]

a b0.1 k (0.012) (0.035) ............(1)a b0.8 k (0.012) (0.070) ............(2)

demide (1) by (2)b1 1

8 2

b 3

Rate does not depent on change in concentration of 0[A] .

Q.33 Ans. (B)

Soln Glycerin 2CH OH

CH OH

2CH OH

Q.34 Ans. (B)

Q.35 Ans. (B)Soln Sodium Nitroprusside 2 5Na [Fe(CN) NO]

Q.36 Ans. (B)

Q.37 Ans. (A)

Q.38 Ans. (A)Soln Pairing would take place in case of 46[Fe(CN) ]

2 5nM : 5 unpaired e

2oC : 3 unpaired 5e

2Fe : 50 unpaired e

Q.39 Ans. (D)



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Q.40 Ans. (D)

SolnCo

Cl

Cl

en

enCo

Cl

Cl

en

enBr Br Co Br

Cl

enen

Cl

2[Co(en) Cl Br]Cl

(i) (ii) (iii) (iv)

Q.41 Ans. (A)

Q.42 Ans. (D)

Q.43 Ans. (D)

Q.44 Ans. (B)

Q.45 Ans. (A)Soln x x y yN N

1/ 2 y 17xy x y x y

y 1/ 2 x

(t )N N i N N ; N 2 10

(t )

Q.46 Ans. (A)

Soln

(ii)27 4 30 '13 2 15 0Al He P n (y)

30 '14 1Al H (X)

30 o14 1Si e (Z)

Q.47 Ans. (A)Soln B B 1 B 2( ) ( )

1 2

2 2 2

1/ 2 B 1/ 2 B 1/ 2 B

n n n(t ) (t ) (t )

B A / 2 A

1 1 1t t 3t

AB

3 tt7

A

B

t 7t 3

Q.48 Ans. (A)

Soln f f bb

RK ; K k k (Pr oducts are stable)k

f bK k k (Reactants are stable)



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Q.49 Ans. (A)

Q.50 Ans. (C)Soln Ea /RTK A e

temperature dependent

Q.51 Ans. (B)Soln

Q.52 Ans. (A)

Soln (a) Fluoro alkene gives mainly Hofmann product due to E1cB type mechanism.

Q.53 Ans. (B)

Soln

nC4H9C C H + n-Bu Li nC4H9 C Li+ + nBuH nC4H9C C CH2CH2 O

H2 Lindlars catalyst

nC4H9C C CH2CH2OH

H+

PBr3 C = C

H

nC4H9

H

CH2CH2OH C = C

H

nC4H9

H

CH2CH2Br

Q.54 Ans. (B)

Soln

22 CHCHBrMg

O

BrMgOCHCH 22

OH2

OHCHCH 22

Q.55 Ans. (A)

Soln C : H : O = 60 13.3 26.7: :12 1 16 =

5 13.3 1.67: :1.67 1.67 1.67

= 3 : 8 : 1The formula of the compound is C3H8O.Since the alcohol gives positive iodoform test, it is 2-propanol.

CH3CHCH3

OH

The answer is (A)



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Q.56 Ans. (B)

SolnH

O

CH3

H O CH3Q.57 Ans. (C)Soln In the first step,

CH3

ONa

Is formed, which under goes Kolbes reaction

Q.58 Ans. (B)

Q.59. Ans. (A)Claises Rearrangement

CH3

O

CH3

CH2 CH CH214

CH3

OH

CH3

H2C CH CH214

(After rearrangement)

Q.60 Ans. (B)

3CH CNNBS

Br

Mg, either

MgBr

3H O

NCH3 MgBr

OH

COMe



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SOLUTIONS PART - III MATHEMATICS SOLUTIONQ.61 Ans. (B)

Soln x 0xlim 1

sin x

x 0

sin xlim 1x

Q.62 Ans. (D)Soln 2 2 2e hence {e } e 7 l

Q.63 Ans. (C)Soln Let z x iy

E z z (z 3)(z 3) (z 6i)(z 6i)

3z z 3(z z) 9 6(z z)i 36 (z z 2x; z z 2iy)

2 23(x y ) 6x 12y 45

2 23[x y 2x 4y 15]

2 2E 3[(x 1) (y 2) 10]

hence minE 30 when x 1 and y 2 i.e. z 1 2i

Q.64 Ans. (C)Soln Let Z a ib, b 0 where Im Z b

5 5 5 5 4 5 3 2 2 5 2 3 3 5 4 4 5 51 2 3 4Z (a ib) a C a bi C a b i C a b i C ab i i b

Im 5 4 2 3 5Z 5a b 10a b b

4 25

5mZ a ay 5 10 1m Z b b

Let 2a x (say),x R

b

2 2 2y 5x 10x 1 5[x 2x] 1 5[(x 1) ] 4

Hence miny 4 Ans.

Q.65 Ans. (B)

Soln 2 2 2z(1 z ) 0 z 0 or z i z 0 or z i Re(z) 0



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Q.66 Ans. (A)Soln z x iy

2 2x iy x y 1 7i

2 2x x y 1 ... (1)

and y 7 ... (2)

2x x 49 1

2 2x 49 1 x 2x

2x 48

x 24

2 2 2| z | x y 625 Ans.

Q.67 Ans. (C)

Soln ; Note that opposite angles are supplementary

Q.68 Ans. (D)

Soln z (3p 7q) i(3q 7p)

for purely imaginary 3p 7q p 7 or q 3 (for least value)

2 2 2 2 2| z | | 3 7i | | p iq | | z | 58(p q ) 58[7 9] 58 (D)

Q.69 Ans. (C)

Soln

13

1 1 1 1 1... 13 9 27 3 2x 9 9 9 3

1 1 1 .....3 9 27y 4

13

1134

144 2

r2 3

r 1

11 1 1 11 iz (1 i) ....... i11 i (1 i) (1 i) i1

1 i

Let x yz 3 2 i (4th quad.) 1 2Arg tan3



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Q.70 Ans. (D)

Soln23

(1, 0)

included

excluded

Q.71 Ans. (A)

Soln x 2(1 2 ) 16 25 ; x 2(1 2 ) 3 0 ; x x(4 2 ) ( 2 2 ) 0 x x(2 4) (2 2) 0

Q.72 Ans. (B)Soln

If | z i | | z i | 8 ,(0, 4)

(0, 4)

P(z)

O

(0, 1)

(0, 1)1 2PF PF 8

max| z | 4 (B) ]

Q.73 Ans. (C)Soln Note that is real

and rz 1 z 1 (z 1) (z 1) (z 1) (z 1) 2zz 2 0z 1 z 1 (z 1) (z 1) D

as 2z z | z | 1 (given) ]

Q.74 Ans. (C)

Soln Let z a ib z a ib

hence, we have 2008z z

2008| z | | z | | z |

2007| z | | z | 1 0

| z | 0 or | z | 1 ; if | z | 0 z 0 (0, 0)

if 2009 2| z | 1 z z z | z | 1 2009 values of z Total 2010 Ans.



IITians TAPASYA

Q.75 Ans. (B)Soln The point of intersection of the two lines are ( 1, 2)

P(1, 2) M(2, 3)

10

Distance PM 10

Hence the required line is one which passes through ( 1, 2) and is to P.M. B ]Q.76 Ans. (A)

Soln 2 2(2y x) (y mx) mx xy(2m 1) 2y 0 the equation to the pair of bisectors are :

2 2x ym 2

2 xy

2m 1

2 212x 7xy 12y m 41/ 38 ]

Q.77 Ans. (A)Soln y 1 m (x 1)

1y 1 (x 1)m

12h 1m

12k 1m

____________ locus is x y 1 ]

Q.78 Ans. (C)Soln equation of any line through (2, 3) is y 3 m(x 2)

y mx 2m 3

with the help of the fig. area of OAB 12

ie.1 2m 3 (3 2m) 122 m

B

AO(0,0)

.(2,3)(0, 32m)

taking sign me get 2(2m 3) 0

this gives one value of m 3 / 2

taking negative sign we get

24m 36m 9 0 (D 0)

quadratic in m gives 2 values of m

3 st. lines are possible. ]



IITians TAPASYA

Q.79 Ans. (B)Soln | PA PB | will be maximum if P, A & B will be collinear . Hence

2a b 5 0 and the determinant a b 14 2 1 02 4 1

]

Q.80 Ans. (D)Soln 1 1 2 2m m & m m equation is

1 2(y m x) (y m x) 0 where 1 22hm mb

& 1 2am mb

]

Q.81 Ans. (D)

Soln2

2n 2

n 1n

n 2 n 2

n 1 n 1n n

1 n 1n 2

n

11nLim

2

12]

Q.82 Ans. (C)Soln

Let f(x) ax b Case-I: f is increasing

f( 1) 3 and f(1) 5 f(x) x 4

Case-II: f is decreasingf( 1) 5 and f(1) 3 f(x) 4 x

Now verify options

Q.83 Ans. (B)

Soln put 1sin x

4

sin cosLim1 tan

4Lim cos 12

]

Q.84 Ans. (D)

Soln2

x k 22

3 51x xLim 4 1 x

x x

l

for existence of limit k 2 0 k 2 Ans. ]



IITians TAPASYA

Q.85 Ans. (C)

Soln

(A) This is false. f(x) x ; g(x) cosec x

now x 0Lim f(x) g(x)

exists 1 also x 0Lim f(x) 0

exist but x cLimg(x)

does not exist.

(B) This is false. Let f be defined as 1 if x 0f(x)2 if x 0

. Let g(x) 0 . Then, f(x) g(x) 0 , and

so x 0Lim f(x)g(x)

exists, while x 0Lim f(x)

does not.

(C) This is true. Notice that g (f g) f . Therefore , by the limit theorem,

x c x c x cLim g(x) Lim f(x) g(x) Limf(x)

(D) This is false.

Q.86 Ans. (A)

Soln1 a

a

xx1

a

log xcotxLimitasec

log x

; aax

log xas 0x

and

x

a

alog x

(using Lopital rule)

/ 2 1/ 2

l ]

Q.87 Ans. (D)

Soln

Limitx 0

tan { x } 1 sin { x }{ x } { x } 1

= 0xLimit f (x) = )1h(h

sinh.)1htan(Limit0h

= 1tan1

)1tan(

x 0

tan((1 h) 1) sin(1 h) sin1Limit(1 h) (1 h 1) 1

sin 1

Hence, limit x 0 f(x) does not exist ]

Q.88 Ans. (D)

Soln

(A) is correct; (B) is correct



IITians TAPASYA

(C) is also correct ]

Q.89 Ans. (A)

Soln f(x) 2 tan3x 5 2 sin3x has a period equal to 3

(A) has a period 3

, (B) has a period 23

, similarly for (C) and (D)]

Q.90 Ans. (C)

Soln 21 1f(x) f xx x

replacing 1xx

; 21 1f f(x) xx x

221 1x xx x

2 21 1x xx x

1 1 1x x xx x x

1xx

1x 1 0x

1 1x ; x 1x x

(rejected)

hence, x 1 or 1 maximum number of elements is {1, 1} (C) ]

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