JEE MAIN ONLINE EXAMINATION - 2017 - Daily Careers · JEE MAIN ONLINE EXAMINATION - 2017 QUESTION...

JEE Online Paper _ Date [9-4-2017] (Page # 1)

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[MATHEMATICS]

1. If the line,

x 3 y 2 z1 1 2

lies in the plane, 2x– 4y +3z=2, then the shortest distance

between this line and the line,

x 1 y z12 9 4

is :

(1) 1 (2) 2 (3) 3 (4) 0Sol. 4

pt (3, –2, ) on pline 2x – 4y +3z –2 = 0= 6 + 8 – 3 –2 = 0

= 3 = 12 4Now

x 31

=

y 21

=

z 42

= k1 ...(1)

x 112

=y9

=z4

= k2 ...(2)

pt on give 1 = (k1+3, – k1 – 2, –2k1 –4)pt on give 2= (12 k2 + 1, 9k2, 4k2)k1 + 3 = 12k2 + 1 | –k1 –2 = 9k2 | –2k1 –4 = 4k2

2

1

k 0k –2

p (1,0,0)gives are ditersech – thortest distance = 0

2. The coefficient of x–5 in the binomial expansion of

10

2 1 13 3 2

x 1 x 1

x xx x 1where x0,1 is :

(1) –1 (2) 4 (3) 1 (4) –4Sol. 3

101/3 2 /3 1/3

2 /3 1 /3(x 1)(x x 1) ( x 1)( x 1)

x( x 1)(x x 1)

= (x1/3 + 1 –1 –1/x1/2)10

= (x1/3 – 1/x1/2) r = 1/3, b = 1/2

r =

10( 5)

31

1 /32

r =

25 /310

1(5 )

2cos. = 10c10 (1) (–1)10 = 1

JEE Online Paper _ Date [9-4-2017](Page # 2)

Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 4 April. 2017

3. The equation

iz 2Im 1 0

z i, zC, z i. represents a part of a circle having radius equal to :

(1) 1 (2) 2 (3) 34

(4) 12

Sol. 3Let = x +y

x y 2 x (y 1)Im 1 0

x (y 1) x (y 1)

+

2

2(y 1)(y 2) x

1 0x2 (y 1)

= 2x2 + 2y2 – y –1 = 0= x2+y2 – 1/2y – 1/2 = 0cot x = 10, 1/4

= 1 /2 1 /2 = 9 /16 =3/4

4. The value of K for which the function

tan4xtan5x4

,0 x5 2f(x)

2k ,x

5 2

is continuous at

x2

, is

(1) 25

(2) 25

(3) 1720

(4) 35

Sol. 4

k 2 /5 (4 /5)

= 2

k 14

K = 35

5. If

2

31 2 4

dx kk 5

(x 2x 4), then k is equal to :

(1) 4 (2) 2 (3) 3 (4) 1Sol. 4

2

3 /21

dx

(x 1) 3)

x 1 3 tanQ

= 23 sec Q

/6

0

3 sec dQ

3 3 sec.3Q



= 13

/6 /600

1cosesQ (tanQ)

3

=

1 kk 5 6k

6 k 5

= k 1

6. For two 3×3 matrices A and B, let A+B =2B' and 3A+2B=I3, where B' is the transpose of B andI3 is 3×3 identity matrix. Then :(1) 10A+5B = 3I3 (2) 3A+6B=2I3 (3) 5A+10B-2I3 (4) B+2A=I3

Sol. 1AT + BT = 2B

T TA BB

2

= T TB AA

2

= 2BT

2A + AT = 2BT

A = T T3B A

2

3A + 2B = I3 ...(i)

T T3B A3

2

+ T TA B2

2

= I3

T T3B 2B

2

+ T T2A 3A

2

= I3

11BT – AT = 2I3 ...(ii)Equation (i) + (ii)35B= 7I3

B = 3I5

3I11 A

5 = 2I3

33

I11 2I

5 = A

3IA5

5A = 5B = I3

10A + 5B = 3I3

7. If y = mx+c is the normal at a point on the parabola y2=8x whose focal distance is 8 units,then |c| is equal to :

(1) 8 3 (2) 10 3 (3) 2 3 (4) 16 3Sol. 2

c = – 29m – 9m3

a = 2Given (at2 – a)2 + 4a2t2 = 64



(a(t2 + 1)) = 8 t2 + 1 = 4 = t2 = 3

t = 3

C = – a [ –2t – t3] = 2at(2 + t2)

= 2 3 5

C = 10 3

8. A line drawn through the point P(4,7) cuts the circle x2+y2 = 9 at the points A and B. ThenPA.PB is equal to :(1) 74 (2) 53 (3) 56 (4) 65

Sol. 3P(4,7) it midpoint the circle

PA PB = 21S = PT2

1S = 16 49 9 = 5621S = 56 ; PA –PB = 56

9. The sum of all the real values of x satisfying the equation 2(x–1)2(x2+5x–50)=1 is :(1) 16 (2) –5 (3) –4 (4) 14

Sol. 3(x – 5)(x2 + 5x – 50) = 0 (x – 5)( x + 10) ( x – 5) = 0 x = 1,5, – 10 sum = – 4

10. The function f :N Ndefined by

xf(x) x 5

5 , where N is the set of natural numbers and [x]

denotes the greatest integer less than or equal to x, is :(1) one-one but not onto (2) one-one and onto(3) neither one-one nor onto (4) onto but not one-one

Sol. 3

f 1 1 5 1/ 5 1Many one

f 6 6 6 6 / 5 1

f(10) = 10 – 5(2) = 0 which is not in codumanmany one + into

11. Let f be a polynomial function such that f(3x)=f'(x). f"(x), for all xR.Then :(1) f(2)+f'(2)=28 (2) f"(2)–f'(2) =0(3) f(2)–f'(2)+f"(2)=10 (4) f"(2)–f(2)=4

Sol. 1f(x) = ax3 + bx2 + cx + lf(3x) = 27ax3 + 9bx2 + 3cx + df'(x) = 3ax2 + 2bx + cf"(x) = 6ax + 2bf(3x) = f'(x) f"(x)27a = 18a2

3a2

, b = 0, c = 0, d = 0



f(x) = 33 x2

f'(x) = 29 x2

, f"(x) = 9x

12. If three positive numbers a, b and c are in A.P. such that abc=8, then the minimum possiblevalue of b is :

(1) 234

(2) 2 (3) 134

(4) 4

Sol. 2a + c = 2b

a ca.c.2

= 8

13. If

a a a

a 1n

1 2 ...n 1lim

60(n 1) [(na 1) (na 2) ... (na n)] for some positive real number a, then a is

equal to

(1) 172

(2) 152

(3) 7 (4) 8

Sol. 3

nlim

a 1 a a 11 2

a 1 2

1.n a n a n .....

1(a 1)1 601n(n 1) .n a

2

nlim

1 22

a

1 a a.... 1a 1 n n1 60

11 n1 an 2

11a 1

1 60a2

(a + 1) (2a + 1) = 120

2a2 + 3a – 119 = 02a2 + 17a – 14a – 119 = 0

(a – 7) (2a + 17) = 0

a = 7, 172

14. If

3x 4f

3x 4 =x+2,x 43

, and f(x)dx A log|1 x | Bx C , then the ordered pair (A,B) is

equal to :(where c is a constant of integration)

(1)

8 2,

3 3 (2)

8 2,

3 3 (3)

8 2,

3 3 (4)

8 2,–

3 3



Sol. 2

3x 4f3x 4

= x + 2, 4x3

Let 3x 43x 4

= t

3x – 4 = 3tx + 4t

x = 4t 43 3t

+ 2

f(t) = 10 2t3 3t

f(x) = 2x 103x 3

f x dx = 2x 10 dx3x 3

= 2x dxdx 10

3x 3 3x 3

= 2 x 1 2 dxdx3 x 1 3 x 1

- 10 dx3 x 1 = 2x 8 ln x 1 C

3 3

15. A square, of each side 2, lies above the x-axis and has one vertex at the origin. If one of the sidespassing through the origin makes an angle 300 with the positive direction of the x-axis, then thesum of the x-coordinates of the vertices of the square is :

(1) 2 3 2 (2) 3 2 (3) 2 3 1 (4) 3 1Sol. 1

xcos30 =

ysin 30 = 2

x = 2 3

2 = 3

y = 1

xcos120 =

ysin120 = 2

x = –1, y = 3

xcos75 =

ysin 75 = 2 2

O(0,0)

3,1C 13,13 B

1,3A45°120°

22

x = 3 1y = 3 1sum = 0 + 3 + 3 –1 + (–1)

= 2 3 2

16. A value of x satisfying the equation sin[cot–1(1+x)]=cos[tan–1x], is :

(1) 12

(2) 0 (3) –1 (4) 12



Sol. 1

sin

1cot (1 x) =

1tan xcos

1

1+x

+ + 22 2x x

cot 1 x x

+ 21 x

1

2

1

x 2x 2=

2

1

1 1 xx2 + 2x + 2 = x2 + 1

x 1 /2

17. From a group of 10 men and 5 women, four member committees are to be formed each of whichmust contain at least one women. Then the probability for these committees to have morewomen than men, is :

(1) 311

(2) 223

(3) 111

(4) 21

220Sol. 3

18. The function f defined by f(x)=x3–3x2+5x+7, is :(1) decreasing in R (2) increasing in R(3) increasing in (0, ) and decreasing in (–,0)(4) decreasing in (0, ) and increasing in (–0)

Sol. 2f(x) = x3 –3x2 + 5x + 7f'(x) = 3x2 – 6x + 5 > 0f'(x) = 3x2 – 6x + 5 < 0

x

19. The ecentricity of an ellipse having centre at the origin, axes along the co-ordinate axes andpassing through the points (4, –1) and (–2,2) is :

(1) 32

(2) 34

(3) 2

5(4)

12

Sol. 1e = ?, centre at (0,0)

2 2

2 2x y

a b = 1

2 2

16 1

a b= 1

16b2 + a2 = a2b2 (1)

2 24 4

a b = 1



4b2 + 4a2 = a2b2 (2)From (1) & (2)

16b2 + a2 = 4a2 + 4b2

3a2 = 12b2 = 2 2a 4b

20. A tangent to the curve, y = f(x) at P(x,y) meets x-axis at A and y-axis at B. If AP : BP = 1 : 3 andf(1)= 1, then the curve also passes through the point :

(1)

1,24

3 (2)

1,4

2 (3)

12,

8 (4)

13,

28

Sol. 3

21. If x = a, y = b, z = c is a solution of the system of linear equationsx + 8y + 7z = 09x + 2y +3z = 0y + y +z = 0such that the point (a,b,c) lies on the plane x + 2y + z = 6, then 2a+b+c equals :(1) 2 (2) –1 (3) 1 (4) 0

Sol. 1

x 8y 7z 0 7y 6z 09x 2y 3z 0

x y z 0 7x z 0

6( 7 )x y z 7

7

x y 6 z 7

12 7 6 2 6 76 6 2

1 2

22. Let Sn = 31

1+ 3 3

1 21 2

+ 3 3 3

1 2 31 2 3

+...+ 3 3 3

1 2 ..., n1 2 ... n

. If 100 Sn = n, then n is equal to :

(1) 200 (2) 199 (3) 99 (4) 19Sol. 2

Tn =

2

n (n 1)2

n (n 1)2

Tn = 2

n(n 1)

Sn = 2

n

n 1

1 1n n 1



= 2

11

21 12 3

1 1n n 1

=

12 1

n 1

n2n

Sn 1

2n100

n 1= n

n + 1 = 200n = 199

23. If the vector ˆ ˆb 3j 4k

is written as the sum of a vector 1b

, parallel to ˆ â i j

and a vector 2b

,

perpendicular to a,then 1 2b b

is equal to :

(1) 9ˆ ˆ ˆ6i 6j k2

(2) – ˆ ˆ ˆ3i 3j 9k (3) –9ˆ ˆ ˆ6i 6j k2

(4) ˆ ˆ ˆ3i 3j 9k

Sol. 1

1b =

ˆ(b.a)a

1

=

ˆˆ ˆ ˆ(3j 4k).(i j)

2

ˆ î j

2

=

ˆ ˆ3(i j)

2 2=

ˆ ˆ3(i j)2

b

b2

b1

1 2b b b

2 1b b b

= 3ˆˆ ˆ ˆ3i 4k i j2

2

3 3 ˆˆ ˆb i j 4k2 2

1 2b b =

ˆˆ î j k3 3

02 23 3

42 2

9 9ˆˆ î(6) j(6) k4 4 ;

9 ˆˆ ˆ6i 6j k2



24. The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 600.

If the area of the quadrilateral is 4 3 , then the perimeter of the quadrilateral is :(1) 12.5 (2) 13 (3) 13.2 (4) 12

Sol. 4

cos 60 = 24 25 c2.2.5

10 = 29 – c2

c2 = 19

c 19

2 21 a b 192 2ab

a2 + b2 – 19 = – aba2 + b2 + ab = 19

Area = 1 1

2 5sin60 absinx 4 32 2

5 3 ab 3

4 32 4

5

a

2

b

1200

600

ab 5 3

44 2 2

ab 6a2+b2 = 13

a 2,b 3

Perimeter= 2 + 5 + 2+ 3= 12

25. The number of ways in which 5 boys and 3 girls can be seated on a round table if a particular boyB1 and a particular girl G1 never sit adjacent to each other, is :(1) 7! (2) 5×6! (3) 6×6! (4) 5×7!

Sol. 24 boy and 2 girls in circle

625! c 2!

6!

5! 2!4!2!

5×6!

26. If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes atA,B and C, then the locus of the centroid of ABC is .

(1) 2 2 21 1 1

1x y z

(2) 2 2 21 1 1

3x y z

(3) 2 2 21 1 1

9x y z

(4) 2 2 21 1 1 1

9x y z



Sol. 1Let Centroid be (h,k,l) x – intp = 3h Y– intp = 3k, 3–int = 3l

Equ.

x y z1

3h 3k 3dist from (0,0,0)

2 2 2

13

1 1 1

9h 9k 9l

2 2 21 1 1

1x y z

27. The sum of 100 observations and the sum of their squares are 400 and 2475, respectively. Lateron, three observations, 3,4 and 5, were found to be incorrect. If the incorrect observations areomitted, then the variance of the remaining observations is :(1) 8.25 (2) 8.50 (3) 9.00 (4) 8.00

Sol. 3

100

i 1

xi 400

100

2

i 1

xi 2475

variance

222 1xxi

N N

2 2 23 4 5

9 16 25 50

=

22475 38897 97

2425

1697

2425 1552 87397 97

= (9)

28. Let E and F be two independent events. The probability that both E and F happen is 1

12and

the probability that neither E nor F happens is 12

, then a value of P(E)P(F) is :

(1) 43

(2) 13

(3) 32

(4) 5

12Sol. 1

P(E F) = P E .P F = 1

12=

1xy

12

P(E F) = P E .P F = 12



(1 – P(E))(1 – P(F) = 12

= 1 – x – y + xy = 12

1 – x – y + 1 1

12 2

1– x – y = 1 1 52 12 12

7

x y12

x + 1

12x=

712

212x 112x

=7

1212x2 – 7x + 1 = 012x2 – 4x – 3x + 1 =04x (3x–1) – 1 (3x–1) = 0

1

x3

, 1

x4

1

y4

, 1

y3

x 1 /3 4y 1 / 4 3 or

1 / 4 31 /3 4

29. If

1 15 52x y y and

22

2d y dy

(x 1) x ky 0dxdx

, then +k is equal to :

(1) 26 (2) –24 (3) –23 (4) –26Sol. 2

y1/5 + y–1/5 = 2x

4 /5 –6 /51 1 dyy y 2

5 5 dx

y' (y1/5 – y–1/5) = 10y

y'

22 x 1 10y

y"

22

2x2 x 1 y '2 y '

2 x 1

y" (x2 –1) + xy' = 25 x 1 y'

2y "(x 1) xy '– 25y 0

= 1, k = –25



30. Contrapositive of the statement'If two numbers are not equal, then their squares are not equal', is :(1) If the squares of two numbers are equal, then the numbers are not equal(2) If the squares of two numbers are not equal, then the numbers are equal(3) If the squares of two numbers are not equal, then the numbers are not equal(4) If the squares of two numbers are equal, then the numbers are equal.

Sol. 4p q

contrapositive is~ q ~ p



[PHYSICS]

1. The electric field component of a monochromatic radiation is given by

0E 2E i coskzcos t

Its magnetic field B is then given by :

(1) tsinkzsinjcE2 0 (2) tcoskzcosj

cE2 0

(3) tcoskzsinjcE2 0 (4) tsinkzsinj

cE2 0

Sol. 1

dzdE

= – dEdt

dzdE

= – 2E0 K sin kz cos t = – dtdB

dB = + 2E0K sin kz cos t dt

B = +2E0K sin kz tdtcos

= + 2E0 k

sin kz sin t

0

0

BE

= k

= C

B = CE2 0 sin kz sin t

2. N moles of a diaotmic gas in a cylinder are at a temperature T. Heat is supplied to the cylinder suchthat the temperature remains constant but n moles of the diatomic gas get converted into monoatomicgas. What is the change in the total kinetic energy of the gas ?

(1) 0 (2) nRT25

(3) nRT23

(4) nRT21

Sol. 4

ui = N 25

RTT

uf = 2n 23

RT + (N – n) 25

RTT

= 21

nRT + 25

nRTT

u = Q = 21

nRTT



3. A circular hole of radius 4R

is made in a thin uniform disc having mass M and radius R, as shown

in figure. The moment of inertia of the remaining portion of the disc about an axis passingthrough the point O and perpendicular to the plane of the disc is -

R

OR/4

O'

3R/4

(1) 256

MR219 2

(2) 512

MR237 2

(3) 256

MR197 2

(4) 512MR19 2

Sol. 2

ID = 2

mr2

Iremoved = 21

16m

16r2

+ 16m

16r9 2

(Im + md)

= 512

mr18mr 22

= 512mr19 2

Iremaining = 2

mr2

– 51219

mr2

= 512237

mr2

4. A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm–1 and oscillates ina damping medium of damping constant 10–2 kg s–1. The system dissipates its energy gradually. Thetime taken for its mechanical energy of vibration to drop to half of its initial value, is closest to -(1) 2 s (2) 5 s (3) 7 s (4) 3.5 s

Sol. 4

E' = 21

b = m

A = A0 e–bt

a' = 2

a = bm

2

1 = 10

t

e



b = 101

2 = 101

e

2ln = 10t

t = 3.5

5. A steel rail of length 5 m and area of cross section 40 cm2 is prevented from expanding along itslength while the temperature rises by 10°C. If coefficient of linear expansion and Young's modulusof steel are 1.2 × 10–5 K–1 and 2 × 1011 Nm–2 respectively, the force developed in the rail isapproximately :(1) 2 × 107 N (2) 2 × 109 N (3) 3 × 10–5 N (4) 1 × 105 N

Sol. 4F = yA t= 2 × 1011 × 40 × 10–4 × 1.2 × 10–5 × 10= 9.6 × 104 = 1 × 105 N

6. In a meter bridge experiment resistances are connected as shown in the figure. Initially resistanceP = 4 and the neutral point N is at 60 cm from A. Now an unknown resistance R is connectedin series to P and the new position of the neutral point is at 80 cm from A. Thevalue of unknownresistance R is -

P Q

G

NB

KRhE

A

(1) 533

(2) 6 (3) 320

(4) 7

Sol. 3

Initially 604

= 40Q

Q = 6

16 =

38

Finally 80

R4 =

20Q

= 608

4 + R = 664

R = 664

– 4 = 6

2464 =

640

= 320



7. A conical pendulum of length 1 m makes an angle = 45° w.r.t. Z-axis and mvoes in a circle inthe XY plane. The radius of the circle is 0.4 m and its center is vertically below O. The speed ofthe pendulum, in its circular path, will be - (Take g = 10 ms–2)

ZO

C

(1) 0.2 m/s (2) 0.4 m/s (3) 2 m/s (4) 4 m/sSol. 3

T sin = r

mv2

T cos = mg

tan = rgv2

T

w = 45°v2 = rg

= rg = 104.0

= 2 m/s

8. A signal is to be transmitted through a wave of wavelength , using a linear antenna. The lengthl of the antenna and effective power radiated Peff will be given respectively as -(K is a constant of proportionality)

(1) 8

, Peff = K

l

(2) 16

, Peff = K 3

1

(3) 5

, Peff = K 21

1

(4) , Peff = K 2

1

Sol. 4Length of antenna = comparable to

Power P = 2

1

here = k



9. A sinusoidal voltageo f peak value 283 V and angular frequency 320/s is applied to a series LCRcircuit. Given that R = 5 , L = 25 mH and C = 1000 F. The total impedance, and phase differencebetween the voltage across the source and the current will respectively be -

(1) 10 and tan–1

35

(2) 7 and 45°

(3) 7 and tan–1

35

(4) 10 and tan–1

38

Sol. 2e0 = 283 volt = 320xL = 320 × 25 × 10–3 = 8

xC = c

1

= 61010003201

= 320

1000 = 3.1

R = 5

Z = 2CL

2 )XX(R = 50 = 7

tan = R

XX CL

= 1 = 45°

10. A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000 Å anddiffraction bands are observed on a screen 0.5 m from the slit. The distance of the third darkband from the central bright band is -(1) 9 mm (2) 3 mm (3) 4.5 mm (4) 1.5 mm

Sol. 1a = 0.1 mm = 10–4

= 6000 × 10–10 = 6 × 10–7

D = 0.5 mfor 3rd darka sin = 3

sin = a3

= Dx

x = aD3

= 4

7

105.01063

= 9 mm

11. In an experiment a convex lens of focal length 15 cm is placed coaxially on an optical bench infront of a convex mirror at a distance of 5 cm from it. It is found that an object and its imagecoincide, if the object is placed at a distance of 20 cm from the lens. The focal length of theconvex mirror is -(1) 20.0 cm (2) 30.5 cm (3) 25.0 cm (4) 27.5 cm



Sol. 4

20 55cm

O

due to this images will from 55 cm behind convex mirror then reflection takes replace due tomirror image will form at v behind mirror (let) this will here as object for lens and final image will

from at a distance of 20 cm from lens (i.e. obext) if focal length in of convex lens = 255

cm

12. The mass density of a spherical body is given by (r) = rk

for r R and (r) = 0 for r > R,

Where r is the distance from the centre. The correct graph that describes qualitatively theacceleration, a, of a test particle as a function of r is -

(1)

a

R r

(2)

a

R r

(3)

a

R r

(4)

a

R r

Sol. 1

VM

= rk

for inside

M = r

kv

g = 3RGmr

= 3RG

. r

kv.r = constant

gout = 2RGM



13. A uniform wire of length I and radius r has a resistance of 100 . It is recast into a wire of radius

2r

. The resistance of new wire will be -

(1) 1600 (2) 100 (3) 200 (4) 400 Sol. 1

R = Al

Al = V

R = 2Av

constantv constant

R 2A1

2r1

R 4r1

R2 = 16 R1 = 1600

14. The figure shows three circuits I, II and III which are connected to a 3V battery. If the powersdissipated by the configurations I, II and III are P1, P2 and P3 respectively, then -

3V

3V

3V

(1) P2 > P1 > P3 (2) P1 > P2 > P3 (3) P3 > P2 > P1 (4) P1 > P3 > P2



Sol. 1

P = RV2

P R1

R1 = 1R2 = 1/2R3 = 2P2 > P1 > R3

15. A standing wave is formed by the superposition of two waves travelling in opposite directions. Thetransverse displacement is given by

y (x,t) = 0.5 sin

x

45

cos (2oo t)

What is the speed of the travelling wave moving in the positive x direction ?(x and t are in meter and second, respectively.)(1) 120 m/s (2) 90 m/s (3) 160 m/s (4) 180 m/s

Sol. 3

V = kw

= 4/5200

= 160

16. In an experiment to determine the period of a simple pendulum of length 1m, it is attached todifferent spherical bobs of radii r1 and r2. The two spherical bobs have uniform mass distribution.If the relative difference in the periods, is found to be 5 × 10–4 s, the difference in radii, |r1 – r2|is best given by -(1) 0.01 cm (2) 0.1 cm (3) 0.5 cm (4) 1 cm

Sol. 2

T l l = 1

lT

= 21

ll

l = r1 – r2

5 × 10–4 = 21

1

rr 21

r1 – r2 = 10 × 10–4

10–3 m = 10–1 cm = 0.1 cm



17. Two particles A and B of equal mass M are moving with the same speed as shown in the figure.They collide completely inelastically and mvoe as a single particle C. The angle that the path ofC makes with the X-axis is given by -

Y

C

X

B45°

30°A

(1) tan = 21

23

(2) tan =

)31(2

21

(3) tan = 21

31

(4) tan =

21

23

Sol. 4

2mv' sin = 2

mv +

23mv

3 mv' cos = 2

mv –

2

mv

sin =

2

121

23

2

1

= 21

32

18. A laser light of wavelength 660 nm is used to weld Retina detachment. If a laser pulse of width 60 msand power 0.5 kW is used the approximate number of photons in the pulse are -[Take Planck's constant h = 6.62 × 10–34 Js](1) 1022 (2) 1019 (3) 1020 (4) 1018

Sol. 3

P = nhc

tn =

hctp

= 5 × 103 × 834

39

103106.6106010660

= 100 × 1018 = 1020



19. The current gain of a common emitter amplifier is 69. If the emitter current is 7.0 mA, collectorcurrent is -(1) 69 mA (2) 0.69 mA (3) 6.9 mA (4) 9.6 mA

Sol. 3

= 69 = B

C

II

=

1 = 7069

= E

C

II

IC = IE × 7069

= 7069

× 7

= 6.9 mA

20. Four closed surfaces and corresponding charge distributions are shown below -

Let the respective electric fluxes through the surfaces be 1, 2, 3 and 4. Then -(1) 1 > 2 > 3 > 4(2) 1 < 2 = 3 > 4(3) 1 > 3 ; 2 > 4(4) 1 = 2 = 3 = 4

Sol. 4Eq = same in all = Same

21. A negative test charge is moving near a long straight wire carrying a current. The force acting onthe test charge is parallel to the direction of the current. The motion of the charge is -(1) away from the wire(2) towards the wire(3) parallel to the wire along the current(4) parallel to the wire opposite to the current



Sol. 2

)Bv(qF

F

ve

22. A uniform magnetic field B of 0.3 T is along the positive Z-direction. A rectangular loop (abcd) ofsides 10 cm × 5 cm carries a current I of 12 A. Out of the following different orientations which onecorresponds to stable equilibrium ?

(1)

X

a b

I

cd

Z

I

Y (2)

X

a

b

Z

IY

B

c

I

(3)

X

a

b

Z

IY

B

d

cI

(4)

X

a

b

Z

IY

B

d

cI

Sol. 3M = NIA

M

and B are on same direction

23. The accelration of an electron in the first orbit of the hydrogen atom (n = 1) is -

(1) 32

2

rm4h

(2) 322

2

rmh

(3) 322

2

rm8h

(4) 322

2

rm4h

Sol. 4

v = 0

2

h2l

r = 20

2

meh

0 = 2

2

hmer

rv2

= 30

4

6

h4ml

= 223

2

mr4h



24. Two tubes of radii r1 and r2, and lengths l1 and l2, respectively, are connected in series and a liquidflows through each of them in stream line conditions. P1 and P2 are pressure differences across

the two tubes. If P2 is 4P1 and l2 is 4l1 , then the radius r2 will be equal to -

(1) 4r1 (2) r1 (3) 2r1 (4) 2r1

Sol. 4

dtv

= 8

nLpr4

1

411

Lrp

= 2

422

Lrp

2

411

lrp

= 4/1

421

lrp4

= r24 =

16r41

r2 = 2r1

25. A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instantbut with different forward accelerations. The bus has acceleration 2 m/s2 and teh car has acceleration4 m/s2. The car will catch up with the bus after at time of -

(1) s120 (2) 15 s (3) s110 (4) s210

Sol. 4

Car Bus2m/sec24m/sec2

200m

aCB = 2 m/sec2

200 = 21

× 2t2

t = 10 2 second

26. The machine as shown has 2 rods of length 1 m connected by a pivot at the top. The end of onerod is connected to the floor by a stationary pivot and the end of the other rod has a roller thatrolls along the floor in a slot. As the roller goes back and forth, a 2 kg weight moves up and down.If the roller is moving towards right at a constant speed, the weight moves up with a -

F

2kg

x Fixed pivot

Movable roller



(1) speed which is 43

th of that of the roller when the weight is 0.4 m above the ground

(2) constant speed(3) decreasing speed(4) increasing speed

Sol. 3

27. A physical quantity P is described by the relationP = a1/2 b2 c3 d–4

If the relative errors in the measurement of a,b,c and d respectively, are 2%, 1%, 3% and 5%,then the relative error in P will be -(1) 12% (2) 8% (3) 25% (4) 32%

Sol. 4

PP

= 21

aa

+ 2 bb

+ 3 cc

+ 4 dd

= 21

× 2 + 2 × 1 + 3 × 3 + 4 × 5

= 32%

28. Imagine that a reactor converts all given mass into energy and that it operates at a power levelof 109 watt. The mass of the fuel consumed per hour in the reactor will be :(velocity of light, c is 3 × 108 m/s)(1) 6.6 × 10–5 gm (2) 0.96 gm (3) 4 × 10–2 gm (4) 0.8 gm

Sol. 3

P = t

E

= t

mc2

tm

= 2cP

= 28

9

)103(10 = 4 × 10–2 gm

29. A combination of parallel plate capacitors is maintained at a certain potential difference.

C1 C C2 C3

E BA D

When a 3 mm thick slab is introduced between all the plates, in order to maintain the samepotential difference, the distance between the plates is increased by 2.4 mm. Find the dielectricconstant of the slab.(1) 6 (2) 4 (3) 3 (4) 5

Sol. 4

C1 = 3A0 before

C1 = 3Ak 0 +

4.2A0 after



3A0 =

4.2A

3A

k

4.2A

.3A

k

00

00

3k = 2.4k + 3

0.6k = 3 k = 6.0

3

k = 630

= 5

30. For the P-V diagram given for an ideal gas, out of the following which one correctly represents theT-P diagram ?

1

P

V

2

VttanCons

P

(1)

T1 2

P

(2)

T2 1

P

(3)

T

1

2

P

(4)

T

1

2

P

Sol. 2

P v1

T = constant



[CHEMISTRY]

1. A compound of molecular formula C8H8O2 reacts with acetophenone to form a single cross-aldolproduct in the presence of base. The same compound on reaction with conc. NaOH forms bezylalcohol as one of the products. The structure of the compound is :

(1) (2)

(3) (4)

Sol. 1

2. Which of the following ions does not liberate hydrogen gas on reaction with dilute acids ?

(1) V2+ (2) Ti2+ (3) Mn2+ (4) Cr2+

Sol. 3

Mn2+

3. The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activationenergy of this reaction is :

(Assume activation energy and pre-exponential factor are independent of temperature; ln2 =0.693; R = 8.314 J mol–1 K–1)

(1) 53.6 kJ mol–1 (2) 214.4 kJ mol–1 (3) 107.2 kJ mol–1 (4) 26.8 kJ mol–1



Sol. 3

4 = Ea 1 1R 300 310e

Ea 10ln(4)

R 300 310

0.693 2 8.314 300 310Ea

10

= 107165.79 J = 107.165 KJ

4. The group having triangular planar structure is :

(1) BF3, NF3, CO32– (2) CO3

2–, NO3–, SO3 (3) NH3, SO3, CO3

2– (4) NCl3, BCl3, SO3

Sol. 2CO3

–2, NO3–, SO3 : SP2 Hybridised

5. The electronic configuration with the highest ionization enthalpy is :

(1) [Ar] 3d10 4s2 4p3 (2) [Ne] 3s2 3p1 (3) [Ne] 3s2 3p3 (4) [Ne] 3s2 3p2

Sol. 3S1 < P1 < S2 < P2 < P4 < P3 < P5 < P6 (IE order)

6. The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius211.6 pm. This transition is associated with :

(1) Lyman series (2) Balmer series (3) Brackett series (4) Paschen series

Sol. 2

R = 211.6 pm = 2.11 Å

R = 0.529 × 2nZ

= 2.11 Å n2 = 4 n = 2

7. Which one of the following is an oxide ?

(1) SiO2 (2) KO2 (3) BaO2 (4) CsO2

Sol. 1SiO2

8. 50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solutionis 4.75, the pH of the mixture will be :

(1) 8.25 (2) 9.25 (3) 3.75 (4) 4.75

Sol. 4

NH3 + HCl NH4Cl

50 0.21000

25 0.21000

5 0 5

Buffer solution

pOH = pkb NH3 + salt

Hgbase

= 4.75



9. Which of the following compounds is most reactive to an aqueous solution of sodium carbonate ?

(1) (2) (3) (4)

Sol. 3

10. The number of P–OH bonds and the oxidation state of phosphorus atom in pyrophosphoric acid(H4P2O7) respectively are :

(1) five and four (2) four and five (3) five and five (4) four and four

Sol. 2

P — O — P

O O

OH OHHO OH

11. The following reaction occurs in the Blast Furnace where iron ore is reduced to iron metal :

2 3 2Fe O (s) 3CO(g) 2Fe( ) 3CO (g)

Using the Le Chatelier’s principle, predict which one of the following will not disturb theequilibrium ?

(1) Addition of CO2 (2) Removal of CO2 (3) Addition of Fe2O3 (4) Removal of CO

Sol. 3

12. A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2and CHCl3 at 298 K are 415 and 200 mm Hg respectively, the mole fraction of CHCl3 in vapourform is : (Molar mass of Cl = 35.5 g mol–1)

(1) 0.162 (2) 0.675 (3) 0.325 (4) 0.486

Sol. 3

13. The increasing order of the boiling points for the following compounds is :

C2H5OH C2H5Cl C2H5CH3 C2H5OCH3

(I) (II) (III) (IV)

(1) (IV) < (III) < (I) < (II) (2) (III) < (II) < (I) < (IV)

(3) (III) < (IV) < (II) < (I) (4) (II) < (III) < (IV) < (I)

Sol. 3

B.P. dipole moment

H-bonding

C2H5CH3 < C2H5OCH3 < C2H5Cl < C2H5OH



14. An ideal gas undergoes isothermal expansion at constant pressure. During the process :

(1) enthalpy remains constant but entropy increases.

(2) enthalpy decreases but entropy increases.

(3) enthalpy increases but entropy decreases.

(4) Both enthalpy and entropy remain constant.

Sol. 1

H = nCpT = 0

S = nRln(Vf/Vi) 0

15. [Co2(CO)8] displays :

(1) no Co–Co bond, six terminal CO and two bridging CO

(2) no Co–Co bond, four terminal CO and four bridging CO

(3) one Co–Co bond, six terminal CO and two bridging CO

(4) one Co-Co bond, four terminal CO and four bridging CO

Sol. 3

OC CO CO CO

CO

COOC

OC

CO

CO

16. The correct sequence of decreasing number of -bonds in the structure of H2SO3, H2SO4 andH2S2O7 is :

(1) H2S2O7 > H2SO3 > H2SO4 (2) H2S2O7 > H2SO4 > H2SO3

(3) H2SO4 > H2S2O7 > H2SO3 (4) H2SO3 > H2SO4 > H2S2O7

Sol. 2

H2S2O7 = HO—S—O—S—OH

O

O O

O

H2SO4 = HO—S—OH

O

O

H2SO3 = —S—OH

O

17. At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen (N2)at 4 bar. The molar mass of gaseous molecule is :

(1) 224 g mol–1 (2) 112 g mol–1 (3) 56 g mol–1 (4) 28 g mol–1

Sol. 4



18. A gas undergoes change from state A to state B. In this process, the heat absorbed and workdone by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another processduring which 3 J of heat is evolved. In this reverse process of B to A :

(1) 6 J of the work will be done by the gas

(2) 6 J of the work will be done by the surrounding on gas.

(3) 10 J of the work will be done by the surrounding on gas.

(4) 10 J of the work will be done by the gas

Sol. 2

19. Which of the following is a biodegradable polymer ?

(1) (2)

(3) (4)

Sol. 2

Nylon–2–Nylon–6Polymer of glycine and amino caproic acid

20. XeF6 on partial hydrolysis with water produces a compound ‘X’. The same compound ‘X’ is formedwhen XeF6 reacts with silica. The compound ‘X’ is :

(1) XeF4 (2) XeF2 (3) XeO3 (4) XeOF4

Sol. 4

21. In the following structure, the double bonds are marked as I, II, III and IV. Geometrical isomerismis not possible at site (s) :

(1) I and III (2) III (3) I (4) III and IVSol. 3

Different group should be attached to each sp2 hybridised c-atom.



22. Adsorption of a gas on a surface follows Freundlich adsorption isotherm. Plot of x

logm

versus log

p gives a straight line with slope equal to 0.5, then : (xm

is the mass of the gas adsorbed per

gram of adsorbent)(1) Adsorption is proportional to the square root of pressure.(2) Adsorption is proportional to the square of pressure.(3) Adsorption is proportional to the pressure.(4) Adsorption is independent of pressure.

Sol. 1

x 1log log P K

m 2

1/2xKP

m

23. The incorrect statement among the following is :(1) -D-glucose and -D-glucose are enantiomers.(2) The penta acetate of glucose does not react with hydroxyl amine.(3) -D-glucose and -D-glucose are anomers.(4) Cellulose is a straight chain polysaccharide made up of only -D-glucose units.

Sol. 1-D-Glucose and -D-glocose are anomer not enantiomer.

24. In the following reaction sequence :

The compound I is :

(1) (2)

(3) (4)

Sol. 1



25. The major product of the following reaction is :

(1) (2) (3) (4)

Sol. 1

26. Among the following compounds, the increasing order of their basic strength is :

(I) (II) (III) (IV)

(1) (II) < (I) < (III) < (IV) (2) (II) < (I) < (IV) < (III)

(3) (I) < (II) < (IV) < (III) (4) (I) < (II) < (III) < (IV)

Sol. 1

Order of basicity



27. What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must be mixed with a20% solution of the same acid to produce 800 mL of a 29.875% acid solution ?

(1) 316 (2) 320 (3) 325 (4) 330

Sol. 1

V 45 (800 V)20 800 29.875100 100 100

9V V160 239

20 5

5V79 V 316 Ans.

20

28. To find the standard potential of M3+/M electrode, the following cell is constituted : Pt/M/M3+

(0.001 mol L–1)/Ag+ (0.01 mol L–1)/Ag

The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3+

+ 3e– M at 298 K will be : (Given Ag /AgE at 298 K = 0.80 volt)

(1) 0.38 volt (2) 1.28 volt (3) 0.32 volt (4) 0.66 volt

Sol. 3

3

0.059 0.0010.421 E log

3 (0.01)

30.059E 0.421 log(10 )

3

E° = 0.480 = 0.8 – 3M /ME

3M /ME 0.32

29. Which of the following is a set of green house gases ?

(1) O3, NO2, SO2, Cl2 (2) CH4, O3, N2, SO2 (3) CO2, CH4, N2O, O3(4) O3, N2, CO2, NO2

Sol. 3

30. Which of the following compounds will show highest dipole moment ?

(I) (II) (III) (IV)

(1) (II) (2) (IV) (3) (III) (4) (I)

Sol. 1

JEE MAIN ONLINE EXAMINATION - 2017 - Daily Careers · JEE MAIN ONLINE EXAMINATION - 2017 QUESTION...

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Transcript of JEE MAIN ONLINE EXAMINATION - 2017 - Daily Careers · JEE MAIN ONLINE EXAMINATION - 2017 QUESTION...