JEE Mains -Previous Year Solved Question Paper with Answer Key -2018 

Answers & Solutions For JEE MAIN- Free PDF Download Link 2018

 

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1. The value of ∫π2π2sin?2x1+2xdx∫2π?2π??1+2xsin2x?dx is:

π44π? π88π? π22π? π4π4ππ?

Solution:

Given∫−π/2π/2sin?2x1+2xdxGiven∫−π/2π/2?1+2xsin2x?dx f(x)+f(−x)=sin?2x1+2x+2x(sin?2x)1+2x=sin?2x=∫0π/2sin?2xdx=∫0π/2sin?2xdx=π4f(x)+f(−x)=1+2xsin2x?+1+2x2x(sin2x)?=sin2x=∫0π/2?sin2xdx=∫0π/2?sin2xdx=4π?

1. Let g(x)=cos?x2g(x)=cosx2 , f(x)=xf(x)=x? , and α,β(α<β)α,β(α<β) be the roots of the quadratic equation 18x2−9πx+π2=018x2−9πx+π2=0 . Then the area (in sq. units) bounded by the curve y=(gof)(x)y=(gof)(x) and the lines x=α,x=βx=α,x=β and y=0y=0 , is:

(1)12(2−1)(1)21?(2?−1) (2)12(3−1)(2)21?(3?−1) (3)12(3+1)(3)21?(3?+1) (4)12(3−2)(4)21?(3?−2?)

Solution:

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1. If sum of all the solution of equation 8cos?x(cos?(π6+x),cos?(π6−x)−12)=18cosx(cos(6π?+x),cos(6π?−x)−21?)=1 in [0,π][0,π] is kπkπ , then k is equal to :

kπ,then k is equal to :kπ,then k is equal to : 209920? 2332? 2332? 8998? 8998?

Solution:

8cos?x[(cos?2π6−sin?2x)−12]=18cosx[(cos26π?−sin2x)−21?]=1 8cos?x(34−12−1+cos?2x)=18cosx(43?−21?−1+cos2x)=1 8cos?x4×(4cos?2x−1−2)=148cosx?×(4cos2x−1−2)=1 cos?3x=4cos?2x−3cos?xcos3x=4cos2x−3cosx 2×cos?3x=12×cos3x=1 cos?3x=12cos3x=21? 3x∈[0,3π]3x∈[0,3π] 3x=π3,2π−π3,2π+π3⇒Sum=13π93x=3π?,2π−3π?,2π+3π?⇒Sum=913π?

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1. Let f(x)=x2+1x2f(x)=x2+x21? and g(x)=x−1x,x∈R−{−1,0,1}g(x)=x−x1?,x∈R−{−1,0,1} . If h(x)=f(x)g(x)h(x)=g(x)f(x)? , then the local minimum value of value of h(x)h(x) is :

(2)3(2)3 (3)−3(3)−3

Solution:

g′(x)=1+1x2>0g′(x)=1+x21?>0 ∴t∈R−{0};t2∈(0,∞)∴t∈R−{0};t2∈(0,∞) ∴f(x)=x2+1x2=(x−1x)2+2=t2+2∈(2,∞)∴f(x)=x2+x21?=(x−x1?)2+2=t2+2∈(2,∞) g′(x)=1+1x2>0g′(x)=1+x21?>0 ∴f(x)g(x)=t2+2t=t+2t∴g(x)f(x)?=tt2+2?=t+t2? Let h(t)=t+2tLet h(t)=t+t2?

∴f(x)g(x)=t2+2t=t+2t∴g(x)f(x)?=tt2+2?=t+t2?Let h(t)=t+2tLet h(t)=t+t2?

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1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 3 1 1 1 1 1 1 1 1 1 1 4 1 1 1 1 1 1 1 1 1 5 1 1 1 1 1 1 1 1 1 6 1 1 1 1 1 1 1 1 1 7 1 1 1 1 1 1 1 1 1 8 1 1 1 1 1 1 1 1 1 9 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 12 1 1 1 1 1 1 1 1 1 13 1 1 1 1 1 1 1 1 1 14 1 1 1 1 1 1 1 1 1 15 1 1 1 1 1 1 1 1 1 16 1 1 1 1 1 1 1 1 1 17 1 1 1 1 1 1 1 1 1 18 1 1 1 1 1 1 1 1 1 19 1 1 1 1 1 1 1 1 1 20 1 1 1 1 1 1 1 1 1 21 1 1 1 1 1 1 1 1 1 22 1 1 1 1 1 1 1 1 1 23 1 1 1 1 1 1 1 1 1 24 1 1 1 1 1 1 1 1 1 25 1 1 1 1 1 1 1 1 1 26 1 1 1 1 1 1 1 1 1 27 1 1 1 1 1 1 1 1 1 28 1 1 1 1 1 1 1 1 1 29 1 1 1 1 1 1 1 1 1 30 1 1 1 1 1 1 1 1 1 31 1 1 1 1 1 1 1 1 1 32 1 1 1 1 1 1 1 1 1 33 1 1 1 1 1 1 1 1 1 34 1 1 1 1 1 1 1 1 1 35 1 1 1 1 1 1 1 1 1 36 1 1 1 1 1 1 1 1 1 37 1 1 1 1 1 1 1 1 1 38 1 1 1 1 1 1 1 1 1 39 1 1 1 1 1 1 1 1 1 40 1 1 1 1 1 1 1 1 1 41 1 1 1 1 1 1 1 1 1 42 1 1 1 1 1 1 1 1 1 43 1 1 1 1 1 1 1 1 1 44 1 1 1 1 1 1 1 1 1 45 1 1 1 1 1 1 1 1 1 46 1 1 1 1 1 1 1 1 1 47 1 1 1 1 1 1 1 1 1 48 1 1 1 1 1 1 1 1 1 49 1 1 1 1 1 1 1 1 1 50 1 1 1 1 1 1 1 1 1 51 1 1 1 1 1 1 1 1 1 52 1 1 1 1 1 1 1 1 1 53 1 1 1 1 1 1 1 1 1 54 1 1 1 1 1 1 1 1 1 55 1 1 1 1 1 1 1 1 1 56 1 1 1 1 1 1 1 1 1 57 1 1 1 1 1 1 1 1 1 58 1 1 1 1 1 1 1 1 1 59 1 1 1 1 1 1 1 1 1 60 1 1 1 1 1 1 1 1 1 61 1 1 1 1 1 1 1 1 1 62 1 1 1 1 1 1 1 1 1 63 1 1 1 1 1 1 1 1 1 64 1 1 1 1 1 1 1 1 1 65 1 1 1 1 1 1 1 1 1 66 1 1 1 1 1 1 1 1 1 67 1 1 1 1 1 1 1 1 1 68 1 1 1 1 1 1 1 1 1 69 1 1 1 1 1 1 1 1 1 70 1 1 1 1 1 1 1 1 1 71 1 1 1 1 1 1 1 1 1 72 1 1 1 1 1 1 1 1 1 73 1 1 1 1 1 1 1 1 1 74 1 1 1 1 1 1 1 1 1 75 1 1 1 1 1 1 1 1 1 76 1 1 1 1 1 1 1 1 1 77 1 1 1 1 1 1 1 1 1 78 1 1 1 1 1 1 1 1 1 79 1 1 1 1 1 1 1 1 1 80 1 1 1 1 1 1 1 1 1 81 1 1 1 1 1 1 1 1 1 82 1 1 1 1 1 1 1 1 1 83 1 1 1 1 1 1 1 1 1 84 1 1 1 1 1 1 1 1 1 85 1 1 1 1 1 1 1 1 1 86 1 1 1 1 1 1 1 1 1 87 1 1 1 1 1 1 1 1 1 88 1 1 1 1 1 1 1 1 1 89 1 1 1 1 1 1 1 1 1 90 1 1 1 1 1 1 1 1 1 91 1 1 1 1 1 1 1 1 1 92 1 1 1 1 1 1 1 1 1 93 1 1 1 1 1 1 1 1 1 94 1 1 1 1 1 1 1 1 1 95 1 1 1 1 1 1 1 1 1 96 1 1 1 1 1 1 1 1 1 97 1 1 1 1 1 1 1 1 1 98 1 1 1 1 1 1 1 1 1 99 1 1 1 1 1 1 1 1 1 100 1 1 1 1 1 1 1 1 1 101 1 1 1 1 1 1 1 1 1 102 1 1 1 1 1 1 1 1 1 103 1 1 1 1 1 1 1 1 1 104 1 1 1 1 1 1 1 1 1 105 1 1 1 1 1 1 1 1 1 106 1 1 1 1 1 1 1 1 1 107 1 1 1 1 1 1 1 1 1 108 1 1 1 1 1 1 1 1 1 109 1 1 1 1 1 1 1 1 1 110 1 1 1 1 1 1 1 1 1 111 1 1 1 1 1 1 1 1 1 112 1 1 1 1 1 1 1 1 1 113 1 1 1 1 1 1 1 1 1 114 1 1 1 1 1 1 1 1 1 115 1 1 1 1 1 1 1 1 1 116 1 1 1 1 1 1 1 1 1 117 1 1 1 1 1 1 1 1 1 118 1 1 1 1 1 1 1 1 1 119 1 1 1 1 1 1 1 1 1 120 1 1 1 1 1 1 1 1 1 121 1 1 1 1 1 1 1 1 1 122 1 1 1 1 1 1 1 1 1 123 1 1 1 1 1 1 1 1 1 124 1 1 1 1 1 1 1 1 1 125 1 1 1 1 1 1 1 1 1 126 1 1 1 1 1 1 1 1 1 127 1 1 1 1 1 1 1 1 1 128 1 1 1 1 1 1 1 1 1 129 1 1 1 1 1 1 1 1 1 130 1 1 1 1 1 1 1 1 1 131 1 1 1 1 1 1 1 1 1 132 1 1 1 1 1 1 1 1 1 133 1 1 1 1 1 1 1 1 1 134 1 1 1 1 1 1 1 1 1 135 1 1 1 1 1 1 1 1 1 136 1 1 1 1 1 1 1 1 1 137 1 1 1 1 1 1 1 1 1 138 1 1 1 1 1 1 1 1 1 139 1 1 1 1 1 1 1 1 1 140 1 1 1 1 1 1 1 1 1 141 1 1 1 1 1 1 1 1 1 142 1 1 1 1 1 1 1 1 1 143 1 1 1 1 1 1 1 1 1 144 1 1 1 1 1 1 1 1 1 145 1 1 1 1 1 1 1 1 1 146 1 1 1 1 1 1 1 1 1 147 1 1 1 1 1 1 1 1 1 148 1 1 1 1 1 1 1 1 1 149 1 1 1 1 1 1 1 1 1 150 1 1 1 1 1 1 1 1 1 151 1 1 1 1 1 1 1 1 1 152 1 1 1 1 1 1 1 1 1 153 1 1 1 1 1 1 1 1 1 154 1 1 1 1 1

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1. A bag contains 4 red and 6 black v balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour is returned to the bag. If now a ball is drawn at random from the bag, then probability that this drawn ball is red, is:

3443? 310103? 2552? 1551? 1551?

Solution:

Total Probability=410⋅12+610⋅13=25Total Probability=104?⋅21?+106?⋅31?=52?

1. Let the orthocenter and centroid of a triangle be A(-3, 5) and B(3, 3) respectively. If C is the circumcentre of this triangle, than the radius of the circle having line segment AC as diameter, is:

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2a- 3 3 2a=12=a=6 2b+5 3 2b=4=b=2 AC=√(6+3)²+3² Diameter=AC=√81+9=√90 Radius= 3√10 3×√10 5 Radius= 2 2×√2 3

1. If the tangent at (1,7) to the curve x2=y−6x2=y−6 touches the circle x2+y2+16x+12y+c=0x2+y2+16x+12y+c=0 than the value of c is :

(1) 95
(2) 195
(3) 185
(4) 85

Solution:

x=y+72−6⇒2x=y+7−12⇒2x=y−5x=2y+7?−6⇒2x=y+7−12⇒2x=y−5

Also, centre of the circle (−8,−6)(−8,−6) and the radius is 64+36−c64+36−c?

⇒(−16+6+55)=100−c⇒5=100−c⇒c=95⇒(5?−16+6+5?)=100−c?⇒5?=100−c?⇒c=95

1. If α,β∈cα,β∈c are the distinct roots, of the equation x2−x+1=0x2−x+1=0 , then α101+β107α101+β107 is equal to :

(1) 2
(2) -1
(3) 0
(4) 1 Solution:

x2−x+1=0x2−x+1=0

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JEE Mains 2018 Mathematics

x=1±32=−ω,−ω2x=21±3??=−ω,−ω2⇒α=−ω and β=−ω2⇒α=−ω and β=−ω2⇒(−ω)101+(−ω2)107=(ω101+ω214)=−(ω2+ω)=1⇒(−ω)101+(−ω2)107=(ω101+ω214)=−(ω2+ω)=1

1. PQR is a triangular park with PQ=PR=200 m. AT.V. tower stands at the mid-point of QR. If the angle of elevation of the top of the tower at P,Q and R are respectively 450,300450,300 and 300300 then the height of tower (in m) is :
   (1) 502502?
   (2) 100
   (3) 50
   (4) 10031003?

Solution:

hx=13xh?=3?1?x=3hx=3?h200=3h2+h2200=3h2+h24h2=(200)24h2=(200)24h2=400004h2=40000h=100h=100

1. If ∑i=19(xi−5)=9∑i=19?(xi?−5)=9 and ∑i=19(xi−5)2=45∑i=19?(xi?−5)2=45 , then the standard deviation of the 9 items x1,x2,…,x9x1?,x2?,…,x9? is:
   (1) 3
   (2) 9
   (3) 4
   (4) 2

Solution:
Variance = 459−(1)2=5−1=4945?−(1)2=5−1=4
σ=Variance=2σ=Variance?=2

Set D

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1. The sum of the co-efficients of all odd degree terms in the expansion of (x+x3−1)5+(x−x3−1)5,(x>1)(x+x3−1?)5+(x−x3−1?)5,(x>1) is:

(1) 2
(2) -1
(3) 0
(4) 1

Solution:

Letx2−1=yLetx2−1?=y (x+y)5+(x−y)5(x+y)5+(x−y)5 =(5C0x5+5C1x4y+……5C5y5)+(5C0x5−5C1x4.y+…?−5C5y5)=(5C0?x5+5C1?x4y+……5C5?y5)+(5C0?x5−5C1?x4.y+…?−5C5?y5) =2[5C0x5+5C2x3y2+5C4xy4]=2[C0x5+5C2x3y2+5C4xy4]=2[5C0?x5+5C2?x3y2+5C4?xy4]=2[C0?x5+5C2?x3y2+5C4?xy4] =2[x5+10x3(x3−1)+5x(x3−1)2]=2[x5+10x6−10x3+5x(x6+1−2x3)]=2[x5+10x3(x3−1)+5x(x3−1)2]=2[x5+10x6−10x3+5x(x6+1−2x3)] =2[x5+10x6−10x3+5x7+5x−10x4]=2[1−10+5+5]=2=2[x5+10x6−10x3+5x7+5x−10x4]=2[1−10+5+5]=2

1. Tangents are drawn to the hyperbola 4x2−y2=364x2−y2=36 at the points P and Q. If these tangents intersects at the point T(0, 3) then the area (in sq. units) of ΔPTQΔPTQ is :

 

Solution:

Equation of PQ, 4x.(0)−3y=364x.(0)−3y=36

Y=−12Y=−12

Area of ΔTPQ=12×15×65=455Area of ΔTPQ=21?×15×65?=455?

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1. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is :

(1) at least 750 but less than 1000
(2) at least 1000
(3) less than 500
(4) at least 500 but less than 750

Solution:

6C4⋅3C1×1×4!6C4?⋅3C1?×1×4! 6×52.3×24=45×24=108026×5?.3×24=45×24=1080?

1. If the system of linear equations

x+ky+3z=0x+ky+3z=0 3x+ky−2z=03x+ky−2z=0 2x+4y−3z=02x+4y−3z=0

has a non- zero solution (x, y, z), then xzy2y2xz? is equal to:

(1) 30
(2) -10
(3) 10
(4) -30

Solution:

?1k33k−224−3?=0?132?kk4?3−2−3??=0 x+ky+3z=0…(i)x+ky+3z=0…(i) 3x+ky−2z=0…(ii)3x+ky−2z=0…(ii) 2x+4y−3z=0…(iii)2x+4y−3z=0…(iii) On Solving (i) and (ii)On Solving (i) and (ii) 4y=−2z…(iv)4y=−2z…(iv) On Solving (iii) and (iv)On Solving (iii) and (iv)

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4y=-2z

xzy2=52z×zz24=10y2xz?=4z2?25?z×z?=10

1. If ?x−42x2x2xx−42x2x2xx−4?=(A+Bx)(x−A)2,?x−42x2x?2xx−42x?2x2xx−4??=(A+Bx)(x−A)2, (1) (4,5) (2) (-4,-5) (3) (-4,3) (4) (-4,5)

Solution:

Put x=0x=0

Put x=1x=1

1. Two sets A and B are as under :

A={(a,b)∈R×R:?a−5?<1and?b−5?<1};A={(a,b)∈R×R:?a−5?<1and?b−5?<1}; B={(a,b)∈R×R:4(a−6)2+9(b−5)2≤36}.Then:B={(a,b)∈R×R:4(a−6)2+9(b−5)2≤36}.Then:

(1) neither A⊂BA⊂B nor B⊂AB⊂A

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1. B⊂AB⊂A (3) A⊂BA⊂B (4) A∩B=ΦA∩B=Φ (an empty set)

Solution:

Since Set A is, ?a−5?<14<a<6?a−5?<14<a<6

And ?b−5?<14<b<6?b−5?<14<b<6

Now, B is

(a−6)29+(b−5)24≤19(a−6)2?+4(b−5)2?≤1

It can be seen that all vertices of rectangle lie inside the ellipse, therefore A⊂BA⊂B

1. Tangent and normal are drawn at P(16, 16) on the parabola y2=16xy2=16x , which intersect the axis of the parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and ∠CPB=θ∠CPB=θ , then a value of tan?θtanθ is :

4312233434(1)34?21?32?43?43??(1)

Solution:

The equation of tangent at P

y−16=12(x−16)⇒A=(−16,0)y−16=21?(x−16)⇒A=(−16,0)

The normal is y=y−16=−2(x−16)y=y−16=−2(x−16)

B=(24,0)B=(24,0)

Since ∠APB=π2∠APB=2π?

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1. The Boolean expression ∼(p∨q)∨(∼p∧q)∼(p∨q)∨(∼p∧q) is equivalent to:

AB is the diameter

Center of the circle C=(4,0)C=(4,0)

Slope of PB=−2=m1PB=−2=m1?

Slope of CP=43=m2⇒tan?θ=?m2−m11+m2m1?=2Slope of CP=34?=m2?⇒tanθ=?1+m2?m1?m2?−m1???=2

1. Lets S={t∈R:f(x)=?x−Π??(e1−1)sin??x?}S={t∈R:f(x)=?x−Π??(e1−1)sin?x?} is not differentiable at t}. Then the set S is equal to:

Solution:

Doubtful point for differentiability are 0 and ππ At x=0x=0

f′(0+)=lim?h→0+?h−π?×(eh−1)×sin??h?−0hf′(0+)=limh→0+?h?h−π?×(eh−1)×sin?h?−0? =lim?h→0+(π−h)×(eh−1)×sin?hh=limh→0+?h(π−h)×(eh−1)×sinh?

∴lim?h→0+sinh?h=1 and lim?h→0+eh−1=0∴h→0+lim?hsinh?=1 and h→0+lim?eh−1=0∴f′(0+)=π×0×1=0∴f′(0+)=π×0×1=0f′(0−)=lim?h→0+?−h−π?×(eh−1)×sin??−h?−0−hf′(0−)=h→0+lim?−h?−h−π?×(eh−1)×sin?−h?−0?=lim?h→0+(π+h)×(eh−1)×sin?h−h=h→0+lim?−h(π+h)×(eh−1)×sinh?∴lim?h→0+sinh?h=1 and lim?h→0+eh−1=0∴h→0+lim?hsinh?=1 and h→0+lim?eh−1=0∴f′(0−)=(−π)×0×1=0∴f′(0−)=(−π)×0×1=0∴f′(0+)=f′(0−)=0∴f′(0+)=f′(0−)=0Similarly f ′(π+)=f′(π−)=0Similarly f ′(π+)=f′(π−)=0

Hence f(x)f(x) is differentiable ∀x∈R∀x∈R

1. The Boolean expression ∼(p∨q)∨(∼p∧q)∼(p∨q)∨(∼p∧q) is equivalent to:

(1) ∼p∼p

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1. ∼q∼q (3) p (4) q Solution: ∼(p∨q)∨(∼p∧q)∼(p∨q)∨(∼p∧q) P q ∼(p∨q)∼(p∨q) ∼p∧q∼p∧q T F F F T F F F F T T F T F F F F F F F F F F F F F F F F F F F F

1. A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is:

3x+2y=6xy3x+2y=6xy 3x+2y=63x+2y=6 2x+3y=xy2x+3y=xy 3x+2y=xy3x+2y=xy

Solution:

Let R=(h,k)R=(h,k)

P=(0,k)P=(0,k)

Q=(h,0)Q=(h,0)

Equation of line would be,

xh+yk=1……(i)hx?+ky?=1……(i)⇒2h+3k=1⇒h2?+k3?=12k+3h=hk2k+3h=hk

Locus of (h,k)(h,k) is 2y+3x=xy2y+3x=xy

1. Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series

12+222+32+242+52+262+…12+222+32+242+52+262+…

If B−2A=100λB−2A=100λ , then λλ is equal to:

(1) 496
(2) 232
(3) 248

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464

Solution:

A=12+2.22+32+2.42+……+A2+2.202A=12+2.22+32+2.42+……+A2+2.202 =(12+2.22+32+42+……+202)+(22+42+……+202)=(12+2.22+32+42+……+202)+(22+42+……+202) =20×21×416+4×10×11×216=2870+1540=4410=2870+1540=4410=620×21×41?+4×610×11×21?=2870+1540=4410=2870+1540=4410 B=40×41×816+4×20×21×416=540×41+41×280=41×820=33620B=640×41×81?+64×20×21×41?=540×41+41×280=41×820=33620 33620−8820=100λ33620−8820=100λ 100λ=24800100λ=24800 λ=248λ=248

1. Let y=y(x) be the solution of the differential equation

sin?xdydx+ycos?x=4x,x∈(0,Π). If y(Π2)=0, then y(Π6) is equal to :sinxdxdy?+ycosx=4x,x∈(0,Π). If y(2Π?)=0, then y(6Π?) is equal to :−49Π2493Π2−893Π2−89Π2(3)−94?Π293?4?Π2−93?8?Π2−98?Π2?(3)

Solution:

dydt+ycot?x=4xcos?ecx⇒d(ysin?x)=4xdxdtdy?+ycotx=4xcosecx⇒d(ysinx)=4xdx

Intragrating both sides we get ysin?x=2x2+Cysinx=2x2+C

Also,y(π2)=0⇒c=−π22Also,y(2π?)=0⇒c=−2π2? ⇒ysin?x=2x2−π22⇒y(π6)=−8π29⇒ysinx=2x2−2π2?⇒y(6π?)=−98π2?

1. The length of the projection of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane, x+y+z=7 is :

(1)23(1)32??

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

Solution:

x−51=y+11=z−41=λ1x−5?=1y+1?=1z−4?=λP(λ+5,λ−1,λ+4)P(λ+5,λ−1,λ+4)

P is foot of perpendicular from A to plane 3λ+8=73λ+8=7

λ=−13λ=−31?P(143,−43,113)P(314?,3−4?,311?)x−41=y+11=z−311x−4?=1y+1?=1z−3?Q(λ+4,λ−1,λ+3)Q(λ+4,λ−1,λ+3)

Q is foot of perpendicular from B to plane

3λ+6=73λ+6=7λ=13λ=31?Q(133,−23,103)Q(313?,3−2?,310?)∴PQ=1+4+13=163=23∴PQ=31+4+1??=316??=32??

1. Let S={x∈R:x≥0S={x∈R:x≥0 and 2x−3]+x(x−6)+6=0}2x?−3]+x?(x?−6)+6=0} . Then S :

(1) contains exactly four elements.

(2) is an empty set.

(3) contains exactly one element.

(4) contains exactly two element.

Solution:

2?x−3?+x(x−6)+6=02?x?−3?+x?(x?−6)+6=0

Case- I x≥3x?≥3

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2(x- 3)+x- 6√x+6=0x- 4√x=0x=0,6 As x≥9⇒x=16x≥9⇒x=16 Case- II x<3⇒−2x+6+x−6x+6=0x?<3⇒−2x?+6+x−6x?+6=0 (x−6)(x−2)=0⇒x=36,4(x?−6)(x?−2)=0⇒x=36,4 As, x<3⇒x=4x?<3⇒x=4 There are exactly two elements in the given set.

There are exactly two elements in the given set.

1. Let a1,a2,a3,…,a49a1?,a2?,a3?,…,a49? be in A.P. such that ∑k=012a4k+1=416∑k=012?a4k+1?=416 and a9+a43=66a9?+a43?=66 . If

a12+a22+?+a172=140m ,then m is equal to:a12?+a22?+?+a172?=140m ,then m is equal to:

(1) 33
(2) 66
(3) 68
(4) 34.

Solution:

a1+a5+a9=416⇒a+24d=32…………………………(i)a1?+a5?+a9?=416⇒a+24d=32…………………………(i) a9+a43=66⇒a+25d=33…………………(ii)a9?+a43?=66⇒a+25d=33…………………(ii)

From (i) and (ii) d=1d=1 and a=8a=8

⇒∑r=117(8+(r−1))⇒∑r=117(7+r)2=140m⇒m=34⇒r=1∑17?(8+(r−1))⇒r=1∑17?(7+r)2=140m⇒m=34

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1. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +σr−σ+σr?−σ and +σ+σ respectively. The potential of shell B is :

σ(b2−c2c+a)σ?0(a2−b2a+c)σ?0(a2+b2b+c)σ?0(b2−c2b+a)(4)σ(cb2−c2?+a)?0?σ?(aa2−b2?+c)?0?σ?(ba2+b2?+c)?0?σ?(bb2−c2?+a)?(4)

Solution:

Charge in sphere

A=qA=σ×4πa2A=qA?=σ×4πa2 B=qB=σ×4πb2B=qB?=σ×4πb2 C=qc=σ×4πc2C=qc?=σ×4πc2

Potential of B:

=qA4π?0b+qb4π?0b+qc4π?0c=σa2?0b+−σb2?0b+σc2?0c=σ?0(a2b−b+c)=σ?0(a2−b2b+c)(3)=4π?0?bqA??+4π?0?bqb??+4π?0?cqc??=?0?bσa2?+?0?b−σb2?+?0?cσc2?=?0?σ?(ba2?−b+c)=?0?σ?(ba2−b2?+c)?(3)

Hence the Solution is Option (3)

1. Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is :

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181 MR2 (2) 19 MR2 (3) 55 MR2 (4) 73 MR2

Solution:

Solution:Moment of inertia of each disc about the given axis is,

I1=12MR2+MR2=32MR2I1?=21?MR2+MR2=23?MR2 I4=12MR2+M(3R)2=192MR2I4?=21?MR2+M(3R)2=219?MR2 I7=12MR2+M(3R)2=512MR2I7?=21?MR2+M(3R)2=251?MR2 Inet=I1+I2+I3+I4+I5+I6+I7Inet?=I1?+I2?+I3?+I4?+I5?+I6?+I7? Inet>I1+I4+I7Inet?>I1?+I4?+I7? Inet>732MR2Inet?>273?MR2

Hence the Solution is Option (4)

1. From a uniform circular disc of radius R and mass 9 M, a small disc of radius R33R? is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is :

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1. 379MR2937?MR2 (2) 4MR24MR2 (3) 409MR2940?MR2 (4) 10MR210MR2

Solution:

Mass of disc == Volume x Density 9M=A×T×ρ9M=A×T×ρ (Area x thickness x density) 9M=πR2×t×ρ……(i)9M=πR2×t×ρ……(i)

For the disc which is cut off.

M′=π(R3)2×t×ρ……(ii)M′=π(3R?)2×t×ρ……(ii) (i)Divided by (ii)(i)Divided by (ii)

9MM=9⇒M′=MM9M?=9⇒M′=M

Moment of inertia of complete disc about an axis passes through O.

I1=12(9M)×R2I1?=21?(9M)×R2

Moment of inertia of cut off disc about an axis passes through O

I2=12M×(R3)2+M×(2R3)2I2?=21?M×(3R?)2+M×(32R?)2 =12MR29+4MR29=21?9MR2?+94MR2?

So, moment of inertia of remaining disc =I1−I2=I1?−I2?

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JEE Mains 2018 Physics

Hence the Solution is Option (2)

1. The reading of the ammeter for a silicon diode in the given circuit is :

(1) 13.5 mA
(2) 0
(3) 15 mA
(4) 11.5 mA

Solution:

The given diode is a forward bias & hence behaves as a perfect conductor i.e., it offers zero resistance.

So,

I=VR=3200=1×10−2AI=RV?=2003?=1×10−2A

=15mA=15mA

Hence the Solution is Option (3)

1. Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be 1221? . Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be 1881? . The angle between polarizer A and C is : (1) 600600 (2) 0000 (3) 300300 (4) 450450

(1) 600600

(2) 0000

(3) 300300

(4) 450450

Solution:

Unpolarized light of intensity I, when passed through a polarizer A, its intensity becomes I22I?

Since intensity of light emerging from polarizer B=I2B=2I?

So, A & B are parallel placed.

Let, C makes angle θθ with A.

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So,

θ=45?θ=45?

Hence the Solution is Option (4)

1. For an RLC circuit driven with voltage of amplitude νmνm? and frequency ω0=1LCω0?=LC?1? the current exhibits resonance. The quality factor, Q is given by :

CRω0ω0?CR? ω0LRRω0?L? ω0RLLω0?R? R(ω0C)(ω0?C)R?

Solution:

Qfactor == (voltage across L or C at Resonance)/(voltage across R)

=I×XLI×R=ω0LR=I×RI×XL??=Rω0?L?

Hence the Solution is Option (2)

1. Two masses m1=5kgm1?=5kg and m2=10kgm2?=10kg , connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2m2? to stop the motion is :

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1. 10.3 kg
   (2) 18.3 g
   (3) 27.3 kg
   (4) 43.3 kg

Solution:

To stop the motion,

m1g=T=μ(m1+m2)gm1?g=T=μ(m1?+m2?)g⇒m1μ−m2=m⇒μm1??−m2?=m⇒50.15−10=m⇒0.155?−10=m⇒m=23.3kg⇒m=23.3kg

Hence the nearest Solution is Option (3)

1. In a collinear collision, a particle with an initial speed ν0ν0? strikes a stationary particle of the same mass. If the final total kinetic energy is 50%50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :

ν022?ν0??ν044ν0??(3)2ν0(3)2?ν0?(4)ν02(4)2ν0??

Solution:

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1. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nthnth power of R. If the period of rotation of the particle is T, then : (1) TαRn/2TαRn/2 (2) TαR3/2TαR3/2 (3) TαRn2+1TαR2n?+1 (4) TαR(n+1)/2TαR(n+1)/2

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Solution:

F=KRn−1F=KRn−1 So,mv2R=KRnSo,Rmv2?=RnK? ⇒ν=KM×1R(n−1)/2⇒ν=MK??×R(n−1)/21? Now,T=2πRν=2πRKM×Rn−12Now,T=ν2πR?=MK??2πR?×R2n−1? ⇒T∝R(n+1)/2⇒T∝R(n+1)/2

Hence the Solution is Option (4)

1. Two batteries with e.m.f. 12V12V and 13V13V are connected in parallel across a load resistor of 10Ω10Ω . The internal resistances of the two batteries are 1Ω1Ω and 2Ω2Ω respectively. The voltage across the load lies between :

(1) 11.7 V and 11.8 V
(2) 11.6 V and 11.7 V
(3) 11.5 V and 11.6 V
(4) 11.4 V and 11.5 V

Solution:

Total

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1. In an a.c. circuit, the instantaneous e.m.f. and current are given by e=100sin?30te=100sin30t i=20sin?(30t−π4)i=20sin(30t−4π?) . In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively : (1) 50, 0 (2) 50, 10 (3) 10002,102?1000?,10 (4) 502,02?50?,0 Solution:

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1. An EM wave from air enters a medium. The electric fields are E1‾=E01x^cos?[2πν(zc−t)]E1??=E01?x^cos[2πν(cz?−t)] in air and E2‾=E02x^cos?[k(2z−ct)]E2??=E02?x^cos[k(2z−ct)] in medium, where the wave number k and frequency νν refer to their values in air. The medium is non- magnetic. If ?r1?r1?? and ?r2?r2?? refer to relative permittivities of air and medium respectively, which of the following options is correct ?

?r1?r2=4?r2???r1???=4 ?r1?r2=2?r2???r1???=2

Solution:

E1‾=E01x^cos?[2πν(zc−t)]E1??=E01?x^cos[2πν(cz?−t)] E2‾=E02x^cos?[k(2z−ct)]E2??=E02?x^cos[k(2z−ct)]

Where

k=2πλ&νc=1λk=λ2π?&cν?=λ1?

So volume in medium 1=C1=C

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Volume in medium 2=C/22=C/2

C=1μ?0?nC=μ?0??n??1? C2=1μ?0?n2C?=μ?0??n??1? ⇒2=?n?n⇒2=?n??n??? ⇒?n?n=14⇒?n??n??=41?

Hence the Solution is Option (4)

1. A telephonic communication service is working at carrier frequency of 10GHz10GHz . Only 10%10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5kHz5kHz ?

(1)2×106(1)2×106 (2)2×103(2)2×103 (3)2×104(3)2×104 (4)2×105(4)2×105

Solution:

Required number of channels

=10100×10×109=10010?×10×109 =2×105=2×105

Hence the Solution is Option (4)

1. A gratitude rod of 60cm60cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7×103kg/m32.7×103kg/m3 and its young's modulus is 9.27×1010Pa9.27×1010Pa What will be the fundamental frequency of the longitudinal vibrations ? (1) 7.5kHz7.5kHz (2) 5kHz5kHz (3) 2.5kHz2.5kHz (4) 10kHz10kHz Solution: Since rod is clamped at centre. So centre it behaves as node & end it behave as antinode.

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4 λ=1.2mλ=1.2m ν=yρ=9.27×10102.7×103ν=ρy??=2.7×1039.27×1010?? =5.85×103m/s=5.85×103m/s S0,ν=5.85×1031.2S0?,ν=1.25.85×103? =5kHz=5kHz Hence the Solution is Option (2)

Hence the Solution is Option (2)

1. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pdpd? ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pdpd? and pcpc? are respectively :

(1) (0, 1)
(2) (.89, .28)
(3) (.28, .89)
(4) (0, 0)

Solution: U1U1? , U2=0U2?=0

Since collision is elastic,

ν1=u1(M1−M2)+2M1M2M1+M2ν1?=M1?+M2?u1?(M1?−M2?)+2M1?M2??

But, u2=0u2?=0

⇒ν1u1=M1−M2M1+M2…………………………(i)⇒u1?ν1??=M1?+M2?M1?−M2??…………………………(i)

Fractional loss in kinetic energy of neutron,

=12M1u12−12M1ν1212M1u12=21?M1?u12?21?M1?u12?−21?M1?ν12?? =1−(ν1u1)2=1−(u1?ν1??)2

So, A.TQ:

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1. The density of a material in the shape of a cube is determines by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5%1.5% and 1%1% , the maximum error in determining the density is :

(1) 6%6% (2) 2.5%2.5% (3) 3.5%3.5% (4) 4.5%4.5%

Solution:

=1.5+3×1=1.5+3×1 =4.5%=4.5%

Hence the Solution is Option (4)

1. Two moles of an ideal monoatomic gas occupies a volume V at 270270 C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

(1) (a) 195 K (b) 2.7 kJ
(2) (a) 189 K (b) 2.7 kJ
(3) (a) 195 K (b) -2.7 kJ
(4) (a) 189 K (b) -2.7 kJ

Solution: Initially n=2n=2 vv T=300kT=300k Finally Vd=2vVd?=2v Gas is monoatomic, SoSo r=5/3r=5/3 So,

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1

⇒Tf=30041/3=3001.6≈189k⇒Tf?=41/3300?=1.6300?≈189k

Since gas undergoes expensed.

So,dw>0So,dw?>0 ⇒du<0.⇒du?<0.

Hence the Solution is Option (4)

1. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, (drr)(rdr?) is :

(1)mgKa(2)Kamg(3)Ka3mg(4)mg3Ka(4)(1)Kamg?(2)mgKa?(3)3mgKa?(4)3Kamg??(4)

Solution: Bulk Modulus,

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1. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1B1? . When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B2B2? . The ratio B1B2B2?B1?? is :

K=ΔP(ΔV/V)K=(ΔV/V)ΔP?Forsphere,V=43πr3Forsphere,V=34?πr3⇒ΔVV=3Δrr⇒VΔV?=r3Δr?⇒Δrr=ΔP3K=Mg3Ka⇒rΔr?=3KΔP?=3KaMg?

Hence the Solution is Option (4)

1. A parallel plate capacitor of capacitance 90pF90pF is connected to a battery of emf 20V20V . If a dielectric material constant K=53K=35? is inserted between the plates, the magnitude of the induced charge will be :

(1) 0.9nC0.9nC

(2) 1.2nC1.2nC

(3) 0.3nC0.3nC

(4) 2.4nC2.4nC

Solution:

qi=C0Vqi?=C0?V =90×10−12×20=90×10−12×20 =1800×10−12=1800×10−12 =1.8nC=1.8nC qf=53×1.8nCqf?=35?×1.8nC =3nC=3nC So,qind=3nC−1.8nCSo,qind?=3nC−1.8nC =1.2nC=1.2nC

Hence the Solution is Option (2)

1. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1B1? . When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B2B2? . The ratio B1B2B2?B1?? is :

122?1?

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1. 33? (4) 22?

Solution:

Magnetic moment, M=IA=I×πr2M=IA=I×πr2 and magnetic field at the centre of circle

=B=μ0I2R=B=2Rμ0?I?Mf=I×πrf2=2MMf?=I×πrf2?=2MMi=I×πri2=MMi?=I×πri2?=M⇒rfri=2⇒ri?rf??=2?S0,B1B2=rfri=2S0?,B2?B1??=ri?rf??=2?

Hence the Solution is Option (4)

1. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λn,λgλn?,λg? be the de Broglie wavelength of the electron in the nthnth state and the ground state respectively. Let ΛnΛn? be the wavelength of the emitted photon in the transition from the nthnth state to the ground state. For large n, (A, B are constants)

Λn2≈λΛn≈A+Bλn2Λn≈A+BλnΛn2≈A+Bλn2(4)Λn2?≈λΛn?≈A+λn2?B?Λn?≈A+Bλn?Λn2?≈A+Bλn2??(4)

Solution:

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1 λ=hmv=h2MEkλ=mvh?=2MEk??h? Ek=h22mλ2Ek?=2mλ2h2? For emitted photon, En−E1En?−E1? =hcΛn=Λn?hc? ⇒h22mλn2−h22mλg2=hcΛn⇒2mλn2?h2?−2mλg2?h2?=Λn?hc? ⇒hcΛn=h22mλn2+K[K=−h22mλg2]⇒Λn?hc?=2mλn2?h2?+K[K=2mλg2?−h2?] ⇒Λn=A+Bλn2⇒Λn?=A+Bλn2? Hence the Solution is Option (4)

Hence the Solution is Option (4)

1. The mass of a hydrogen molecule is 3.32×10−27kg3.32×10−27kg . If 10231023 hydrogen molecules strike, per second, a fixed wall of area 2cm22cm2 at an angle of 450450 to the normal, and rebound elastically with a speed of 103m/s103m/s , then the pressure on the wall is nearly :

(1)4.70×102N/m2(1)4.70×102N/m2 (2)2.35×103N/m2(2)2.35×103N/m2 (3)4.70×103N/m2(3)4.70×103N/m2 (4)2.35×102N/m2(4)2.35×102N/m2

Solution:

Change in momentum normal to the wall

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2mvcos45o So, Force=2mvcos45N 2x3.32x10- 27x10- 3x10- 3 Pr essure= 2x10- 4x1 P=2.35x10N/m2 Hence the Solution is Option (2)

1. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

Solution:

(1),(2),(4) uniform retardation & then uniform acceleration. (3) Normal uniform acceleration & retardation. Hence the Solution is Option (3)

1. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii rere? , rprp? , rαrα? respectively in a uniform magnetic field B. The relation between rere? , rprp? , rαrα? is:

(1)re<rp<rα(1)re?<rp?<rα? (2)re>rp=rα(2)re?>rp?=rα?

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1. re<rp=rare?<rp?=ra? (4) re<rp<rare?<rp?<ra?

Solution:

For change moving in circular orbit is a uniform magnetic field,

r=mvBq=2mEkBqr=Bqmv?=Bq2mEk??? ⇒Ek=B2q2r22m⇒Ek?=2mB2q2r2?

Since all particles came same Ek&BEk?&B

So,

q2r22m=Cons tant2mq2r2?=Cons tantqe=qp&me<mpqe?=qp?&me?<mp?⇒re<rp⇒re?<rp?

For proton & αα - particle

12×rp22×1=22×ra22×42×112×rp2??=2×422×ra2?? ⇒rp=ra⇒rp?=ra?

Hence the Solution is Option (3)

1. On interchanging the resistances, the balane point of a meter bridge shifts to the left by 10 cm. The resistance of their series combinations is 1kΩ1kΩ . How much was the resistance on the left slot before interchanging the resistances ?

(1) 910Ω (2) 990Ω (3) 505Ω (4) 550Ω

Solution:

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1.0 1.0 1.0 1.0 1.0 1.1 1.1 1.1 1.1 1.1 1.2 1.2 1.2 1.2 1.2 1.3 1.3 1.3 1.3 1.3 1.4 1.4 1.4 1.4 1.4 1.5 1.5 1.5 1.5 1.5 1.6 1.6 1.6 1.6 1.6 1.7 1.7 1.7 1.7 1.7 1.8 1.8 1.8 1.8 1.8 1.9 1.9 1.9 1.9 1.9 2.0 2.0 2.0 2.0 2.0 2.1 2.1 2.1 2.1 2.1 2.2 2.2 2.2 2.2 2.2 2.3 2.3 2.3 2.3 2.3 2.4 2.4 2.4 2.4 2.4 2.5 2.5 2.5 2.5 2.5 2.6 2.6 2.6 2.6 2.6 2.7 2.7 2.7 2.7 2.8 2.8 2.8 2.8 2.9 3.0 3.0 3.0 3.0 3.1 3.1 3.1 3.1 3.2 3.2 3.2 3.2 3.3 3.3 3.3 3.3 3.4 3.4 3.4 3.4 3.5 3.5 3.5 3.5 3.6 3.6 3.6 3.6 3.7 3.7 3.7 3.7 3.8 3.8 3.8 3.8 3.9 4.0 4.0 4.0 4.0 4.1 4.1 4.1 4.1 4.2 4.2 4.2 4.2 4.3 4.3 4.3 4.3 4.4 4.4 4.4 4.4 4.5 4.5 4.5 4.5 4.6 4.6 4.6 4.6 4.7 4.7 4.7 4.7 4.8 4.8 4.8 4.8 4.9 5.0 5.0 5.0 5.0 5.1 5.1 5.1 5.1 5.2 5.2 5.2 5.2 5.3 5.3 5.3 5.3 5.4 5.4 5.4 5.4 5.5 5.5 5.5 5.5 5.6 5.6 5.6 5.6 5.7 5.7 5.7 5.7 5.8 5.8 5.8 5.8 5.9 6.0 6.0 6.0 6.0 6.1 6.1 6.1 6.1 6.2 6.2 6.2 6.2 6.3 6.3 6.3 6.3 6.4 6.4 6.4 6.4 6.5 6.5 6.5 6.5 6.6 6.6 6.6 6.6 6.7 6.7 6.7 6.7 6.8 6.8 6.8 6.8 6.9 7.0 7.0 7.0 7.0 7.1 7.1 7.1 7.1 7.2 7.2 7.2 7.2 7.3 7.3 7.3 7.3 7.4 7.4 7.4 7.4 7.5 7.5 7.5 7.5 7.6 7.6 7.6 7.6 7.7 7.7 7.7 7.7 7.8 7.8 7.8 7.8 7.9 8.0 8.0 8.0 8.0 8.1 8.1 8.1 8.1 8.2 8.2 8.2 8.2 8.3 8.3 8.3 8.3 8.4 8.4 8.4 8.4 8.5 8.5 8.5 8.5 8.6 8.6 8.6 8.6 8.7 8.7 8.7 8.7 8.8 8.8 8.8 8.8 8.9 9.0 9.0 9.0 9.0 9.1 9.1 9.1 9.1 9.2 9.2 9.2 9.2 9.3 9.3 9.3 9.3 9.4 9.4 9.4 9.4 9.5 9.5 9.5 9.5 9.6 9.6 9.6 9.6 9.7 9.7 9.7 9.7 9.8 9.8 9.8 9.8 9.9 10.0 10.0 10.0 10.0 10.1 10.1 10.1 10.1 10.2 10.2 10.2 10.2 10.3 10.3 10.3 10.3 10.4 10.4 10.4 10.4 10.5 10.5 10.5 10.5 10.6 10.6 10.6 10.6 10.7 10.7 10.7 10.7 10.8 10.8 10.8 10.8 10.9 11.0 11.0 11.0 11.0 11.1 11.1 11.1 11.1 11.2 11.2 11.2 11.2 11.3 11.3 11.3 11.3 11.4 11.4 11.4 11.4 11.5 11.5 11.5 11.5 11.6 11.6 11.6 11.6 11.7 11.7 11.7 11.7 11.8 11.8 11.8 11.8 11.9 12.0 12.0 12.0 12.0 12.1 12.1 12.1 12.1 12.2 12.2 12.2 12.2 12.3 12.3 12.3 12.3 12.4 12.4 12.4 12.4 12.5 12.5 12.5 12.5 12.6 12.6 12.6 12.6 12.7 12.7 12.7 12.7 12.8 12.8 12.8 12.8 12.9 13.0 13.0 13.0 13.0 13.1 13.1 13.1 13.1 13.2 13.2 13.2 13.2 13.3 13.3 13.3 13.3 13.4 13.4 13.4 13.4 13.5 13.5 13.5 13.5 13.6 13.6 13.6 13.6 13.7 13.7 13.7 13.7 13.8 13.8 13.8 13.8 13.9 14.0 14.0 14.0 14.0 14.1 14.1 14.1 14.1 14.2 14.2 14.2 14.2 14.3 14.3 14.3 14.3 14.4 14.4 14.4 14.4 14.5 14.5 14.5 14.5 14.6 14.6 14.6 14.6 14.7 14.7 14.7 14.7 14.8 14.8 14.8 14.8 14.9 15.0 15.0 15.0 15.0 15.1 15.1 15.1 15.1 15.2 15.2 15.2 15.2 15.3 15.3 15.3 15.3 15.4 15.4 15.4 15.4 15.5 15.5 15.5 15.5 15.6 15.6 15.6 15.6 15.7 15.7 15.7 15.7 15.8 15.8 15.8 15.8 15.9 16.0 16.0 16.0 16.0 16.1 16.1 16.1 16.1 16.2 16.2 16.2 16.2 16.3 16.3 16.3 16.3 16.4 16.4 16.4 16.4 16.5 16.5 16.5 16.5 16.6 16.6 16.6 16.6 16.7 16.7 16.7 16.7 16.8 16.8 16.8 16.8 16.9 17.0 17.0 17.0 17.0 17.1 17.1 17.1 17.1 17.2 17.2 17.2 17.2 17.3 17.3 17.3 17.3 17.4 17.4 17.4 17.4 17.5 17.5 17.5 17.5 17.6 17.6 17.6 17.6 17.7 17.7 17.7 17.7 17.8 17.8 17.8 17.8 17.9 18.0 18.0 18.0 18.0 18.1 18.1 18.1 18.1 18.2 18.2 18.2 18.2 18.3 18.3 18.3 18.3 18.4 18.4 18.4 18.4 18.5 18.5 18.5 18.5 18.6 18.6 18.6 18.6 18.7 18.7 18.7 18.7 18.8 18.8 18.8 18.8 18.9 19.0 19.0 19.0 19.0 19.1 19.1 19.1 19.1 19.2 19.2 19.2 19.2 19.3 19.3 19.3 19.3 19.4 19.4 19.4 19.4 19.5 19.5 19.5 19.5 19.6 19.6 19.6 19.6 19.7 19.7 19.7 19.7 19.8 19.8 19.8 19.8 19.9 20.0 20.0 20.0 20.0 20.1 20.1 20.1 20.1 20.2 20.2 20.2 20.2 20.3 20.3 20.3 20.3 20.4 20.4 20.4 20.4 20.5 20.5 20.5 20.5 20.6 20.6 20.6 20.6 20.7 20.7 20.7 20.7 20.8 20.8 20.8 20.8 20.9 21.0 21.0 21.0 21.0 21.1 21.1 21.1 21.1

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1.5Ω Hence the Solution is Option (3)

1. If the series limit frequency of the Lyman series is νLνL? , then the series limit frequency of the Pfund series is :

 

Solution:

1λ=R(1n12−1n22)∝νλ1?=R(n12?1?−n22?1?)∝ν R(112)∝νl&R(152)∝νpR(121?)∝νl?&R(521?)∝νp?

Hence the Solution is Option (1)

1. The angular width of the central maximum in a single slit diffraction pattern is 600600 . The width of the slit is 1μm1μm . The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance of 50cm50cm from the slits. If the observed fringe width is 1cm1cm , what is slit separation distance ? (i.e. distance between the centres of each slit.)

(1) 100μm100μm (2) 25μm25μm (3) 50μm50μm (4) 75μm75μm

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2 2λd=600×π180d2λ?=600×180π? λ=(π3×12)μmλ=(3π?×21?)μm In Ydse, B′=λDd′×10−6B′=d′λD?×10−6 ⇒10−2=(π2×12)×12d′×10−6⇒10−2=d′(2π?×21?)×21??×10−6 ⇒d′=π10−2×12×10−6⇒d′=10−2×12π?×10−6 =26.16≈25μm=26.16≈25μm Hence the Solution is Option (2)

Solution:

2λd=600×π180d2λ?=600×180π? λ=(π3×12)μmλ=(3π?×21?)μm

In Ydse,

B′=λDd′×10−6B′=d′λD?×10−6 ⇒10−2=(π2×12)×12d′×10−6⇒10−2=d′(2π?×21?)×21??×10−6 ⇒d′=π10−2×12×10−6⇒d′=10−2×12π?×10−6 =26.16≈25μm=26.16≈25μm

Hence the Solution is Option (2)

1. A particle is moving in a circular path of radius a under the action of an attractive potential

U=−k2r2U=−2r2k? . Its total energy is :

(1)−32ka2(2)−k4a2(3)k2a2(4)Zero(4)(1)2−3?a2k?(2)4a2−k?(3)2a2k?(4)Zero?(4)

Solution:

u=−K2r2u=2r2−K? ButF=−dudr=−Kr3ButF=dr−du?=r3−K?

' - ' sign of force implies attractive force So,

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1. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec1012/sec . What is the force constant of the bonds connecting one atom with the other ? (Mole wt. of silver =108=108 and Avagadro number =6.02×1023gm=6.02×1023gm mole −1−1 )

(1) 5.5 N/m5.5 N/m (2) 6.44 N/m6.44 N/m (3) 7.1 N/m7.1 N/m (4) 2.2 N/m2.2 N/m

Solution:

T=2πMKT=2πKM?? K=4π2×MT2K=T24π2×M? =4π2×108×10−36.023×1023×(1012)2=4π2×6.023×1023108×10−3?×(1012)2 =4π2×10.86.023×10−2=4π2×6.02310.8?×10−2 =7.1N/m=7.1N/m

Hence the Solution is Option (3)

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1. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

(1)[Co(H2O)3Cl3].3H2O(1)[Co(H2?O)3?Cl3?].3H2?O (2)[Co(H2O)6]Cl3(2)[Co(H2?O)6?]Cl3? (3)[Co(H2O)5Cl]Cl2H2O(3)[Co(H2?O)5?Cl]Cl2?H2?O (4)[Co(H2O)4Cl2]Cl.2H2O(4)[Co(H2?O)4?Cl2?]Cl.2H2?O

Solution:

ΔTf=i×kf×mΔTf=i×kf×m ΔTf∝iΔTf∝i [Co(H2O)3Cl3].3H2O;i=1[Co(H2?O)3?Cl3?].3H2?O;i=1 [Co(H2O)6]Cl3;i=2[Co(H2?O)6?]Cl3?;i=2 [Co(H2O)5Cl]Cl2.H2O;i=4[Co(H2?O)5?Cl]Cl2?.H2?O;i=4 [Co(H2O)4Cl2]Cl.2H2O;i=3[Co(H2?O)4?Cl2?]Cl.2H2?O;i=3

Hence the answer is option (1).

1. Hydrogen peroxide oxides [Fe(CN)6]4−[Fe(CN)6?]4− to [Fe(CN)6]3−[Fe(CN)6?]3− in acidic medium but reduces [Fe(CN)6]3−[Fe(CN)6?]3− to [Fe(CN)6]4−[Fe(CN)6?]4− in alkaline medium. The other products formed are, respectively:

(1)H2O and (H2O+OH−)(1)H2?O and (H2?O+OH−) (2)(H2O+O2)and H2O(2)(H2?O+O2?)and H2?O (3)(H2O+O2)and (H2O+OH−)(3)(H2?O+O2?)and (H2?O+OH−) (4)H2O and (H2O+O2)(4)H2?O and (H2?O+O2?)

Solution:

[Fe(+2)(CN)6]4−→H2O2[Fe(+3)(CN)6]3−+H2O[Fe(+2)(CN)6?]4−H2?O2??[Fe(+3)(CN)6?]3−+H2?O

reducing agent

(oxidisin gagent)H2O2−1→H2O−2(Reduction)(oxidisin gagent)H2?O2−1?→H2?O−2(Reduction) (Reducing agent)H2O2−1→O2(oxidation)(Reducing agent)H2?O2−1?→O2?(oxidation)

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1. Which of the following compounds will be suitable for Kjeldahl's method for nitrogen estimation?

Solution: Kjeldahl's method:-

Organic compounds nitrogen + Conc.H2SO4→(NH4)2SO4Conc.H2?SO4?→(NH4?)2?SO4? ↓↓ alkali NH3NH3?

Nitro compounds, A2OA2?O compounds & Nitrogen part of the aromatic ring will not give positive result for Kjeldahl's method.

Aniline is the best suitable to estimate nitrogen using Kjeldahl's method.

Hence the answer is option (3).

1. Glucose on prolonged heating with HI gives :

(1) 6-iodohexanal
(2) n-Hexane
(3) 1-Hexane
(4) Hexanoic acid

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1. An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination?

Base Acid End point (1) Strong Strong Pink to colourless (2) Weak Strong Colourless to pink (3) Strong Strong Pinkish red to yellow (4) Weak Strong Yellow to pinkish red

Solution:

When a weak base is titrated with string acid with methyl orange as an indicator then at end point The colour change will be yellow to pinkish red. Hence the answer is option (4).

1. The predominant form of histamine present in human blood is (pka(pka? , Histidine =6.0=6.0

Solution:

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JEE Mains 2018 Chemistry

Hence the answer is option (1).

1. The increasing order of basicity of the following compounds is :

(1) (d)<(b)<(a)<(c)(d)<(b)<(a)<(c) (2) (a)<(b)<(c)<(d)(a)<(b)<(c)<(d) (3) (b)<(a)<(c)<(d)(b)<(a)<(c)<(d) (4) (b)<(a)<(d)<(c)(b)<(a)<(d)<(c)

Solution:

The lone pair on the nitrogen atom is in conjugation with the ππ bond hence it can involve in resonance.

during resonance the other nitrogen attains a negative charge.

Hence 'c' is a strong base.

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JEE Mains 2018 Chemistry

The +1+1 group of CH3CH3? and C2H5C2?H5? makes 'd' more basic but less basic than 'c'

Due to - I of ππ bond, basic character decreases.

The nitrogen involves sp2sp2 hybridization which is highly electro negative have least basic. Hence the answer is option (4).

1. Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction?

(1) A and D
(2) A and B
(3) B and C
(4) C and D

Solution:

ln?K=−ΔHoRT+ΔSoR(RelationbetweenRateconstant)lnK=RT−ΔHo?+RΔSo?(RelationbetweenRateconstant)

For exothermic reaction ΔH=ΔH= - ve.

Slope=−ΔHoR>0Slope=R−ΔHo?>0

Hence the answer is option (2).

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1. How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B=10.8uB=10.8u ) (1) 1.6 hours (2) 6.4 hours (3) 0.8 hours (4) 3.2 hours

Solution:

Cathode:H2O+2e−→H2+2OH−(Reduction)Cathode:H2?O+2e−→H2?+2OH−(Reduction)

Anode:

2H2O→4H++O2+4e−(oxidation)2H2?O→4H++O2?+4e−(oxidation) B2H6+3O2→B2O3+3H2OB2?H6?+3O2?→B2?O3?+3H2?O 27.66g27.66g n=27.66276≈1n=27627.66?≈1

Moles of O2O2? required =3=3

Gram equivalent of O2=3×4=12O2?=3×4=12

Gram equivalent

w=it96500w=96500it? 12=100×t9650012=96500100×t? t=12×96560×60sec?t=60×6012×965?sec t=3.2hrt=3.2hr

Hence the answer is option (4).

1. Consider the following reaction and statements:

[Co(NH3)4Br2]++Br−→[Co(NH3)3Br3]++NH3[Co(NH3?)4?Br2?]++Br−→[Co(NH3?)3?Br3?]++NH3?

(I) Two isomers are produced if the reactant complex ion is a cis-isomer.

(II) Two isomers are produced if the reactant complex ion is a trans-isomer.

(III) Only one isomer is produced if the reactant complex ion is a trans-isomer.

(IV) Only one isomer is produced if the reactant complex ion is a cis-isomer.

The correct statements are:

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

Solution:

one product can be formed.

Hence the answer is option (3).

1. Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with Br2Br2? to form product B. A and B are respectively :

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3 Solution:

Hence the answer is option (4).

1. An aqueous solution contains an unknown concentration of Ba2+Ba2+ . When 50 mL50 mL of a 1M solution of Na2SO4Na2?SO4? is added, BaSO4BaSO4? just begins to precipitate. The final volume is 500 mL500 mL . The solubility product of BaSO4BaSO4? is 1×10−101×10−10 . What is the original concentration of Ba2+Ba2+ ?

(1) 1.0×10−10M1.0×10−10M (2) 5×10−9M5×10−9M (3) 2×10−9M2×10−9M (4) 1.1×10−9M1.1×10−9M

Solution:

KSp(BaSO4)=10−10KSp??(BaSO4?)=10−10 QSp=KSpQSp?=KSp? (Ba2+)(SO42−)=10−10(Ba2+)(SO42−?)=10−10 Ba2+50500=10−10Ba2+50050?=10−10 ∴(Ba2+)=10−9∴(Ba2+)=10−9 m1ν1=m2ν2m1?ν1?=m2?ν2? m×450=10−9×500m×450=10−9×500 ∴m=500450×10−9∴m=450500?×10−9 m=1.1×10−9m=1.1×10−9

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1. At 5180C5180C , the rate of decomposition of a sample of gaseous acetaldehyde, initially ar a pressure of 363 Torr, was 1.00 Torr s−1s−1 when 5%5% had reacted and 0.5 Torr s−1s−1 when 33%33% had reacted. The order of the reaction is :

(1) 0 (2) 2 (3) 3 (4) 1

Solution:

1=k[363×95100]m……………………………(1)1=k[363×10095?]m……………………………(1) 0.5=k(362×67100)…………………(2)0.5=k(362×10067?)…………………(2)

equation−1/Equation−2equation−1/Equation−210.5=[9567]m0.51?=[6795?]m2=(9567)m2=(6795?)m2=(0.4)m2=(0.4)mlog?2=mlog?1.4log2=mlog1.40.3010log?1.4=mlog1.40.3010?=mm≈0.30100.15m≈0.150.3010?m=2m=2

Hence the answer is option (2).

1. The combustion of benzene (1) gives CO2(g)CO2?(g) and H2O(I)H2?O(I) . Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol−1−1 at 250250 C; heat of combustion (in kJ mol−1−1 ) of benzene at constant pressure will be:

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1. -3267.6
   (2) 4152.6
   (3) -452.46
   (4) 3260

Solution:

C6H6(l)+152O2(g)→6CO2+3H2O(l)C6?H6?(l)+215?O2?(g)→6CO2?+3H2?O(l) Δng=6−7.5=−1.5Δng=6−7.5=−1.5 ΔH=ΔU+ΔngRTΔH=ΔU+ΔngRT =−3263.9−1.5×8.134×10−3×298=−3263.9−1.5×8.134×10−3×298 =−3267.6=−3267.6

Hence the answer is option (1).

1. The ratio of mass present of C and H of an organic compound (CXHYOZ)(CX?HY?OZ?) is 6:16:1 . If one molecule of the above compound (CXHYOZ)(CX?HY?OZ?) contains half as much oxygen as required to burn one molecule of compound CXHYCX?HY? completely to CO2CO2? and H2OH2?O . The empirical formula of compound CXHYOZCX?HY?OZ? is :

(1) C2H4O3C2?H4?O3? (2) C3H6O3C3?H6?O3? (3) C2H4OC2?H4?O (4) C3H4O2C3?H4?O2?

Solution:

CXHYOZCX?HY?OZ? - Organic compound mass of CC mass of HH mass of C/12=612:12C/12=126?:21? mass of H/1=612:12H/1=126?:21? Noof moles of C=12C=21? Noof moles of HH CXHYOZ=molesofCMolesofHCX?HY?OZ?=MolesofHmolesofC?

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1 CxHy+(x+yx)o2→xco2+y2H2ocxhyoz(x+y4−x2)o2→xco2+yxH2O12(x+y4)=x+y4−z2x+y4=2x+y2−ZZ=x+y4c2h4o3Z=2+44=3Cx?Hy?+(x+xy?)o2?→xco2?+2y?H2?ocx?hy?oz?(x+4y?−2x?)o2?→xco2?+xy?H2?O21?(x+4y?)=x+4y?−2z?4x+y?=2x+2y?−ZZ=x+4y?c2?h4?o3?Z=2+44?=3?

Hence the answer is option (1).

1. The trans-alkenes are formed by the reduction of alkynes with :

(1) Sn-HCl
(2) H2−pd/C,BaSO4H2?−pd/C,BaSO4? (3) NaBH4
(4) Na/liqu. NH3NH3?

Solution:

Trans alkenes are formed by the reaction of alkynes with Na/liqu. NH3NH3? (birch Reduction) Hence the answer is option (4).

1. Which of the following are Lewis acids ?

(1) BCl3BCl3? and AlCl3AlCl3? (2) PH3PH3? and BCl3BCl3? (3) AlCl3AlCl3? and SiCl4SiCl4? (4) PH3PH3? and SiCl4SiCl4?

Solution:

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JEE Mains 2018 Chemistry

BCl3BCl3? and AlCl3AlCl3? are electron deficient compounds. Boron and aluminium has 6 electrons in their valence shell in BCl3BCl3? & AlCl3AlCl3? , PH3PH3? , SiCl4SiCl4? have 8 electrons in their valence shell. They are not Lewis acids.

Hence the answer is option (1).

1. When metal 'M' is treated with NaOH, a white gelatinous precipitate 'X' is obtained, which is soluble in excess of NaOH. Compound 'X' when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal 'M' is :

(1) Fe
(2) Zn
(3) Ca
(4) Al

Solution:

The Gelatinous precipitate formed in Al(OH)3Al(OH)3? ; Al(OH)3Al(OH)3? on strong heating gives Al2O3Al2?O3? which is used in chromatography as an adsorbent. So the metal is Al. Hence the answer is option (4).

1. According to molecular orbital theory, which of the following will not be a viable molecule ?

(1) H22−H22−? (2) H22+H22+? (3) He2+He2+? (4) H2−H2−?

Solution :

H22− e l e c t r o n i c i c f o r i g r a t i o n σ1S2,σ1S2H22−? e l e c t r o n i c i c f o r i g r a t i o n σ1S2,σ1S2  N u m b e r o f e l e c t r o n s =4 ; Bo=12(Nb−Na)=12(2−2)=0 N u m b e r o f e l e c t r o n s =4 ; Bo=21?(Nb?−Na?)=21?(2−2)=0

Molecule does not exist.

Hence the answer is option (1).

1. The major product formed in the following reaction is :

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JEE Mains 2018 Chemistry

Hence the answer is option (1).

1. Phenol on treatment with CO2CO2? in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with (CH3CO)2O(CH3?CO)2?O in the presence of catalytic amount of H2SO4H2?SO4? produces :(1)

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1

Hence the answer is option (2).

1. Which of the following compounds contain(s) no covalent bond(s) ?

KCl,PH3,O2,B2H6,H2SO4KCl,PH3?,O2?,B2?H6?,H2?SO4?

(1) KCl, B2H6B2?H6? (2) KCl, B2H6,PH3B2?H6?,PH3? (3) KCl, H2SO4H2?SO4? (4) KCl

Solution:

Solution:KCl is an ionic compound. It cannot from covalent bond. Elements of s - block & p - block combine to form ionic compounds. Hence the answer is option (4).

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1. Which type of 'defect' has the presence of cations in the interstitial sites ? (1) Metal deficiency defect (2) Schottky defect (3) Vacancy defect (4) Frenkel defect

Solution: Frankel defect has the presence of cation in interstitial site. Hence the answer is option (4).

1. The major product of the following reaction is :

Solution:

Hence the answer is option (3).

1. The compound that does not produce nitrogen gas by the thermal decomposition is :

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1. (NH4)2SO4(NH4?)2?SO4? (2) Ba(N3)2Ba(N3?)2? (3) (NH4)2Cr2O7(NH4?)2?Cr2?O7? (4) NH4NO2NH4?NO2?

Solution:

NH4NO2→ΔN2+H2ONH4?NO2?Δ?N2?+H2?O (NH4)2Cr2O7→ΔN2+H2O+Cr2O3(NH4?)2?Cr2?O7?Δ?N2?+H2?O+Cr2?O3? Ba(N3)2→ΔBa+N2Ba(N3?)2?Δ?Ba+N2?

Hence the answer is option (1).

1. An aqueous solution contains 0.10MH2S0.10MH2?S and 0.20MHCl0.20MHCl . If the equilibrium constants for the formation of HS−HS− from H2SH2?S is 1.0x10−71.0x10−7 and that of S2−S2− from HS−HS− ions is 1.2x10−131.2x10−13 then the concentration of S2−S2− ions in aqueous solution is :

(1) 5x10−195x10−19 (2) 5x10−85x10−8 (3) 3x10−203x10−20 (4) 6x10−216x10−21

Solution:

H2S→ka1,ka22H++S2−H2?S2H++S2−ka1??,ka2??? 0.1−x0.2x0.1−x0.2x ka1,ka2=(0.2)2×520.1ka1??,ka2??=0.1(0.2)2×52? 1.2×10−2×0.10.04=[S2−]0.041.2×10−2×0.1?=[S2−] [S2−]=3×10−20[S2−]=3×10−20

Hence the answer is option (3).

1. The oxidation states of CrCr in [Cr(H2O)6]Cl3,[Cr(C6H6)2][Cr(H2?O)6?]Cl3?,[Cr(C6?H6?)2?] , and K2[Cr(CN)2(O)2(O2)NH3]K2?[Cr(CN)2?(O)2?(O2?)NH3?] respectively are :

(1) +3+3 ,0,and +4+4 (2) +3+3 +4+4 ,and +6+6 (3) +3+3 +2+2 ,and +4+4 (4) +3+3 ,0,and +6+6

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1. The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required to make teeth enamel harder by converting [3Ca3(PO4)2.Ca(OH)2][3Ca3?(PO4?)2?.Ca(OH)2?] :

(1) [3{Ca(OH)2}.CaF2][3{Ca(OH)2?}.CaF2?] (2) [CaF2][CaF2?] (3) [3(CaF2).Ca(OH)2][3(CaF2?).Ca(OH)2?] (4) [3Ca3(PO4)2.CaF2][3Ca3?(PO4?)2?.CaF2?]

Solution:

[3Ca3(PO4)2.CaF2][3Ca3?(PO4?)2?.CaF2?]

Hence the answer is option (4).

1. Which of the following salts is the most basic in aqueous solution?

(1) Pb(CH3COO)2Pb(CH3?COO)2? (2) Al(CN)3Al(CN)3? (3) CH3COOKCH3?COOK (4) FeCl3FeCl3?

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1. Total number of lone pair of electrons in I3−I3−? ion is :

(1) 12 (2) 3 (3) 6 (4) 9

Solution:

The total number of lone pair of electrons is I3−I3−? is 9

Hence the answer is option (4).

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