JEE Main Previous Year Question paper-solved

JEE Main exam with PDF Download link-2025

1. Let α,β,γα,β,γ and δδ be the coefficients of x?,x?,x?x?,x?,x? and xx respectively in the expansion of (x+x3−1)5+(x−x3−1)5(x+x3−1?)5+(x−x3−1?)5 x>1x>1 If u and v satisfy the equations αu+βv=18αu+βv=18 γu+δv=20γu+δv=20 then u+vu+v equals :
(1) 5
(2) 4
(3) 3
(4) 8
Ans. (1)
Sol. (x+x3−1)5+(x−x3−1)5(x+x3−1?)5+(x−x3−1?)5
=2{5Cox5+5Cox3(x3−1)+5Cox(x3−1)5}=2{5Co?x5+5Co?x3(x3−1)+5Co?x(x3−1)5}
=2{5x7+10x6+x5−10x4−10x3+5x}=2{5x7+10x6+x5−10x4−10x3+5x}
⇒α=10,β=2,γ=−20,δ=10⇒α=10,β=2,γ=−20,δ=10
Now, 10u+2v=1810u+2v=18
−20u+10v=20−20u+10v=20
⇒u=1,v=4⇒u=1,v=4
u+v=5u+v=5

2. In a group of 3 girls and 4 boys, there are two boys B1B1? and B2B2? . The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but B1B1? and B2B2? are not adjacent to each other, is :
(1) 144
(2) 72
(3) 96
(4) 120
Ans. (1)
Sol. Total - when B1B1? and B2B2? are together
=2!(3!4!)−2!(3!(3!2!))=144=2!(3!4!)−2!(3!(3!2!))=144

3. Let P(4,43)P(4,43?) be a point on the parabola y2=4xy2=4x and PQ be a focal chord of the parabola. If M and N are the foot of perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to:
(1) 2633882633??
(2) 173173?
(3) 3433883433??
(4) 34333343??
Ans. (3)
Sol. (x+x3−1)5+(x−x3−1)5(x+x3−1?)5+(x−x3−1?)5
=2{5Cox5+5Cox3(x3−1)+5Cox(x3−1)5}=2{5Co?x5+5Co?x3(x3−1)+5Co?x(x3−1)5}
=2{5x7+10x6+x5−10x4−10x3+5x}=2{5x7+10x6+x5−10x4−10x3+5x}
⇒α=10,β=2,γ=−20,δ=10⇒α=10,β=2,γ=−20,δ=10
Now, 10u+2v=1810u+2v=18
−20u+10v=20−20u+10v=20
⇒u=1,v=4⇒u=1,v=4
u+v=5u+v=5

JEE main Previous Year Question Paper 

4. For a 3×33×3 matrix M, let trace (M) denote the sum of all the diagonal elements of M. Let A be a 3×33×3 matrix such that ?A?=12?A?=21? and trace (A)=3(A)=3 . If B=adj?(adj?(2A))B=adj(adj(2A)) , then the value of ?B?+trace?(B)?B?+trace(B) equals:
(1) 56
(2) 132
(3) 174
(4) 280
Ans. (4)
Sol. ?A?=12?A?=21? , trace?(A)=3trace(A)=3 , B=adj?(adj?(2A))=?2A?2(2A)B=adj(adj(2A))=?2A?2(2A)
n=3n=3 , B=?2A?(2A)=23,?A?(2A)=8AB=?2A?(2A)=23,?A?(2A)=8A
?B?=?8A?=83,?A?=28=256?B?=?8A?=83,?A?=28=256
trace?(B)=8trace?(A)=24trace(B)=8trace(A)=24
?B?+trace?(B)=280?B?+trace(B)=280

5. Suppose that the number of terms in an A.P. is 2k2k k∈Nk∈N . If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then kk is equal to
(1) 5
(2) 8
(3) 6
(4) 4
Ans. (1)
Sol. a1,a2,a3,…,a2k→a2ka1?,a2?,a3?,…,a2k?→a2k?
∑r=1ka2r−1=40,∑r=1ka2r=55,a2k−a1=27∑r=1k?a2r−1?=40,∑r=1k?a2r?=55,a2k?−a1?=27
k2[2a1+(k−1)2d]=40,k2[2a2+(k−1)2d]=55,2k?[2a1?+(k−1)2d]=40,2k?[2a2?+(k−1)2d]=55,
d=272k−1d=2k−127?
a1=40k−(k−1)d=55k−kda1?=k40?−(k−1)d=k55?−kd
d=15k⇒272k−1=15k⇒9k=10k−5d=k15?⇒2k−127?=k15?⇒9k=10k−5
∴k=5∴k=5

6. Let a line pass through two distinct points P(−2,−1,3)P(−2,−1,3) and QQ , and be parallel to the vector 3i^+2j^+2k3i^+2j^?+2k . If the distance of the point Q from the point R(1,3,3)R(1,3,3) is 5, then the square of the area of ΔPQRΔPQR is equal to:
(1) 136
(2) 140
(3) 144
(4) 148
Ans. (1)
Sol. PQ‾PQ? parallel to 3i^+2j^+2k^3i^+2j^?+2k^
R(1,3,3)R(1,3,3)
⇒Q(3λ−2,2λ−1,2λ+3),λ∈R−{0}⇒Q(3λ−2,2λ−1,2λ+3),λ∈R−{0}
?QR‾?=5=(3λ−3)2+(2λ−4)2+(2λ)2?QR??=5=(3λ−3)2+(2λ−4)2+(2λ)2?
∴17λ2−34λ+25=25⇒λ=2(?λ≠0)∴17λ2−34λ+25=25⇒λ=2(?λ?=0)
∴Q(4,3,7),P(−2,−1,3),R(1,3,3)∴Q(4,3,7),P(−2,−1,3),R(1,3,3)
Area of ΔPQR=[PQR]=12[PQ‾×PR‾]ΔPQR=[PQR]=21?[PQ?×PR]
[PQR]=12?i^j^k^644340?=?i^j^k^322340?[PQR]=21??i^63?j^?44?k^40??=?i^33?j^?24?k^20??
\left[\mathrm{PQR}\right] = \left|\begin{array}{lll}- 8\hat{\mathbf{i}} +6\hat{\mathbf{j}} +6\hat{\mathbf{k}}\right| = \sqrt{136}
∴[PQR]2=136∴[PQR]2=136

7. If lim?x→∞((e1−e)(1e−x1+x))x=αlimx→∞?((1−ee?)(e1?−1+xx?))x=α , then the value of log?eα1+log?eα1+loge?αloge?α? equals :
(1) e
(2) e2e2
(3) e2e2
(4) e−1e−1
Ans. (1)
Sol. α=lim?x→∞((e1−e)(1e−x1+x))xα=limx→∞?((1−ee?)(e1?−1+xx?))x (1° form)
∴α=et∴α=et
Where L=lim?x→∞((e1−e)(1e−x1+x)−1)L=limx→∞?((1−ee?)(e1?−1+xx?)−1)
⇒L=lim?x→∞(e1−e)x(1e−x1+x−(1−ee))⇒L=limx→∞?(1−ee?)x(e1?−1+xx?−(e1−e?))
⇒L=e1−elim?x→∞(1−x1+x)⇒L=1−ee?limx→∞?(1−1+xx?)
⇒L=e1−elim?x→∞xx+1⇒L=1−ee?limx→∞?x+1x?
⇒L=e1−e⋅1⇒L=1−ee?⋅1
⇒L=e1−e⇒L=1−ee?
∴α=et−e⇒log?α=e1−e∴α=et−e⇒logα=1−ee?
∴∴ Required value =1−e1+e1−e=e=1+1−ee?1−e?=e

8. Let f(x)=∫0x2t2−8t+15dt,x∈R.f(x)=∫0x2?t2−8t+15dt,x∈R. Then the numbers of local maximum and local minimum points of f, respectively, are :
(1) 2 and 3
(2) 3 and 2
(3) 1 and 3
(4) 2 and 2
Ans. (1)
Sol. f′(x)=(x4−8x2+15ex2)(2x)f′(x)=(ex2x4−8x2+15?)(2x)
=(x2−3)(x2−5)(2x)ex2=(x−3)(x+3)(x−5)(x+5)xex2=− + + + + + + +  −5−3 0 3 5 =ex2(x2−3)(x2−5)(2x)?=ex2(x−3?)(x+3?)(x−5?)(x+5?)x?= −5?−3? 0 3? 5? − + + + + + + + ??
Maxima at x∈{−3,3}x∈{−3?,3?}
Minima at x∈{−5,0,5}x∈{−5?,0,5?}
2 points of maxima and 3 points of minima.

9. The perpendicular distance, of the line x−12=y+2−1=z+322x−1?=−1y+2?=2z+3? from the point P(2,−10,1)P(2,−10,1) , is:
(1) 6
(2) 5252?
(3) 3535?
(4) 4343?
Ans. (3)
Sol. P(2,−10,1)n?=2i?−j?+2k?P(2,−10,1)n?=2i?−j??+2k??
x−12=y+2−1=z+32=λ2x−1?=−1y+2?=2z+3?=λ (let)
(2) λ+1,−λ−2,2λ−3)λ+1,−λ−2,2λ−3)
∴PA.n?=0∴PA.n?=0
⇒(2λ−1)2+(−λ+8)(−1)+(2λ−4)2=0⇒(2λ−1)2+(−λ+8)(−1)+(2λ−4)2=0
⇒4λ−2+λ−8+4λ−8=0⇒4λ−2+λ−8+4λ−8=0
⇒9λ−18=0⇒λ=2⇒9λ−18=0⇒λ=2
∴A(5,−4,1)∴A(5,−4,1)
∴AP=32+62+02=45=35∴AP=32+62+02?=45?=35?

10. If x=f(y)x=f(y) is the solution of the differential equation

(1+y2)+(x−2etan?−1y)dydx=0,y∈(−π2,π2)(1+y2)+(x−2etan−1y)dxdy?=0,y∈(−2π?,2π?)

with f(0)=1f(0)=1 , then f(13)f(3?1?) is equal to :
(1) eπ/4eπ/4
(2) eπ/12eπ/12
(3) eπ/3eπ/3
(4) eπ/6eπ/6
Ans. (4)
Sol. dxdy+x1+y2=2etan?−1y1+y2dydx?+1+y2x?=1+y22etan−1y?
I.F.=etan?−1yI.F.=etan−1y
xetan?−1y=∫2(etan?−1y)2dy1+y2xetan−1y=∫1+y22(etan−1y)2dy?
Puttan?−1y=t,dy1+y2=dtPuttan−1y=t,1+y2dy?=dt
xetan?−1y=∫2e2tdtxetan−1y=∫2e2tdt
xetan?−1y=e2tan?−1y+cxetan−1y=e2tan−1y+c
x=etan?−1y+ce−tan?−1yx=etan−1y+ce−tan−1y
?y=0,x=1?y=0,x=1
1=1+c⇒c=01=1+c⇒c=0
y=13,x=eπ/6y=3?1?,x=eπ/6

11. If ∫ex(xsin?−1x1−x2+sin?−1x(1−x2)3/2+x1−x2)dx=g(x)+C,∫ex(1−x2?xsin−1x?+(1−x2)3/2sin−1x?+1−x2x?)dx=g(x)+C, where C is the constant of integration, then g(12)g(21?) equals :
(1) π6c26π?2c??
(2) π4c24π?2c??
(3) π6c36π?3c??
(4) π4c34π?3c??
Ans. (3)
Sol. ?d(xsin?−1x)dx(1−x2)=sin?−1x(1−x2)3/2+x1−x2?dx(1−x2?)d(xsin−1x)?=(1−x2)3/2sin−1x?+1−x2x?
⇒∫ex(xsin?−1x1−x2+sin?−1x(1−x2)3/2+x1−x2)dx⇒∫ex(1−x2?xsin−1x?+(1−x2)3/2sin−1x?+1−x2x?)dx
=ex⋅xsin?−1x1−x2+c=g(x)+C=ex⋅1−x2?xsin−1x?+c=g(x)+C
Note:assumingg(x)=xexsin?−1x1−x2Note:assumingg(x)=1−x2?xexsin−1x?
g(1/2)=e1/22⋅63=π6e3g(1/2)=2e1/2?⋅3?6?=6π?3e??

Comment : In this question we will not get a unique function g(x)g(x) , but in order to match the answer we will have to assume g(x)=xexsin?−1x1−x2g(x)=1−x2?xexsin−1x? .

12. Let α0α0? and β0β0? be the distinct roots of 2x2+(cos?θ)x−1=02x2+(cosθ)x−1=0 , θ∈(0,2π)θ∈(0,2π) . If m and M are the minimum and the maximum values of α04+β04α04?+β04? , then 16(M+m)16(M+m) equals :
(1) 24
(2) 25
(3) 27
(4) 17
Ans. (2)
Sol. (α2+β2)2−2α2β2(α2+β2)2−2α2β2
[(α+β)2−2αβ]2−2(αβ)2[(α+β)2−2αβ]2−2(αβ)2
[cos?2θ4+1]2−214[4cos2θ?+1]2−241?
(cos?2θ4+1)2−12(4cos2θ?+1)2−21?
M=2516−12=1716M=1625?−21?=1617?
m=12,16(M+m)=25m=21?,16(M+m)=25

13. Let A={1,2,3,4}A={1,2,3,4} and B={1,4,9,16}B={1,4,9,16} . Then the number of many- one functions f:A→Bf:A→B such that 1∈f(A)1∈f(A) is equal to :
(1) 127
(2) 151
(3) 163
(4) 139
Ans. (2)
Sol. Total =44=44
One- one =4!=4!
Many- one =256−24=232=256−24=232
Many- one which 1∉f(A)1∈/f(A)
=3.3.3.3=81=3.3.3.3=81
232−81=151232−81=151

14. If the system of linear equations : x+y+2z=6,x+y+2z=6, 2x+3y+az=a+1,2x+3y+az=a+1, −x−3y+bz=2b,−x−3y+bz=2b, where a, b∈R,b∈R, has infinitely many solutions, then 7a+3b7a+3b is equal to :
(1) 9
(2) 12
(3) 16
(4) 22
Ans. (3)
Sol. Δ=?11223a−1−3b?=0Δ=?12−1?13−3?2ab??=0
⇒2a+b−6=0⇒2a+b−6=0
∴…(1)∴…(1)
Δ1=?11623a+1−1−32b?=0Δ1?=?12−1?13−3?6a+12b??=0
⇒a+b−8=0⇒a+b−8=0
∴…(2)∴…(2)
Solving (1)+(2)(1)+(2)
a=−2,b=10a=−2,b=10
⇒7a+3b=16⇒7a+3b=16

15. Let a?a? and b?b? be two unit vectors such that the angle between them is π33π? . If λa?+2b?λa?+2b? and 3a?−λb?3a?−λb? are perpendicular to each other, then the number of values of λλ in [−1,3][−1,3] is :
(1) 3
(2) 2
(3) 1
(4) 0
Ans. (4)
Sol. a^.b^=12a^.b^=21?
Now (λa^+2b^)(3a^−λb^)=0(λa^+2b^)(3a^−λb^)=0
3λa^.a^−λ2a^.b^+6a^.b^−2λb^.b^=03λa^.a^−λ2a^.b^+6a^.b^−2λb^.b^=0
3λ−λ22+3−2λ=03λ−2λ2?+3−2λ=0
λ2−2λ−6=0λ2−2λ−6=0
λ=1±7λ=1±7?
⇒⇒ number of values =0=0

16. Let E:x2a2+y2b2=1E:a2x2?+b2y2?=1 a>b and H: x2A2−y2B2=1A2x2?−B2y2?=1 Let the distance between the foci of E and the foci of H be 2323? .If a- A=2,and the ratio of the eccentricities of E and H is 1331? then the sum of the lengths of their latus rectums is equal to :
(1) 10
(2) 7
(3) 8
(4) 9
Ans. (3)
Sol. x2a2+y2b2=1a2x2?+b2y2?=1 foci are (ae, 0) and (- ae, 0)
x2A2+y2B2=1A2x2?+B2y2?=1 foci are (Ae', 0) and (- Ae', 0)
⇒2ae=23⇒ae=3⇒2ae=23?⇒ae=3?
and 2Ae′=23⇒Ae′=32Ae′=23?⇒Ae′=3?
⇒ae=Ae′⇒ee′=Aa⇒ae=Ae′⇒e′e?=aA?
⇒13=Aa⇒a=3A⇒31?=aA?⇒a=3A
Now a−A=2⇒a−a3−2⇒a=3a−A=2⇒a−3a?−2⇒a=3 and A=1A=1
Ae=3⇒e=13Ae=3?⇒e=3?1? and e′=3e′=3?
b2=a2(1−e2)b2=a2(1−e2)
b2=6b2=6 and B2=A2((e′)2−1)=(2)⇒B2=2B2=A2((e′)2−1)=(2)⇒B2=2
sum of LR=2b2a+2B2A=8LR=a2b2?+A2B2?=8

17. If A and B are two events such that P(A∩B)=0.1P(A∩B)=0.1 and P(A?B)P(A?B) and P(B?A)P(B?A) are the roots of the equation 12x2−7x+1=012x2−7x+1=0 then the value of P(A‾∪B‾)P(A‾∩B‾)P(A∩B)P(A∪B)? is:
(1) 5335?
(2) 4334?
(3) 9449?
(4) 7447?
Ans. (3)
Sol. 12x2−7x+1=012x2−7x+1=0
x=13,14x=31?,41?
Let P(AB)=13P(BA?)=31? & P(BA)=14P(AB?)=41?
P(A∩B)P(B)=13P(B)P(A∩B)?=31? & P(A∩B)P(A)=14P(A)P(A∩B)?=41?
⇒P(B)=0.3⇒P(B)=0.3 & P(A)=0.4P(A)=0.4
P(A∪B)=P(A)+P(B)−P(A∩B)P(A∪B)=P(A)+P(B)−P(A∩B)
=0.3+0.4−0.1=0.6=0.3+0.4−0.1=0.6
Now P(A‾∪B‾)P(A‾∩B‾)=P(A‾∩B‾)P(A∪B)P(A∩B)P(A∪B)?=P(A∪B)P(A∩B)?
=1−P(A∩B)1−P(A∪B)=1−0.11−0.6=94=1−P(A∪B)1−P(A∩B)?=1−0.61−0.1?=49?

18. The sum of all values of θ∈[0,2π]θ∈[0,2π] satisfying 2sin?2θ=cos?2θ2sin2θ=cos2θ and 2cos?2θ=3sin?θ2cos2θ=3sinθ is
(1) π22π?
(2) 4π4π
(3) 5π665π?
(4) ππ
Ans. (4)
Sol. 2sin?2θ=cos?2θ2sin2θ=cos2θ
2sin?2θ=1−2sin?2θ2sin2θ=1−2sin2θ
4sin?2θ=14sin2θ=1
sin?2θ=14sin2θ=41?
sin?θ=±12sinθ=±21?
2cos?2θ=3sin?θ2cos2θ=3sinθ
2−2sin?2θ+3sin?θ−2=02−2sin2θ+3sinθ−2=0
(2sin?θ−1)(2sin?θ−2)=0(2sinθ−1)(2sinθ−2)=0
sin?θ=12sinθ=21? so common equation which satisfy both equations is sin?θ=12sinθ=21?
θ=π6,5π6(θ∈[0,2π])θ=6π?,65π?(θ∈[0,2π])
Sum=πSum=π

19. Let the curve z(1+i)+z?(1−i)=4z(1+i)+z?(1−i)=4 z∈Cz∈C divide the region ?z−3?≤1?z−3?≤1 into two parts of areas αα and ββ .Then ?α−β??α−β? equals :
(1) 1+π21+2π?
(2) 1+π31+3π?
(3) 1+π41+4π?
(4) 1+π61+6π?
Ans. (1)
Sol. Let z=x+iyz=x+iy
(x+iy)(1+i)+(x−iy)(1−i)=4(x+iy)(1+i)+(x−iy)(1−i)=4
x+ix+iy−y+x−ix−iy−y=4x+ix+iy−y+x−ix−iy−y=4
2x−2y=42x−2y=4
x−y=2x−y=2
?z−3?≤1?z−3?≤1
(x−3)2+y2≤1(x−3)2+y2≤1
Area of shaded region =π.124−12.1.1=π4−12=4π.12?−21?.1.1=4π?−21?
Area of unshaded region inside the circle
=34π.12+12.1.1=3π4+12=43?π.12+21?.1.1=43π?+21?
∴difference of area=(3π4+12)−(π4−12)∴difference of area=(43π?+21?)−(4π?−21?)
=π2+1=2π?+1

20. The area of the region enclosed by the curves y=x2−4x+4y=x2−4x+4 and y2=16−8xy2=16−8x is :
(1) 8338?
(2) 4334?
Ans. (1)
Sol.
y=(x−2)2,y2=8(x−2)y=(x−2)2,y2=8(x−2)
y=x2,y2=−8xy=x2,y2=−8x
=16ab3=16×14×23=83=316ab?=316×41?×2?=38?

SECTION-B JEE main Previous Year Question Paper 

21. Let y=f(x)y=f(x) be the solution of the differential

dydx+dydx=x6+4x1−x2,−1

that f(0)=0f(0)=0 .If 6∫11f(x)dx=2π−α6∫11?f(x)dx=2π−α then α2α2 is equal to
Ans. (27)
Sol. I.F.e−12∫1−x22xdx=e−12∫1−x2xdx=1−x2I.F.e−21?∫1−x22x?dx=e−21?∫1−x2x?dx=1−x2?
y×1−x2=∫(x6+4x)dx=x77+2x2+cy×1−x2?=∫(x6+4x)dx=7x7?+2x2+c
Given y(0)=0⇒c=0y(0)=0⇒c=0
y=x77+2x21−x2y=1−x2?7x7?+2x2?
Now,6∫11x71−x2dx=6∫112x21−x2dxNow,6∫11?1−x2x7?dx=6∫11?1−x22x2?dx
=24∫01x21−x2dx=24∫01?1−x2x2?dx
Put x=sin?θx=sinθ
dx=cos?θdθdx=cosθdθ
=24∫01sin?2θcos?θcos?θdθ=24∫01?cosθsin2θ?cosθdθ
=24∫01(1−cos?2θ2)dθ=12∫01(sin?2θ2)6=24∫01?(21−cos2θ?)dθ=12∫01?(2sin2θ?)6
=12(π6−34)=12(6π?−43??)
=2π−33=2π−33?
α2=(33)2=27α2=(33?)2=27

22. Let A(6,8),B(10 cosα,-10 sinα) and C(-10 sinα,10 cosα),be the vertices of a triangle. If L(a,9) and G(h,k) be its orthocenter and centroid respectively,then (5a−3h+6k+100sin?2α)(5a−3h+6k+100sin2α) is equal to
Ans. (145)

23. Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then (QR) is equal to
Ans. (28)
Sol.
PR = cosecθ, PQ = 4secθ
For equilateral
d = PR = PQ
⇒ cos(θ + 30°) = 4sinθ
⇒ 32cos?θ−12sin?θ=4sin?θ23??cosθ−21?sinθ=4sinθ
⇒ tan?θ=133tanθ=33?1?
QR² = d² = cosec²θ = 28

24. If ∑r=130r2(30Cr)2Cr−1=α×229∑r=130?Cr−1?r2(30Cr?)2?=α×229 , then αα is equal to
Ans. (465)
Sol. ∑r=130r2(30Cr)2Cr−1∑r=130?Cr−1?r2(30Cr?)2?
=∑r=130r2(31−rr)30!r!(30−r)!(?30CrCr−1=30−r+1r=31−rr)=∑r=130(31−r)30!(r−1)!(30−r)!=30∑r=130(31−r)29!(r−1)!(30−r)!=30∑r=130(30−r+1)29C30−r=30(∑r=130(31−r)29C30−r+∑r=13029C30−r)=30(29×228+229)=30(29+2)228=15×31×229=465(229)α=465=∑r=130?r2(r31−r?)r!(30−r)!30!?(?Cr−1?30Cr??=r30−r+1?=r31−r?)=∑r=130?(r−1)!(30−r)!(31−r)30!?=30∑r=130?(r−1)!(30−r)!(31−r)29!?=30∑r=130?(30−r+1)29C30−r?=30(∑r=130?(31−r)29C30−r?+∑r=130?29C30−r?)=30(29×228+229)=30(29+2)228=15×31×229=465(229)α=465?

25. Let A={1,2,3}A={1,2,3} . The number of relations on A, containing (1, 2) and (2, 3), which are reflexive and transitive but not symmetric, is
Ans. (3)
Sol. Transitivity
(1,2)∈R,(2,3)∈R⇒(1,3)∈R For reflexive (1,1),(2,2)(3,3)∈R Now (2,1),(3,2),(3,1) (3,1) cannot be taken (1) (2, 1) taken and (3, 2) not taken (2) (3, 2) taken and (2, 1) not taken (3) Both not taken therefore 3 relations are possible.

26. A symmetric thin biconvex lens is cut into four equal parts by two planes AB and CD as shown in figure. If the power of original lens is 4D then the power of a part of the divided lens is
(1)8D
(2)4D
(3)D
(4)2D
Ans. (4)
\mathrm{Sol.}\left\{ \begin{array}{l}\frac{1}{\mathrm{f}_1} = (\mu -1)\frac{2}{\mathrm{R}} = \mathrm{P} = 4\mathrm{D}\\ \displaystyle \left(\frac{1}{\mathrm{f}_2} = (\mu -1)\frac{1}{\mathrm{R}} = \frac{\mathrm{P}}{2} = 2\mathrm{D} \end{array} \right.

27. A small rigid spherical ball of mass M is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball, then the viscous force acting on the ball will be (consider g as acceleration due to gravity)
(1) 32Mg23?Mg
(2) Mg22Mg?
(3) MgMg
(4) 2Mg2Mg
Ans. (2)
mg−Fb−f=0mg−Fb?−f=0
⇒mg−mg2−f=0⇒mg−2mg?−f=0
∴f=mg2∴f=2mg?

28. The maximum percentage error in the measurement of density of a wire is [Given, mass of wire =(0.60±0.003)g=(0.60±0.003)g radius of wire =(0.50±0.01)cm=(0.50±0.01)cm length of wire (10.00±0.05)cm](10.00±0.05)cm]
(1)4
(2)5
(3)8
(4)7
Ans. (2)
Sol.d=mvol.=mπR2?⇒dρρ=dmm+2dRR+d??Sol.d=vol.m?=πR2?m?⇒ρdρ?=mdm?+R2dR?+?d??
⇒dρρ=(0.0030.6+2×0.010.5+0.0510)100=5%⇒ρdρ?=(0.60.003?+0.52×0.01?+100.05?)100=5%

29. A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is IoIo? If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be
(1) IoIo?
(2) Io22Io??
(3) Io22?Io??
(4) 2Io2Io?
Ans. (2)
Sol. Initially, Io=emRIo?=Rem?? Finally, Iol=em2R=Io2Iol?=2Rem??=2Io??

30. For a short dipole placed at origin O, the dipole moment P is along x-axis, as shown in the figure. If the electric potential and electric field at A are V0V0? and E0E0? respectively, then the correct combination of the electric potential and electric field, respectively, at point B on the y-axis is given by
(1) V022V0?? and E01616E0??
(2) zero and E088E0??
(3) zero and E01616E0??
(4) V0V0? and E044E0??
Ans. (3)
EA=2kPr3=E0&VA=kPr2=V0EA?=r32kP?=E0?&VA?=r2kP?=V0?
EB=kP(2r)3=E016&VB=kPP^r2=0EB?=(2r)3kP?=16E0??&VB?=r2kPP^?=0

31. Which one of the following is the correct dimensional formula for the capacitance in F ? MF ? M L, T and C stand for unit of mass, length, time and charge,
(1) [F]=[C2M−2L2T2][F]=[C2M−2L2T2]
(2) [F]=[CM−2L−2T−2][F]=[CM−2L−2T−2]
(3) [F]=[CM−1L−2T2][F]=[CM−1L−2T2]
(4) [F]=[C2M−1L−2T2][F]=[C2M−1L−2T2]
Ans. (4)
C=qV=qqVq=q2WD=C2ML2T−2=C2M−1L−2T2C=Vq?=Vqqq?=WDq2?=ML2T−2C2?=C2M−1L−2T2

32. An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :
(1) 2hπeBπeB2h??
(3) h2πeB2πeBh??
Ans. (4)
r=mveB&mvr=nh2π⇒(eBr)r=nh2πr=eBmv?&mvr=2πnh?⇒(eBr)r?=2πnh?
⇒r=nh2πeB⇒r=2πeBnh??
first excited state : n=2∴r=hπeBn=2∴r=πeBh??

33. For a diatomic gas, if γ1=(CpCv)γ1?=(CvCp?) for rigid molecules and γ2=(CpCv)γ2?=(CvCp?) for another diatomic molecules, but also having vibrational modes. Then, which one of the following options is correct? (Cp and Cv are specific heats of the gas at constant pressure and volume)
(1) γ2>γ1γ2?>γ1?
(2) γ2=γ1γ2?=γ1?
(3) 2γ2=γ12γ2?=γ1?
(4) γ2<γ1γ2?<γ1?
Ans. (4)
γ=2f+1γ=f2?+1
without vibration : f=5:γ1=1.4f=5:γ1?=1.4
without vibration : f=7:γ2=1.14f=7:γ2?=1.14
∴γ2<γ1∴γ2?<γ1?

34. A rectangular metallic loop is moving out of a uniform magnetic field region to a field free region with a constant speed. When the loop is partially inside the magnate field, the plot of magnitude of induced emf (ε) with time (t) is given by
Ans. (4)

35. A light source of wavelength λλ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2eV2eV . If the same surface is illuminated by a light source of wavelength λ22λ? , then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1eV1eV )
(1) 2eV2eV
(2) 6eV6eV
(3) 5eV5eV
(4) 3eV3eV
Ans. (3)
Sol.hcλ=?+eV⇒hcλ=1+2=3eV……(1)Sol.λhc?=?+eV⇒λhc?=1+2=3eV……(1)
hcλ/2=6=1+kmax∴kmax=5eVλ/2hc?=6=1+kmax?∴kmax?=5eV

36. Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet. Reason (R): The mass of the pendulum remains unchanged at Earth and the other planet. In the light of the above statements, choose the correct answer from the options given below:
(1) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
(2) (A) is true but (R) is false
(3) (A) is false but (R) is true
(4) Both (A) and (R) are true and (R) is the correct explanation of (A)
Ans. (1)
Sol.g=GMR2Sol.g=R2GM?
g′=G(4M)(2R)2=gg′=(2R)2G(4M)?=g
A is correct,R is correct;but since T=2π?gA is correct,R is correct;but since T=2πg???
doesn′t depend on mass;R doesn′t explain A.doesn′t depend on mass;R doesn′t explain A.

JEE main Previous Year Question Paper 

37. The torque due to the force (2i^+j^+2k^)(2i^+j^?+2k^) about the origin, acting on a particle whose position vector is (i^+j^+k^)(i^+j^?+k^) , would be
(1) i^−j^+k^i^−j^?+k^
(2) i^+k^i^+k^
(3) i^−k^i^−k^
(4) j^−k^j^?−k^
Ans. (3)

38. To obtain the given truth table, following logic gate should be placed at G:
(1) NOR Gate
(2) AND Gate
(3) NAND Gate
(4) OR Gate
NTA Ans. (1)
For NOR gate: AB‾=A‾+BAB=A+B

39. A force F?=2i?+b?j?+k?F=2i+bj?+k is applied on a particle and it undergoes a displacement i?−2j?−k?i−2j?−k . What will be the value of b, if work done on the particle is zero.
(1) 0
(2) 1221?
(3) 1331?
(4) 2
Ans. (2)
Sol. WD=F‾.S‾=2−2b−1=0WD=F.S=2−2b−1=0
∴b=12∴b=21?

40. Given below are two statements. On is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In Young's double slit experiment, the fringes produced by red light are closer as compared to those produced by blue light. Reason (R) : The fringe width is directly proportional to the wavelength of light. In the light of above statements, choose the correct answer from the options given below :
(1) Both (A) and (R) are true and (R) is the correct explanation of (A)
(2) (A) is false but (R) is true.
(3) Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4) (A) is true but (R) is false.
Ans. (2)
Sol. β=λDd&λR>λbβ=dλD?&λR?>λb?
∴βR>βb∴βR?>βb?

41. A ball of mass 100g100g is projected with velocity 20m/s20m/s at 60?60? with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is :
(1) 20J20J
(2) 15J15J
(3) zero
(4) 5J5J
Ans. (2)
Sol.
ki=12mv2ki?=21?mv2
kf=12m(vcos?60?)2=18mv2kf?=21?m(vcos60?)2=81?mv2
Δk=ki−kf=38mv2=38×0.1×400=15JΔk=ki?−kf?=83?mv2=83?×0.1×400=15J

42. A transparent film of refractive index, 2.0 is coated on a glass slab of refractive index, 1.45. What is the minimum thickness of transparent film to be coated for the maximum transmission of Green light of wavelength 550nm550nm . [Assume that the light is incident nearly perpendicular to the glass surface.]
(1) 94.8nm94.8nm
(2) 68.7nm68.7nm
(3) 137.5nm137.5nm
(4) 275nm275nm
Ans. (3)
Sol.
For transmitted green light to be maxima, reflected green should be minima.
ΔP=2μ0t=nλΔP=2μ0?t=nλ
⇒t=nλ2μ0∴tmin=λ2μ0=5502×2=137.5⇒t=2μ0?nλ?∴tmin?=2μ0?λ?=2×2550?=137.5

43. The tube of length L is shown in the figure. The radius of cross section at the point (1) is 2cm2cm and at the point (2) is 1cm1cm respectively. If the velocity of water entering at point (1) is 2m/s2m/s then velocity of water leaving the point (2) will be :
(1) 2m/s2m/s
(2) 4m/s4m/s
(3) 6m/s6m/s
(4) 8m/s8m/s
Ans. (4)
Aivi=A2V2⇒2π(2R)2=V2πR2Ai?vi?=A2?V2?⇒2π(2R)2=V2?πR2
∴V2=8m/s∴V2?=8m/s

JEE main Previous Year Question Paper 

44. Given are statements for certain thermodynamic variables, (A) Internal energy, volume (V) and mass (M) are extensive variables. (B) Pressure (P), temperature (T) and density (p) are intensive variables. (C) Volume (V), temperature (T) and density (p) are intensive variables. (D) Mass (M), temperature (T) and internal energy are extensive variables. Choose the correct answer from the points given below :
(1) (C) and (D) only
(2) (D) and (A) only
(3) (A) and (B) only
(4) (B) and (C) only
Ans. (3)
Sol. Extensive variables depends on size or mass of system ex : internal energy, volume, mass

45. A body of mass 100g100g is moving in circular path of radius 2m2m on vertical plane as shown in figure. The velocity of the body at point A is 10m/s10m/s . The ratio of its kinetic energies at point B and C is :
(Take acceleration due to gravity as 10m/s210m/s2
(1) 2+3332+3??
(2) 2+2332+2??
(3) 3+3223+3??
(4) 3−2223−2??
Ans. (3)
Sol.
12m×100+0=12mVB2+mg(R−R32)21?m×100+0=21?mVB2?+mg(R−2R3??)
100=VB2+2gR(1−32)100=VB2?+2gR(1−23??)
VB2=100−20(2−3)VB2?=100−20(2−3?)
VB2=60+203VB2?=60+203?
K.EB=12mVB2=m2(60+203)K.EB?=21?mVB2?=2m?(60+203?)
12m(100)=12mVC2+mg(3R2)21?m(100)=21?mVC2?+mg(23R?)
100=VC2=60100=VC2?=60
VC2=40VC2?=40
K.EC=12mVC2=12m(40)K.EC?=21?mVC2?=21?m(40)
K.EB=60+20340=32+32=3+32K.EB?=4060+203??=23?+23??=23+3??

46. A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2×105 ms−12×105 ms−1 . When the electric field is switched off, the proton moves along a circular path of radius 2 cm2 cm . The magnitude of electric field is x×104 N/Cx×104 N/C . The value of xx is ______. Take the mass of the proton =1.6×10−27 kg=1.6×10−27 kg .
Ans. (2)
Sol. For uniform speed V=EBV=BE?
R=mVeBR=eBmV?
=mV2eE=eEmV2?
⇒E=mV2eR⇒E=eRmV2?
=1.6×10−27×4×10101.6×10−19×2×10−2=1.6×10−19×2×10−21.6×10−27×4×1010?
=2×104N/C.=2×104N/C.

47. Two long parallel wires X and Y, separated by a distance of 6 cm6 cm , carry currents of 5 A5 A and 4 A4 A , respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm4 cm from wire Y is x×10−5 Tx×10−5 T . The value of xx is ______. Take permeability of free space as μ0=4π×10−7 SIμ0?=4π×10−7 SI units.
Ans. (1)
Sol.
B=μ0(5)2π×.01−μ042π×.04B=2π×.01μ0?(5)?−2π×.04μ0?4?
=−100μ04π=−4π100μ0??
=−100×10−7=−100×10−7
=−1×10−5T=−1×10−5T

JEE main Previous Year Question Paper 

48. A parallel plate capacitor of area A=16 cm2A=16 cm2 and separation between the plates 10 cm10 cm , is charged by a DC current. Consider a hypothetical plane surface of area A0=3.2 cm2A0?=3.2 cm2 inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6 A6 A . At the same instant the displacement current through A0A0? is ______ mA.
Ans. (1200)
Jd=1A=616Jd?=A1?=166?
I through small area Jd×A′=616×3.2=1.2A=1200mAJd?×A′=166?×3.2=1.2A=1200mA

49. A tube of length 1 m1 m is filled completely with an ideal liquid of mass 2 M2 M , and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is FF then angular velocity of the tube is FαMαMF?? in SI unit. The value of αα is ______.
Ans. (1)
Sol.
F=2Mω2?2=Mw2?F=2Mω22??=Mw2?
ω=FM?ω=M?F??

50. The net current flowing in the given circuit is ______.
Ans. (1)
Sol.
Req=2ΩReq?=2Ω
I=22=1AI=22?=1A

CHEMISTRY

SECTION-A

51. Arrange the following compounds in increasing order of their dipole moment : HBr, H?S, NF? and CHCl?
(1) NF? < HBr < H?S < CHCl?
(2) HBr < H?S < NF? < CHCl?
(3) H?S < HBr < NF? < CHCl?
(4) CHCl? < NF? < HBr < H?S
Ans. (1)
Sol. Increasing order of Dipole moment
NF? < HBr < H?S < CHCl?
μ = 0.24D 0.79D 0.95D 1.04D
It is NCERT Data Based

52. Identify the number of structure/s from the following which can be correlated to D-glyceraldehyde.
(1) three
(2) two
(3) four
(4) one
Ans. (1)
Sol.
In A, B, D – OH group in right hand side then D-configuration is assigned

53. The maximum covalency of a non-metallic group 15 element 'E' with weakest E–E bond is :
(1) 5
(2) 3
(3) 6
(4) 4
Ans. (4)
Sol. N – N < P – P : single (σ) bond strength
Due to L.P.-L.P. repulsion
and maximum possible covalency of nitrogen is 4.

54. Consider the given figure and choose the correct option :
(1) Activation energy of backward reaction is E1E1? and product is more stable than reactant.
(2) Activation energy of forward reaction is E1+E2E1?+E2? and product is more stable than reactant.
(3) Activation energy of forward reaction is E1+E2andE1?+E2?and product is less stable than reactant.
(4) Activation energy of both forward and backward reaction is E1+E2E1?+E2? and reactant is more stable than product.
Ans. (3)
Sol. Activation energy of forward reaction =E1+E2=E1?+E2? Energy of product >> Energy of reactant Stability Reactant >> Product

JEE main Previous Year Question Paper 

55. When sec-butylcyclohexane reacts with bromine in the presence of sunlight, the major product is :
(1)
(2)
(3)
(4)
Ans. (4)
Sol.
Formation of more stable free radical intermediate

56. The species which does not undergo disproportionation reaction is :
(1) ClO2−ClO2−?
(2) ClO4−ClO4−?
(3) ClO−ClO−
(4) ClO3−ClO3−?
Ans. (2)
Sol. ClO4−→x+{(−2)×4}=−1⇒x=+7ClO4−?→x+{(−2)×4}=−1⇒x=+7
Chlorine is in its maximum oxidation state, so disproportionation not possible in ClO4−ClO4−?

57. Match the Compounds (List- I) with the appropriate Catalyst/Reagents (List- II) for their reduction into corresponding amines.
List- I List- II
(Compounds) (Catalyst/Reagents)
(A) O (I) NaOH (aqueous) R- C- NH?
(B) C6H5NO2C6?H5?NO2? (II) H2/NiH2?/Ni
(C) R- C=N (III) LiAlH?, H?O
(D) C6H4(CO)2NRC6?H4?(CO)2?NR (IV)
Choose the correct answer from the options given below :
(1) (A)- (III), (B)- (II), (C)- (IV), (D)- (I)
(2) (A)- (II), (B)- (IV), (C)- (III), (D)- (I)
(3) (A)- (II), (B)- (I), (C)- (III), (D)- (IV)
(4) (A)- (III), (B)- (IV), (C)- (II), (D)- (I)
Ans. (4)

61. Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is :
(1) 1.79m1.79m
(2) 2m2m
(3) 3m3m
(4) 2.79m2.79m
Ans. (4)
Sol. 3M NaCl, dsol=1.25g/moldsol?=1.25g/mol
Molality=M×10001000d−M×MwMolality=1000d−M×Mw?M×1000?
=30001250−175.5=2.79=1250−175.53000?=2.79

62. The molar solubility(s) of zirconium phosphate with molecular formula (Zr4+)3(PO43−)4(Zr4+)3?(PO43−?)4? is given by relation :
(1) (Ksp6912)17(6912Ksp??)71?
(2) (Ksp5348)16(5348Ksp??)61?
(3) (Ksp8435)17(8435Ksp??)71?
(4) (Ksp9612)13(9612Ksp??)31?
Ans. (1)
Sol. Zr3(PO4)4(s)?3Zr4+(aq)+4PO4−3(aq)Zr3?(PO4?)4?(s)?3Zr4+(aq)+4PO4−3?(aq)
Ksp=(3s)3(4s)4=6912s7Ksp?=(3s)3(4s)4=6912s7
s=(Ksp6912)1/7s=(6912Ksp??)1/7

63. The most stable carbocation from the following is :
(1)
(2)
(3)
(4)
Ans. (1)

66. Given below are two statements:
Statement (I) : Nitrogen, sulphur, halogen and phosphorus present in an organic compound are detected by Lassaigne's Test.
Statement (II) : The elements present in the compound are converted from covalent form into ionic form by fusing the compound with Magnesium in Lassaigne's test.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Ans. (3)
Sol. The elements present in the compound are converted from covalent form into ionic form by fusing the compound with sodium in Lassigne's test

67. Identify the homoleptic complex(es) that is/are low spin.
(A) [Fe(CN)5NO]2−[Fe(CN)5?NO]2−
(B) [CoF6]3−[CoF6?]3−
(C) [Fe(CN)6]4−[Fe(CN)6?]4−
(D) [Co(NH3)6]3+[Co(NH3?)6?]3+
(E) [Cr(H2O)6]2+[Cr(H2?O)6?]2+
Choose the correct answer from the options given below :
(1) (B) and (E) only
(2) (A) and (C) only
(3) (C) and (D) only
(4) (C) only
Ans. (3)
Sol. (A) [Fe(CN)5NO]2−→[Fe(CN)5?NO]2−→ Heteroleptic, Fe2+Fe2+ 3d6, t2g6eg0t2g6?eg0? d2sp3d2sp3 , Low spin (3d series ++ SFL)
(B) [CoF6]3−→[CoF6?]3−→ Homoleptic, sp3d2sp3d2 High spin, Co3+Co3+ 3d6 (3d series ++ WFL)
(C) [Fe(CN)6]4−→[Fe(CN)6?]4−→ Homoleptic
Fe2+Fe2+ 3d6, d2sp3d2sp3 t2g6eg0t2g6?eg0? Low spin
(3d series ++ SFL)
(D) [Co(NH3)6]3−→[Co(NH3?)6?]3−→ Homoleptic, Co3+Co3+ 3d6, d2sp3d2sp3 t2g6eg0t2g6?eg0? Low spin (3d series ++ SFL)
(E) [Cr(H2O)6]2−→[Cr(H2?O)6?]2−→ Homoleptic
Cr2+Cr2+ 3d4, d2sp3d2sp3 High spin t2g3eg1t2g3?eg1?
(3d series ++ WFL)

68. Toluene (excess) →(iii)NaHSO3(i)CrO2Cl2,CS2,(ii)H3O+(i)CrO2?Cl2?,CS2?,(ii)H3?O+(iii)NaHSO3?? Filter ?? Residue (A)
Residue (A) ++ HCl (dil.) →→ Compound (B)
Structure of residue (A) and compound (B) Formed respectively is :
(1)
(2)
(3)
(4)
Ans. (4)
Sol.

69. Given below are two statements :
Statement (I) : Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure metal as a cathode.
Statement (II) : The rate of corrosion is more in alkaline medium than in acidic medium.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both Statement I and Statement II are false
(2) Statement I is false but Statement II is true
(3) Both Statement I and Statement II are true
(4) Statement I is true but Statement II is false
Ans. (4)
Sol. Statement I :
Corrosion is an example of electrochemical phenomenon
In which pure metal act as anode and impure metal (rusted metal) act as cathode.
Statement II :
Corrosion is more favourable in acid medium than alkaline so rate of corrosion is high is acid medium then alkaline.

70. The alkane from below having two secondary hydrogens is :
(1) 4-Ethyl-3,4-dimethyloctane
(2) 2,2,4,4-Tetramethylhexane
(3) 2,2,3,3-Tetramethylpentane
(4) 2,2,4,5-Tetramethylheptane
Ans. (3)
Sol. Alkane 2?H

SECTION-B JEE main Previous Year Question Paper 

71. The compound with molecular formula C6H6C6?H6? which gives only one monobromo derivative and takes up four moles of hydrogen per mole for complete hydrogenation has ππ electrons.
Ans. (8)
Sol.
No. of π e? = 8

72. Niobium (Nb) and ruthenium (Ru) have "x" and "y" number of electrons in their respective 4d orbitals. The value of x+yx+y is
Ans. (11)
Sol. Z=41→NbZ=41→Nb (Niobium) : [Kr]364d45s1[Kr]36?4d45s1 Number of electron in 4d=4=x4d=4=x
Z=44→RuZ=44→Ru (Ruthenium) [Kr]364d75s1[Kr]36?4d75s1 Number of electron in 4d=7=y4d=7=y
x+y=11

73. The complex of Ni2+Ni2+ ion and dimethyl glyoxime contains number of Hydrogen (H) atoms.
Ans. (14)
Sol. [Ni(dmg)2]
Number of H- atom =14

74. Consider the following cases of standard enthalpy of reaction (ΔHi?(ΔHi?? in kJ mol-1)
C2H6(g)+72O2(g)→2CO2(g)+3H2O(?)ΔHio=−1550C2?H6?(g)+27?O2?(g)→2CO2?(g)+3H2?O(?)ΔHio?=−1550
C(graphtic)+O2(g)→CO2(g)ΔHio=−393.5C(graphtic)+O2?(g)→CO2?(g)ΔHio?=−393.5
H2(g)+12O2(g)→H2O(?)ΔHio=−286H2?(g)+21?O2?(g)→H2?O(?)ΔHio?=−286
The magnitude of ΔHi?C2H6(g)ΔHi??C2?H6?(g) is kJ mol- 1 (Nearest integer).
Ans. (95)
2C(graphtic)+3H2(g)→C2H6(g)ΔHio=?2C(graphtic)?+3H2?(g)→C2?H6?(g)ΔHio?=?
C2H6(g)+72O2(g)→2CO2(g)+3H2O(l)C2?H6?(g)+27?O2?(g)→2CO2?(g)+3H2?O(l)
ΔHi=−1550ΔHi?=−1550
C(graphtic)+O2(g)→CO2(g)ΔH2=−393.5C(graphtic)?+O2?(g)→CO2?(g)ΔH2?=−393.5
H2(g)+12O2(g)→H2O(l)ΔH3=−286H2?(g)+21?O2?(g)→H2?O(l)ΔH3?=−286
ΔHi=2ΔH2+3ΔH3−ΔH1ΔHi?=2ΔH2?+3ΔH3?−ΔH1?
=95kJ/mole.

75. 20 mL20 mL of 2M2M NaOH solution is added to 400 mL400 mL of 0.5M0.5M NaOH solution. The final concentration of the solution is ×10−2M×10−2M .(Nearest integer).
Ans. (57)
MF=MiVi+M2V2Vi+V2MF?=Vi?+V2?Mi?Vi?+M2?V2??
=2×20+0.5×400420=0.571M=4202×20+0.5×400?=0.571M
=57.1×10−2M=57.1×10−2M
=57

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