1. The largest n∈N such that 3 divides 50! is:
(1) 21
(2) 22
(3) 20
(4) 23

Ans. (2)
Sol. 2a3b572a3b57
B=[503]+[5032]+[5033]+[5034]B=[350?]+[3250?]+[3350?]+[3450?]
=16+5+1=16+5+1
=22=22
Maximum value of n is 22

2. Let one focus of the hyperbola H: x2a2−y2b2=1a2x2?−b2y2?=1 be at (10,0)(10?,0) and the corresponding directrix be x=910x=10?9?. If e and ll respectively are the eccentricity and the length of the latus rectum of H, then 9(e2+l)9(e2+l) is equal to:
(1) 14
(2) 15
(3) 16
(4) 12

Ans. (3)
Sol. ae=10ae=10? and ae=910ea?=109?
⇒a2=9⇒a2=9 and e=103e=310??
Now (ae)2=a2+b2(ae)2=a2+b2
10=9+b2⇒b2=110=9+b2⇒b2=1
?=2b2a=2(1)3?=a2b2?=32(1)?
⇒9(e2+?)⇒9(e2+?)
=9(109+23)=9(910?+32?)
=10+6=10+6
=16=16

JEE Previous Year Question Paper

3. The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to
(1) 360
(2) 45
(3) 2520
(4) 1820

Ans. (3)
Sol. 11111 222 0 0
No. of sequences =10!5!3!2!=2520=5!3!2!10!?=2520
Note : Sequence can start with 0.

4. Let f:R→Rf:R→R be a twice differentiable function such that (sin?xcos?y)(f(2x+2y)−f(2x−2y))=(cos?xsin?y)(f(2x+2y)+f(2x−2y))(sinxcosy)(f(2x+2y)−f(2x−2y))=(cosxsiny)(f(2x+2y)+f(2x−2y)), for all x,y∈Rx,y∈R. If f′(0)=12f′(0)=21? then the value of 24f′′(5π3)24f′′(35π?) is:
(1) 2
(2) -3
(3) 3
(4) -2

Ans. (2)
Sol. (sin?xcos?y)(f(2x+2y)−f(2x−2y))=(cos?xsin?y)(f(2x+2y)+f(2x−2y))(sinxcosy)(f(2x+2y)−f(2x−2y))=(cosxsiny)(f(2x+2y)+f(2x−2y))
f(2x+2y)sin?(x+y)=f(2x−2y)sin?(x−y)sin(x+y)f(2x+2y)?=sin(x−y)f(2x−2y)?
f(m)sin?(m2)=f(n)sin?(n2)=Ksin(2m?)f(m)?=sin(2n?)f(n)?=K
⇒f(m)=Ksin?(m2)⇒f(m)=Ksin(2m?)
∴f(x)=Ksin?(x2)∴f(x)=Ksin(2x?)
f′(x)=K2cos?(x2)f′(x)=2K?cos(2x?)
Put x=0;12=K2⇒K=1x=0;21?=2K?⇒K=1
f′(x)=12cos?x2f′(x)=21?cos2x?

JEE Previous Year Question Paper

5. Let A=[α−16β],α>0A=[α6?−1β?],α>0, such that det(A) = 0 and α+β=1α+β=1. If I denotes 2×22×2 identity matrix, then the matrix (1+A)8(1+A)8 is:
(1) [766−2551530−509][7661530?−255−509?]
(2) [7662551530−509][7661530?255−509?]
(3) [766−2551530509][7661530?−255509?]
(4) [7662551530509][7661530?255509?]

Ans. (1)
Sol. ?A?=0?A?=0
αβ+6=0αβ+6=0
αβ=−6αβ=−6
α+β=1α+β=1
⇒α=3,β=−2⇒α=3,β=−2
A=[3−16−2]A=[36?−1−2?]
A2=[3−16−2][3−16−2]=[3−16−2]A2=[36?−1−2?][36?−1−2?]=[36?−1−2?]
∴A2=A∴A2=A
A=A2=A3=A4=A5A=A2=A3=A4=A5
(I+A)8(I+A)8
I+8C1A+8C2A+?+8C8AI+8C1?A+8C2?A+?+8C8?A
I+A(8C1+8C2+?+8C8)I+A(8C1?+8C2?+?+8C8?)
I+A(28−1)I+A(28−1)
=[1001]+[765−2551530−510]=[10?01?]+[7651530?−255−510?]
=[766−2551530−509]=[7661530?−255−509?]

6. The term independent of x in the expansion of ((x+1)(x2/3+1−x1/3)−(x+1)(x−x1/2))10,x>1((x2/3+1−x1/3)(x+1)?−(x−x1/2)(x+1)?)10,x>1 is:
(1) 210
(2) 150
(3) 240
(4) 120

Ans. (1)
Sol. ((x+1)(x2/3+1−x1/3)−(x−1)(x−x1/2))10((x2/3+1−x1/3)(x+1)?−(x−x1/2)(x−1)?)10
=((x1/3+1)−(x+1x))10=((x1/3+1)−(x?x?+1?))10
=(x1/3−1x)10=(x1/3−x?1?)10
Tr+1=10Cr(x1/3)10−r(−1)r(x−1/2)rTr+1?=10Cr?(x1/3)10−r(−1)r(x−1/2)r
10−r3−r2=0310−r?−2r?=0
(20−2r)−3r=0(20−2r)−3r=0
r=4r=4
⇒10C4(−1)4=210⇒10C4?(−1)4=210

JEE Previous Year Question Paper

7. If θ∈[−2π,2π]θ∈[−2π,2π], then the number of solutions of 22cos?2θ+(2−6)cos?θ−3=022?cos2θ+(2−6?)cosθ−3?=0, is equal to:
(1) 12
(2) 6
(3) 8
(4) 10

Ans. (3)
Sol. 22cos?2θ+2cos?θ−6cos?θ−3=022?cos2θ+2cosθ−6?cosθ−3?=0
(2cos?θ−3)(2cos?θ+1)=0(2cosθ−3?)(2?cosθ+1)=0
cos?θ=32,−12cosθ=23??,2?−1?
Number of solution =8=8

8. Let a1,a2,a3a1?,a2?,a3? be in an A.P. such that ∑k=112a2k−1=−725a1,a1≠0∑k=112?a2k−1?=−572?a1?,a1??=0. If ∑k=1nak=0∑k=1n?ak?=0, then n is:
(1) 11
(2) 10
(3) 18
(4) 17

Ans. (1)
Sol. Let a1=aa1?=a, common difference =d=d
a1+a3+a5+…+a23=−725aa1?+a3?+a5?+…+a23?=−572?a
122[2a+11×2d]=−725a212?[2a+11×2d]=−572?a
12a+132d=−725a12a+132d=−572?a
132a+132×5d=0132a+132×5d=0
a=−5da=−5d
n2(2a+(n−1)d)=0⇒−10d+nd−d=02n?(2a+(n−1)d)=0⇒−10d+nd−d=0
n=11n=11

JEE Previous Year Question Paper

9. If the function f(x)=2x3−9ax2+12a2x+1f(x)=2x3−9ax2+12a2x+1, where a>0a>0, attains its local maximum and local minimum values at p and q, respectively, such that p2=qp2=q, then f(3)f(3) is equal to:
(1) 55
(2) 10
(3) 23
(4) 37

Ans. (4)
Sol. f′(x)=6x2−18ax+12a2f′(x)=6x2−18ax+12a2
f′(x)=6(x2−3ax+2a2)f′(x)=6(x2−3ax+2a2)
roots are a, 2a
p2=q⇒a2=2ap2=q⇒a2=2a
a=2a=2
f(x)=2x3−18x2+48x+1f(x)=2x3−18x2+48x+1
f(3)=37f(3)=37

10. Let z be a complex number such that ?z?=1?z?=1. If 2+k2zk+z?=kz,k∈Rk+z?2+k2z?=kz,k∈R, then the maximum distance of k+ik2k+ik2 from the circle ?z−(1+2i)?=1?z−(1+2i)?=1 is:
(1) 5+15?+1
(2) 2
(3) 3
(4) 3+13?+1

Ans. (1)
Sol. 2+k2zk+z?=kzk+z?2+k2z?=kz
?z?2k=2?z?2k=2
k=2k=2
point P(2,4)P(2,4) center (1,2)(1,2)
distance from circle (x−1)2+(y−2)2=1(x−1)2+(y−2)2=1 is max. if (OP+r)=1+4+1=5+1(OP+r)=1+4?+1=5?+1

JEE Previous Year Question Paper

11. If a?a? is nonzero vector such that its projections on the vectors 2i^−j^+2k^2i^−j^?+2k^, i^+2j^−2k^i^+2j^?−2k^ and k^k^ are equal, then a unit vector along a?a? is:
(1) 1155(−7i^+9j^+5k^)155?1?(−7i^+9j^?+5k^)
(2) 1155(−7i^+9j^−5k^)155?1?(−7i^+9j^?−5k^)
(3) 1155(7i^+9j^+5k^)155?1?(7i^+9j^?+5k^)
(4) 1155(7i^+9j^−5k^)155?1?(7i^+9j^?−5k^)

Ans. (3)
Sol. Let a?=a1i^+a2j^+a3k^a?=a1?i^+a2?j^?+a3?k^
a12+a22+a32=1a12?+a22?+a32?=1
Let b?=2i^−j^+2k^b?=2i^−j^?+2k^
c?=i^+2j^−2k^c?=i^+2j^?−2k^
d?=k^d?=k^
a?.b??b?=a?.c??c?=a?.d??d??b?a?.b??=?c?a?.c??=?d?a?.d??
2a1−a2+2a33=a1+2a2−2a33=a332a1?−a2?+2a3??=3a1?+2a2?−2a3??=a3?
By solving
a1=7155a1?=155?7?
a2=9155a2?=155?9?
a3=5155a3?=155?5?

12. Let A be the set of all functions f:Z→Zf:Z→Z and R be a relation on A such that R={(f,g):f(0)=g(1)R={(f,g):f(0)=g(1) and f(1)=g(0)}f(1)=g(0)}. Then R is:
(1) Symmetric and transitive but not reflexive
(2) Symmetric but neither reflexive nor transitive
(3) Reflexive but neither symmetric nor transitive
(4) Transitive but neither reflexive nor symmetric

Ans. (2)
Sol. R={(f,g):f(0)=g(1) and f(1)=g(0)}R={(f,g):f(0)=g(1) and f(1)=g(0)}
Reflexive: (f,f)∈R(f,f)∈R
=f(0)=f(1) and f(1)=f(0)→must hold=f(0)=f(1) and f(1)=f(0)→must hold
⇒⇒ but this is not true for all function
so not reflexive
Symmetric: (f,g)∈R⇒(g,f)∈R(f,g)∈R⇒(g,f)∈R
Now, g(0)=f(1) and g(1)=f(0)→trueg(0)=f(1) and g(1)=f(0)→true
∴∴ symmetric
Transitive: (f,g)∈R and (g,h)∈R(f,g)∈R and (g,h)∈R
⇒(f,h)∈R⇒(f,h)∈R
Now (f,g)∈R⇒f(0)=g(1) and f(1)=g(0)(f,g)∈R⇒f(0)=g(1) and f(1)=g(0)
(g,h)∈R⇒g(0)=h(1) and g(1)=h(0)(g,h)∈R⇒g(0)=h(1) and g(1)=h(0)
For (f,h)∈R(f,h)∈R we need f(0)=h(1) and f(1)=h(0)f(0)=h(1) and f(1)=h(0)
Now f(0)=g(1)=h(0) and f(1)=g(0)=h(1)f(0)=g(1)=h(0) and f(1)=g(0)=h(1)
Hence not transitive

JEE Previous Year Question Paper

13. For α,β,γ,∈Rα,β,γ,∈R if lim?x→0x2sin?αx+(γ−1)ex2sin?2x−βx=3limx→0?sin2x−βxx2sinαx+(γ−1)ex2?=3 then β+γ−αβ+γ−α is equal to:
(1) 7
(2) 4
(3) 6
(4) -1

Ans. (1)
Sol. lim?x→0x2(αx)+(γ−1)(1+x21)2x−8x36−βx=3limx→0?2x−68x3?−βxx2(αx)+(γ−1)(1+1x2?)?=3
lim?x→0(γ−1)+(γ−1)x2+αx3(2−β)x−43x3=3limx→0?(2−β)x−34?x3(γ−1)+(γ−1)x2+αx3?=3
γ−1=0,β=2,−3α4=3⇒α=−4γ−1=0,β=2,4−3α?=3⇒α=−4
β+γ−α=7β+γ−α=7

14. If the system of linear equations
3x+y+βz=33x+y+βz=3
2x+αy−z=−32x+αy−z=−3
x+2y+z=4x+2y+z=4
has infinitely many solutions, then the value of 22β−9α22β−9α is:
(1) 49
(2) 31
(3) 43
(4) 37

Ans. (2)
Sol. 3α+4β−αβ+3=03α+4β−αβ+3=0
9α+19=09α+19=0
α=−199,β=611α=9−19?,β=116?
⇒22β−9α=31⇒22β−9α=31

JEE Previous Year Question Paper

15. Let Pn=αn+βn,n∈NPn?=αn+βn,n∈N. If P10=123,P9=76,P8=47P10?=123,P9?=76,P8?=47 and P1=1P1?=1 then the quadratic equation having roots 1αα1? and 1ββ1? is:
(1) x2−x+1=0x2−x+1=0
(2) x2+x−1=0x2+x−1=0
(3) x2−x−1=0x2−x−1=0
(4) x2+x+1=0x2+x+1=0

Ans. (2)
Sol. α10+β10=123α10+β10=123
α+β=1α+β=1
α9+β9=76α9+β9=76
α8+β8=47α8+β8=47
P10=P9+P8P10?=P9?+P8?
x2=x+1⇒x2−x−1=0x2=x+1⇒x2−x−1=0
α+β=1,αβ=−1α+β=1,αβ=−1
1α+1β=α+βαβ=1−1=−1,1αβ=−1α1?+β1?=αβα+β?=−11?=−1,αβ1?=−1

16. If S and S′S′ are the foci of the ellipse x218+y29=118x2?+9y2?=1 and P be a point on the ellipse, then min?(SP.S′P)+max?(SP.S′P)min(SP.S′P)+max(SP.S′P) is equal to:
(1) 3(1+2)3(1+2?)
(2) 3(6+2)3(6+2?)
(3) 9
(4) 27

Ans. (4)
Sol. PS+PS′=2×32PS+PS′=2×32?
b2=a2(1−e2)⇒9=18(1−e2)b2=a2(1−e2)⇒9=18(1−e2)
⇒e=12⇒e=2?1?
Directrix x=ae=3212=6x=ea?=2?1?32??=6
PS⋅PS′=?12(32cos?θ−6)12(32cos?θ+6)?PS⋅PS′=?2?1?(32?cosθ−6)2?1?(32?cosθ+6)?
=12?18cos?2θ−36?=21??18cos2θ−36?
(PS⋅PS′)max?=18;(PS⋅PS′)min?=9(PS⋅PS′)max?=18;(PS⋅PS′)min?=9
sum = 27

JEE Previous Year Question Paper

17. Let the vertices Q and R of the triangle PQR lie on the line x+35=y−12=z+435x+3?=2y−1?=3z+4?, QR = 5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is mnnm? then:
(1) m−521n=0m−521?n=0
(2) 2m−521n=02m−521?n=0
(3) 5m−221n=05m−221?n=0
(4) 5m−212n=05m−212?n=0

Ans. (2)
Sol. M(5λ−3,2λ+1,3λ−4)M(5λ−3,2λ+1,3λ−4)
Drs of PM ⇒5λ−3,2λ−1,3λ−7⇒5λ−3,2λ−1,3λ−7
Drs of line L ⇒5,2,3⇒5,2,3
PM⊥LPM⊥L
⇒(5λ−3)5+(2λ−1)2+(3λ−7)3=0⇒(5λ−3)5+(2λ−1)2+(3λ−7)3=0
⇒λ=1⇒λ=1
∴M(2,3,−1)∴M(2,3,−1)
PM=4+1+16=21PM=4+1+16?=21?
Area =12×5×21=mn=21?×5×21?=nm?
2m−521n=02m−521?n=0

18. Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the BCD is equal to:
(1) 340340?
(2) 12
(3) 110110?
(4) 7373?

Ans. (3)
Sol. Ar(ΔBCD)Ar(ΔBCD)
=(Ar(ΔABC))2+(Ar(ACD))2+(Ar(ΔADB))2=(Ar(ΔABC))2+(Ar(ACD))2+(Ar(ΔADB))2?
=52+62+72=52+62+72?
=110=110?

JEE Previous Year Question Paper

19. Let a ∈ R and A be a matrix of order 3×3 such that det(A) = -4 and A+I=[1a1210a12]A+I=?12a?a11?102??, where I is the identity matrix of order 3×3. If det((a + 1)adj((a - 1)A)) is 2m3n2m3n, m, n ∈ {0,1,2,....20}, then m + n is equal to:
(1) 14
(2) 17
(3) 15
(4) 16

Ans. (4)
Sol. A=[0a1200a11]A=?02a?a01?101??
?A?=−4⇒2−2a=−4⇒a=3?A?=−4⇒2−2a=−4⇒a=3
?(a+1)adj(a−1)A?=?4adj3A??(a+1)adj(a−1)A?=?4adj3A?
=43?adj3A?=43?adj3A?
=43×?3A?3−1=64?3A?2=43×?3A?3−1=64?3A?2
=64×(33)2?A?2=64×(33)2?A?2
=26×36×16=26×36×16
2m×3n=210×362m×3n=210×36
∴m=10,n=6∴m=10,n=6
⇒m+n=16⇒m+n=16

20. Let the focal chord PQ of the parabola y2=4xy2=4x make an angle of 60?60? with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, S being the focus of the parabola, touches the y-axis at the point (0,α)(0,α), then 5α25α2 is equal to:
(1) 15
(2) 25
(3) 30
(4) 20

Ans. (1)
Sol. tan?60?=2t−0t2−1=3⇒t=3tan60?=t2−12t−0?=3?⇒t=3?
∴P(3,23)∴P(3,23?)
Circle:
(x−1)(x−3)+(y−0)(y−23)=0(x−1)(x−3)+(y−0)(y−23?)=0
at x=0x=0
⇒3+y2−23y=0⇒3+y2−23?y=0
⇒y=3=α⇒y=3?=α
5α2=155α2=15

JEE Previous Year Question Paper

SECTION-B

21. Let [] denote the greatest integer function. If ∫0e3[1ex−1]dx=α−log?e2∫0e3?[ex−11?]dx=α−loge?2, then α3α3 is equal to

Ans. (8)
Sol. f(x)=1ex−1=e1−xf(x)=ex−11?=e1−x
f(x)=2f(x)=1f(x)=2f(x)=1
1ex−1=2x=1ex−11?=2x=1
x=1−?n2x=1−?n2
f(0)=e1=2.71f(0)=e1=2.71
f(e3)=e1−e3∈(0,1)f(e3)=e1−e3∈(0,1)
1=∫01−?n22dx+∫1−?n21dx+∫1e30dx1=∫01−?n2?2dx+∫1−?n21?dx+∫1e3?0dx
=2(1−?n2−0)+1(1−1+?n2)+0=2(1−?n2−0)+1(1−1+?n2)+0
α−?n2=2−?n2α−?n2=2−?n2
α=2α=2
α3=8α3=8

22. Let f:R→Rf:R→R be a thrice differentiable odd function satisfying f′′(x)≥0,f′′(x)=f(x),f(0)=0,f′(0)=3f′′(x)≥0,f′′(x)=f(x),f(0)=0,f′(0)=3. Then 9f(log?23)9f(log2?3) is

Ans. (36)
Sol. f′′(x)=f(x)f′′(x)=f(x)
⇒f′(x)⋅f′′(x)=f′(x)⋅f(x)⇒f′(x)⋅f′′(x)=f′(x)⋅f(x)
⇒(f′(x))22=(f(x))22+C⇒2(f′(x))2?=2(f(x))2?+C
⇒(f′(x))2=(f(x))2+C′⇒(f′(x))2=(f(x))2+C′
f(0)=0,f′(0)=3⇒C′=9f(0)=0,f′(0)=3⇒C′=9
∴(f′(x))2=(f(x))2+9∴(f′(x))2=(f(x))2+9
f′(x)=(f(x))2+9?f′(x)≥0f′(x)=(f(x))2+9??f′(x)≥0
∫dyy2+9=∫dx⇒?n?y+y2+9?=x+C∫y2+9?dy?=∫dx⇒?n?y+y2+9??=x+C
⇒f(0)=0⇒C=?n3⇒f(0)=0⇒C=?n3
⇒y+y2+9=3ex⇒y+y2+9?=3ex
at x=?n3;y=4x=?n3;y=4
∴9f(?n3)=36∴9f(?n3)=36

JEE Previous Year Question Paper

23. If the area of the region {(x,y):?4−x2?≤y≤x2,y≤4,x≥0}{(x,y):?4−x2?≤y≤x2,y≤4,x≥0} is (802α−β)(α802??−β), α,β∈Nα,β∈N, then α+βα+β is equal to

Ans. (22)
Sol. A=∫044+ydy−∫024−ydy−∫24ydyA=∫04?4+y?dy−∫02?4−y?dy−∫24?y?dy
=((4+y)3/23/2+(4−y)3/23/2−y3/23/2)02−(y3/23/2)02=(3/2(4+y)3/2?+3/2(4−y)3/2?−3/2y3/2?)02?−(3/2y3/2?)02?
=8023−16=8023−16=3802??−16=3802??−16
α=6,β=16α=6,β=16
α+β=22α+β=22

24. Three distinct numbers are selected randomly from the set {1,2,3,…,40}{1,2,3,…,40}. If the probability, that the selected numbers are in an increasing G.P. is mnnm?, gcd?(m,n)=1gcd(m,n)=1, then m + n is equal to

Ans. (4949)
Sol. 1≤a<ar<ar2≤401≤a<ar<ar2≤40
(If r∈Nr∈N)
If r=2r=2
1≤a<2a<4a≤401≤a<2a<4a≤40
a∈{1,…,10}(10GP)a∈{1,…,10}(10GP)
If r=3r=3
1≤a<3a<9a≤401≤a<3a<9a≤40
a∈{1,2,3,4}a∈{1,2,3,4}
(4GP)(4GP)
If r=4r=4
1≤a<4a<16a≤401≤a<4a<16a≤40
a∈{1,2}(2GP)a∈{1,2}(2GP)
If r=5r=5
1≤a<5a<25a≤401≤a<5a<25a≤40
a∈{1}(1GP)a∈{1}(1GP)
If r=6r=6
1≤a<6a<36a≤401≤a<6a<36a≤40
a∈{1}(1GP)a∈{1}(1GP)
(P=189880=94940)(P=988018?=49409?) as per NTA for r∈Nr∈N
m+n=4949m+n=4949
If r∈Nr∈N (also possible)
r=32r=23?
ar2=9a4;a=4kar2=49a?;a=4k
(4,6,9)(4,6,9)
(8,12,18)(8,12,18)
(12,18,27)(12,18,27)
(16,24,36)(16,24,36)
r=52ar2=25a4;a=4kr=25?ar2=425a?;a=4k
(4,10,25)……(1)GP(4,10,25)……(1)GP
r=43ar2=16a9→a=9kr=34?ar2=916a?→a=9k
(9,12,16),(18,24,32)……(2)GP(9,12,16),(18,24,32)……(2)GP
r=53ar2=25a9;a=9kr=35?ar2=925a?;a=9k
(9,15,25)……(1)GP(9,15,25)……(1)GP
r=54ar2=25a16;a=16kr=45?ar2=1625a?;a=16k
r=65ar2=36a25;a=25kr=56?ar2=2536a?;a=25k
(25,30,36)……(1)GP(25,30,36)……(1)GP
Total =18+10=28=18+10=28
P=2840C3=289880=72470P=40C3?28?=988028?=24707?
m+n=2477m+n=2477

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25. The absolute difference between the squares of the radii of the two circles passing through the point (−9,4)(−9,4) and touching the lines x+y=3x+y=3 and x−y=3x−y=3 is equal to

Ans. (768)
Sol. Centre (a, 0)
r=?a−0−32?r=?2?a−0−3??
circle (x−a)2+y2=(a−32)2(x−a)2+y2=(2?a−3?)2
passes through (−9,4)(−9,4)
2(a2+18a+81+16)=(a2−6a+9)2(a2+18a+81+16)=(a2−6a+9)
a2+42a+185=0a2+42a+185=0
(a+37)(a+5)=0(a+37)(a+5)=0
⇒a=−37,−5⇒a=−37,−5
r1=?−37−32?=202r1?=?2?−37−3??=202?
r2=?−5−32?=42r2?=?2?−5−3??=42?
?r12−r22?=?800−32?=768?r12?−r22??=?800−32?=768

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PHYSICS

SECTION-A

26. A light wave is propagating with plane wave fronts of the type x + y + z = constant. The angle made by the direction of wave propagation with the x-axis is:
(1) cos?−1(13)cos−1(3?1?)
(2) cos?−1(23)cos−1(32?)
(3) cos?−1(13)cos−1(31?)
(4) cos?−1(23)cos−1(32??)

Ans. (1)
Sol. The direction of propagation of light is perpendicular to the wave front and is symmetric about x, y and z axis.
∴∴ Angle made by the light with x, y & z axis is same.
∴cos?α=cos?β=cos?γ∴cosα=cosβ=cosγ (α,βα,β & γγ are angle made by light with x, y & z axis respectively)
Also cos?2α+cos?2β+cos?2γ=1cos2α+cos2β+cos2γ=1 [Sum of direction cosines]
∴α=cos?−113∴α=cos−13?1?

27. The equation for real gas is given by (P+aV2)(V−b)=RT(P+V2a?)(V−b)=RT, where P,V,T and R are the pressure, volume, temperature and gas constant, respectively. The dimension of ab2ab2 is equivalent to that of:
(1) Planck's constant
(2) Compressibility
(3) Strain
(4) Energy density

Ans. (4)
Sol. [P+aV2](V−b)=RT[P+V2a?](V−b)=RT
∴[a]=[P][V2]=ML−1T−2L6=ML5T−2∴[a]=[P][V2]=ML−1T−2L6=ML5T−2
[b]=[V]=L3[b]=[V]=L3
[ab2]=ML5T−2L6=ML11T−2[ab2]=ML5T−2L6=ML11T−2
Dimension of energy density.

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28. A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10 kg and radius is 10 cm and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20 N is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1 m, would be:
(1) 20 rad/s
(2) 30 rad/s
(3) 10 rad/s
(4) 0 rad/s

Ans. (1)
Sol. WF=20×1=20JWF?=20×1=20J
∴ΔKE=20J=12Iω2∴ΔKE=20J=21?Iω2
I=MR2=10×0.12=0.1kgm2I=MR2=10×0.12=0.1kgm2
∴20=12×0.1×ω2∴20=21?×0.1×ω2
⇒ω=20rad/sec⇒ω=20rad/sec

29. A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is:
(1) −α2−2α?
(2) −45?−45?
(3) +45?+45?
(4) −α−α

Ans. (2)
Sol. Location of image of A:
1v−1u=1f⇒1v−1−30=120⇒1v=160⇒v=60cmv1?−u1?=f1?⇒v1?−−301?=201?⇒v1?=601?⇒v=60cm
∴m=2∴m=2
Since size of object is small wrt the location hence
dv=m2du⇒dv=4×1=4cmdv=m2du⇒dv=4×1=4cm
hi=mho⇒hi(dv)=2×2=4cmhi?=mho?⇒hi?(dv)=2×2=4cm
Angle made with principle axis =−45?=−45?

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30. Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density +σ+σ and −2σ−2σ. The force experienced by a point charge +q+q placed at the mid point between two plates will be:
(1) σq4?04?0?σq?
(2) 3σq2?02?0?3σq?
(3) 3σq4?04?0?3σq?
(4) σq2?02?0?σq?

Ans. (2)
Sol. Final charge distribution will be
∴Fnet=3σ2?0q∴Fnet?=2?0?3σ?q

31. A river is flowing from west to east direction with speed of 9kmh−19kmh−1. If a boat capable of moving at a maximum speed of 27kmh−127kmh−1 in still water, crosses the river in half a minute, while moving with maximum speed at an angle of 150?150? to direction of river flow, then the width of the river is:
(1) 300 m
(2) 112.5 m
(3) 75 m
(4) 112.5×3m112.5×3?m

Ans. (2)
Sol. V⊥=27×cos?60?=272km/hrV⊥?=27×cos60?=227?km/hr
Time taken = 30 sec
∴S=Vt=272×518×30m=112.5m∴S=Vt=227?×185?×30m=112.5m

32. A point charge +q+q is placed at the origin. A second point charge +9q+9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is:
(1) (4d/3,0,0)(4d/3,0,0)
(2) (d/4,0,0)(d/4,0,0)
(3) (3d/4,0,0)(3d/4,0,0)
(4) (4d/3,0,0)(4d/3,0,0)

Ans. (2)
Sol. Let Ep=0Ep?=0
∴kqx2=k9q(d−x)2∴x2kq?=(d−x)2k9q?
⇒d−xx=3⇒x=d4⇒xd−x?=3⇒x=4d?
co-ordinate of P is (d4,0,0)(4d?,0,0)

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33. The battery of a mobile phone is rated as 4.2 V, 5800 mAh. How much energy is stored in it when fully charged?
(1) 43.8 kJ
(2) 48.7 kJ
(3) 87.7 kJ
(4) 24.4 kJ

Ans. (3)
Sol. Given V = 4.2 volt
Energy supplied by battery
=vq=4.2×5800×3600×10−3J=87.696kJ=vq=4.2×5800×3600×10−3J=87.696kJ
Energy stored in the battery when fully charged =87.696kJ≈87.7kJ=87.696kJ≈87.7kJ

34. A particle is subjected two simple harmonic motions as:
x1=7sin?5tcmx1?=7?sin5tcm
and x2=27sin?(5t+π3)cmx2?=27?sin(5t+3π?)cm
where x is displacement and t is time in seconds.
The maximum acceleration of the particle is x×10−3ms−2x×10−3ms−2. The value of x is:
(1) 175
(2) 257257?
(3) 5757?
(4) 125

Ans. (1)
Sol. x1=7sin?5tx1?=7?sin5t
x2=27sin?(5t+π3)x2?=27?sin(5t+3π?)
From phasor,
Amplitude of resultant SHM = 7
?=tan?−127×3/27+27×12=tan?−12127=tan?−132?=tan−17?+27?×21?27?×3?/2?=tan−127?21??=tan−123??
∴XR=7sin?(5t+?)∴XR?=7sin(5t+?)
aR=−7×25sin?(5t+?)aR?=−7×25sin(5t+?)
∴amax=175cm/sec=175×10−2m/sec∴amax?=175cm/sec=175×10−2m/sec

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35. The relationship between the magnetic susceptibility (χ)(χ) and the magnetic permeability (μ)(μ) is given by:
(μ0μ0? is the permeability of free space and μrμr? is relative permeability)
(1) χ=μμ0−1χ=μ0?μ?−1
(2) χ=μrμ0+1χ=μ0?μr??+1
(3) χ=μr+1χ=μr?+1
(4) χ=1−μμ0χ=1−μ0?μ?

Ans. (1)
Sol. We have
μr=(1+χ)⇒χ=(μr−1)μr?=(1+χ)⇒χ=(μr?−1)
μ=μ0μr⇒μr=μμ0μ=μ0?μr?⇒μr?=μ0?μ?
∴χ=(μμ0−1)∴χ=(μ0?μ?−1)

36. A zener diode with 5V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400Ω400Ω resistor connected in series, the zener current is found to be 4 times load current. The load current (IL)(IL?) and load resistance (RL)(RL?) are:
(1) IL=20mA;RL=250ΩIL?=20mA;RL?=250Ω
(2) IL=10A;RL=0.5ΩIL?=10A;RL?=0.5Ω
(3) IL=0.02mA;RL=250ΩIL?=0.02mA;RL?=250Ω
(4) IL=10mA;RL=500ΩIL?=10mA;RL?=500Ω

Ans. (4)
Sol. From the circuit diagram,
5i=20400=120A5i=40020?=201?A
∴i=1100A=10mA=Loadcurrent∴i=1001?A=10mA=Loadcurrent
Also, VL=5VVL?=5V
∴RL=510×10−3Ω=500Ω∴RL?=10×10−35?Ω=500Ω

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37. In an adiabatic process, which of the following statements is true?
(1) The molar heat capacity is infinite
(2) Work done by the gas equals the increase in internal energy
(3) The molar heat capacity is zero
(4) The internal energy of the gas decreases as the temperature increases

Ans. (3)
Sol. For adiabatic process, dQ=0dQ=0
Molar heat capacity = 0
∴dQ=0⇒dU=−dW∴dQ=0⇒dU=−dW
Also dU=f2nRdTdU=2f?nRdT
Only option (3) is correct.

38. A square Lamina OABC of length 10cm10cm is pivoted at 'O'. Forces act at Lamina as shown in figure. If Lamina remains stationary, then the magnitude of F is:
(1) 20N20N
(2) 0 (zero)
(3) 10N10N
(4) 102N102?N

Ans. (3)
Sol. Since the lamina is equilibrium.
∴Fnet=0∴Fnet?=0 & τnet=0τnet?=0
To=10?−F?⇒F=10NTo?=10?−F?⇒F=10N

39. Let B1B1? be the magnitude of magnetic field at center of a circular coil of radius R carrying current I. Let B2B2? be the magnitude of magnetic field at an axial distance 'x' from the center. For x:R=3:4x:R=3:4, B2B1B1?B2?? is:
(1) 4:54:5
(2) 16:2516:25
(3) 64:12564:125
(4) 25:1625:16

Ans. (3)
Sol. B1=μ0i2RB1?=2Rμ0?i?
B2=B1sin?3θB2?=B1?sin3θ
∴B2B1=sin?3θ=(45)3=64125∴B1?B2??=sin3θ=(54?)3=12564?

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40. Considering Bohr's atomic model for hydrogen atom:
(A) the energy of H atom in ground state is same as energy of He? ion in its first excited state.
(B) the energy of H atom in ground state is same as that for Li²? ion in its second excited state.
(C) the energy of H atom in its ground state is same as that of He? ion for its ground state.
(D) the energy of He? ion in its first excited state is same as that for Li²? ion in its ground state
Choose the correct answer from the options given below:
(1) (B), (D) only
(2) (A), (B) only
(3) (A), (D) only
(4) (A), (C) only

Ans. (2)
Sol. E∝Zn2E∝n2Z?
ZHe=2ZHe+=2ZLi2+=3ZHe?=2ZHe+?=2ZLi2+?=3
1st excited state ⇒n=2⇒n=2
2nd excited state ⇒n=3⇒n=3
From the given statements only A & B are correct.

41. Moment of inertia of a rod of mass 'M' and length 'L' about an axis passing through its center and normal to its length is 'α'. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is:
(1) αα
(2) α/4α/4
(3) α/8α/8
(4) α/2α/2

Ans. (2)
Sol. α=M?212…(i)α=12M?2?…(i)
α′=2[M2(?2)212]α′=2[122M?(2??)2?]
α′=M?248=α4α′=48M?2?=4α?
Correct option is (2)

42. A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is:
(C is the center of curvature of the spherical surface and R is the radius of curvature)
(1) 0.24 m right to the spherical surface
(2) 0.4 m left to the spherical surface
(3) 0.24 m left to the spherical surface
(4) 0.4 m right to the spherical surface

Ans. (2)
Sol. μ2v−μ1u=μ2−μ1Rvμ2??−uμ1??=Rμ2?−μ1??
1.5v−1(−0.2)=1.5−10.4v1.5?−(−0.2)1?=0.41.5−1?
1.5v=0.50.4−10.2v1.5?=0.40.5?−0.21?
1.5v=1.50.4v1.5?=0.41.5?
v=−0.4mv=−0.4m

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43. Match List-I with List-II.
List-I
(A) Coefficient of viscosity
(B) Intensity of wave
(C) Pressure gradient
(D) Compressibility
List-II
(I) [ML−1T−1][ML−1T−1]
(II) [ML0T−3][ML0T−3]
(III) [ML−2T−2][ML−2T−2]
(IV) [M−1L0T2][M−1L0T2]
Choose the correct answer from the options given below:
(1) (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
(2) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(3) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
(4) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Ans. (2)
Sol. (A) Coefficient of viscosity
[η]=[ML−1T−1][η]=[ML−1T−1]
(B) Intensity [I]=[ML0T−3][I]=[ML0T−3]
(C) Pressure gradient = [ML−2T−2][ML−2T−2]
(D) Compressibility [K]=[M−1L0T2][K]=[M−1L0T2]

44. A small bob of mass 100mg100mg and charge +10μC+10μC is connected to an insulating string of length 1m1m. It is brought near to an infinitely long nonconducting sheet of charge density 'σ' as shown in figure. If string subtends an angle of 45?45? with the sheet at equilibrium the charge density of sheet will be:
(Given, ?0=8.85×10−12F/m?0?=8.85×10−12F/m and acceleration due to gravity, g=10m/s2g=10m/s2)
(1) 0.885nC/m20.885nC/m2
(2) 17.7nC/m217.7nC/m2
(3) 885nC/m2885nC/m2
(4) 1.77nC/m21.77nC/m2

Ans. (4)
Sol. qE=mgqE=mg
q[σ2?0]=mgq[2?0?σ?]=mg
σ=2?0mgqσ=q2?0?mg?
σ=2×8.85×10−12×100×10−6×1010×10−6σ=10×10−62×8.85×10−12×100×10−6×10?
σ=17.7×10−10C/m2σ=17.7×10−10C/m2
σ=1.77nC/m2σ=1.77nC/m2

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45. A monochromatic light is incident on a metallic plate having work function ??. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is:
(Given: The magnitude of charge of an electron is e and mass is m, h is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)
(1) 2m(hcλ−?)/eB2m(λhc?−?)?/eB
(2) m(hcλ−?)/eBm(λhc?−?)?/eB
(3) 8m(hcλ−?)/eB8m(λhc?−?)?/eB
(4) 2m(hcλ−?)/eB2m(λhc?−?)?/eB

Ans. (3)
Sol. KEmax=hcλ−?KEmax?=λhc?−?
p=2mKmaxp=2mKmax??
p=2m(hcλ−?)p=2m(λhc?−?)?
dA−B=2RdA−B?=2R
=2[pqB]=2[qBp?]
dAB=22m(hcλ−?)eB=8m(hcλ−?)eBdAB?=eB22m(λhc?−?)??=eB8m(λhc?−?)??

SECTION-B

46. A vessel with square cross-section and height of 6m6m is vertically partitioned. A small window of 100cm3100cm3 with hinged door is fitted at a depth of 3m3m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density 1.5×103kg/m31.5×103kg/m3. What force one needs to apply on the hinged door so that it does not get opened? (Acceleration due to gravity =10m/s2=10m/s2)

Ans. (150)
Sol. in equilibrium
Fext+Fw=F?Fext?+Fw?=F??
⇒Fext=F?−Fw⇒Fext?=F??−Fw?
=(P0+ρ?gh)A−(P0+ρwgh)A=(P0?+ρ??gh)A−(P0?+ρw?gh)A
=(ρ?−ρw)ghA=(ρ??−ρw?)ghA
=(1500−1000)×10×3×(100×10−4)=(1500−1000)×10×3×(100×10−4)
=150N=150N

47. A steel wire of length 2m2m and Young's modulus 2.0×1011Nm−22.0×1011Nm−2 is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10−310−3 respectively, then the elastic potential energy density of the wire is ×105×105 (in SI units)

Ans. (25)
Sol. ?=2m;Y=2×1011N/m2?=2m;Y=2×1011N/m2
μ=−(Δrr)(Δ??)⇒Δ??=1μ×(Δrr)μ=−(?Δ??)(rΔr?)?⇒?Δ??=μ1?×(rΔr?)
=10.2×(10−3)=0.21?×(10−3)
⇒Δ??=5×10−3⇒?Δ??=5×10−3
u=12Yet2=12×2×1011×[5×10−3]2u=21?Yet2?=21?×2×1011×[5×10−3]2
=25=25

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48. If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is 30?30? in a single slit diffraction pattern recorded using 628nm628nm light, then the width of the slit is ______ mm.

Ans. (6)
Sol. θ1=sin?−1(2λa)θ1?=sin−1(a2λ?)
θ2=sin?−1(3λa)θ2?=sin−1(a3λ?)
∴θ1+θ2=30?∴θ1?+θ2?=30?
⇒sin?−1(2λa)+sin?−1(3λa)=π6⇒sin−1(a2λ?)+sin−1(a3λ?)=6π?
⇒2λa1−(3λa)2+3λa1+(2λa)2=sin?π6⇒a2λ?1−(a3λ?)2?+a3λ?1+(a2λ?)2?=sin6π?
Here λ=628nmλ=628nm
After solving A=6.07μmA=6.07μm
Approximate Method:
θ=θ1+θ2θ=θ1?+θ2?
⇒π6=2λa+3λa⇒6π?=a2λ?+a3λ?
⇒π6=5a(628nm)⇒6π?=a5?(628nm)
⇒a=6μm⇒a=6μm

49. γAγA? is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. γBγB? is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If γAγB=(1+1n)γB?γA??=(1+n1?), then the value of n is

Ans. (3)
Sol. γAγB=fA+2fA×fBfB+2γB?γA??=fA?fA?+2?×fB?+2fB??
=3+23×(6+2)(6+2)+2=33+2?×(6+2)+2(6+2)?
=53×810=4030=35?×108?=3040?
∴4030=1+1n∴3040?=1+n1?
⇒4030−1=1n⇒3040?−1=n1?
n=3n=3

50. A person travelling on a straight line moves with a uniform velocity v1v1? for a distance x and with a uniform velocity v2v2? for the next 32x23?x distance. The average velocity in this motion is 507m/s750?m/s. If v1v1? is 5m/s5m/s then v2=v2?= ______ m/s.

Ans. (10)
Sol. Vavg=x1+x2t1+t2Vavg?=t1?+t2?x1?+x2??
⇒507=x+3x2x5+3x2v2⇒750?=5x?+2v2?3x?x+23x??
⇒507=5/215+32v2⇒750?=51?+2v2?3?5/2?
⇒15+32v2=720⇒51?+2v2?3?=207?
⇒32v2=720−15=7−420⇒2v2?3?=207?−51?=207−4?
⇒32v2=320⇒2v2?3?=203?
⇒v2=10m/s⇒v2?=10m/s

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CHEMISTRY

SECTION-A

51. Designate whether each of the following compounds is aromatic or not aromatic.
(1) e, g aromatic and a, b, c, d, f, h not aromatic
(2) b, e, f, g aromatic and a, c, d, h not aromatic
(3) a, b, c, d aromatic and e, f, g, h not aromatic
(4) a, c, d, e, h aromatic and b, f, g not aromatic

Ans. (4)
Sol. Aromatic compounds: a, c, d, e, h follow Huckel's rule.
b, f, g are not aromatic, these compounds do not follow Huckel's rule.

52. An optically active alkyl halide C?H?Br [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic NaNH?. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E]. The IUPAC name of compound [E] is:
(1) But-2-yne
(2) Butan-2-ol
(3) Butan-2-one
(4) Butan-1-al

Ans. (3)
Sol. Butan-2-one

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53. The property/properties that show irregularity in first four elements of group-17 is/are:
(A) Covalent radius
(B) Electron affinity
(C) Ionic radius
(D) First ionization energy
Choose the correct answer from the options given below:
(1) B and D only
(2) A and C only
(3) B only
(4) A, B, C and D

Ans. (3)
Sol. The order of first four elements of group-17 are as follows.
F < Cl < Br < I (Covalent radius)
Cl > F > Br > I (Electron affinity)
F < Cl < Br < I (Ionic radius)
F > Cl > Br > I (I ionization energy)
Electron affinity order is irregular.

54. Which of the following graph correctly represents the plots of KHKH? at 1 bar gases in water versus temperature?
(1) graph
(2) graph
(3) graph
(4) graph

Ans. (4)
Sol. As temperature increases solubility first decrease then increase hence KHKH? first increase than decrease also at moderate temperature KHKH? value H > N? > CH?.

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55. According to Bohr's model of hydrogen atom, which of the following statement is incorrect?
(1) Radius of 3rd orbit is nine times larger than that of 1st orbit.
(2) Radius of 8th orbit is four times larger than that of 4th orbit.
(3) Radius of 6th orbit is three time larger than that of 4th orbit.
(4) Radius of 4th orbit is four times larger than that of 2nd orbit.

Ans. (3)
Sol. r∝n2r∝n2
r6/r4=(6/4)2=9/4r6?/r4?=(6/4)2=9/4

56. Two vessels A and B are connected via stopcock. The vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and is allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true?
(1) dw ≠ 0
(2) dq ≠ 0
(3) dU ≠ 0
(4) The pressure in the vessel B before opening the stopcock is zero.

Ans. (4)
Sol. It is free expansion of gas ⇒Pext=0⇒Pext?=0
Where w=0w=0, q=0q=0 and ΔU=0ΔU=0

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57. A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200mmHg200mmHg and that of the solution is 500mmHg500mmHg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are:
(1) 1400mmHg1400mmHg A
(2) 1400mmHg1400mmHg B
(3) 600mmHg600mmHg B
(4) 600mmHg600mmHg A

Ans. (4)
Sol. PS=PAoXA+PBoXBPS?=PAo?XA?+PBo?XB?
500=200×14+PBo34500=200×41?+PBo?43?
PBo=600mmHgPBo?=600mmHg
As PAo<PBo⇒APAo?<PBo?⇒A is least volatile.

58. Consider the reaction:
CaCO?(s) + 2HCl(aq) → CaCl?(aq) + CO?(g) + H?O(l)
What mass of CaCl? will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of CaCO??
(Given: Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g mol?¹, respectively)
(1) 3.908 g
(2) 2.636 g
(3) 10.545 g
(4) 5.272 g

Ans. (3)
Sol. Moles of CaCO? = 1000/100 = 10
Moles of HCl = 0.76 × 250/1000 = 0.19 (L.R.)
Moles of CaCl? formed = 0.19/2
Mass of CaCl? = (0.19/2) × 111 = 10.545 gm

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59. If equal volumes of AB? and XY (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of AY? at 300 K? (Given Ksp (at 300 K) for AY? = 5.2 × 10??)
(1) 3.6×10−3MAB2,5.0×10−4MXY3.6×10−3MAB2?,5.0×10−4MXY
(2) 2.0×10−4MAB2,0.8×10−3MXY2.0×10−4MAB2?,0.8×10−3MXY
(3) 2.0×10−2MAB2,2.0×10−2MXY2.0×10−2MAB2?,2.0×10−2MXY
(4) 1.5×10−4MAB2,1.5×10−3MXY1.5×10−4MAB2?,1.5×10−3MXY

Ans. (3)
Sol. When equal volumes are mixed molarity reduce to half.
For precipitation Qsp=[A3+][Y−]2>KspQsp?=[A3+][Y−]2>Ksp?
For option (3): Qsp=(10−3)(10−3)2>KspQsp?=(10−3)(10−3)2>Ksp?

60. Among SO?, NF?, NH?, XeF?, ClF? and SF?, the hybridization of the molecule with non-zero dipole moment and highest number of lone-pairs of electrons on the central atom is
(1) sp3sp3
(2) dsp2dsp2
(3) sp3d2sp3d2
(4) sp3dsp3d

Ans. (4)
Sol.

Molecule	Hybridisation	Dipole Moment	Lone pair on central atom
SO?	sp²	Non-zero	1
NF?	sp³	Non-zero	1
NH?	sp³	Non-zero	1
XeF?	sp³d	zero	3
ClF?	sp³d	Non-zero	2
SF?	sp³d	Non-zero	1

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61. Given below are two statements:
Statement (I): Vanillin will react with NaOH and also with Tollen's reagent.
Statement (II): Vanillin will undergo self aldol condensation very easily.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is incorrect but Statement II is correct
(2) Statement I is correct but Statement II is incorrect
(3) Both Statement I and Statement II are incorrect
(4) Both Statement I and Statement II are correct

Ans. (2)
Sol. Vanillin has phenolic group soluble in NaOH and benzaldehyde derivative reacts with Tollen's reagent. Vanillin does not give self-aldol reaction due to lack of acidic H for condensation.

62. Identify the correct statement among the following:
(1) All naturally occurring amino acids except glycine contain one chiral centre.
(2) All naturally occurring amino acids are optically active.
(3) Glutamic acid is the only amino acid that contains a –COOH group at the side chain.
(4) Amino acid, cysteine easily undergo dimerization due to the presence of free SH group.

Ans. (4)
Sol. Isoleucine has 2 chiral centre. Glycine is optically inactive. Aspartic acid also contains COOH group at side chain. Cysteine easily dimerises due to free SH group.

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63. The correct order of basic nature on aqueous solution for the bases NH?, H?N–NH?, CH?CH?NH?, (CH?CH?)?NH and (CH?CH?)?N is:
(1) NH? < H?N–NH? < (CH?CH?)?N < CH?CH?NH? < (CH?CH?)?NH
(2) NH? < H?N–NH? < CH?CH?NH? < (CH?CH?)?NH < (CH?CH?)?N
(3) H?N–NH? < NH? < (CH?CH?)?N < CH?CH?NH? < (CH?CH?)?NH
(4) NH? < H?N–NH? < CH?CH?NH? < (CH?CH?)?N < (CH?CH?)?NH

Ans. (4)
Sol. Basic strength of amine depends on hydrogen bonding and electronic inductive effect.
NH(Et)? > N(Et)? > NH?Et > NH? > NH?–NH?

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64. Given below are two statements:
Statement (I): The metallic radius of Al is less than that of Ga.
Statement (II): The ionic radius of Al³? is less than that of Ga³?.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Both Statement I and Statement II are incorrect
(2) Statement I is incorrect but Statement II is correct
(3) Statement I is correct but Statement II is incorrect
(4) Both Statement I and Statement II are correct

Ans. (2)
Sol. Metallic radius order: B < Ga < Al < In < Tl (due to poor shielding of d-subshell electrons).
Ionic radius order: B³? < Al³? < Ga³? < In³? < Tl³?.

65. Given below are two statements:
Statement (I): In octahedral complexes, when Δ? < P high spin complexes are formed. When Δ? > P low spin complexes are formed.
Statement (II): In tetrahedral complexes because of Δ? < P low spin complexes are rarely formed.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is correct but Statement II is incorrect.
(2) Both Statement I and Statement II are incorrect
(3) Statement I is incorrect but Statement II is correct
(4) Both Statement I and Statement II are correct

Ans. (4)
Sol. In octahedral complex CN = 6. If Δ? < P.E., high spin complexes are formed. If Δ? > P.E., low spin complexes are formed. But in tetrahedral complex CN = 4, Δ? < P.E., mainly high spin complexes are formed and rarely low spin complexes are formed.

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66. Choose the correct tests with respective observations.
(A) CuSO? (acidified with acetic acid) + K?[Fe(CN)?] → Chocolate brown precipitate.
(B) FeCl? + K?[Fe(CN)?] → Prussian blue precipitate.
(C) ZnCl? + K?[Fe(CN)?], neutralised with NH?OH → White or bluish white precipitate.
(D) MgCl? + K?[Fe(CN)?] → Blue precipitate.
(E) BaCl? + K?[Fe(CN)?], neutralised with NaOH → White precipitate.
Choose the correct answer from the options given below:
(1) A, D and E only
(2) B, D and E only
(3) A, B and C only
(4) C, D and E only

Ans. (3)
Sol.
2CuSO? + K?[Fe(CN)?] → Cu?[Fe(CN)?] + 2K?SO? (chocolate brown ppt)
4FeCl? + 3K?[Fe(CN)?] → Fe?[Fe(CN)?]? + 12KCl (Prussian blue ppt)
3ZnCl? + 2K?[Fe(CN)?] → K?Zn?[Fe(CN)?]? + 6KCl (white or bluish white ppt)

67. On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of CO? and 0.567 g of H?O. The empirical formula mass of compound (X) is ______ g.
(Given molar mass in g mol?¹ C:12, H:1, O:16)
(1) 30
(2) 45
(3) 60
(4) 15

Ans. (1)
Sol. Moles of C = n(CO?) = 1.46/44 = 0.033
Mass of C = 0.033 × 12
Moles of H = 2 × n(H?O) = 2 × 0.567/18 = 0.063
Mass of H = 0.063
Mass of Oxygen = 1 − (0.033×12 + 0.063×1) = 0.541 g
Moles of O = 0.541/16 = 0.033
Empirical formula = CH?O
Empirical formula mass = 30

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68. Consider the following compound (X). The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding C–H bond are:
(1) II, IV
(2) III, II
(3) I, IV
(4) II, I

Ans. (4)
Sol. II most stable carbon radical due to resonance stabilisation. I least stable carbon radical due to no stabilising factor.

69. Consider the following molecules. The correct order of rate of hydrolysis is:
(1) r > q > p > s
(2) q > p > r > s
(3) p > r > q > s
(4) p > q > r > s

Ans. (4)
Sol. Rate of hydrolysis ∝ Leaving group ability
CH?CH?COCl > EtCOOCOCH? > EtCOOEt > EtCONH?

70. A molecule with the formula AX?Y has all its elements from p-block. Element A is rarest, monoatomic, non-radioactive from its group and has the lowest ionization enthalpy value among A, X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is:
(1) Square pyramidal
(2) Octahedral
(3) Pentagonal planar
(4) Trigonal bipyramidal

Ans. (1)
Sol. A is Xe, X is F, Y is O. Compound is XeOF? with square pyramidal shape.

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71. A transition metal (M) among Mn, Cr, Co and Fe has the highest standard electrode potential (M³?/M²?). It forms a metal complex of the type [M(CN)?]³?. The number of electrons present in the e_g orbital of the complex is

Ans. (1)
Sol. Co has highest standard electrode potential (M³?/M²?) among Mn, Cr, Co, Fe. Complex is [Co(CN)?]³? and its splitting shows 1 electron in e_g orbital.

72. Consider the following electrochemical cell at standard condition.
Au(s)|QH?,Q|NH?X(0.01M)||Ag?(1M)|Ag(s)
E_cell = +0.4 V
The couple QH?/Q represents quinhydrone electrode, the half cell reaction is given below
(Given: E°(Ag?/Ag) = +0.8 V and 2.303RT/F = 0.06 V)
The pK_b value of the ammonium halide salt (NH?X) used here is (nearest integer)

Ans. (6)
Sol. QH? + 2Ag? → 2Ag + Q + 2H?
E = E° − (0.06/2) log[H?]²
E = E° − 0.06 × log[H?]
pH = −log(H?) = (E − E°)/0.06 = (0.4 − 0.1)/0.06 = 5
pH + NH?X = 7 − ½ pK_b − ½ log C
5 = 7 − ½ pK_b − ½ log(10?²)
pK_b = 6

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73. 0.1 mol of the following given antiviral compound (P) will weigh ______ × 10?¹ g.
(Given: molar mass in g mol?¹ H:1, C:12, N:14, O:16, F:19, I:127)

Ans. (372)
Sol. Molar mass = 372 g
0.1 mole has = 372 × 10?¹ g

74. Consider the following equilibrium, CO(g) + 2H?(g) ? CH?OH(g). 0.1 mol of CO along with a catalyst is present in a 2 dm³ flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of CH?OH is formed. The K_p° is ______ × 10?³ (nearest integer).
Given: R = 0.08 dm³ bar K?¹ mol?¹
Assume only methanol is formed as the product and the system follows ideal gas behaviour.

Ans. (74)
Sol. CO(g) + 2H?(g) ? CH?OH(g)
t = 0: 0.1 mol, a mol
t_eq: 0.1 − x, a − 2x, x = 0.04
= 0.06, a − 0.08, 0.04
a = 0.23, n_total = 0.25 = 1/4 mol
P_total = n_total × RT/V
5 = (0.06 + a − 0.08 + 0.04) × 0.08×500/2
a = 0.23 mol
K_p = [X_CH3OH/(X_CO × X_H2²)] × (1/P_f²)
= [0.04/(0.06×(0.15)²)] × [ (1/4)/5 ]²
= 0.074 = 74 × 10?³

75. For the reaction A → products.
The concentration of A at 10 minutes is ______ × 10?³ mol L?¹ (nearest integer).
The reaction was started with 2.5 mol L?¹ of A.

Ans. (2435)
Sol. t?/? ∝ [A]? ⇒ Order = zero
t?/? = A?/2K ⇒ Slope = 1/2K = 76.92
K = 1/(2×76.92)
[A] = −Kt + A? = −(1/(2×76.92))×10 + 2.5 = 2.435
= 2435 × 10?³ mol/L

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