MATHEMATICS

SECTION-A

1. If the domain of the function

f(x)=cos?−1(2x−511−3x)+sin?−1(2x2−3x+1)f(x)=cos−1(11−3x2x−5?)+sin−1(2x2−3x+1)

is the interval [α,β][α,β], then α+2βα+2β is equal to :
(1) 1
(2) 3
(3) 5
(4) 2

Ans. (2)

Sol. f(x)=cos?−1(2x−511−3x)+sin?−1(2x2−3x+1)f(x)=cos−1(11−3x2x−5?)+sin−1(2x2−3x+1)

−1≤2x−511−3x≤1−1≤11−3x2x−5?≤1−1≤2x2−3x+1≤1−1≤2x2−3x+1≤12x2−3x+2≥0,2x2−3x≤02x2−3x+2≥0,2x2−3x≤0x∈[0,32]……(i)x∈[0,23?]……(i)2x−511−3x+1≥02x−511−3x−1≤011−3x2x−5?+1≥011−3x2x−5?−1≤02x−5+11−3x11−3x≥05x−1611−3x≤011−3x2x−5+11−3x?≥011−3x5x−16?≤06−x11−3x≥011−3x6−x?≥0x∈(−∞,165]∪(113,∞)x∈(−∞,516?]∪(311?,∞)x∈(−∞,113)∪[6,∞)x∈(−∞,311?)∪[6,∞)

intersection

x∈(−∞,165]∪[6,∞)…(i)x∈(−∞,516?]∪[6,∞)…(i)

Intersection of (i) & (ii) x∈[0,32]x∈[0,23?]

α=0,β=32⇒α+2β=3α=0,β=23?⇒α+2β=3

JEE Previous Year Question Paper

2. The area of the region, inside the ellipse x2+4y2=4x2+4y2=4 and outside the region bounded by the curves y=?x?−1y=?x?−1 and y=1−?x?y=1−?x?, is :
(1) 2(π−1)2(π−1)
(2) 2π−122π−21?
(3) 3(π−1)3(π−1)
(4) 2π−12π−1

Ans. (1)

Sol. Required area = area of ellipse - shaded area
=π×2×1−4(12×1×1)=2π−2=π×2×1−4(21?×1×1)=2π−2

3. The number of relations, defined on the set {a,b,c,d}, which are both reflexive and symmetric, is equal to:
(1) 256
(2) 16
(3) 1024
(4) 64

Ans. (4)

Sol. Number of relation which are reflexive and symmetric both
=14×26=64=14×26=64
(a,a) (a,b) (a,c) (a,d)
(b,a) (b,b) (b,c) (b,d)
(c,a) (c,b) (c,c) (c,d)
(d,a) (d,b) (d,c) (d,d)

4. Let a point A lie between the parallel lines L1L1? and L2L2? such that its distances from L1L1? and L2L2? are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines L1L1? and L2L2? respectively, is :
(1) 156156?
(2) 27
(3) 213213?
(4) 122122?

Ans. (3)

Sol.

sin?θ=3asinθ=a3?sin?(60?+θ)=9asin(60?+θ)=a9?32cos?θ+12sin?θ=9a23??cosθ+21?sinθ=a9?31−9a2+3a=18a3?1−a29??+a3?=a18?

a=84a=84?
Area of ΔABC=34a2=34×84=213ΔABC=43??a2=43??×84=213?

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5. Let a?=−i^+2j^+2k^,b?=8i^+7j^−3k^a=−i^+2j^?+2k^,b=8i^+7j^?−3k^ and c?c be a vector such that a?×c?=b?a×c=b. If c?⋅(i^+j^+k^)=4c⋅(i^+j^?+k^)=4, then ?a?+c??2?a+c?2 is equal to :
(1) 33
(2) 30
(3) 35
(4) 27

Ans. (4)

Sol. a?=−i^+2j^+2k^a=−i^+2j^?+2k^
b?=8i^+7j^−3k^b=8i^+7j^?−3k^
c?=c1i^+c2j^+c3k^c=c1?i^+c2?j^?+c3?k^
a?×c?=b?⇒(2c3−2c2)i^+(c1+2c3)j^−(c2+2c1)k^=8i^+7j^−3k^a×c=b⇒(2c3?−2c2?)i^+(c1?+2c3?)j^?−(c2?+2c1?)k^=8i^+7j^?−3k^
2c3−2c2=8,c1+2c3=7,c2+2c1=32c3?−2c2?=8,c1?+2c3?=7,c2?+2c1?=3
(c1i^+c2j^+c3k^)⋅(i^+j^+k^)=4(c1?i^+c2?j^?+c3?k^)⋅(i^+j^?+k^)=4
⇒c1+c2+c3=4,c1=2,c2=−1,c3=3⇒c1?+c2?+c3?=4,c1?=2,c2?=−1,c3?=3
?a?+c??2=?i^+j^+5k^?2=27?a+c?2=?i^+j^?+5k^?2=27

6. Let a1,a2,a3,…a1?,a2?,a3?,… be a G.P. of increasing positive terms such that a3a5a7a9=64a3?a5?a7?a9?=64 and a1+a3+a5=8137a1?+a3?+a5?=7813?. Then a7+a9+a11a7?+a9?+a11? is equal to :
(1) 3256
(2) 3252
(3) 3244
(4) 3248

Ans. (2)

Sol. ar,ar3,ar5,ar7=64ar,ar3,ar5,ar7=64
a4r16=64⇒ar4=4a4r16=64⇒ar4=4
a+ar2+ar4=8137a+ar2+ar4=7813?
r2=28r2=28
ar6+ar8+ar10=?ar6+ar8+ar10=?
ar6(1+r2+r4)=4(1+28+784)=3252ar6(1+r2+r4)=4(1+28+784)=3252

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7. Let c?c and d?d be vectors such that ?c?+d??=29?c+d?=29? and c?×(2i^+3j^+4k^)=(2i^+3j^+4k^)×d?c×(2i^+3j^?+4k^)=(2i^+3j^?+4k^)×d. If λ1,λ2(λ1>λ2)λ1?,λ2?(λ1?>λ2?) are the possible values of (c?+d?)⋅(−7i^+2j^+3k^)(c+d)⋅(−7i^+2j^?+3k^), then the equation k2x2+(k2−5k+λ1)xy+(3k+λ22)y2−8x+12y+λ2=0k2x2+(k2−5k+λ1?)xy+(3k+2λ2??)y2−8x+12y+λ2?=0 represents a circle, for k equal to :
(1) 4
(2) 1
(3) -1
(4) 2

Ans. (2)

Sol. ?c?+d??=29?c+d?=29?
c?+d?=λ(2i^+3j^+4k^)c+d=λ(2i^+3j^?+4k^)
λ=±1λ=±1
λ(−14+6+12)=4λ,λ1=4,λ2=−4λ(−14+6+12)=4λ,λ1?=4,λ2?=−4
k2x2+(k2−5k+4)xy+(3k−2)y2−8x+12y−4=0k2x2+(k2−5k+4)xy+(3k−2)y2−8x+12y−4=0 is circle
k2−5k+4=0⇒k=1,4k2−5k+4=0⇒k=1,4
k2=3k−2⇒k=1,2k2=3k−2⇒k=1,2
k=1k=1

8. Let y=y(x)y=y(x) be the solution curve of the differential equation (1+x2)dy+(y−tan?−1x)dx=0(1+x2)dy+(y−tan−1x)dx=0, y(0)=1y(0)=1. Then the value of y(1)y(1) is :
(1) 2e4+π4−1e42?+4π?−1
(2) 2e4−π4−1e42?−4π?−1
(3) 4e4+π2−1e44?+2π?−1
(4) 4e4−π2−1e44?−2π?−1

Ans. (1)

Sol. dydx+yx2+1=tan?−1xx2+1dxdy?+x2+1y?=x2+1tan−1x?
I.F. =etan?−1x=etan−1x
y×etan?−1x=∫etan?−1xtan?−1x1+x2dxy×etan−1x=∫etan−1x1+x2tan−1x?dx
y×etan?−1x=tan?−1x(etan?−1x)−etan?−1x+cy×etan−1x=tan−1x(etan−1x)−etan−1x+c
y(0)=1⇒c=2y(0)=1⇒c=2
y(1)=2e3/4+π4−1y(1)=e3/42?+4π?−1

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9. The number of strictly increasing functions f from the set {1,2,3,4,5,6}{1,2,3,4,5,6} to the set {1,2,3,…,9}{1,2,3,…,9} such that f(i)≠if(i)?=i for 1≤i≤61≤i≤6 is equal to :
(1) 21
(2) 27
(3) 22
(4) 28

Ans. (4)

Sol. f(i)≠if(i)?=i
f(x)f(x) is strictly increasing function
f:A→Bf:A→B, where A {1,2,3,…,6}{1,2,3,…,6}
B {1,2,3,…,9}{1,2,3,…,9} then number of function
f:A→B is equal to
Case-i f(1)=2⇒5C5=21f(1)=2⇒5C5?=21
Case-ii f(1)=3⇒5C4=6f(1)=3⇒5C4?=6
Case-iii f(1)=4⇒5C3=1f(1)=4⇒5C3?=1
No of function A to B = 21 + 6 + 1 = 28

10. Let f:R→(0,∞)f:R→(0,∞) be a twice differentiable function such that f(3)=18f(3)=18, f′(3)=0f′(3)=0 and f′′(3)=4f′′(3)=4. Then lim?x→1(log?e(f(2+x)f(3))18f(3))limx→1?(loge?(f(3)f(2+x)?)f(3)18?) is equal to :
(1) 1
(2) 9
(3) 2
(4) 18

Ans. (3)

Sol. Let T=lim?x→1(f(x+2)f(3))18f(3);1?T=limx→1?(f(3)f(x+2)?)f(3)18?;1? form
⇒T=e18f(3)ln?(f(x+2)f(3))⇒T=ef(3)18?ln(f(3)f(x+2)?)
⇒T=e18f(3)(f(x+2)−f(3)f(3))⇒T=ef(3)18?(f(3)f(x+2)−f(3)?) form
⇒T=e18f(3)⋅f(x+2)−f(3)x−1⋅x−1f(3)⇒T=ef(3)18?⋅x−1f(x+2)−f(3)?⋅f(3)x−1?
⇒T=e1818⋅f′(3)⋅1=e2⇒T=e1818?⋅f′(3)⋅1=e2
⇒log?e(T)=2⇒loge?(T)=2

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11. Let the foci of hyperbola coincide with the foci of the ellipse x236+y216=136x2?+16y2?=1. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is:
(1) 12
(2) 16
(3) 9655?96?
(4) 245245?

Ans. (3)

Sol. Let e1e1? be eccentricity of ellipse
⇒e1=1−1636=1−49=53⇒e1?=1−3616??=1−94??=35??
So ae1=6⋅53=25ae1?=6⋅35??=25?
Now H:x2p2−y2q2=1H:p2x2?−q2y2?=1
p⋅e=ae1p⋅e=ae1?
p=25⇒e2=1+q2p2⇒25=1+5q24⇒q2=965p=5?2?⇒e2=1+p2q2?⇒25=1+45q2?⇒q2=596?
So length of LR 2q2p=965p2q2?=5?96?

12. The value of ∫−π/6π/6(π+4x111−sin?(?x?+π/6))dx∫−π/6π/6?(1−sin(?x?+π/6)π+4x11?)dx is equal to
(1) 2π2π
(2) 4π4π
(3) 8π8π
(4) 6π6π

Ans. (2)

Sol. =2π∫0π/611−sin?(x+π6)dx=2π∫0π/6?1−sin(x+6π?)1?dx let x+π6=tx+6π?=t, dx=dtdx=dt
=2π∫π/6π/3dt1−sin?t=2π∫π/6π/31+sin?tcos?2tdt=2π∫π/6π/3?1−sintdt?=2π∫π/6π/3?cos2t1+sint?dt
=2π[∫π/6π/3sec?2tdt+∫π/6π/3sec?ttan?tdt]=2π[∫π/6π/3?sec2tdt+∫π/6π/3?secttantdt]
=2π[(tan?t)π/6π/3+(sec?t)π/6π/3]=2π[(tant)π/6π/3?+(sect)π/6π/3?]
=2π[(3−13)+(2−23)]=2π[(3?−3?1?)+(2−3?2?)]
=2π[3+2−3]=4π=2π[3?+2−3?]=4π

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13. Let the mean and variance of 7 observations 2,4,10,x,12,14,y, x>y, be 8 and 16 respectively. Two numbers are chosen from {1,2,3,x-4,y,5} one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is:
(1) 3553?
(2) 4554?
(3) 2552?
(4) 1331?

Ans. (2)

Sol. Mean (x?)=8(x?)=8 (Given)
⇒2+4+10+x+12+14+y7=8⇒72+4+10+x+12+14+y?=8
⇒x+y=14…(1)⇒x+y=14…(1)
Variance (σ2)=16(σ2)=16 (Given)
⇒16=22+42+102+x2+122+142+y27−82⇒16=722+42+102+x2+122+142+y2?−82
⇒x2+y2=100…(2)⇒x2+y2=100…(2)
?(x+y)2=x2+y2+2xy?(x+y)2=x2+y2+2xy
⇒xy=48⇒xy=48
Since problem states x>yx>y
∴x=8∴x=8 and y=6y=6
Now set X={1,2,3,4,6,5}X={1,2,3,4,6,5}
Now we choose two numbers one after another without replacement total outcomes =6×5=30=6×5=30
We want the prob. that the smaller number among the two is less than 4
P(smaller<4)=1−P(smaller≥4)P(smaller<4)=1−P(smaller≥4)
=1−630=45=1−306?=54?

14. Let (α,β,γ)(α,β,γ) be the co-ordinates of the foot of the perpendicular drawn from the point (5,4,2) on the line r?=(−i^+3j^+k^)+λ(2i^+3j^−k^)r=(−i^+3j^?+k^)+λ(2i^+3j^?−k^). Then the length of the projection of the vector αi^+βj^+γk^αi^+βj^?+γk^ on the vector 6i^+2j^+3k^6i^+2j^?+3k^ is :
(1) 157715?
(2) 167716?
(3) 187718?
(4) 3

Ans. (3)

Sol. r?=(−i^+3j^+k^)+λ(2i^+3j^−k^)r=(−i^+3j^?+k^)+λ(2i^+3j^?−k^)
x+12=y−33=z−1−1=λ2x+1?=3y−3?=−1z−1?=λ
Any general point P on the line is
(2λ−1,3λ+3,−λ+1)(2λ−1,3λ+3,−λ+1)
Let the given point is A (5,4,2)
AP→=(2λ−6)i^+(3λ−1)j^+(−λ−1)k^AP=(2λ−6)i^+(3λ−1)j^?+(−λ−1)k^
∴AP→⊥∴AP⊥ Line (L)
∴AP→⋅(2i^+3j^−k^)=0∴AP⋅(2i^+3j^?−k^)=0
2(2λ−6)+3(3λ−1)−1(−λ−1)=02(2λ−6)+3(3λ−1)−1(−λ−1)=0
⇒λ=1⇒λ=1
∴α=1,β=6,γ=0∴α=1,β=6,γ=0
Let the vector u?=αi^+βj^+γk^u=αi^+βj^?+γk^
u?=i^+6j^+0k^u=i^+6j^?+0k^
& w?=6i^+2j^+3k^w=6i^+2j^?+3k^
So projection =?u?⋅w???w??=187=?w??u⋅w??=718?

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15. Let PQ and MN be two straight lines touching the circle x2+y2−4x−6y−3=0x2+y2−4x−6y−3=0 at the points A and B respectively. Let O be the centre of the circle and ∠AOB=π/3∠AOB=π/3. Then the locus of the point of intersection of the lines PQ and MN is:
(1) 3(x2+y2)−18x−12y+25=03(x2+y2)−18x−12y+25=0
(2) x2+y2−12x−18y−25=0x2+y2−12x−18y−25=0
(3) x2+y2−18x−12y−25=0x2+y2−18x−12y−25=0
(4) 3(x2+y2)−12x−18y−25=03(x2+y2)−12x−18y−25=0

Ans. (4)

Sol. Given circle
x2+y2−4x−6y−3=0x2+y2−4x−6y−3=0
C(2,3)C(2,3) & r=4r=4
cos?30?=rOR=4ORcos30?=ORr?=OR4?
⇒OR=83⇒OR=3?8?
Now
OR2=(h−2)2+(k−3)2OR2=(h−2)2+(k−3)2
⇒3(x2+y2)−12x−18y−25=0⇒3(x2+y2)−12x−18y−25=0

16. If the coefficient of x in the expansion of (ax2+bx+c)(1−2x)26(ax2+bx+c)(1−2x)26 is −56−56 and the coefficients of x2x2 and x3x3 are both zero, then a+b+ca+b+c is equal to
(1) 1300
(2) 1500
(3) 1403
(4) 1483

Ans. (3)

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PHYSICS

SECTION-A

26. Potential energy (V) versus distance (x) is given by the graph. Rank various regions as per the magnitudes of the force (F) acting on a particle from high to low.
(1) FBC>FCD>FDE>FABFBC?>FCD?>FDE?>FAB?
(2) FCD>FAB>FBC>FDEFCD?>FAB?>FBC?>FDE?
(3) FCD>FDE>FAB>FBCFCD?>FDE?>FAB?>FBC?
(4) FBC>FAB>FDE>FCDFBC?>FAB?>FDE?>FCD?

Ans. (4)

Sol. Slope of potential energy v/s position curve gives negative of force.
∴FBC>FAB>FDE>FCD∴FBC?>FAB?>FDE?>FCD?
Correct option (4)

27. A gas based geyser heats water flowing at the rate of 5.0 litres per minute from 27?C27?C to 87?C87?C. The rate of consumption of the gas is ____ g/s. (Take heat of combustion of gas = 5.0×1045.0×104 J/g; specific heat capacity of water = 4200 J/kg.°C)
(1) 2.1
(2) 4.2
(3) 0.42
(4) 0.21

Ans. (3)

Sol. Water flow rate = 5 ?/min = 560605? kg/s
∴∴ Power of heater = dmdtSΔT=112×4200×60Wdtdm?SΔT=121?×4200×60W
∴∴ Let rate of consumption of gas be x g/s.
∴x×5.0×104=112×4200×60∴x×5.0×104=121?×4200×60
⇒x=4200×10−4=0.42⇒x=4200×10−4=0.42 g/s
Correct option (3)

28. A conducting circular loop of area 1.0 m² is placed perpendicular to a magnetic field which varies as B = sin(100 t) Tesla. If the resistance of the loop is 100 Ω, then the average thermal energy dissipated in the loop in one period is ____ J.
(1) π22π?
(2) 2π2π
(3) ππ
(4) π2π2

Ans. (3)

Sol. Area of the loop = 1 m²
B = sin(100 t)
∴?=BA=sin?(100t)∴?=BA=sin(100t)
∴d?dt=100cos?(100t)∴dtd??=100cos(100t)
∴P=V2R=104cos?2(100t)100∴P=RV2?=100104cos2(100t)?
∴∴ Thermal energy dissipated in 1 time period
=∫0TPdt=∫0T100cos?2(100t)dt=∫0T?Pdt=∫0T?100cos2(100t)dt
T=2π100=π50sec?T=1002π?=50π?sec
∴Q=100∫0π/50cos?2(100t)dt∴Q=100∫0π/50?cos2(100t)dt
=100∫0π/501+cos?200t2dt=100∫0π/50?21+cos200t?dt
=100[π100]=π=100[100π?]=π
Correct option (3)

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29. Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at A and B are 6 cm² and 3 cm² respectively. The rate of flow will be ____ cm³/s. (take g = 10 m/s²)
(1) 20033?200?
(2) 20062006?
(3) 20032003?
(4) 10031003?

Ans. (3)

Sol. From continuity equation
AAVA=ABVB⇒6VA=3VB⇒VB=2VAAA?VA?=AB?VB?⇒6VA?=3VB?⇒VB?=2VA?
Applying Bernoulli's equation between A & B,
PA+12ρVA2=PB+12ρVB2PA?+21?ρVA2?=PB?+21?ρVB2?
⇒ρg×0.05=12ρ[VB2−VA2]=12ρ(3VA2)⇒ρg×0.05=21?ρ[VB2?−VA2?]=21?ρ(3VA2?)
⇒VA=2g×0.053m/s=13m/s=1003cm/s⇒VA?=32g×0.05??m/s=3?1?m/s=3?100?cm/s
⇒⇒ Volume flow rate = AAVA=6×1003cm3/sec=2003cm3/secAA?VA?=3?6×100?cm3/sec=2003?cm3/sec
Correct option (3)

30. In an experiment the values of two spring constants were measured as k1=(10±0.2)k1?=(10±0.2) N/m and k2=(20±0.3)k2?=(20±0.3) N/m. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :
(1) 2.67%
(2) 2.33%
(3) 1.33%
(4) 1.67%

Ans. (4)

Sol. For parallel combination of spring,
Keq=K1+K2=30Keq?=K1?+K2?=30 N/m
ΔKeq=ΔK1+ΔK2=0.2+0.3=0.5ΔKeq?=ΔK1?+ΔK2?=0.2+0.3=0.5 N/m
∴%Error in K=0.530×100=1.67%∴%Error in K=300.5?×100=1.67%
Correct option (4)

31. A 4 kg mass moves under the influence of a force F?=(4t2i^−3tj^)F=(4t2i^−3tj^?) N where t is the time in second. If mass starts from origin at t = 0, the velocity and position after t = 2s will be :
(1) v?=3i^+32j^,r?=65i^+j^v=3i^+23?j^?,r=56?i^+j^?
(2) v?=4i^−32j^,r?=85i^−j^v=4i^−23?j^?,r=58?i^−j^?
(3) v?=4i^+52j^,r?=85i^+2j^v=4i^+25?j^?,r=58?i^+2j^?
(4) v?=4i^−32j^,r?=65i^−j^v=4i^−23?j^?,r=56?i^−j^?

Ans. (2)

Sol. F?=4t2i^−3tj^F=4t2i^−3tj^?
a?=F?m=t2i^−34tj^a=mF?=t2i^−43?tj^?
ax=t2,ay=−34tax?=t2,ay?=−43?t
dvxdt=t2,dvydt=−34tdtdvx??=t2,dtdvy??=−43?t
∫0vxdvx=∫0tt2dt,∫0vydvy=∫0t−34tdt∫0vx??dvx?=∫0t?t2dt,∫0vy??dvy?=∫0t?−43?tdt
vx=t33,vy=−38t2vx?=3t3?,vy?=−83?t2
v?2=4i^−32j^v2?=4i^−23?j^?
x2=85,y2=−1x2?=58?,y2?=−1
r?=85i^−j^r=58?i^−j^?
Correct option (2)

JEE Previous Year Question Paper

32. Consider a modified Bernoulli equation. (P+ABt2)+ρg(h+Bt)+12ρV2=constant(P+Bt2A?)+ρg(h+Bt)+21?ρV2=constant If t has the dimension of time then the dimensions of A and B are ____, ____ respectively.
(1) [ML−1T−1][ML−1T−1] and [M0LT1][M0LT1]
(2) [ML0T−1][ML0T−1] and [M0LT−1][M0LT−1]
(3) [ML0T−2][ML0T−2] and [M0LT−1][M0LT−1]
(4) [ML0T−3][ML0T−3] and [M0LT−1][M0LT−1]

Ans. (2)

Sol. ⇒[P]=[ABt2]…(1)⇒[P]=[Bt2A?]…(1)
⇒[h]=[Bt]…(2)⇒[h]=[Bt]…(2)
⇒[B]=[ht]=[LT]=[LT−1]⇒[B]=[th?]=[TL?]=[LT−1]
Putting B in equation (1)
[ML−1T−2]=[ALT−1×T2][ML−1T−2]=[LT−1×T2A?]
[A]=[ML0T−1][A]=[ML0T−1]
Correct option (2)

33. A current carrying is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy :
(1) a=ga=g
(2) a>ga>g
(3) a=0a=0
(4) 0<a<g0<a<g

Ans. (1)

Sol. Since the solenoid is placed vertically, the magnetic field inside the solenoid will be either along –y or +y axis.
⇒⇒ Particle will gain velocity along –y axis.
⇒F?B=q(v?×B?)⇒FB?=q(v×B)
⇒F?B=0⇒FB?=0
⇒F?net=mg?⇒Fnet?=mg?
⇒anet=g⇒anet?=g
Correct option (1)

JEE Previous Year Question Paper

34. A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness (13)rd(31?)rd of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :
(1) 3KC2K+12K+13KC?
(2) CK2+K2+KCK?
(3) 3CK2(2K+1)2(2K+1)23CK2?
(4) 4KC3K−13K−14KC?

Ans. (1)

Sol. C=A?0dC=dA?0??
C1=3A?02d=32CC1?=2d3A?0??=23?C
C2=3A?0×Kd=3KCC2?=d3A?0?×K?=3KC
Ceq=C1C2C1+C2=32C×3KC32C+3KCCeq?=C1?+C2?C1?C2??=23?C+3KC23?C×3KC?
Ceq=92KC232C(2K+1)=3KC2K+1Ceq?=23?C(2K+1)29?KC2?=2K+13KC?
Correct option (1)

35. The electric field a plane electromagnetic wave is given by : Ey=69sin?[0.6×103x−1.8×1011t]V/mEy?=69sin[0.6×103x−1.8×1011t]V/m. The expression for magnetic field associated with this electromagnetic wave is ____ T.
(1) Bz=2.3×10−7sin?[0.6×103x−1.8×1011t]Bz?=2.3×10−7sin[0.6×103x−1.8×1011t]
(2) Bz=2.3×10−7sin?[0.6×103x+1.8×1011t]Bz?=2.3×10−7sin[0.6×103x+1.8×1011t]
(3) Bz=69sin?[0.6×103x+1.8×1011t]Bz?=69sin[0.6×103x+1.8×1011t]
(4) By=2.3×10−7sin?[0.6×103x−1.8×1011t]By?=2.3×10−7sin[0.6×103x−1.8×1011t]

Ans. (1)

Sol. B^=c^×E^B^=c^×E^
⇒c^=i^⇒c^=i^ because phase of electric field is function of x.
⇒E?=j^⇒E=j^? (given)
⇒B^=i^×j^=k^⇒B^=i^×j^?=k^
?B?=?E?c=69×0.6×1031.8×1011=693×108?B?=c?E??=1.8×101169×0.6×103?=3×10869?
?B?=2.9×10−7?B?=2.9×10−7
B?z=2.9×10−7sin?(0.6×103x−1.8×1011t)Bz?=2.9×10−7sin(0.6×103x−1.8×1011t)
(phase is same as that of electric field)
Correct option (1)

JEE Previous Year Question Paper

36. In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index μ=1.5μ=1.5, the central fringe shifts by 0.2 cm. The value of t is ____ cm.
(1) 8×10−48×10−4
(2) 6.0×10−36.0×10−3
(3) 5.6×10−45.6×10−4
(4) 5.0×10−35.0×10−3

Ans. (1)

Sol. dsin?θ=(μ−1)tdsinθ=(μ−1)t
d[xD]=(μ−1)td[Dx?]=(μ−1)t
t=xdD(μ−1)=(0.2)(0.1)50(1.5−1)t=D(μ−1)xd?=50(1.5−1)(0.2)(0.1)?
t=8×10−4cmt=8×10−4cm
Correct option (1)

37. A light wave described by E=60sin?(3×1015t)+sin?(12×1015t)E=60sin(3×1015t)+sin(12×1015t) (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) ____ eV. (h = 6.6×10−346.6×10−34 J-s and e = 1.6×10−191.6×10−19 C)
(1) 5.1
(2) 3.8
(3) 6.0
(4) 7.8

Ans. (1)

Sol. ω1=3×1015ω1?=3×1015 rad/sec
ω2=12×1015ω2?=12×1015 rad/sec
∴ν=ω2π∴ν=2πω?
Ephoton=hν=6.6×10−34×1.91×1015Ephoton?=hν=6.6×10−34×1.91×1015
=1.26×10−18J=1.26×10−18J
Emax=1.26×10−181.6×10−19=7.9eVEmax?=1.6×10−191.26×10−18?=7.9eV
Kmax=Emax−?0Kmax?=Emax?−?0?
=7.9−2.8=7.9−2.8
Kmax=5.1eVKmax?=5.1eV
Correct option (1)

38. If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and 14π?0=9×1094π?0?1?=9×109 in SI units)
(1) 2.95×10−142.95×10−14
(2) 2.95×10−162.95×10−16
(3) 3.85×10−163.85×10−16
(4) 3.85×10−143.85×10−14

Ans. (1)

Sol. Energy conservation
Ki+Ui=Kf+UfKi?+Ui?=Kf?+Uf?
7.7×106×1.6×10−19+0=0+9×109(1.6×10−19)(79×1.6×10−19)r7.7×106×1.6×10−19+0=0+r9×109(1.6×10−19)(79×1.6×10−19)?
r=2.95×10−14r=2.95×10−14
Correct option (1)

JEE Previous Year Question Paper

39. A uniform rod of mass m and length l suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is ____. (g acceleration due to gravity)
(1) mg/2mg/2
(2) mg/4mg/4
(3) mg/3mg/3
(4) mgmg

Ans. (2)

Sol. mgl2=ml23αmg2l?=3ml2?α
α=3g2l…(1)α=2l3g?…(1)
mg−T=macmg−T=mac?
T=mg−mac=mg−m(l2α)T=mg−mac?=mg−m(2l?α)
=mg−m(l2⋅3g2l)=mg−m(2l?⋅2l3g?)
T=mg4T=4mg?
Correct option (2)

40. An aluminium and steel rods having same lengths and cross-sections are joined to make total length of 120 cm at 30?C30?C. The coefficient of linear expansion of aluminium and steel are 24×10−6/?C24×10−6/?C and 1.2×10−6/?C1.2×10−6/?C respectively. The length of this composite rod when its temperature is raised to 100?C100?C is ____ cm.
(1) 120.20
(2) 120.15
(3) 120.03
(4) 120.06

Ans. (2)

Sol. ?final=?s(1+αsΔT)+?s(1+αnΔT)?final?=?s?(1+αs?ΔT)+?s?(1+αn?ΔT)
=?s[2+(αs+αn)ΔT]=?s?[2+(αs?+αn?)ΔT]
=60[2+(36×10−6)×70]=60[2+(36×10−6)×70]
=60[2+0.0025]=60[2+0.0025]
=120.15cm=120.15cm
Correct option (2)

41. A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2 Ω then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is ____ N.
(1) 7.5×10−27.5×10−2
(2) 5.7×10−35.7×10−3
(3) 5.7×10−25.7×10−2
(4) 7.5×10−37.5×10−3

Ans. (4)

Sol. To maintain constant speed
Fext=FB=iLBFext?=FB?=iLB
Fext=B2L2vRFext?=RB2L2v?
=(0.1)2(1)2(1.5)2=7.5×10−3N=2(0.1)2(1)2(1.5)?=7.5×10−3N
Correct option (4)

42. The given circuit works as :
(1) AND gate
(2) NOR gate
(3) NAND gate
(4) OR gate

Ans. (3)

JEE Previous Year Question Paper

43. Two strings (A,B) having linear densities μA=2×10−4μA?=2×10−4 kg/m and μB=4×10−4μB?=4×10−4 kg/m and lengths LA=2.5LA?=2.5 m and LB=1.5LB?=1.5 m respectively are joined. Free ends of A and B are tied to two rigid supports C and D, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from C and D ends, take time t1t1? and t2t2? respectively, to reach the joint. The ratio t1/t2t1?/t2? is :
(1) 1.08
(2) 1.90
(3) 1.67
(4) 1.18

Ans. (4)

Sol. Given LA=2.5mLA?=2.5m, LB=1.5mLB?=1.5m, T=500NT=500N
vA=TμA=5002×10−4=510×102m/svA?=μA?T??=2×10−4500??=510?×102m/s
vB=TμB=5004×10−4=55×102m/svB?=μB?T??=4×10−4500??=55?×102m/s
t1=LAvA=2.5510×10−2st1?=vA?LA??=510?2.5?×10−2s
t2=LBvB=1.555×10−2st2?=vB?LB??=55?1.5?×10−2s
∴t1t2=2.5510×551.5=53×12=1.661.41=1.18∴t2?t1??=510?2.5?×1.555??=35?×2?1?=1.411.66?=1.18
Correct Option (4)

44. Initially a satellite of 100 kg is in a circular orbit of radius 1.5RE1.5RE?. This satellite can be moved to a circular orbit of radius 3RE3RE? by supplying α×105Jα×105J of energy. The value of αα is ____. (Take Radius of Earth RE=6×106mRE?=6×106m and g=10m/s2g=10m/s2)
(1) 150
(2) 500
(3) 100
(4) 1000

Ans. (4)

Sol. Energy of a satellite in a circular orbit is given as
E=−GMEm2r;r=E=2r−GME?m?;r= radius of circular orbit
Required energy to be supplied =ΔE=Ef−Ei=ΔE=Ef?−Ei?
ΔE=(−GMEm2(3RE))−(−GMEm2(1.5RE))ΔE=(2(3RE?)−GME?m?)−(2(1.5RE?)−GME?m?)
=GMEm6RE=6RE?GME?m?
Now, g=GMERE2⇒GMERE=gREg=RE2?GME??⇒RE?GME??=gRE?
∴ΔE=16gmRE∴ΔE=61?gmRE?
=16×10×100×6×106=61?×10×100×6×106
=1000×106=1000×106
α=1000α=1000
Correct option (4)

JEE Previous Year Question Paper

45. A point charge of 10−8C10−8C is placed at origin. The work done in moving a point charge 2μC2μC from point A(4,4,2) m to B(2,2,1) m is ____ J. (14π?0=9×105 in SI units)(4π?0?1?=9×105 in SI units)
(1) 45×10−645×10−6
(2) 0
(3) 30×10−630×10−6
(4) 15×10−615×10−6

Ans. (3)

Sol. Work done by external agent :
Wext=ΔUWext?=ΔU
ΔU→ΔU→ Change in potential energy in taking the charge from initial to final configuration
⇒Wext=14π?0q1q2ri−14π?0q1q2rf⇒Wext?=4π?0?1?ri?q1?q2??−4π?0?1?rf?q1?q2??
Now, rf=(2−0)2+(2−0)2+(1−0)2=3mrf?=(2−0)2+(2−0)2+(1−0)2?=3m
ri=(4−0)2+(4−0)2+(2−0)2=6mri?=(4−0)2+(4−0)2+(2−0)2?=6m
Wext=(9×109)×(10−8×2×10−6)[13−16]Wext?=(9×109)×(10−8×2×10−6)[31?−61?]
=3×10−5=30×10−6J=3×10−5=30×10−6J
Correct Option (3)

SECTION-B

46. A collimated beam of light of diameter 2 mm is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm, then the focal length of second lens is ____ mm.

Ans. (280)

Sol. 402=f14240?=14f?
⇒f=280mm⇒f=280mm
Correct Answer : 280

47. The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9V with internal resistance of 1 Ω is connected across these points is ____ J.

Ans. (1080)

Sol. Balanced Wheatstone bridge
i=93=3Ai=39?=3A
HAB=i2RABt=(3)2×2×60=1080JHAB?=i2RAB?t=(3)2×2×60=1080J
Correct Answer : 1080

48. Two identical thin rods of mass M kg and length L m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is x2ML2kgm22x?ML2kgm2. The value of x is ____.

Ans. (17)

Sol. I=ML23+(ML212+ML2)I=3ML2?+(12ML2?+ML2)
=4ML2+ML2+12ML212=124ML2+ML2+12ML2?
I=1712ML2I=1217?ML2
∴x=17∴x=17
Correct Answer : 17

49. 10 mole of oxygen is heated at constant volume from 30?C30?C to 40?C40?C. The change in the internal energy of the gas is ____ cal. (The molecular specific heat of oxygen at constant pressure, Cp=7Cp?=7 cal./mol °C and R=2R=2 cal./mol °C.)

Ans. (500)

Sol. ΔU=nCvΔTΔU=nCv?ΔT
=n(Cp−R)ΔT=n(Cp?−R)ΔT
=10(7−2)(40−30)=10(7−2)(40−30)
ΔU=500ΔU=500
Correct Answer: 500

50. In a microscope the objective is having focal length f0=2f0?=2 cm and eye-piece is having focal length fe=4fe?=4 cm. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ____.

Ans. (100)

Sol. m=Df0fem=f0?fe?D?
=322×254=232?×425?
m=100m=100
Correct Answer: 100

JEE Previous Year Question Paper

CHEMISTRY

SECTION-A

51. Consider the following reactions.
PbCl2+K2CrO4→A+2KClPbCl2?+K2?CrO4?→A+2KCl (Hot solution)
A+NaOH?B+Na2CrO4A+NaOH?B+Na2?CrO4?
PbSO4+4CH3COONH4→(NH4)2SO4+XPbSO4?+4CH3?COONH4?→(NH4?)2?SO4?+X
In the above reactions, A, B and X are respectively.
(1) Na2[Pb(OH)4],PbCrO4Na2?[Pb(OH)4?],PbCrO4? and (NH4)2[Pb(CH3COO)4](NH4?)2?[Pb(CH3?COO)4?]
(2) PbCrO4,Na2[Pb(OH)4]PbCrO4?,Na2?[Pb(OH)4?] and [Pb(NH3)4]SO4[Pb(NH3?)4?]SO4?
(3) Na2[Pb(OH)4],PbCrO4Na2?[Pb(OH)4?],PbCrO4? and [Pb(NH3)4]SO4[Pb(NH3?)4?]SO4?
(4) PbCrO4,Na2[Pb(OH)4]PbCrO4?,Na2?[Pb(OH)4?] and (NH4)2[Pb(CH3COO)4](NH4?)2?[Pb(CH3?COO)4?]

Ans. (4)

Sol. PbCl2+K2CrO4→PbCrO4+2KClPbCl2?+K2?CrO4?→PbCrO4?+2KCl (Hot solution)
PbCrO4+4NaOH(excess)→Na2[Pb(OH)4]+Na2CrO4PbCrO4?+4NaOH(excess)→Na2?[Pb(OH)4?]+Na2?CrO4?
PbSO4+4CH3COONH4→(NH4)2[Pb(CH3COO)4]+(NH4)2SO4PbSO4?+4CH3?COONH4?→(NH4?)2?[Pb(CH3?COO)4?]+(NH4?)2?SO4?

52. Which of the following represents the correct trend for the mentioned property?
A. F > P > S > B – First Ionization Energy
B. Cl > F > S > P – Electron Affinity
C. K > Al > Mg > B – Metallic character
D. K2O>Na2O>MgO>Al2O3K2?O>Na2?O>MgO>Al2?O3? – Basic character
Choose the correct answer from the option given below.
(1) A, B and D only
(2) A, B, C and D
(3) A and B only
(4) B and C only

Ans. (1)

Sol. ⇒⇒ On moving left to right in a period IE increases and from top to bottom in a group IE decreases.
F > P > S > B (IE order)
⇒⇒ On moving left to right in a period metallic and basic character decreases.
K > Mg > Al > B (Metallic character order)
⇒⇒ On moving top to bottom in a group metallic and basic character increases.
K2O>Na2O>MgO>Al2O3K2?O>Na2?O>MgO>Al2?O3?
⇒⇒ EA : Group 17 > Group 16 > Group 15
Cl > F > S > P

JEE Previous Year Question Paper

53. Identify A in the following reaction.
(1)
(2)
(3)
(4)

Ans. (3)

Sol. Naphthalene reacts with 2H2/Pt2H2?/Pt to give tetralin, which on oxidation with KMnO4KMnO4? gives phthalic acid.

54. A hydrocarbon ‘P’ (C4H8)(C4?H8?) on reaction with HCl gives an optically active compound ‘Q’ (C4H9Cl)(C4?H9?Cl) which on reaction with one mole of ammonia gives compound ‘R’ (C4H11N)(C4?H11?N). ‘R’ on diazotization followed by hydrolysis gives ‘S’. Identify P, Q, R and S.
(1) P=CH3−CH2−CH=CH2,Q=CH3−CH2−CH2−CH2ClP=CH3?−CH2?−CH=CH2?,Q=CH3?−CH2?−CH2?−CH2?Cl
R=CH3−CH2−CH2−CH2NH2,S=CH3−CH2−CH2−CH2OHR=CH3?−CH2?−CH2?−CH2?NH2?,S=CH3?−CH2?−CH2?−CH2?OH
(2) P=CH3−?,Q=Cl−CH2−?,R=H2N−CH2−?,S=?OHP=CH3?−?,Q=Cl−CH2?−?,R=H2?N−CH2?−?,S=?OH
(3) P=CH3−CH=CH−CH3,Q=CH3−CH2−CH(Cl)−CH3P=CH3?−CH=CH−CH3?,Q=CH3?−CH2?−CH(Cl)−CH3?
R=CH3−CH2−CH(NH2)−CH3,S=CH3−CH2−CH(OH)−CH3R=CH3?−CH2?−CH(NH2?)−CH3?,S=CH3?−CH2?−CH(OH)−CH3?
(4) P=CH3−CH=CH−CH3,Q=CH3−CH2−CH2−CH2ClP=CH3?−CH=CH−CH3?,Q=CH3?−CH2?−CH2?−CH2?Cl
R=CH3−CH2−CH2−CH2NH2,S=CH3−CH2−CH2−CH2OHR=CH3?−CH2?−CH2?−CH2?NH2?,S=CH3?−CH2?−CH2?−CH2?OH

Ans. (3)

Sol. But-2-ene reacts with HCl to give 2-chlorobutane (optically active), which with NH3NH3? gives butan-2-amine, and diazotization-hydrolysis gives butan-2-ol.

55. Given below are two statements :
Statement I : The number of pairs among [SiO3,CO2][SiO3?,CO2?], [SnO,SnO2][SnO,SnO2?], [PbO,PbO2][PbO,PbO2?] and [GeO,GeO2][GeO,GeO2?], which contain oxides that are both amphoteric is 2.
Statement II : BF3BF3? is an electron deficient molecule can act as a lewis acid, forms adduct with NH3NH3? and has a trigonal planar geometry.
In the light of the above statement, choose the correct answer from the option 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. (1)

Sol. ⇒⇒ SiO2,CO2,GeO,GeO2SiO2?,CO2?,GeO,GeO2? are acidic in nature.
SnO,SnO2,PbO,PbO2SnO,SnO2?,PbO,PbO2? are amphoteric in nature.
⇒⇒ BF3BF3? is lewis acid according to lewis octet theory and has sp2sp2 hybridization with trigonal planar geometry and it can accept lone pair form ammonia to form adduct.

JEE Previous Year Question Paper

56. 80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :
(1) C2H4C2?H4?
(2) C4H10C4?H10?
(3) C2H2C2?H2?
(4) C2H6C2?H6?

Ans. (3)

Sol. CxHy(g)+(x+y4)O2(g)→xCO2(g)+y2H2O(l)Cx?Hy?(g)+(x+4y?)O2?(g)→xCO2?(g)+2y?H2?O(l)
t=0 80 264 0 -
t=t_final - 264 - 80(x + y/4) 80x -
264−80(x+y4)+80x=224264−80(x+4y?)+80x=224
264−80y4=224264−480y?=224
40=80y4⇒y=240=480y?⇒y=2
264−80(x+y4)=64264−80(x+4y?)=64
264−80(x+12)=64264−80(x+21?)=64
264−80x−40=64264−80x−40=64
x=2x=2

57. 14.0 g of calcium metal is allowed to react with excess HCl at 1.0 atm pressure and 273 K. Which of the following statements is incorrect ? [Given : Molar mass in g mol?¹ of Ca-40, Cl-35.5, H-1]
(1) 0.35 mol of H2H2? gas is evolved.
(2) 7.84 L of H2H2? gas is evolved.
(3) 33.3 g of CaCl2CaCl2? is produced.
(4) The limiting reagent is calcium metal.

Ans. (3)

Sol. Ca(s)+2HCl(g)?CaCl2(s)+H2(g)Ca(s)+2HCl(g)?CaCl2?(s)+H2?(g)
0.35 mole 0.35 mole
Volume of H2(g)H2?(g) evolved =0.35×22.4=7.84L=0.35×22.4=7.84L
(3) is wrong because weight of CaCl2=0.35×111=38.85gmCaCl2?=0.35×111=38.85gm

58. In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass 32gmol−132gmol−1). Molar mass of barium sulphate is 233gmol−1233gmol−1
(1) 4.55%
(2) 10.30%
(3) 21.97%
(4) 16.48%

Ans. (3)

Sol. nBaSO4×32W(unknowncomp.)×100W(unknowncomp.)?nBaSO4??×32?×100
=1.2×32233×1000.75=21.97%=2331.2×32?×0.75100?=21.97%

JEE Previous Year Question Paper

59. Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ2PQ2?. When 1 g of PQ is dissolved in 50 g of solvent ‘A’, ΔTbΔTb? was 1.176 K while when 1 g of PQ2PQ2? is dissolved in 50 g of solvent ‘A’, ΔTbΔTb? was 0.689 K. KbKb? of ‘A’ = 5 K kg mol?¹. The molar masses of elements P and Q (in g mol?¹) respectively, are :
(1) 70, 110
(2) 65, 145
(3) 60, 25
(4) 25, 60

Ans. (4)

Sol. (ΔTb)PQ=Kbm(ΔTb?)PQ?=Kb?m
1.176=5×1M1×1000501.176=5×M1?1?×501000?
M1=85.03M1?=85.03
(ΔTb)PQ2=5×1M2×100050=0.689(ΔTb?)PQ2??=5×M2?1?×501000?=0.689
M2=145.13M2?=145.13
Let molar mass of P & Q are MpMp? and MQMQ? respectively
Mp+MQ=85.03Mp?+MQ?=85.03
Mp+2MQ=145.13Mp?+2MQ?=145.13
Mp=24.93≈25Mp?=24.93≈25
MQ=60.1≈60MQ?=60.1≈60

60. An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with Br2Br2? and KOH forms compound (R) having molecular formula C4H7NC4?H7?N. Names of P, Q and R respectively are.
(1) Benzoic acid, benzamide, aniline
(2) Toluic acid, methylbenzamide, 2-methylaniline
(3) Benzoic acid, 4-methylbenzamide, 4-methylaniline.
(4) Phenylethanoic acid, phenylethanamide, benzamine

Ans. (1)

Sol. Ph−COOH→NH3Ph−CONH2Ph−COOHNH3??Ph−CONH2?
(P) (Q)
→Br2/KOHPh−NH2Br2?/KOH?Ph−NH2?
(R)

JEE Previous Year Question Paper

61. An organic compound "P" of molecular formula C6H12O3C6?H12?O3? gives positive Iodoform test but negative Tollen's test. When "P" is treated with dilute acid, it produces "Q". "Q" gives positive Tollen's test and also Iodoform test. The structure of "P" is :
(1) CH3−CO−CH2−CH(OCH3)2CH3?−CO−CH2?−CH(OCH3?)2?
(2) CH3−CO−CH2−CH(OCH3)2CH3?−CO−CH2?−CH(OCH3?)2?
(3) H−CO−CH2−CH2−CH(OCH3)2H−CO−CH2?−CH2?−CH(OCH3?)2?
(4) CH3−C(OCH3)2−CH3CH3?−C(OCH3?)2?−CH3?

Ans. (2)

Sol. Structure with ketone gives +ve iodoform, acetal gives -ve Tollen's, hydrolysis gives aldehyde.

62. From the following, the least stable structure is :
(1)
(2)
(3)
(4)

Ans. (3)

Sol. This resonating structure having +ve charge on adjacent atoms so it is least stable.

63. MnO42−MnO42−?, in acidic medium, disproportionates to :
(1) Mn2O4Mn2?O4? and MnO2MnO2?
(2) MnO4−MnO4−? and MnOMnO
(3) MnO4−MnO4−? and MnO2MnO2?
(4) Mn2O4Mn2?O4? and MnOMnO

Ans. (3)

Sol. 3MnO42−+4H+→2MnO4−+MnO2+2H2O3MnO42−?+4H+→2MnO4−?+MnO2?+2H2?O

64. Given below are two statements:
Statement I: The number of species among SF4SF4?, NH4+NH4+?, [NiCl4][NiCl4?], XeF4XeF4?, [PtCl4]2−[PtCl4?]2−, SeF4SeF4? and [Ni(CN)4]2−[Ni(CN)4?]2−, that have tetrahedral geometry is 3.
Statement II: In the set [NO2,BeH2,BF3,AlCl3][NO2?,BeH2?,BF3?,AlCl3?], all the molecules have incomplete octet around central atom.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is true but Statement II is false
(2) Both Statement I and Statement II are false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true

Ans. (3)

Sol. Statement-I
SF4SF4? (See-saw), XeF4XeF4? (square planar), [PtCl4]2−[PtCl4?]2− (square planar), [NiCl4]2−[NiCl4?]2− (Tetrahedral), [Ni(CN)4]2−[Ni(CN)4?]2− (square planar), SeF4SeF4? (See-saw), NH4+NH4+? (Tetrahedral)
Statement-II
NO (seven electrons on N)
BeH2BeH2? (four electrons on Be)
BF3BF3? (six electrons on B)
AlCl3AlCl3? (six electrons on Al)

JEE Previous Year Question Paper

65. Given below are two statements:
Statement I: Among [Cu(NH3)4]2+[Cu(NH3?)4?]2+, [Ni(en)3]2+[Ni(en)3?]2+, [Ni(NH3)4]2+[Ni(NH3?)4?]2+ and [Mn(H2O)4]2+[Mn(H2?O)4?]2+, [Mn(H2O)4]2+[Mn(H2?O)4?]2+ has the maximum number of unpaired electrons.
Statement II: The number of pairs among {[NiCl4]2−,[Ni(CO)4]}{[NiCl4?]2−,[Ni(CO)4?]}, {[NiCl4]2−,[Ni(CN)4]2−}{[NiCl4?]2−,[Ni(CN)4?]2−} and {[Ni(CO)4],[Ni(CN)4]2−}{[Ni(CO)4?],[Ni(CN)4?]2−} that contain only diamagnetic species is two.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false

Ans. (4)

Sol. [Cu(NH3)4]2+⇒d9,dsp2[Cu(NH3?)4?]2+⇒d9,dsp2, one unpaired electron
[Ni(en)3]2+⇒d8,sp3d2[Ni(en)3?]2+⇒d8,sp3d2, two unpaired electrons
[Ni(NH3)4]2+⇒d8,sp3d2[Ni(NH3?)4?]2+⇒d8,sp3d2, two unpaired electrons
[Mn(H2O)4]2+⇒d5,sp3d2[Mn(H2?O)4?]2+⇒d5,sp3d2, five unpaired electrons
[Ni(CO)4][Ni(CO)4?] (diamagnetic)
[NiCl4]2−[NiCl4?]2− (paramagnetic)
[Ni(CN)4]2−[Ni(CN)4?]2− (diamagnetic)

66. Identify correct statement from the following :
A. Propanal and propanone are functional isomers.
B. Ethoxyethane and methoxypropane are metameres.
C. But-2-ene shows optical isomerism.
D. But-1-ene and but-2-ene are functional isomers.
E. Pentane and 2,2-dimethyl propane are chain isomers.
Choose the correct answer from the options given below:
(1) B,C and D only
(2) A,B and C only
(3) A,B and E only
(4) C,D and E only

Ans. (3)

Sol. Propanal and propanone are functional isomers.
CH3−CH=CH−CH3CH3?−CH=CH−CH3? & CH3−CH2−CH=CH2CH3?−CH2?−CH=CH2? are positional isomers.
CH3−CH2−O−CH2−CH3CH3?−CH2?−O−CH2?−CH3? & CH3−CH2−CH2−O−CH3CH3?−CH2?−CH2?−O−CH3? are metamers.
CH3−CH=CH−CH3CH3?−CH=CH−CH3? does not have optical isomerism.
Pentane and 2,2-dimethyl propane are chain isomers.

JEE Previous Year Question Paper

67. Identify the correct statements.
A. Arginine and Tryptophan are essential amino acids.
B. Histidine does not contain heterocyclic ring in its structure.
C. Proline is a six membered cyclic ring amino acid.
D. Glycine does not have chiral centre.
E. Cysteine has characteristic feature of side chain as MeS−CH2−CH2−MeS−CH2?−CH2?−
Choose the correct answer from the options given below:
(1) C and E Only
(2) B and E Only
(3) C and D Only
(4) A and D Only

Ans. (4)

Sol. Histidine does contain heterocyclic ring.
Proline is a five membered cyclic ring amino acid.
Cysteine has characteristic feature of side chain as CH2−SHCH2?−SH.

68. Which of the following graphs between pressure 'P' versus volume 'V' represent the maximum work done?
(1)
(2)
(3)
(4)

Ans. (4)

Sol. Area under the P v/s V curve, is equal to magnitude of work.
In option (2) work done is zero while in remaining options net work done is negative due to expansion.
NTA has given the answer without considering the negative sign that is considered only magnitude.

69. For the reaction, N2O4?2NO2N2?O4??2NO2?, graph is plotted as shown below. Identify correct statements.
A. Standard free energy change for the reaction is −5.40kJmol−1−5.40kJmol−1
B. As ΔG?ΔG? in graph is positive, N2O4N2?O4? will not dissociate into NO2NO2? at all.
C. Reverse reaction will go to completion.
D. When 1 mole of N2O4N2?O4? changes into equilibrium mixture, value of ΔG?=−0.84kJmol−1ΔG?=−0.84kJmol−1
E. When 2 mole of NO2NO2? changes into equilibrium mixture, ΔG?ΔG? for equilibrium mixture is −6.24kJmol−1−6.24kJmol−1
Choose the correct answer from the options given below:
(1) D and E only
(2) C and E only
(3) A and D only
(4) B and C only

Ans. (1)

Sol.
(A) ΔG?=GB?−GA?=+veΔG?=GB??−GA??=+ve
(B) ΔG?=+veΔG?=+ve, N2O4N2?O4? will partially dissociate into NO2NO2?
(C) For reverse reaction, it is partially completed as there is equilibrium at E.
(D) For 1 mole N2O4N2?O4?, ΔG?=−0.84kJmol−1ΔG?=−0.84kJmol−1
(E) For 2 mole NO2NO2?, ΔG?=−5.4−0.84=−6.24kJmol −1ΔG?=−5.4−0.84=−6.24kJmol−1

JEE Previous Year Question Paper

70. Given below are two statements:
Statement I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies.
Statement II: The frequency of second line of Balmer series obtained from He+He+ is equal to that of first line of Lyman series obtained from hydrogen atom.
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 false but Statement II is true
(4) Statement I is true but Statement II is false

Ans. (1)

Sol. 1λ=RZ2(1n12−1n22)λ1?=RZ2(n12?1?−n22?1?)
For Ist line of Lyman series in H-atom
1λ=R(1)2(112−122)λ1?=R(1)2(121?−221?)
1λ=3R4λ1?=43R?
For 2nd2nd line of Balmer series of He+He+
1λ′=R(2)2(122−142)λ′1?=R(2)2(221?−421?)
1λ′=3R4λ′1?=43R?
As λλ and λ′λ′ is equal so frequency of these lines will be also equal.

SECTION-B

71. Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20kJmol−120kJmol−1. If k1k1? and k2k2? are the rate constants of first and second reaction respectively at 300 K then ln?k2k1lnk1?k2?? will be ____ (nearest integer) [R = 8.3 J K?¹ mol?¹]

Ans. (8)

Sol. A→RX2(1)product E1ARX2(1)?product E1?
B→RX2(2)product E2BRX2(2)?product E2?
Assuming A′A′ same for both reaction.
ln?k1=ln?A−E1300Rlnk1?=lnA−300RE1??
ln?k2=ln?A−E2300Rlnk2?=lnA−300RE2??
ln?(k2k1)=E1−E2300R=20×1000300R

ln(k1?k2??)=300RE1?−E2??=300R20×1000?
=8.032

JEE Previous Year Question Paper

72. The pH and conductance of a weak acid (HX) was found to be 5 and 4×10−5S4×10−5S respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1 cm² were at a distance of 15 cm apart. The value of the limiting molar conductivity is ____ S m² mol?¹ (nearest integer) (Given: degree of dissociation of the weak acid (αα) ?1?1)

Ans. (6)

Sol. pH = 5
[H+]=10−5=[HX]⋅α[H+]=10−5=[HX]⋅α
=[HX]⋅ΛmΛm∞=[HX]⋅Λm∞?Λm??
Λm=k×1000[HX]Λm?=[HX]k×1000?
K=G⋅G∗=4×10−5×151=6×10−4Scm−1K=G⋅G∗=4×10−5×115?=6×10−4Scm−1
[H+]=10−5=[HX]×6×10−4×1000Λm∞×[HX][H+]=10−5=[HX]×Λm∞?×[HX]6×10−4×1000?
Λm∞=60000Scm2mol−1Λm∞?=60000Scm2mol−1
Λm∞=6Sm2mol−1

Λm∞?=6Sm2mol−1

73. Use the following data :

Substance	ΔH°(500K) kJ mol?¹	S°(500K) J K?¹ mol?¹
AB(g)	32	222
A?(g)	6	146
B?(g)	X	280

One mole each of A2(g)A2?(g) and B2(g)B2?(g) are taken in a 1L closed flask and allowed to establish the equilibrium at 500 K
A2(g)+B2(g)?2AB(g)A2?(g)+B2?(g)?2AB(g)
The value of x (in kJ mol) is ____ (Nearest integer)
(Given: log K = 2.2, R = 8.3 J K?¹ mol?¹)

Ans. (70)

Sol. A2+B2→500K2ABA2?+B2?500K?2AB, log?K=2.2logK=2.2
ΔH?=(2×32)−(6+x)=(58−x)kJΔH?=(2×32)−(6+x)=(58−x)kJ
ΔS?=(2×222)−(146+280)=18JΔS?=(2×222)−(146+280)=18J
ΔG?=−RTln?KΔG?=−RTlnK
ΔG?=−8.314×500×2.2×2.3031000ΔG?=−10008.314×500×2.2×2.303?
ΔG?=−21.06ΔG?=−21.06
ΔH?−TΔS?=−21.06ΔH?−TΔS?=−21.06
58−x−500(181000)=−21.0658−x−500(100018?)=−21.06
x=70.06kJ/mol

x=70.06kJ/mol

JEE Previous Year Question Paper

74. Consider the following reaction sequence
The percentage of nitrogen in product "T" formed is ____ %. (Nearest integer)
(Given molar mass in g mol?¹ H:1, C:12, N:14, O:16)

Ans. (20)

Sol. C6H6→HNO3/H2SO4Ph−NO2→Sn/HClPh−NH2C6?H6?HNO3?/H2?SO4??Ph−NO2?Sn/HCl?Ph−NH2?
→(CH3CO)2OPh−NH−COCH3→HNO3/H2SO4p−NO2−C6H4−NH−COCH3(CH3?CO)2?O?Ph−NH−COCH3?HNO3?/H2?SO4??p−NO2?−C6?H4?−NH−COCH3?
→HCl/EtOHp−NO2−C6H4−NH2HCl/EtOH?p−NO2?−C6?H4?−NH2?
Mol. wt =6×12+(6×1)+(2×14)+(2×16)=138=6×12+(6×1)+(2×14)+(2×16)=138
%N=28138×100=20.29%

%N=13828?×100=20.29%

75. Consider the following reactions:
NaCl+K2Cr2O7+H2SO4→A+KHSO4+NaHSO4+H2ONaCl+K2?Cr2?O7?+H2?SO4?→A+KHSO4?+NaHSO4?+H2?O
A+NaOH→B+NaCl+H2OA+NaOH→B+NaCl+H2?O
B+H2SO4+H2O2→C+Na2SO4+H2O

B+H2?SO4?+H2?O2?→C+Na2?SO4?+H2?O
In the product ‘C’, ‘X’ is the number of O22−O22−? units, ‘Y’ is the total number oxygen atoms present and ‘Z’ is the oxidation state of Cr. The value of X + Y + Z is ____.

Ans. (13)

Sol. A→CrO2Cl2A→CrO2?Cl2?
B→Na2CrO4B→Na2?CrO4?
C→CrO3C→CrO3?
Structure of CrO3CrO3?:
X=2,Y=5X=2,Y=5 and Z=6Z=6
Thus X+Y+Z=2+5+6=13

X+Y+Z=2+5+6=13

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