JEE-MAIN EXAMINATION - JANUARY 2025
(HELD ON FRIDAY 24th JANUARY 2025)
TIME: 9:00 AM TO 12:00 PM
MATHEMATICS

**1. Let `a?=i^+2j^+3k^` `b?=3i^+j^−k^` and `c?` be three vectors such that `c?` is coplanar with `a?` and `b?` . If the vector `c?` is perpendicular to `b?` and `a?.c?=5` then `?c??` is equal to**

(1) 132
(3) 16
(4)
Ans. (4)
Sol. c?=λ(b?×(a?×b?))=λ(b?b?)a?−(a?b?)b?)=λ(1a?−2b?)=λ(11a?+22b?+33b?−6a?−2b?+2b?)=λ(5a?+20b?+35b?)=5λ(5a?+4b?+7b?)= Given c?.a?=5=5λ(1+8+21)=5=λ=130⇒c?=16(a?+4b?+7b?)?c??=1+16+496=116(2)
**2. In `I(m,n)=∫01xm−1(1−x)n−1dx,m,n>0,` then `I(9,14)+I(10,13)` is**
(1) `I(9,1)`
(2) `I(19,27)`
(3) `I(1,13)`
(4) `I(9,13)`
Ans. (4)
Sol. `I(m,m)=∫01xm−1(1−x)n−1dx`
Let `x=sin?2θ`
`dx=2sin?θcos?θdθ`
`I(m,n)=2∫0π/2(sin?θ)2m−1(cos?θ)2n−1dθ`
`I(9,14)+I(10,13)=2∫0π/2(sin?θ)17(cos?θ)27dθ`
`+2∫0π/2(sin?θ)19(cos?θ)25dθ`
`=2∫0π/2(sin?θ)17(cos?θ)25dθ`
`=I(9,13)`

**3. Let `f:R−{0}→R` be a function such that**

f(x)−6f(1x)=353x−52. If the lim?x→0(1αx+f(x))=β;
**`α,β∈R` then `α+2β` is equal to**

(1)3
(2)5
(3)4
(4)6
Ans. (3)
Sol. `F(x)−6f(1/x)=353x−52……(1)`

Replace `x→1x`

F(1/x)−6(x)=35x3−52……(2)
Using (1) & (2)

f(x)=−2x−13x+12
B=lim?x→0(1αx+f(x))
=lim?x→0(1αx−2x−13x+12)
α=3,B=12
So,α+2B=3+1=4
**4. Let `Sn=12+16+112+120+…` upto n terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is `2026S2025` , then the absolute difference between `20n` and `15n` terms of the A.P. is**

(1) 25
(2) 90
(3) 20
(4) 45
Ans. (1)
Sol. `Sn=12+16+112+120…N` terms

S2025=∑n=120251n(n+1)=∑n=12025(1n−1n+1)
=(11−12)+(12−13)……(12025−12026)
2025 2026 `2026.S2025=2025=45`
Given: `62[−2p+(6−1)p]=45`
`9p=45`
`p=5`
`?A20−A15?=?−5+19×5?−[−5+14×5]`
`=?90−65?`
`=25`

**5. Let `f(x)=2x+2+1622x+1+2x+4+32` . Then the value of `8(f(115)+f(215)+…+f(5915))` is equal to**
(1) 118
(2) 92
(3) 102
(4) 108
Ans. (1)

f(x)=42x+162.22x+16.2x+32
f(x)=2(2x+4)22x+8.2x+16
f(x)=22x+4
f(4−x)=2x2(2x+4)
f(x)+f(4−x)=12
So,f(115)+f(5915)=12
Similarly=f(2915)+f(3115)=12
f(3015)=f(2)=222+4=28=14
⇒8(29×12+14)
Ans. 118
Option (4)

**6. If `α` and `β` are the roots of the equation `2z2−3z−2i=0` , where `i=−1` , then `16.Re(α19+β19+α11+β11α15+β15).Im(α19+β19+α11+β11α15+β15)` is equal to**
(1) 398
(2) 312
(3) 409
(4) 441
Ans. (4)
Sol.2z2−32−2i=0
2(z−iz)=3
α−iα=32
⇒α2−1α2−2i=94
⇒α2−1α2−2i=94
⇒94+2i=α2−1α2
⇒8116−4+9i=α4+1α4−2
⇒4916+9i=α4+1α4
Similarly

⇒4916+9i=β4+1β4
⇒α19+β19+α11+β11α15+β15=α15(α4+1α4)+β15(β4+1β4)α15+β15
=(α15+β15)(4916+9i)(α15+β15)
Real=4916
Im=9
Ans. 441

**7. `lim?x→0coscex(2cos2x+3cosx−cos2x+sin?x+4)` is**

(1)0(2)125
(3)115(4)−125
Ans. (4)
Sol. `lim?x→0cosecx(2cos?2x+3cos?x−cos?2x+sin?x+4)`

lim?x→0cosecx(cos?2x+3cos?x−sin?x−4)2cos?2x+3cos?x+cos?2x+sin?x+4
lim?x→01sin?x(cos?2x+3cos?x−4)−sin?x2cos?2x+3cos?x+cos?2x+sin?x+4
lim?x→0(cos?x+4)(cos?x−1)−sin?xsin?x(2cos?2x+3cos?x+cos?2x+sin?x+4)
lim?x→0−2sin?2x2(cos?x+4)−2sin?x2cos?x22sin?x2cos?x2(2cos?2x+3cos?x+cos?2x+sin?x+4)
lim?x→0−(sin?x2(cos?x+4)+cos?x2)cos?x2(2cos?2x+3cos?x+cos?2x+sin?x+4)
125
**8. Let in a `ΔABC` , the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line `x−63=y−72=z−7−2` . Then the area (in sq. units) of `ΔABC` is**
(1) 42
(2) 21
(3) 56
(4) 17
Ans. (2)
Sol.

Let M `(3λ+6,2λ+7,−2λ+7)`

`BM‾=(3λ+5)i^+(2λ+5)j^+(−2λ+4)k^`

`AC‾.BM‾=0=3(3λ+5)+2(2λ+5)−2(−2λ+4)`

`BM‾=2i^+3j^+6k^`

`?BM‾?=7`

`Area=12×6×7=21`
Option (2)

**9. Let `y=y(x)` be the solution of the differential equation `(xy−5x21+x2)dx+(1+x2)dy=0,` `y(0)=0` . Then `y(3)` is equal to**
(1) `532`
(2) `143`
(3) `22`
(4) `152`
Ans. (1)
Sol. `(1+x2)dydx+xy=5x1+x2`

dydx+xy1+x2=5x21+x2
∴I.F.=e∫x1+x2dx=eln?(1+x2)2=1+x2
∴y1+x2=∫5x21+x21+x2dx
∴y1+x2=∫5x21+x21+x2dx
y1+x2=5x33+C
?y(0)=0⇒0=0+C⇒C=0
∴y=5x331+x2
y(3)=15332=532
Option (1)

**10. Let the product of the focal distances of the point `(3,12)` on the ellipse `x2a2+y2b2=1,` `(a>b)`, be `74` . Then the absolute difference of the eccentricities of two such ellipses is**
(1) `3−2232`
(2) `1−32`
(3) `3−2223`
(4) `1−223`
Ans. (3)

1. Product of focal distances `Ψ=a+ex1` a- ex)
   `Ψ=a2−e2x12=a2−e2(3)`
   `Ψ=a2−3e2=74⇒a2=74+3e2`
   `⇒4a2=7+12e2`
   & `(3,12)` lines on `x2a2+y2b2=1`
   `∴3a2+14b2=1`
   `3a2+14(a2)(1−e2)=1`
   `12(1−e2)+1=4a2(1−e2)`
   `13−12e2=(7+12e2)(1−e2)`
   `⇒13−12e2=7−7e2+12e2−12e4`
   `⇒12e4−17e2+6=0`
   `∴e2=17±289−28824=17±124=34×23`
   `∴e=32×23`
   `∴difference=32−23=3−2223`
Option (3)

**11. A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is**

(1) 917
(3) 817
(4) 819
(4)
Ans. (2)
Sol. `p(S3)=19`
`p(S3)=536`
required prob `=19+89⋅3136⋅19+(89⋅3136)2⋅19+…∞`
`=19−6281=919`
Option(2)

**12. Consider the region**

R={(x,y):x≤y≤9−113x2,x≥0}. The area, of
**the largest rectangle of sides parallel to the coordinate axes and inscribed in `R` is :**

(1)625111
(2)730119
(3)567121
(4)821123
Ans. (3)
Sol. `t(9−11t23−t)`

A=9t−t2−113t3
dAdt=9−2t−11t2
⇒11t2+2t−9=0
11t2+11t−9t−9=0
t=−1 & t=911
∴dAdt=9−1119
Δ  Δ  Δ  Δ maxima at t  =911
∴ largest area =911(9−113,81121−911)
=911⋅6311=567121
Option (3)

**13. The area of the region `{(x,y):x2+4x+2≤y≤?x+2?}` is equal to**

(1)7
(2)24/5
(3)20/3
(4)5
Ans. (3)
Sol. `x2+4x+2≤y≤?x+2?`
The area bounded between
`y=x2+4x+2=(x+2)2−2`
and `y=?x+2?` is same as
area bounded between `y=x2−2` and `y=?x?`
For P.O.I `?x?2−2=?x?`
`⇒?x?=2⇒x=±2`
`∴` Required area `=−∫−22(x2−2)dx+∫−22?x?dx`
`=−2∫02(x2−2)dx+2∫02x.dx`
`=−2[x33−2x]02+2[x22]02`
`=−2[83−4]+2[42]`
`=−2×(−43)+4`
`=203`

**14. For a statistical data `x1,x2,…,x10` of 10 values, a student obtained the mean as 5.5 and `∑i=110xi2=371` . He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is**
(1) 7
(2) 4
(3) 9
(4) 5
Ans. (1)
Sol. Mean `x?=5.5`
`=∑i=110xi=5.5×10=55`
`=∑i=110xi2=371`
`(∑xi)new=55−(4+5)+(6+8)=60`
`(∑xi)new=371−(42+52)+(62+82)=430`
Variance `σ2=∑xi210−(∑xi10)2`
`σ2=43010−(6010)2`
`σ2=43−36`
`σ2=7`

**15. Let circle C be the image of `x2+y2−2x+4y−4=0` in the line `2x−3y+5=0` and A be the point on C such that OA is parallel to x-axis and A lies on the right hand side of the centre O of C. If B(α,β), with β < 4, lies on C such that the length of the arc AB is (1/6)th of the perimeter of C, then β−√3α is equal to**
(1) 3
(2) 3 + √3
(3) 4 − √3
(4) 4
Ans. (4)
Sol.

`x2+y2−2x+4y−4=0`
Centre (1, −2), r = 3
Reflection of (1, −2) about 2x − 3y + 5 = 0
`x−12=y+2−3=−2(2+6+5)13=−2`
`x=−3,y=4`
Equation of circle ‘C’
C : (x+3)² + (y−4)² = 9
A.T.Q.
B(α,β)
AB = 3√3
`?(arcAB)=16×2πr`
`rθ=16×2πr`
`θ=π3`
`(α+6)2+(β−4)2=27`
`(α+3)2±(β−4)2=9`
`(α+6)2−(α+3)2=18`
`⇒6α=−9`
`⇒α=−32,β=(4−332)`
`∴β−3α`
`(4−332)+332`
`=4`

**16. For some n ≠ 10, let the coefficients of the 5th, 6th and 7th terms in the binomial expansion of (1 + x)^(n+4) be in A.P. Then the largest coefficient in the expansion of (1 + x)^(n+4) is :**
(1) 70
(2) 35
(3) 20
(4) 10
Ans. (2)
Sol. (1 + x)^(n+4)
???C?, ???C?, ???C? → A.P.
⇒ 2 × ???C? = ???C? + ???C?
⇒ 4 × ???C? = (???C? + ???C?) + (???C? + ???C?)
⇒ 4 × ???C? = ???C? + ???C?
⇒ 4 × `(n+4)!5!(n−1)!=(n+6)!6!n!`
⇒ 4 = `(n+6)(n+5)6n`
⇒ n² + 11n + 30 = 24n
⇒ n² − 13n + 30 = 0
⇒ n = 3, 10 (rejected)
? n ≠ 10
∴ Largest binomial coefficient in expansion of (1 + x)?
(? n + 4 = 7)
is coeff. of middle term
⇒ ?C? = ?C? = 35
N.T.A. Ans
Option (2)

**17. The product of all the rational roots of the equation (x² − 9x + 11)² − (x − 4)(x − 5) = 3, is equal to :**
(1) 14
(2) 7
(3) 28
(4) 21
Ans. (1)
Sol. (x² − 9x + 11)² − (x² − 9x + 20) = 3
Let x² − 9x = t
⇒ t² + 22t + 121 − t − 20 − 3 = 0
⇒ t² + 21t + 98 = 0
⇒ (t + 14)(t + 7) = 0
⇒ t = −7, −14

So, x² − 9x = −7, −14
x² − 9x + 7 = 0 or x² − 9x + 14 = 0
x = `9±81−4(7)2` or x = `9±81−4(14)2`
\= `9±532` = `9±52`
Product of all rational roots = 7 × 2 = 14
Option (1)

**18. Let the line passing through the points (−1, 2, 1) and parallel to the line `x−12=y+13=z4` intersect the line `x+23=y−32=z−41` at the point P. Then the distance of P from the point Q(4, −5, 1) is :**
(1) 5
(2) 10
(3) 5√6
(4) 5√5
Ans. (4)
Sol. Equation of line through point (−1, 2, 1) is →
⇒ `x+12=y−23=z−14=λ` ...(1)
So, x = 2λ − 1, y = 3λ + 2, z = 4λ + 1
By (1) → `x+23=y−32=z−41=μ` (Let)
So, x = 3μ − 2, y = 2μ + 3, z = μ + 4
For intersection point ‘P’
x = 2λ − 1 = 3μ − 2
y = 3λ + 2 = 2μ + 3
z = 4λ + 1 = μ + 4
`[λ=1μ=1]`
So, point P(x, y, z) = (1, 5, 5)
& Q(4, −5, 1)
∴ PQ = `9+100+16` = `125` = 5√5
Option (4)

**19. Let the lines `3x−4y−=0,8x−11y−33=0,` and `2x−3y+λ=0` be concurrent. If the image of the point (1, 2) in the line `2x−3y+λ=0` is `(5713,−4013)` , then `?αλ?` is equal to :**
(1) 84
(2) 91
(3) 113
(4) 101
Ans. (2)
Sol.

`PM=QM`
So, M `(5713+i−4013+2)`
` 12  12 `
` 12  13 `
` 12  −713 `

M lies on the time

`2x−3y+λ=0`
`2(3513)−3(−713)+λ=0`
`λ=−7013+2113`
`=−−9113=−7`

`⇒3(−11λ−99)+4(8λ+66)−α(−24+22)=0`
`⇒33λ−297+32λ+264+24α−22α=0`
`⇒−λ+2α−33=0……(1)`
`∴λ=−7`
`−(−7)+2α−33=0`
`2α=26`
`α=13`
`∴?αλ?=?13×(−7)?`

**20. If the system of equations `2x−y+z=4,` `5x+λy+3z=12,` `100x−47y+μz=212,` has infinitely many solutions, then `μ−2λ` is equal to**
(1) 56
(2) 59
(3) 55
(4) 57
Ans. (4)
Sol. `Δ=0⇒?2−115λ3100−47μ?=0`

`2(λμ+141)+(5μ−300)−235−100λ=0…(1)`

`Δ3=0⇒?2−145λ12100−47212?=0`

`6λ=−12⇒λ=−2`

Put `λ=2` in (1)

`2(−2μ+141)+5μ−300−235+200=0`

`μ=53`

:57

## SECTION-B

**21. Let f be a differentiable function such that `2(x+2)2f(x)−3(x+2)2=10∫0x(t+2)f(t)dt,` `x≥0` . Then `f(2)` is equal to**
Ans. (19)
Sol. Differentiate both sides

`4(x+2)f(x)+2(x+2)2f′(x)−6(x+2)=10(x+2)f(x)`

`2(x+2)2f′(x)−6(x+2)f(x)=6(x+2)`

`(x+2)dydx−3y=3`

`∫dydx=3∫dxx+2`

`ln?(y+1)=3ln?(x+2)+C`

`(y+1)=C(x+2)3`

`f(0)=32`
`f(2)=19`

**22. If for some `α,β;α≤β,α+β=8` and `sec?2(tan?−1α)+cosec?2(cot?−1β)=36,` then `α2+β` is \_\_\_\_\_\_.**
Ans. (14)
Sol. If `tan?(tan?−1(α))+1(cot?(cot?−1β))2=36`
`α2+β2=34`
`αβ=15`
`α=3,β=5`
`∴α2+β=9+5=14`

**23. The number of 3-digit numbers, that are divisible by 2 and 3, but not divisible by 4 and 9, is**
Ans. (125)
Sol. No. of 3 digits = 999 − 99 = 900
No. of 3 digit numbers divisible by 2 & 3 i.e. by 6
`9006=150`
No. of 3 digit numbers divisible by 4 & 9 i.e. by 36
`90036=25`
∴ No of 3 digit numbers divisible by 2 & 3 but not by 4 & 9
150 − 25 = 125

**24. Let be a 3 × 3 matrix such that `XTAX=O` for all nonzero 3 × 1 matrices `X=[xyz]`. If `A[111]=[14−5],A[211]=[04−8],` and det(adj(2(A + I))) = 2^α 3^β 5^γ, α, β, γ, ∈ N, then α² + β² + γ² is**
Ans. (44)
Sol. `XTAX=0`
`(xyz)(a1a2a3b1b2b3c1c2c3)(xyz)=0`
`(xyz)(a1x+a2y+a3zb1x+b2y+b3zc1x+c2y+c3z)=0`
`x(a1x+a2y+a3z)+y(b1x+b2y+b3z)+z(c1x+c2y+c3z)=0`
`a1=0,b2=0,c3=0`
`a2+b1=0,a3+c1=0,b3=c2=0`
A = skew symm matrix
`A=(0xy−x0z−y−z0);A(111)=(14−5)`
`⇒A=(0xy−x0z−y−z0)(111)=(14−5)`
`x+y=1`
`−x+z=4`
`y+z=5`
`(0xy−x0z−y−z0)(121)=(14−8)`
`2x+y=0x=−1`
`−x+z=4y=2`
`−y−2z=−8z=3`
`A=(0−12103−2−30)`
`2(A+I)=(2−24226−2−62)`
`2(A+I)=120⇒det?(adj(2(A+I)))=1202=263252`
`α=6,β=2,γ=2`

**25. Let S = {p?, p?, …, p??} be the set of first ten prime numbers. Let A = S ∪ P, where P is the set of all possible products of distinct element of S. Then the number of all ordered pairs (x, y), x ∈ S, y ∈ A, such that x divides y, is \_\_\_\_\_\_.**
Ans. (5120)
Sol. Let `yx=λ`
`y=λx`
\= 10 × (C? + ?C? + ?C? + ?C? + … + ?C?)
\= 10 × (2?)
10 × 512
5120

**26. Consider a parallel plate capacitor of area A (of each plate) and separation 'd' between the plates. If E is the electric field and `?0` is the permittivity of free space between the plates, then potential energy stored in the capacitor is :-**

112?0E2Ad114?0E2Ad134?0E2Ad1(4)
Ans. (1)
Sol. `UV=12?0E2`
`U=12?0E2V`
`=12?0E2(Ad)`
Ans. (1)

**27. What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from `2.5D` ? ['D' stands for dioptre]**

(1)0.04(3)0.1(2)0.40(4)0.01
Ans. (1)
Sol. When `P=2.5D`
`F=1P=12.5`
When `P′=2.6D`
`F′=1P′=12.6`
Relative decrease in focal length
`F−F′F=25=513=1−2526=126=0.04`
Ans. (1)

**28. An air bubble of radius `0.1cm` lies at a depth of `20cm` below the free surface of a liquid of density `1000kg/m3` . If the pressure inside the bubble is `2100N/m2` greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use `g=10m/s2`)**

(1)0.02(2)0.1
(3)0.25(4)0.05
Ans. (4)
Sol. T is surface tension
`Pinairbubble=P0+ρgh+2TR`
`Pin−P0=ρgh+2TR=2100`
`2TR=2100−ρgh`
`T=R2(2100−103×10×0.2)`
`=120(2100−2000)×10−2`
`=0.05`
`Ans.(4)`
Ans. (4)

**29. For an experimental expression `y=32.3×112527.4` where all the digits are significant. Then to report the value of y we should write :-**

(1)y=1326.2(2)y=1326.19
(3)y=1326.186
(4)y=1330
Ans. (4)
Sol. `y=32.3×112527.4=1326.186`
Last significant digits are 3 in operands so results should rounded off to 3 digits.
`∴y=1330`
Ans. (4)

**30. During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is `2000A?` and it becomes `6000A?` when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is :-**

(1) `3000A?`
(2) `6000A?`
(3) `4000A?`
(4) `2000A?`
Ans. (1)

EA−EC=hc2000A?…(i)
and EB−EC=hc6000A?…(ii)
Now EA−EB=(EA−EC)−(EB−EC)
hcλAB=hc2000−hc6000
1λAB=13000A?
λAB=3000A?
Ans. (1)

**31. Consider the following statements :**

A. The junction area of solar cell is made very narrow compared to a photo diode.
B. Solar cells are not connected with any external bias.
C. LED is made of lightly doped p-n junction.
D. Increase of forward current results in continuous increase of LED light intensity.
E. LEDs have to be connected in forward bias for emission of light.

(1) B, D, E Only
(2) A, C Only
(3) A, C, E Only
(4) B, E Only
Ans. (4)
Sol. Conceptual
Ans. (4)

**32. The amount of work done to break a big water drop of radius 'R' into 27 small drops of equal radius is `10J` . The work done required to break the same big drop into 64 small drops of equal radius will be :-**

(1) `15J`
(2) `10J`
(3) `20J`
(4) `5J`
Ans. (1)
Sol. `W=ΔU=SΔA`
One drop to n drop
`43λR3=n43λr3`
`r=Rn3`
`So W=S(n4πr2−4πR2)`
`=S4πR2(n3−1)`
For on drop to 27 drops
`W=S4πR2(273−1)=10……(i)`
For one drop to 64 drops
`W′=S4πR2(643−1)…(ii)`
(ii)(i)
`W′W=4−13−1=32`
`W′=32W=155`
Ans. (1)

**33. An object of mass 'm' is projected from origin in a vertical xy plane at an angle `45?` with the x-axis with an initial velocity `v0` . The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [g is acceleration due to gravity]**

(1) `mv0322g` along negative z-axis
(2) `mv0322g` along positive z-axis
(3) `mv0342g` along positive z-axis
(4) `mv0342g` along negative z-axis
Ans. (4)
Sol.
`H=(v02)22g=v024g`
L = mvh
L = m `v02v024g`
Ans. (4)

**34. The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is :-**
(1) 4
(2) 8
(3) 6
(4) 5
Ans. (4)
Sol. `n1λ1Dd=n2λ2Dd`
n 480 = m 600
`nmin?=5`
Ans. (4)

**35. A car of mass 'm' moves on a banked road having radius 'r' and banking angle θ. To avoid slipping from banked road, the maximum permissible speed of the car is `v0`. The coefficient of friction μ between the wheels of the car and the banked road is :-**

(1) `μ=v02+rgtan?θrg−v02tan?θ`
(2) `μ=v02+rgtan?θrg+v02tan?θ`
(3) `μ=v02−rgtan?θrg+v02tan?θ`
(4) `μ=v02−rgtan?θrg−v02tan?θ`
Ans. (3)
Sol.
`Nsin?θ+fcos?θ=mv2R`
`Ncos?θ−fsin?θ=mg`
`sin?θ+μcos?θcos?θ−μsin?θ=v2Rg`
`Rgtan?θ+μRg=v2−v2μtan?θ`
`μ=v2−Rgtan?θRg+v2tan?θ`
Ans. (3)

**36. A uniform solid cylinder of mass 'm' and radius 'r' rolls along an inclined rough plane of inclination 45°. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder axis will be :-**
(1) `12g`
(2) `132g`
(3) `2g3`
(4) `2g`
Ans. (3)
Sol. `a=gsin?θ1+1mR2`
`a=g21+12=2g32=2g3`
Ans. (3)

**37. A thin plano convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m. The radius of curvature of the curved surface of the lens is :-**
(1) 0.15 m
(2) 0.10 m
(3) 0.20 m
(4) 0.25 m
Ans. (2)
Sol.
`1.5v=1.5−1.2R`
`v=1.5R0.3=5R`
`1.2f=1.55R=1.2−1.5−R`
`1.2f=0.3R×2⇒f=2R⇒R=0.1`
Ans (2)

**38. A particle is executing simple harmonic motion with time period 2s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then `Dd` is :-**
(1) `154`
(2) 25
(3) 10
(4) `165`
Ans. (2)
Sol. A = 1 cm
`n=12.52=6.25 cycles`
`∴D=4×6+1=25`
d = 1
`Dd=25`
Ans. (2)

**39. A satellite is launched into a circular orbit of radius 'R' around the earth. A second satellite is launched into an orbit of radius 1.03 R. The time period of revolution of the second satellite is larger than the first one approximately by :-**
(1) 3 %
(2) 4.5 %
(3) 9 %
(4) 2.5 %
Ans. (2)
Sol. `T2=KR3`
`2ΔTT=3ΔRR`
`2ΔTT=3×0.03RR`
`ΔTT=3×0.032×100=4.5%`
Ans. (2)

**40. A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of `f1` in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of `f2` when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5, the ratio of `f1` and `f2` will be :-**
(1) 3 : 5
(2) 1 : 3
(3) 1 : 2
(4) 2 : 3
Ans. (2)
Sol. `1f1=(1.5−1)[12−0]⇒f1=4cm`
`1f2=(1.51.2−1)(13−0)`
`1f2=0.31.2×13`
`f2=12`
`f1:f2=4:12=1:3`
Ans. (2)

**41. An alternating current is given by `I=IAsin?ωt+IBcos?ωt`. The r.m.s. current will be :-**
(1) `IA2+IB2`
(2) `IA2+IB22`
(3) `IA2+IB22`
(4) `?IA+IB?2`
Ans. (3)
Sol. `irms=∫I2dt∫dt`
`IA2+IB22=irms`

**42. An electron of mass 'm' with an initial velocity `v?=v0i^(v0>0)` enters an electric field `E?=−E0k^` . If the initial de Broglie wavelength is `λ0` , the value after time t would be :-**
(1) `λ01+e2E02t2m2v02`
(2) `λ01−e2E02t2m2v02`
(3) `λ0`
(4) `λ01+e2E02t2m2v02`
Ans. (1)
Sol. `v?=v0i^−E0emtk^`
`?v??=v02+E02e2t2m2`
`λ0=hmv0`
`λ′=hmv01+E02e2t2v02m2`
`λ′=λ01+E02e2t2v02m2`
Ans. (1)

**43. A parallel plate capacitor was made with two rectangular plates, each with a length of `l=3cm` and breath of `b=1cm` . The distance between the plates is `3μm` . Out of the following, which are the ways to increase the capacitance by a factor of `10?`**
A. `l=30cm` `b=1cm` `d=1μm`
B. `l=3cm` `b=1cm` `d=30μm`
C. `l=6cm` `b=5cm` `d=3μm`
D. `l=1cm` `b=1cm` `d=10μm`
E. `l=5cm` `b=2cm` `d=1μm`
Choose the correct answer from the options given below :
(1) C and E only
(2) B and D only
(3) A only
(4) C only
Ans. (1)
Sol. `C=Ae0d`
A : plate area
d : distance between the plates.
Capacitance initial
`=?0lbd=?0 unitsOption ′C′?=6 cmb=5 cmd=3 cmCapacitance=10 e0 unitsOption ′E′?=5 cmb=2 cmd=1 cmCapacitance=10 e0 units∴Ans is option (1)`

**44. A force `F=α+βx2` acts on an object in the `x` - direction. The work done by the force is `5J` when the object is displaced by `1m` . If the constant `α=1N` then `β` will be**
(1) `15N/m2`
(2) `10N/m2`
(3) `12N/m2`
(4) `8N/m2`
Ans. (3)
Sol. `F=α+βx2`
Work done `=∫Fdx`
`5=∫(α+βx2)dx5=αx+βx33?015=α+β3[α=1]4=β3⇒β=12N/m2`

**45. An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature.**
A. The work done by gas during the process is zero.
B. The heat added to gas is different from change in its internal energy.
C. The volume of the gas is increased.
D. The internal energy of the gas is increased.
E. The process is isochoric (constant volume process)
Choose the correct answer from the options given below:-
(1) A, B, C, D Only
(2) A, D, E Only
(3) E Only
(4) A, C Only
Ans. (2)
Sol. Given that `P=kT`
`PT=` constant : Volume is constant or isochoric process.
`∴WD=0`
`∴Q=ΔU`
Also temperature increases hence internal energy increases.

## SECTION-B

**46. A square loop of sides `a=1m` is held normally in front of a point charge `q=1C` . The flux of the electric field through the shaded region is `5p×1Nm2e0C` , where the value of `p` is \_\_\_\_\_\_.**
Ans. (48)
Sol.

Total flux through square `a=qe0(16)`
Lets divide square is 8 equal parts. Flux is same for each part.
Flux through shaded portion is `58` (Total flux)
`=58×qe016=5481e0`
required
Ans. is 48
Note : Distance of charge from square loop is not mentioned we have assume it as `a2`

**47. The least count of a screw guage is `0.01mm` . If the pitch is increased by `75%` and number of divisions on the circular scale is reduced by `50%` , the new least count will be `_×10−3mm` .**
Ans. (35)
Sol. Given least count of Screw Gauge `=0.01mm`
`L.C=(pitch)No. of circular turn=PN=0.01mm`
`New pitch=P(1+0.75)N(1−0.5)=P[1.75]N[0.5]`
`=(0.01)3.5`
`=0.035mm`
`=35×10−3mm`
Ans. is 35

**48. A wire of resistance 9 is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be \_ohm.**
Ans. (2)
Sol.

`9Ω` is the resistance of whole wire : resistance of each wire `=3Ω`
Equivalent resistance `=2Ω`

**49. A current of 5A exists in a square loop of side `12m` . Then the magnitude of the magnetic field B at the centre of the square loop will be `p×10−6T` . where, value of p is [Take `μ0=4π×10−7TmA−1]`**
Ans. (8)

Let B be the magnetic field due to single side
`then B=μ0i4πd(sin?θ1+sin?θ2)`
`=10−7×5×2122×12=2×10−6`
`∴ Bnet at centre O=4B`
`=8×10−6`
`∴ P=8
Ans.`

**50. The temperature of 1 mole of an ideal monoatomic gas is increased by `50?C` at constant pressure. The total heat added and change in internal energy are `E1` and `E2` respectively. If `E1E2=x9` then the value of `x` is**
Ans. (15)
Sol. Given that process is isobaric `ΔT=50?C`
`Q` in isobaric process `=nCpΔT=E1`
`ΔU` in isobaric process `=nCvΔT=E2`
`∴E1E2=CpCv=γ`
Given, gas is monoatomic
`∴γ=1+2f`
`=1+23`
`=53`

Now, as per question.
`53=x9`
`x=15`

**51. For the given cell `Fe2+(eq)+Ag+(aq)→Fe3+(aq)+Ag(s)` The standard cell potential of the above reaction is Given : `Ag++e−→AgE0=xV` `Fe2++2e−→FeE0=yV` `Fe3++3e−→FeE0=zV`**
(1) `x+y−z`
(2) `x+2y−3z`
(3) `y−2x`
(4) `x+2y`
Ans. (2)
Sol. `Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)`
`Fe→ΔG10Fe2+→ΔG20Fe3+`
`ΔG30=ΔG10+ΔG20`
`−3F(−z)=−2F(−y)+ΔG20`
`ΔG20=3Fz−2Fy`
Also `ΔG30=−nFE00e−z/Fe3`
`3Fz−2Fy=−1F(E0e2/Fe30)`
`E0e2/Fe31=2y−3z`
`E0cell0` for reaction will be
`EAg+/Ag0+EFe2/Fe30`
`=x+2y−3z`
Option (2)

## SECTION-A

**51. For the given cell**

Fe2+(eq)+Ag+(aq)→Fe3+(aq)+Ag(s)
**The standard cell potential of the above reaction is**

Given :

Ag++e−→AgE0=xV
Fe2++2e−→FeE0=yV
Fe3++3e−→FeE0=zV
(1)x+y−z
(2)x+2y−3z
(3)y−2x
(4)x+2y
Ans. (2)
Sol. `Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)`

ΔG30=ΔG10+ΔG20
−3F(−z)=−2F(−y)+ΔG20
ΔG20=3Fz−2Fy
ΔG30=−nFE00e−z/Fe3
3Fz−2Fy=−1F(E0e2/Fe30)
E0e2/Fe30=2y−3z
E0cell0for reaction will be
EAg+/Ag0+EFe2/Fe30
=x+2y−3z
Option (2)

**52. Following are the four molecules "P", "Q", "R" and "S". Which one among the four molecules will react with H- Br(aq) at the fastest rate ?**

(1) S
(2) Q
(3) R
(4) P
Ans. (2)
Sol. Addition of `H−Br(aq)` to alkene follows electrophilic addition mechanism. In the rate determining step a carbonation intermediate is formed. Among P, Q, R & S compound Q will form most stable carbonation intermediate since it is resonance stabilized.

**53. One mole of the octahedral complex compound `Co(NH3)5Cl3` gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with excess of `AgNO3` solution to yield two moles of `AgCl(s)` . The structure of the complex is :**

(1)[Co(NH3)5Cl]Cl2
(2)[Co(NH3)4Cl].Cl2.NH3
(3)[Co(NH3)4Cl2]Cl.NH3
(4)[Co(NH3)5Cl3].2NH3
Ans. (1)
Sol.

[Co(NH3)5Cl]Cl2→[Co(NH3)5Cl]2+(aq.)+2Cl−(aq.)3 ions in water[Co(NH3)5Cl]Cl2(aq.)+2AgNO3(aq.)→[Co(NH3)5Cl](NO3)2(aq.)+2AgCl(s)(3)
**54. Which one of the carbocations from the following is most stable ?**
Ans. (2)
Sol. Carbocation intermediate is stabilised by `+I` `+M` & hyperconjugation effect. Since in option 2 carbocation is in conjugation with stronger `+M` group- OCH, hence it will be most stable.

**55. Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [intermuclear axis in z-direction] ?**
A. `2pz` and `2px`
B. 2s and `2px`
C. `3dxy` and `3dx2−y2`
D. 2s and `2pz`
E. `2pz` and `3dx2−y2`

(1) E Only
(2) A and B Only
(3) D Only
(4) C and D Only
Ans. (3)
Sol.

**56. Which of the following ions is the strongest oxidizing agent ? (Atomic Number of Ce `=58` `Eu=63` `Tb=65` `Lu=71`)**
(1) `Lu3+`
(2) `Eu2+`
(3) `Tb4+`
(4) `Ce3+`
Ans. (3)
Sol. `Tb4+` is strongest oxidising agent

**57. Ksp for `Cr(OH)3` is `1.6×10−30` . What is the molar solubility of this salt in water?**

(1)1.6×10−30274(2)1.8×10−3027
(3)1.8×10−305(4)1.6×10−302
Ans. (1)
Sol. `Cr(OH)3(s)?Craq3++3OHaq−`
At eq : s 3s
`Ksp=(s).(3s)3=27s4`
`27s4=1.6×10−30`
`s=(1.627×10−30)1/4`
Option (1)

**58. Let us consider an endothermic reaction which is non- spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.**
(1) Both `ΔH` and `ΔS` are `(+ve)`
(2) `ΔH` is `(−ve)` but `ΔS` is `(+ve)`
(3) `ΔH` is `(+ve)` but `ΔS` is `(−ve)`
(4) Both `ΔH` and `ΔS` are `(−ve)`
Ans. (1)
Sol. Reaction is spontaneous at relatively high temperature and non- spontaneous at low temperature
`ΔG=ΔH−TΔS`
It is only possible when `ΔH` and `ΔS` both are positive.
Option (1)

**59. Given below are two statements I and II. Statement I : Dumas method is used for estimation of "Nitrogen" in an organic compound. Statement II : Dumas method involves the formation of ammonium sulphate by heating the organic compound with conc `H2SO4` 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) Statement I is false but Statement II is true
(3) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Ans. (4)
Sol. In Dumas method nitrogen present in organic compound is converted into `N2` gas whose volumetric analysis gives the percentage of nitrogen atom in the organic compound.

**60. Which of the following Statements are NOT true about the periodic table?**
A. The properties of elements are function of atomic weights.
B. The properties of elements are function of atomic numbers.
C. Elements having similar outer electronic configuration are arranged in same period.
D. An element's location reflects the quantum numbers of the last filled orbital.
E. The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled.
Choose the correct answer from the options given below:
(1) A, C and E Only
(2) D and E Only
(3) A and E Only
(4) B, C and E Only
Ans. (1)
Sol. Properties of elements are periodic function of their atomic number. Elements having similar outer electronic configuration are arranged in same group. Number of elements in a period is not equal to number of atomic orbitals available in energy level that is being filled. Hence, A, C & E are incorrect

**61. The carbohydrates "Ribose" present in DNA, is**
A. A pentose sugar
B. present in pyranose from
C. in "D" configuration
D. a reducing sugar, when free
E. in `α` -anomeric form
Choose the correct answer from the options given below:
(1) A, C and D Only
(2) A, B and E Only
(3) B, D and E Only
(4) A, D and E Only
Ans. (1)
Sol. In Ribose carbohydrate present in DNA is `β−2` - Deoxy- D- Ribose whose structure is which is a reducing D- sugar in `β` anomeric form & it is a pentose sugar.

**62. Preparation of potassium permanganate from `MnO2` involves two step process in which the `1st` step is a reaction with KOH and `KNO3` to produce**
(1) `K4[Mn(OH)6]`
(2) `K3MnO4`
(3) `KMnO4`
(4) `K2MnO4`
Ans. (4)
Sol. `MnO2→KOHK2MnO4`

**63. The large difference between the melting and boiling points of oxygen and sulphur may be explained on the basis of**
(1) Atomic size
(2) Atomicity
(3) Electronegativity
(4) Electron gain enthalpy
Ans. (2)
Sol. Oxygen exists as `O2` (Atomicity = 2)
Sulphur exists as `S8` (Atomicity = 8)
Hence, Melting point & Boiling point of sulphur are significantly large compared to oxygen.

**64. For a reaction, `N2O5(g)→2NO2(g)+12O2(g)` in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is**
(1) 7/2 times of initial pressure
(2) 5 times of initial pressure
(3) 5/2 times of initial pressure
(4) 7/4 times of initial pressure
Ans. (4)

Sol
`N2O5(g)?2NO2(g)+12O2(g)`
t = 0 `P0` – –
t = t `P0−x` `2x` `x/2`
`x=P02`
`Ptotal=P0−P02+P0+P04=74P0`
Option (4)

**65. Which of the following arrangements with respect to their reactivity in nucleophilic addition reaction is correct?**
(1) benzaldehyde < acetophenone < p-nitrobenzaldehyde < p-tolualdehyde
(2) acetophenone < benzaldehyde < p-tolualdehyde < p-nitrobenzaldehyde
(3) acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde
(4) p-nitrobenzaldehyde < benzaldehyde < p-tolualdehyde < acetophenone
Ans. (3)
Sol. The rate of nucleophilic addition decreased due to steric crowding around carbonyl carbon & increased by electron withdrawing group if the steric crowding is same hence the reactivity towards nucleophilic addition will be

**66. Aman has been asked to synthesise the molecule (x). He thought of preparing the molecule using an aldol condensation reaction. He found a few cyclic alkenes in his laboratory. He thought of performing ozonolysis reaction on alkene to produce a dicarbonyl compound followed by aldol reaction to prepare "x". Predict the suitable alkene that can lead to the formation of "x".**
Ans. (1)
Sol.

**67. Consider the given plots of vapour pressure (VP) vs temperature (T/K) Which amongst the following options is correct graphical representation showing `ΔTf` depression in the freezing point of solvent in a solution ?**
Ans. (3)
Sol. On adding non- volatile solute in a solvent, the freezing point of solution decreases.
`Tf<Tf0`
F.P. of solution < F.P. of pure solvent
Also V.P. of solution decreases on adding nonvolatile solute in a solvent.

**68. Which of the following statement is true with respect to `H2O` `NH3` and `CH4` ?**
A. The central atoms of all the molecules are `sp3` hybridized.
B. The `H−O−H` `H−N−H` and `H−C−H` angles in the above molecules are `104.5?` `107.5?` and `109.5?` respectively.
C. The increasing order of dipole moment is `CH4<NH3<H2O`
D. Both `H2O` and `NH3` are Lewis acids and `CH4` is a Lewis base
E. A solution of `NH3` in `H2O` is basic. In this solution `NH3` and `H2O` act as Lowry- Bronsted acid and base respectively.
Choose the correct answer from the options given below :
(1) A, B and C only
(2) C, D and E only
(3) A, D and E only
(4) A, B, C and E only
Ans. (1)
Sol.

Dipole moment `H2O>NH3>CH4`
`H2O` and `NH3` are Lewis Bases
`NH3` act as Lowry- Bronsted base
Hence, A, B & C are correct

**69. Given below are two statements : Statement- I : The conversion proceeds well in the less polar medium. `CH3−CH2−CH2−CH2−Cl→HOCH3−CH2−CH2−CH2−OH+Cl−` Statement- II : The conversion proceeds well in the more polar medium. `CH3−CH2−CH2−CH2−Cl→R1N−CH3−CH2−CH2−CH2−N−RCl+` 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. `CH3−CH2−CH2−CH2−Cl+OH`
Reactant (higher charge density)
Transition state (less charge density)
`⇒` This reaction will proceed faster in less polar medium which will not increase the activation energy value.
`CH3−CH2−CH2−CH2−Cl+R3N`
Reactant (low charge density)
Transition state (Higher charge density)
`⇒` This reaction will proceed faster in more polar medium which will decrease the activation energy value.

**70. The product (A) formed in the following reaction sequence is :**
Ans. (2)
Sol. `CH3−C≡CH→Hg2+,H2SO4CH3−C−CH3`
`CH3−C−CH3→H2/NiCH3−C−CH3`
`CH2−NH2`
`CH2−NH2`

## SECTION-B

**71. `37.8gN2O5` was taken in a 1 L reaction vessel and allowed to undergo the following reaction at `500K`**

2N2O5(g)→2N2O4(g)+O2(g)
**The total pressure at equilibrium was found to be 18.65 bar. Then, `Kp=ΩΩ×10−2` [nearest integer] Assume `N2O5` to behave ideally under these conditions Given : `R=0.082barLmol−1K−1`**

**73. Xg of benzoic acid on reaction with aq. NaHCO3 release `CO2` that occupied 11.2 L volume at STP. X is g.**
Ans. (61)
Sol. `C6H5COOH+NaHCO3→C6H5COONa+`
`+H2O+CO2`
x gm 11.2 L
mole of `C6H5COOH=` mole of `CO2=11.222.4=0.5`
mass of `C6H5COOH=x=0.5×122=61` gm
Ans. 61

**74. Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with `K4[Fe(CN)6]` is : `Cu2+` `Fe3+` `Ba2+` `Ca2+` `NH4+` `Mg2+` `Zn2+`**

NTA
Ans. (3)
Sol. Only `Cu2+` `Fe3+` `Ca2+` `Zn2+` form precipitate with `K4[Fe(CN)6]`

**75. Consider the following reaction occurring in the blast furnace. `Fe3O4(s)+4CO(g)→3Fe(0)+4CO(g)` 'x' kg of iron is produced when `2.32×103` kg `Fe3O4` and `2.8×102` kg CO are brought together in the furnace. The value of 'x' is (nearest integer) {Given : Molar mass of `Fe3O4=232gmol−1` Molar mass of `CO=28gmol−1` Molar mass of `Fe=56gmol−1`**
Ans. (420)
Sol. moles of `Fe3O4=2.32×103×103232=10000mol`
moles of `CO=2.8×102×10328=10000mol`
`Fe3O4+4CO?3Fe+4CO2`
`104mol` `104mol`
CO is L.R.
mole of `Fe=34×104`
mass of `Fe=34×104×561000kg=420kg`
Ans. 420
Ans. (962)
Sol. Initial pressure of `N2O5`
`37.8108×0.082×500`
`=14.35bar1`
`2N2O5?2N2O4+O2`
`t=014.35`
`t=eq14.35−2P2PP`
`PTotalateqb=14.35+P=18.65`
`P=4.3`
`PN2O5=5.75bar`
`PN2O4=8.6bar`
`PO2=4.3bar`
`kp=(8.6)2×(4.3)(5.75)2=9.619=x×10−2`
`x=961.9≈962`
Ans. 962

**72. Standard entropies of `X2` `Y2` and `XY5` are 70, 50 and 110 J `K−1` mol `−1` respectively. The temperature in Kelvin at which the reaction**

12X2+52Y2?XY5 Δ ΔH=−35 kJ mol−1
**Will be at equilibrium is (Nearest integer)**
Ans. (700)
Sol.12X2+52Y2?XY5
ΔSRin0=110−[(12×70)+(52×50)]
=110−160=−50JK−1mol−1
ΔG0=0ateqb
ΔG0=ΔH0−TΔS0
0=−35000−T(−50)
T=700Kelvin
Ans. 700
 

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