JEE Mains -Previous Year Solved Question Paper with Answer Key -CODE D-2017
Answers & Solutions For JEE MAIN- Free PDF Download Link 2017
Answers & Solutions For JEE MAIN 2017
(Mathematics, Physics and Chemistry)
PART-A : MATHEMATICS
-
If S is the set of distinct values of b for which the following system of linear equations
x + y + z = 1
x + ay + z = 1
ax + by + z = 0
has no solution, then S is
(1) An empty set
(2) An infinite set
(3) A finite set containing two or more elements
(4) A singleton
Answer (4)
Sol. |1 1 1|
|1 a 1| = 0
|a b 1|
⇒ -(1 - a)² = 0
⇒ a = 1
For a = 1
Eq. (1) & (2) are identical i.e., x + y + z = 1
To have no solution with x + by + z = 0.
b = 1
-
The following statement (p → q) → [(~p → q) → q] is
(1) A tautology
(2) Equivalent to ~p → q
(3) Equivalent to p → ~q
(4) A fallacy
Answer (1)
Sol.
p q p→q ~p (~p→q) ((~p→q)→q) (p→q)→((~p→q)→q)
T T T F T T T
T F F F T F T
F T T T T T T
F F T T F T T
(a tautology)
-
If 5(tan²x - cos²x) = 2cos2x + 9, then the value of cos4x is
(1) -3/5
(2) 1/3
(3) 2/9
(4) -7/9
Answer (4)
Sol. 5 tan²x = 9 cos²x + 7
5 sec²x - 5 = 9 cos²x + 7
Let cos²x = t
5/t = 9t + 12
9t² + 12t - 5 = 0
t = 1/3 as t ≠ -5/3
cos²x = 1/3, cos2x = 2cos²x - 1 = -1/3
cos4x = 2cos²2x - 1 = 2/9 - 1 = -7/9
-
For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = 1/4 and P(All the three events occur simultaneously) = 1/16. Then the probability that at least one of the events occurs, is
(1) 7/32
(2) 7/16
(3) 7/64
(4) 3/16
Answer (2)
Sol. P(A) + P(B) - P(A∩B) = 1/4
P(B) + P(C) - P(B∩C) = 1/4
P(C) + P(A) - P(A∩C) = 1/4
P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(A∩C) = 3/8
∴ P(A∩B∩C) = 1/16
∴ P(A∪B∪C) = 3/8 + 1/16 = 7/16
===== Page 3 =====
-
Let ω be a complex number such that 2ω + 1 = z where z = √-3. If
Answer (1)
Sol. 2ω + 1 = z, z = √3 i
ω = (-1 + √3 i)/2 → Cube root of unity.
C1 → C1 + C2 + C3
-
Let k be an integer such that the triangle with vertices (k, -3k), (5, k) and (-k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point
Answer (4)
(k² - 7k + 10) + 4k² + 20k = ±56
5k² + 13k + 10 = ±56
5k² + 13k - 46 = 0
5k² + 13k - 46 = 0
k = (-13 ± √(169 + 920))/10
= 2, -4.6
reject
For k = 2
Equation of AD
x = 2 ...(i)
Also equation of BE
y - 2 = 1/2 (x - 5)
2y - 4 = x - 5
x - 2y - 1 = 0
Solving (i) & (ii), 2y = 1
y = 1/2
Orthocentre is (2, 1/2)
-
Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is
(1) 12.5
(2) 10
(3) 25
(4) 30
Answer (3)
Sol.
2r + θr = 20 ...(i)
===== Page 4 =====
-
If the image of the point P(1, -2, 3) in the plane, 2x + 3y - 4z + 22 = 0 measured parallel to the line, x/1 = y/4 = z/5 is Q, then PQ is equal to
(1) 3√5
(2) 2√42
(3) √42
(4) 6√5
A to be maximum
dA/dr = 10 - 2r = 0 ⇒ r = 5
d²A/dr² = -2 < 0
Hence for r = 5 A is maximum
Now, 10 + θ·5 = 20 ⇒ θ = 2 (radian)
Area = 2/(2π) × π(5)² = 25 sq m
-
The area (in sq. units) of the region
{(x,y): x ≥ 0, x + y ≤ 3, x² ≤ 4y and y ≤ 1 + √x}
is
(1) 59/12
(2) 3/2
(3) 7/3
(4) 5/2
Answer (4)
Sol.
Area of shaded region
= ∫?¹(√x + 1 - x²/4)dx + ∫?²((3 - x) - x²/4)dx
= 5/2 sq. unit
Answer (2)
Sol. Equation of PQ, (x - 1)/1 = (y + 2)/4 = (z - 3)/5
Let M be (λ + 1, 4λ - 2, 5λ + 3)
As it lies on 2x + 3y - 4z + 22 = 0
λ = 1
For Q, λ = 2
Distance PQ = 2√(1² + 4² + 5²) = 2√42
-
If for x ∈ (0, 1/4), the derivative of tan?¹(6x√x/(1 - 9x³)) is √x · g(x), then g(x) equals
(1) 9/(1 + 9x³)
(2) 3x/(1 - 9x³)
(3) 3x/(1 - 9x³)
(4)
Answer (1)
Sol. f(x) = 2 tan?¹(3x√x) For x ∈ (0, 1/4)
f'(x) = 9√x/(1 + 9x³)
g(x) = 9/(1 + 9x³)
===== Page 5 =====
-
If (2 + sin x)dy/dx + (y + 1)cos x = 0 and y(0) = 1, then y(π/2) is equal to
(1) 1/3
(2) -2/3
(3) 1/3
(4) 4/3
Answer (1)
Sol. (2 + sin x)dy/dx + (y + 1)cos x = 0
y(0) = 1, y(π/2) = ?
1/(y + 1) dy + cos x/(2 + sin x) dx = 0
ln|y + 1| + ln(2 + sin x) = ln C
(y + 1)(2 + sin x) = C
Put x = 0, y = 1
(1 + 1)·2 = C ⇒ C = 4
Now, (y + 1)(2 + sin x) = 4
For, x = π/2
(y + 1)(2 + 1) = 4
y + 1 = 4/3
y = 4/3 - 1 = 1/3
-
Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If ∠BPC = β then tan β is equal to
(1) 6/7
(2) 1/4
(3) 2/9
(4) 4/9
Answer (3)
Sol.
tan θ = 1/4
tan(θ + β) = 1/2
∴ (1/4 + tanβ)/(1 - 1/4 tanβ) = 1/2
Solving tanβ = 2/9
-
If A = [2 -3; -4 1], then adj(3A² + 12A) is equal to
(1) [72 -84; -63 51]
(2)
(3) [51 84; 63 72]
(4)
Answer (2)
Sol. A = [2 -3; -4 1]
|A - λI| = |2 - λ -3; -4 1 - λ|
= (2 - 2λ - λ + λ²) - 12
f(λ) = λ² - 3λ - 10
∴ A satisfies f(λ)
∴ A² - 3A - 10I = 0
A² - 3A = 10I
3A² - 9A = 30I
3A² + 12A = 30I + 21A
= [30 0; 0 30] + [42 -63; -84 21]
= [72 -63; -84 51]
adj(3A² + 12A) = [51 63; 84 72]
===== Page 6 =====
-
For any three positive real numbers a, b and c, 9(25a² + b²) + 25(c² - 3ac) = 15b(3a + c). Then
(1) b, c and a are in G.P.
(2) b, c and a are in A.P.
(3) a, b and c are in A.P.
(4) a, b and c are in G.P.
Answer (2)
9(25a² + b²) + 25(c² - 3ac) = 15b(3a + c)
⇒ (15a)² + (3b)² + (5c)² - 45ab - 15bc - 75ac = 0
⇒ (15a - 3b)² + (3b - 5c)² + (15a - 5c)² = 0
It is possible when
15a - 3b = 0 and 3b - 5c = 0 and 15a - 5c = 0
15a = 3b = 5c
a/1 = b/5 = c/3
∴ b, c, a are in A.P.
-
The distance of the point (1, 3, -7) from the plane passing through the point (1, -1, -1), having normal perpendicular to both the lines (x - 1)/1 = (y + 2)/-2 = (z - 4)/3 and (x - 2)/2 = (y + 1)/-1 = (z + 7)/-1, is
(1) 20/√74
(2) 10/√83
(3) 5/√83
(4) 10/√74
Answer (2)
Sol. Let the plane be
a(x - 1) + b(y + 1) + c(z + 1) = 0
It is perpendicular to the given lines
a - 2b + 3c = 0
2a - b - c = 0
Solving, a:b:c = 5:7:3
The plane is 5x + 7y + 3z + 5 = 0
Distance of (1, 3, -7) from this plane = 10/√83
-
Let I_n = ∫tan?x dx, (n > 1). If I? + I? = a tan?x + bx? + C, where C is a constant of integration, then the ordered pair (a,b) is equal to
(1) (-1/5, 1)
(2) (1/5, 0)
(3) (1/5, -1)
(4) (-1/5, 0)
Answer (2)
Sol. I_n = ∫tan?x dx, n > 1
I? + I? = ∫(tan?x + tan?x)dx
= ∫tan?x sec²x dx
Let tan x = t
sec²x dx = dt
= ∫t?dt
= t?/5 + C
= 1/5 tan?x + C
a = 1/5, b = 0
-
The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of its directrices is x = -4, then the equation of the normal to it at (1, 3/2) is
(1) 2y - x = 2
(2) 4x - 2y = 1
(3) 4x + 2y = 7
(4) x + 2y = 4
Answer (2)
Sol.
e = 1/2
-a/e = -4
-a = -4 × e
===== Page 7 =====
a = 2
Now, b² = a²(1 - e²) = 3
Equation to ellipse
x²/4 + y²/3 = 1
Equation of normal is
(x - 1)/(1/4) = (y - 3/2)/(2/(2×3)) ⇒ 4x - 2y - 1 = 0
-
A hyperbola passes through the point P(√2, √3) and has foci at (±2, 0). Then the tangent to this hyperbola at P also passes through the point
(1) (3√2, 2√3)
(2) (2√2, 3√3)
(3) (√3, √2)
(4) (-√2, -√3)
Answer (2)
Sol. x²/a² - y²/b² = 1
and 2/a² - 3/b² = 1
and 2/(4 - b²) - 3/b² = 1
⇒ b² = 3
∴ a² = 1
∴ x² - y²/3 = 1
∴ Tangent at P(√2, √3) is √2 x - y/√3 = 1
Clearly it passes through (2√2, 3√3)
-
The function f: R → [-1/2, 1/2] defined as
f(x) = x/(1 + x²), is
(1) Invertible
(2) Injective but not surjective
(3) Surjective but not injective
(4) Neither injective nor surjective
Answer (3)
Sol. f(x) = x/(1 + x²)
f'(x) = ((1 + x²)·1 - x·2x)/(1 + x²)² = (1 - x²)/(1 + x²)²
f(x) changes sign in different intervals.
Not injective.
y = x/(1 + x²)
yx² - x + y = 0
For y ≠ 0
D = 1 - 4y² ≥ 0 ⇒ y ∈ [-1/2, 1/2] - {0}
For, y = 0 ⇒ x = 0
Part of range
∴ Range: [-1/2, 1/2]
Surjective but not injective.
-
lim_{x→π/2} (cot x - cos x)/(π - 2x)³ equals
(1) 1/24
(2) 1/16
(3) 1/8
(4) 1/4
Answer (2)
Sol. lim_{x→π/2} (cot x - cos x)/(π - 2x)³
Put, π/2 - x = t
lim_{t→0} (tan t - sin t)/(8t³)
= lim_{t→0} (sin t · 2sin²(t/2))/(8t³)
= 1/16
===== Page 8 =====
-
Let a = 2i + j - 2k and b = i + j. Let c be a vector such that |c - a| = 3, |(a × b) × c| = 3 and the angle between c and a × b be 30°. Then a·c is equal to
(1) 25/8
(2) 2
(3) 5
(4) 1/8
Answer (2)
|(a × b) × c| = 3, a × b = 2i - 2j + k
⇒ |a × b||c| sin 30° = 3, |a| = 3 = |a × b|
⇒ |c| = 2
|c - a| = 3
⇒ |c|² + |a|² - 2(a·c) = 9
a·c = (9 - 3 - 2)/2 = 2
-
The normal to the curve y(x - 2)(x - 3) = x + 6 at the point where the curve intersects the y-axis passes through the point
(1) (-1/2, -1/2)
(2) (1/2, 1/2)
(3) (1/2, -1/3)
(4) (1/2, 1/3)
Answer (2)
Sol. y(x - 2)(x - 3) = x + 6
At y-axis, x = 0, y = 1
Now, on differentiation.
dy/dx (x - 2)(x - 3) + y(2x - 5) = 1
dy/dx (6) + 1(-5) = 1
dy/dx = 6/6 = 1
Now slope of normal = -1
Equation of normal y - 1 = -1(x - 0)
y + x - 1 = 0 ...(i)
Line (i) passes through (1/2, 1/2)
-
If two different numbers are taken from the set {0,1,2,3,...,10} then the probability that their sum as well as absolute difference are both multiple of 4, is
(1) 6/55
(2) 12/55
(3) 14/45
(4) 7/55
Answer (1)
Sol. Total number of ways = ¹¹C? = 55
Favourable ways are
(0,4), (0,8), (4,8), (2,6), (2,10), (6,10)
Probability = 6/55
-
A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is
(1) 485
(2) 468
(3) 469
(4) 484
Answer (1)
Sol. X(4L3G) Y(3L4G)
3L0G 0L3G
2L1G 1L2G
1L2G 2L1G
0L3G 3L0G
Required number of ways
= ?C?·?C? + (?C?·³C?)² + (?C?·³C?)² + (³C?)²
= 16 + 324 + 144 + 1
= 485
-
The value of
(²C? - ¹?C?) + (²C? - ¹?C?) + (²C? - ¹?C?) + (²C? - ¹?C?) + ... + (²C?? - ¹?C??) is
(1) 2²¹ - 2¹¹
(2) 2²¹ - 2¹?
(3) 2²? - 2?
(4) 2²? - 2¹?
===== Page 9 =====
Answer (4)
Sol. ²C? + ²C? + ... + ²C?? = 1/2[²C? + ²C? + ... + ²C??] - 1
= 2²? - 1
(¹?C? + ¹?C? + ... + ¹?C??) = 2¹? - 1
∴ Required sum = (2²? - 1) - (2¹? - 1)
= 2²? - 2¹?
-
A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is
(1) 12/5
(2) 6
(3) 4
(4) 6/25
Answer (1)
Sol. n = 10
p(Probability of drawing a green ball) = 15/25
∴ p = 3/5, q = 2/5
var(X) = n.p.q
= 10·6/25 = 12/5
-
Let a, b, c ∈ R. If f(x) = ax² + bx + c is such that a + b + c = 3 and
f(x + y) = f(x) + f(y) + xy, ∀ x, y ∈ R,
then Σ_{n=1}^{10} f(n) is equal to
(1) 330
(2) 165
(3) 190
(4) 255
Answer (1)
Sol. As, f(x + y) = f(x) + f(y) + xy
Given, f(1) = 3
Putting, x = y = 1 ⇒ f(2) = 2f(1) + 1 = 7
Similarly, x = 1, y = 2 ⇒ f(3) = f(1) + f(2) + 2 = 12
Now, Σ_{n=1}^{10} f(n) = f(1) + f(2) + f(3) + ... + f(10)
= 3 + 7 + 12 + 18 + ... = S (let)
Now, S_n = 3 + 7 + 12 + 18 + ... + t_n
Again, S_n = 3 + 7 + 12 + ... + t_{n-1} + t_n
We get, t_n = 3 + 4 + 5 + ... n terms
= n(n + 5)/2
i.e., S_n = Σ_{n=1}^{n} t_n
= 1/2[Σn² + 5Σn]
= n(n + 1)(n + 8)/6
So, S?? = 10×11×18/6 = 330
-
The radius of a circle, having minimum area, which touches the curve y = 4 - x² and the lines, y = |x| is
(1) 2(√2 + 1)
(2) 2(√2 - 1)
(3) 4(√2 - 1)
(4) 4(√2 + 1)
Answer (3)
Sol.
x² = -(y - 4)
Let a point on the parabola P(t/2, 4 - t²/4)
Equation of normal at P is
y + t²/4 - 4 = 1/t(x - t/2)
⇒ x - ty - t³/4 + 7/2 t = 0
It passes through centre of circle, say (0, k)
-tk - t³/4 + 7/2 t = 0 ...(i)
t = 0, t² = 14 - 4k
===== Page 10 =====
Radius = r = 0 - k (Length of perpendicular from (0, k) to y = x)
⇒ r = k/√2
Equation of circle is x² + (y - k)² = k²/2
It passes through point P
t²/4 + (4 - t²/4 - k)² = k²/2
t? + t²(8k - 28) + 8k² - 128k + 256 = 0 (ii)
For t = 0 ⇒ k² - 16k + 32 = 0
k = 8 ± 4√2
∴ r = k/√2 = 4(√2 - 1) (discarding 4(√2 + 1)) (iii)
For t = ±√(14 - 4k)
(14 - 4k)² + (14 - 4k)(8k - 28) + 8k² - 128k + 256 = 0
2k² + 4k - 15 = 0
k = (-2 ± √34)/2
∴ r = k/√2 = (√17 - √2)/2 (Ignoring negative value of r)
From (iii) & (iv), r_min = (√17 - √2)/2
But from options, r = 4(√2 - 1)
-
If, for a positive integer n, the quadratic equation, x(x + 1) + (x + 1)(x + 2) + ... + (x + n - 1)(x + n) = 10n has two consecutive integral solutions, then n is equal to
(1) 12
(2) 9
(3) 10
(4) 11
Answer (4)
Sol. Rearranging equation, we get
n x² + {1 + 3 + 5 + ... + (2n - 1)}x + {1·2 + 2·3 + ... + (n - 1)n} = 10n
⇒ n x² + n²x + (n - 1)n(n + 1)/3 = 10n
⇒ x² + n x + (n² - 31)/3 = 0 ...(i)
Given difference of roots = 1
⇒ |α - β| = 1
⇒ D = 1
⇒ n² - 4/3(n² - 31) = 1
So, n = 11
-
The integral ∫_{π/4}^{3π/4} dx/(1 + cos x) is equal to
(1) -2
(2) 2
(3) 4
(4) -1
Answer (2)
Sol. ∫{π/4}^{3π/4} dx/(2cos²(x/2)) dx = 1/2 ∫{π/4}^{3π/4} sec²(x/2) dx
= 1/2 [tan(x/2)/(1/2)]_{π/4}^{3π/4}
= tan(3π/8) - tan(π/8)
tan(π/8) = √(1 - cosπ/4)/(1 + cosπ/4) = (√2 - 1)/(√2 + 1) = √2 - 1
tan(3π/8) = √(1 - cos3π/4)/(1 + cos3π/4) = (√2 + 1)/(√2 - 1) = √2 + 1
= (√2 + 1) - (√2 - 1) = 2
===== Page 11 =====
-
A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0 there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by
(1) t = T/log(1.3)
(2) t = T/2 log2/log1.3
(3) t = T log1.3/log2
(4) t = T log(1.3)
Answer (3)
Sol. (N? - N?e^{-λt})/(N?e^{-λt}) = 0.3
⇒ e^{λt} = 1.3
∴ λt = ln 1.3
∴ (ln2)/T t = ln 1.3
t = T·ln(1.3)/ln2
t = T·log(1.3)/log2
-
The following observations were taken for determining surface tension T of water by capillary method:
diameter of capillary, D = 1.25 × 10?² m
rise of water, h = 1.45 × 10?² m
Using g = 9.80 m/s² and the simplified relation T = rhg/2 × 10³ N/m, the possible error in surface tension is closest to
(1) 10%
(2) 0.15%
(3) 1.5%
(4) 2.4%
Answer (3)
Sol. ΔT/T × 100 = ΔD/D × 100 + Δh/h × 100
= 0.01/1.25 × 100 + 0.01/1.45 × 100
= 100/125 + 100/145
= 0.8 + 0.689
= 1.489 ≈ 1.5%
-
An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If λ_min is the smallest possible wavelength of X-ray in the spectrum, the variation of log λ_min with log V is correctly represented in
Answer (2)
Sol. In X-ray tube
λ_min = hc/eV
ln λ_min = ln(hc/e) - ln V
Slope is negative
Intercept on y-axis is positive
===== Page 12 =====
-
The moment of inertia of a uniform cylinder of length ? and radius R about its perpendicular bisector is I. What is the ratio ?/R such that the moment of inertia is minimum?
(1) 3/√2
(2) √3/2
(3) √3/2
(4) 1
Answer (2)
Sol.
I = mR²/4 + m?²/12
I = m/4 [R² + ?²/3]
= m/4 [V/(π?) + ?²/3]
dI/d? = m/4 [-V/(π?²) + 2?/3] = 0
V = 2?/π?² = 2?/3
πR²? = 2π?³/3
?²/R² = 3/2
?/R = √(3/2)
-
A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is
(1) 2g/(3?) cosθ
(2) 3g/(2?) sinθ
(3) 2g/(3?) sinθ
(4) 3g/(2?) cosθ
Answer (2)
Sol. Torque at angle θ
τ = Mg sinθ · ?/2
τ = Iα
Iα = Mg sinθ ?/2 ∴ I = M?²/3
M?²/3 · α = Mg sinθ ?/2
===== Page 13 =====
-
C_p and C_v are specific heats at constant pressure and constant volume respectively. It is observed that
C_p - C_v = a for hydrogen gas
C_p - C_v = b for nitrogen gas
The correct relation between a and b is
(1) a = 28b
(2) a = 1/14 b
(3) a = b
(4) a = 14b
Answer (4)
Sol. Let molar heat capacity at constant pressure = X_p and molar heat capacity at constant volume = X_v
X_p - X_v = R
MC_p - MC_v = R
C_p - C_v = R/M
For hydrogen; a = R/2
For N?; b = R/28
a/b = 14
a = 14b
-
A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by:
(Given: room temperature = 30°C, specific heat of copper = 0.1 cal/gm°C)
(1) 825°C
(2) 800°C
(3) 885°C
(4) 1250°C
Answer (3)
Sol. 100 × 0.1 × (t - 75) = 100 × 0.1 × 45 + 170 × 1 × 45
10t - 750 = 450 + 7650
10t = 1200 + 7650
10t = 8850
t = 885°C
-
In amplitude modulation, sinusoidal carrier frequency used is denoted by ω_c and the signal frequency is denoted by ω_m. The bandwidth (Δω_m) of the signal is such that Δω_m << ω_c. Which of the following frequencies is not contained in the modulated wave?
(1) ω_c - ω_m
(2) ω_m
(3) ω_c
(4) ω_m + ω_c
Answer (2)
Sol. Modulated wave has frequency range.
ω_c ± ω_m
∴ Since ω_c >> ω_m
∴ ω_m is excluded.
-
The temperature of an open room of volume 30 m³ increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 × 10? Pa. If n_i and n_f are the number of molecules in the room before and after heating, then n_f - n_i will be
(1) -2.5 × 10²?
(2) -1.61 × 10²³
(3) 1.38 × 10²³
(4) 2.5 × 10²?
Answer (1)
Sol. n? = initial number of moles
n? = P?V?/RT? = 10?×30/(8.3×290) ≈ 1.24×10³
n? = final number of moles
= P?V?/RT? = 10?×30/(8.3×300) ≈ 1.20×10³
Change of number of molecules:
n_f - n_i = (n? - n?)×6.023×10²³
= -2.5×10²?
===== Page 14 =====
-
In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is
(1) 15.6 mm
(2) 1.56 mm
(3) 7.8 mm
(4) 9.75 mm
Answer (3)
Sol. For λ?
y = mλ?D/d
For λ?
y = nλ?D/d
⇒ m/n = λ?/λ? = 4/5
For λ?
y = mλ?D/d, λ? = 650 nm
= 7.8 mm
-
A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λ_A to λ_B after the collision is
(1) λ_A/λ_B = 1/2
(2) λ_A/λ_B = 1/3
(3) λ_A/λ_B = 2
(4) λ_A/λ_B = 2/3
Answer (3)
Sol. v? = (m? - m?)v/(m? + m?) + 0, m? = m, m? = m/2
= v/3
∴ p? = m[v/3]
v? = 2m?v/(m? + m?) + 0
= 4v/3
∴ p? = m/2[4v/3] = 2mv/3
∴ de-Broglie wavelength λ_A/λ_B = p_B/p? = 2:1
-
A magnetic needle of magnetic moment 6.7 × 10?² Am² and moment of inertia 7.5 × 10?? kg m² is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is
(1) 8.76 s
(2) 6.65 s
(3) 8.89 s
(4) 6.98 s
Answer (2)
Sol. T = 2π√(I/MB)
= 2π√(7.5×10??/(6.7×10?²×0.01)) = 2π/10 × 1.06
For 10 oscillations,
t = 10T = 2π × 1.06
= 6.6568 ≈ 6.65 s
-
An electric dipole has a fixed dipole moment p, which makes angle θ with respect to x-axis. When subjected to an electric field E? = Ei, it experiences a torque T? = τk. When subjected to another electric field E? = √3E? j it experiences a torque T? = -T?. The angle θ is
(1) 90°
(2) 30°
(3) 45°
(4) 60°
===== Page 15 =====
Answer (4)
Sol.
p = p cosθ i + p sinθ j
E? = Ei
T? = p × E?
= (p cosθ i + p sinθ j) × E(i)
τk = pE sinθ (-k)
-
In a coil of resistance 100Ω a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is
(1) 275 Wb
(2) 200 Wb
(3) 225 Wb
(4) 250 Wb
Answer (4)
Sol. ε = dφ/dt
iR = dφ/dt
∫dφ = R∫i dt
Magnitude of change in flux = R × (area under current vs time graph)
= 100 × 1/2 × 1/2 × 10
= 250 Wb
-
A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 second will be
(1) 18 J
(2) 4.5 J
(3) 22 J
(4) 9 J
Answer (2)
Sol. 6t = 1·dv/dt
∫dv = ∫6t dt
v = 6[t²/2]?¹
= 3 ms?¹
W = ΔKE = 1/2 × 1 × 9 = 4.5 J
-
Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = λ?/λ?, is given by
(1) r = 1/3
(2) r = 4/3
(3) r = 2/3
(4) r = 3/4
===== Page 16 =====
-
Answer (1)
Sol. From energy level diagram
λ? = hc/E
λ? = hc/(E/3)
∴ λ?/λ? = 1/3
-
In the above circuit the current in each resistance is
(1) 0 A
(2) 1 A
(3) 0.25 A
(4) 0.5 A
Answer (1)
Sol. The potential difference in each loop is zero.
No current will flow.
-
A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?
Answer (4)
Sol. Acceleration is constant and negative
-
A capacitance of 2μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1μF capacitors are available which can withstand a potential difference of not more than 300 V.
The minimum number of capacitors required to achieve this is
(1) 32
(2) 2
(3) 16
(4) 24
Answer (1)
Sol. Following arrangement will do the needful:
8 capacitors of 1μF in parallel with four such branches in series.
===== Page 17 =====
-
In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be:
(1) CE r?/(r? + r)
(2) CE
(3) CE r?/(r? + r)
(4) CE r?/(r + r?)
Answer (4)
Sol. In steady state, flow of current through capacitor will be zero.
i = E/(r + r?)
V_C = i r?C = Er?C/(r + r?)
V_C = CE r?/(r + r?)
-
In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be
(1) 180°
(2) 45°
(3) 90°
(4) 135°
Answer (1)
Sol. In common emitter configuration for n-p-n transistor, phase difference between output and input voltage is 180°.
-
Which of the following statements is false?
(1) Kirchhoff's second law represents energy conservation
(2) Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude
(3) In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed
(4) A rheostat can be used as a potential divider
Answer (3)
Sol. In a balanced Wheatstone bridge, the null point remains unchanged even if cell and galvanometer are interchanged.
-
A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like:
Answer (1)
Sol. K.E = 1/2 mω²A² cos²ωt
===== Page 18 =====
-
An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 × 10? ms?¹)
(1) 15.3 GHz
(2) 10.1 GHz
(3) 12.1 GHz
(4) 17.3 GHz
Answer (4)
Sol. For relativistic motion
f = f?√((c + v)/(c - v)); v = relative speed of approach
f = 10√((c + c/2)/(c - c/2)) = 10√3 = 17.3 GHz
-
A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of
(1) 1/81
(2) 9
(3) 1/9
(4) 81
Answer (2)
Sol. v_f/v_i = 9³
Density remains same
So, mass ∝ Volume
m_f/m_i = 9³
(Area)_f = 9²
Stress = (Mass)×g/Area
σ?/σ? = (m_f/m_i)(A_i/A_f)
= 9³/9² = 9
-
When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 - 10 V is
(1) 4.005 × 10³ Ω
(2) 1.985 × 10³ Ω
(3) 2.045 × 10³ Ω
(4) 2.535 × 10³ Ω
Answer (2)
Sol. i_g = 5 × 10?³ A
G = 15 Ω
Let series resistance be R.
V = i_g(R + G)
10 = 5 × 10?³(R + 15)
R = 2000 - 15 = 1985 = 1.985 × 10³ Ω
-
The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius)
Answer (1)
Sol.
Variation of g inside earth surface
d < R = g = Gm/R² · d
===== Page 19 =====
-
An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by
(1) 3PKα
(2) P/(3αK)
(3)
(4)
Answer (2)
Sol. K = ΔP/(-ΔV/V)
ΔV/V = P/K
∴ V = V?(1 + γΔt)
ΔV/V? = γΔt
∴ P/K = γΔt ⇒ Δt = P/(γK) = P/(3αK)
-
A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is
(1) Real and at a distance of 6 cm from the convergent lens
(2) Real and at a distance of 40 cm from convergent lens
(3) Virtual and at a distance of 40 cm from convergent lens
(4) Real and at a distance of 40 cm from the divergent lens
Answer (2)
Sol. For converging lens
u = -40 cm which is equal to 2f
Image will be real and at a distance of 40 cm from convergent lens.
-
A body of mass m = 10?² kg is moving in a medium and experiences a frictional force F = -kv². Its initial speed is v? = 10 ms?¹. If, after 10 s, its energy is 1/8 mv?², the value of k will be
(1) 10?¹ kg m?¹ s?¹
(2) 10?³ kg m?¹
(3) 10?³ kg s?¹
(4) 10?? kg m?¹
Answer (4)
Sol. k_f/k_i = (1/8 mv?²)/(1/2 mv?²) = 1/4
v_f/v_i = 1/2
v_f = v?/2
-kv² = m dv/dt
∫{v?}^{v?/2} dv/v² = ∫?^{t?} -k dt
[-1/v]{v?}^{v?/2} = -k/m t?
1/v? - 2/v? = -k/m t?
1/v? = k/m t?
k = m/(v?t?)
= 10?²/(10×10) = 10?? kg m?¹
===== Page 20 =====
-
1 gram of a carbonate (M?CO?) on treatment with excess HCl produces 0.01186 mole of CO?. The molar mass of M?CO? in g mol?¹ is
(1) 84.3
(2) 118.6
(3) 11.86
(4) 1186
Answer (1)
Sol. M?CO? + 2HCl → 2MCl + H?O + CO?
n_{M?CO?} = n_{CO?}
1/M_{M?CO?} = 0.01186
M_{M?CO?} = 1/0.01186
= 84.3 g/mol
-
Given
C(graphite) + O?(g) → CO?(g)
Δ_rH° = -393.5 kJ mol?¹
H?(g) + 1/2 O?(g) → H?O(l)
Δ_rH° = -285.8 kJ mol?¹
CO?(g) + 2H?O(l) → CH?(g) + 2O?(g)
Δ_rH° = +890.3 kJ mol?¹
Based on the above thermochemical equations, the value of Δ_rH° at 298 K for the reaction
C(graphite) + 2H?(g) → CH?(g) will be
(1) +144.0 kJ mol?¹
(2) -74.8 kJ mol?¹
(3) -144.0 kJ mol?¹
(4) +74.8 kJ mol?¹
Answer (2)
Sol. C(graphite) + O?(g) → CO?(g)
Δ_rH° = -393.5 kJ mol?¹ ...(i)
H?(g) + 1/2 O?(g) → H?O(l)
Δ_rH° = -285.8 kJ mol?¹ ...(ii)
CO?(g) + 2H?O(l) → CH?(g) + 2O?(g)
Δ_rH° = 890.3 kJ mol?¹ ...(iii)
By applying the operation
(i) + 2×(ii) + (iii), we get
C(graphite) + 2H?(g) → CH?(g)
Δ_rH° = -393.5 - 285.8×2 + 890.3
= -74.8 kJ mol?¹
-
The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be
K_f for benzene = 5.12 K kg mol?¹
(1) 80.4%
(2) 74.6%
(3) 94.6%
(4) 64.6%
Answer (3)
Sol. ΔT_f = i K_f m
0.45 = i × 5.12 × (0.2/60)/0.02
i = 0.527
2CH?COOH ? (CH?COOH)?
1 - α
i = 1 - α/2
i = 0.527
i = 1 - α/2
⇒ α/2 = 0.473
⇒ α = 0.946
∴ % association = 94.6%
-
The most abundant elements by mass in the body of a healthy human adult are:
Oxygen (61.4%) Carbon (22.9%) Hydrogen (10.0%) and Nitrogen (2.6%)
The weight which a 75 kg person would gain if all ¹H atoms are replaced by ²H atoms is
(1) 37.5 kg
(2) 7.5 kg
(3) 10 kg
(4) 15 kg
Answer (2)
Sol. Mass of hydrogen = 10/100 × 75 = 7.5 kg
Replacing ¹H by ²H would replace 7.5 kg with 15 kg
Net gain = 7.5 kg
-
ΔU is equal to
(1) Isobaric work
(2) Adiabatic work
(3) Isothermal work
(4) Isochoric work
Answer (2)
Sol. For adiabatic process, q = 0
As per 1st law of thermodynamics,
ΔU = W
===== Page 21 =====
-
The formation of which of the following polymers involves hydrolysis reaction?
(1) Bakelite
(2) Nylon 6, 6
(3) Terylene
(4) Nylon 6
Answer (4)
Sol. Caprolactam is hydrolysed to produce caproic acid which undergoes condensation to produce Nylon-6.
-
Given
E°{Cl?/Cl?} = 1.36 V, E°{Cr³?/Cr} = -0.74 V
E°{Cr?O?²?/Cr³?} = 1.33 V, E°{MnO??/Mn²?} = 1.51 V
Among the following, the strongest reducing agent is
(1) Mn²?
(2) Cr³?
(3) Cl?
(4) Cr
Answer (4)
Sol. For Cr³?, E°{Cr³?/Cr?O?²?} = -1.33 V
For Cl?, E°{Cl?/Cl?} = -1.36 V
For Cr, E°{Cr/Cr³?} = 0.74 V
For Mn²?, E°{Mn²?/MnO??} = -1.51 V
Positive E° is for Cr hence it is strongest reducing agent.
-
The Tyndall effect is observed only when following conditions are satisfied
(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used.
(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used
(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude
(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude
(1) (b) and (d)
(2) (a) and (c)
(3) (b) and (c)
(4) (a) and (d)
Answer (1)
Sol. For Tyndall effect refractive index of dispersion phase and dispersion medium must differ significantly. Secondly, size of dispersed phase should not differ much from wavelength used.
-
In the following reactions, ZnO is respectively acting as a/an
ZnO + Na?O → Na?ZnO?
ZnO + CO? → ZnCO?
(1) Base and base
(2) Acid and acid
(3) Acid and base
(4) Base and acid
Answer (3)
Sol. In (a), ZnO acts as acidic oxide as Na?O is basic oxide.
In (b), ZnO acts as basic oxide as CO? is acidic oxide.
-
Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?
===== Page 22 =====
-
On treatment of 100 mL of 0.1 M solution of CoCl?·6H?O with excess AgNO?, 1.2×10²² ions are precipitated. The complex is
Sol. Sugars in which there is free anomeric -OH group are reducing sugars
-
The major product obtained in the following reaction is
(1) C?H?CH = CHC?H?
(2) (+)C?H?CH(O'Bu)CH?C?H?
(3) (-)C?H?CH(O'Bu)CH?C?
(4) (±)C?H?CH(O'Bu)CH?C?H?
Answer (1)
Sol.
(E-2)
-
Which of the following species is not paramagnetic?
(1) CO
(2) O?
(3) B?
(4) NO
Answer (1)
Sol. CO has 14 electrons (even): it is diamagnetic
NO has 15e? (odd): it is paramagnetic and has 1 unpaired electron in π²p molecular orbital.
B? has 10e? (even) but still paramagnetic and has two unpaired electrons in π2p_x and π2p_y (s-p mixing).
O? has 16e? (even) but still paramagnetic and has two unpaired electrons in π2p_x and π2p_y molecular orbitals.
(1) [Co(H?O)?Cl?]·3H?O
(2) [Co(H?O)?]Cl?
(3) [Co(H?O)?Cl]Cl?·H?O
(4) [Co(H?O)?Cl?]Cl·2H?O
Answer (3)
Sol. Millimoles of AgNO? = (1.2×10²²)/(6×10²³) × 1000 = 20
Millimoles of CoCl?·6H?O = 0.1×100 = 10
Each mole of CoCl?·6H?O gives two chloride ions.
[Co(H?O)?Cl]Cl?·H?O
-
pK_a of a weak acid (HA) and pK_b of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB) solution is
(1) 6.9
(2) 7.0
(3) 1.0
(4) 7.2
Answer (1)
Sol. pH = 7 + 1/2(pK_a - pK_b)
= 7 + 1/2(3.2 - 3.4)
= 6.9
-
The increasing order of the reactivity of the following halides for the S_N1 reaction is
I. CH?CH(Cl)CH?CH?
II. CH?CH?CH?Cl
III. p-H?CO-C?H?-CH?Cl
(1) (II) < (I) < (III)
(2) (I) < (III) < (II)
(3) (II) < (III) < (I)
(4) (III) < (II) < (I)
===== Page 23 =====
Answer (1)
Sol. Rate of S_N1 reaction ∝ stability of carbocation
I. CH?-CH(Cl)-CH?-CH? → CH?-CH?-CH?-CH?
II. CH?-CH?-CH?-Cl → CH?-CH?-CH??
III. CH?-Cl → CH??
OCH? OCH?
So, II < I < III
Increase stability of carbocation and hence increase reactivity of halides.
-
Both lithium and magnesium display several similar properties due to the diagonal relationship, however, the one which is incorrect, is
(1) Both form soluble bicarbonates
(2) Both form nitrides
(3) Nitrates of both Li and Mg yield NO? and O? on heating
(4) Both form basic carbonates
Answer (4)
Sol. Mg forms basic carbonate
3MgCO?·Mg(OH)?·3H?O but no such basic carbonate is formed by Li.
-
The correct sequence of reagents for the following conversion will be
(1) CH?MgBr, H?/CH?OH, [Ag(NH?)?]?OH?
(2) CH?MgBr, [Ag(NH?)?]?OH?, H?/CH?OH
(3) [Ag(NH?)?]?OH?, CH?MgBr, H?/CH?OH
(4) [Ag(NH?)?]?OH?, H?/CH?OH, CH?MgBr
Answer (4)
Sol.
(i) CH?MgBr (3 moles)
(ii) H?O
-
The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are
(1) ClO?? and ClO??
(2) Cl? and ClO?
(3) Cl? and ClO??
(4) ClO? and ClO??
Answer (2)
Sol. Cl? + 2NaOH → NaCl + NaOCl + H?O
Cold & dilute Sodium hypochlorite
-
Which of the following compounds will form significant amount of meta product during mono-nitration reaction?
(1) OCOCH?
(2) NH?
(3) NHCOCH?
(4) OH
Answer (2)
Sol.
NH? → NH?? → NO??
NH?? NH?? NH??
===== Page 24 =====
-
3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is
(1) Zero
(2) Two
(3) Four
(4) Six
Answer (3)
Sol. CH?-CH=C(CH?)-CH?-CH? --HBr/R?O?--> product
3-methyl pent-2-ene
CH?-CH(Br)-C(CH?)(H)-CH?-CH?
Product (X)
Since product (X) contains two chiral centres and it is unsymmetrical.
So, its total stereoisomers = 2² = 4
-
Two reactions R? and R? have identical preexponential factors. Activation energy of R? exceeds that of R? by 10 kJ mol?¹. If k? and k? are rate constants for reactions R? and R? respectively at 300 K then ln(k?/k?) is equal to
(R = 8.314 J mole?¹ K?¹)
(1) 12
(2) 6
(3) 4
(4) 8
Answer (3)
Sol. k? = Ae^{-Ea?/RT}
k? = Ae^{-Ea?/RT}
k?/k? = e^{1/RT(Ea? - Ea?)}
ln k?/k? = (Ea? - Ea?)/RT
= (10×10³)/(8.314×300) = 4
-
Which of the following molecules is least resonance stabilized?
Answer (3)
Sol. However, all molecules given in options are stabilised by resonance but compound given in option (3) is least resonance stabilised (other three are aromatic)
-
The group having isoelectronic species is
(1) O?, F?, Na, Mg?
(2) O²?, F?, Na, Mg²?
(3) O?, F?, Na?, Mg²?
(4) O²?, F?, Na?, Mg²?
Answer (4)
Sol. Mg²?, Na?, O²? and F? all have 10 electrons each.
===== Page 25 =====
Sol. DIBAL-H reduces esters and carboxylic acids into aldehydes
-
Which of the following reactions is an example of a redox reaction?
(1) XeF? + PF? → [XeF]?PF??
(2) XeF? + H?O → XeOF? + 2HF
(3) XeF? + 2H?O → XeO?F? + 4HF
(4) XeF? + O?F? → XeF? + O?
Answer (4)
Sol. Xe is oxidised from +4 (in XeF?) to +6 (in XeF?)
Oxygen is reduced from +1 (in O?F?) to zero (in O?)
-
A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in metallic crystal will be
(1) 2√2a
(2) √2a
(3) a/√2
(4) 2a
Answer (3)
Sol. In FCC, one of the face is like
By ΔABC
2a² = 16r²
⇒ r² = 1/8 a²
⇒ r = 1/(2√2) a
Distance of closest approach = 2r = a/√2
-
Sodium salt of an organic acid 'X' produces effervescence with conc. H?SO?. 'X' reacts with the acidified aqueous CaCl? solution to give a white precipitate which decolourises acidic solution of KMnO?. 'X' is
(1) HCOONa
(2) CH?COONa
(3) Na?C?O?
(4) C?H?COONa
Answer (3)
Sol. Na?C?O? + H?SO? → Na?SO? + H?C?O?
Conc.
H?C?O? --Conc. H?SO?--> CO↑ + CO?↑
(effervescence)
Na?C?O? + CaCl? → CaC?O?↓ + 2NaCl
white ppt.
2MnO?? + 5C?O?²? + 16H? → 2Mn²? + 10CO? + 8H?O
-
A water sample has ppm level concentration of following anions
F? = 10; SO?²? = 100; NO?? = 50
The anion/anions that make/makes the water sample unsuitable for drinking is/are
(1) Both SO?²? and NO??
(2) Only F?
(3) Only SO?²?
(4) Only NO??
Answer (2)
Sol. Permissible limit of F? in drinking water is upto 1 ppm. Excess concentration of F? > 10 ppm causes decay of bones.
-
Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to decolourize the colour of bromine?
Answer (4)
Sol.
The above product does not have any C=C or C≡C bond, so, it will not give Br?-water test.
NOTE : For complete Question paper with image or diagram please refer page no . 413 to 437 from given PDF
JEE MAINS 2012 to 2018 Previous Year Question Paper FREE PDF DOWNLOAD LINK
JEE MAINS 2016 PREVIOUS YEAR QUESTION PAPER SOLWED WITH FREE PDF DOWNLOAD LINK
JEE MAINS Previous Year Solwed Question Paper 2015
