JEE MAIN
CLASS 11th
APNI KAKSHA STAR BATCH
Time : 3 Hours
Maximum Marks : 180
ANSWER KEY
&
SOLUTION WITH EXPLANATION
Solution Sheet Available Only for 24 Hours
PART – 1: MATHEMATICS
•
This question paper contains two sections , section A & B.
•
Section A contains 20 multiple choice questions (MCQs) with four options (A),(B),(C),(D) out
of which only one option is correct.
•
Section B contains 10 Numerical Value Type questions, out of which candidate have to
attempt only 5 questions.
Section A
•
This Section contain 20 questions (Q.No. 1 to Q.No. 20)
•
Answer to each question in Section A will be evaluated according to the following marking
scheme:
Full Marks
: +𝟒 for correct answer
Zero Marks
:
0 If the question is unanswered;
Negative Marks : −𝟏 for incorrect answer
1.
Let (x0 , y0 ) be solution of following equations :
x+y =3
x ln2 + y ln3 = ln27
Then y0 is
(A) 1
(B) 0
(C) 3
(D) 2
Ans.
(C)
2.
ABCD is a square of length 10 metres. x, y, z, w are mid-points of sides. Shyam walks along the
path AxyCzwA. Distance covered by Shyam is____ metres. (xy, zw are circular arcs with centres
at B and D respectively)
(A) 20 + 5π
Ans.
(B) 10 + 10π
(C) 40 − 5π
(D) 40 − 10π
(A)
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3.
APNI KAKSHA STAR BATCH
sin50∘ + sin10∘
The value of cos67.5∘ ⋅cos22.5∘ is
(A) 2√2 sin20∘
(B) 2√2 cos20∘
(C) √2 sin20∘
Ans.
(B)
4.
If log y x + log x y = 7, then the value of (log y x) + (log x y)3 is
(D) √2 cos20∘
3
(A) 322
(B) 342
(C) 343
(D) 344
Ans.
(A)
5.
In an equilateral triangle, 3 coins of radius 1 unit are kept so that they touch each other and
also the sides of the triangle, then area of the triangle is
(A) 4 + 2√3
(B) 6 + 4√3
Ans.
(B)
6.
The value of sin24∘ + cos6∘ is
(A)
√3+√5
4
(B)
√3+√5
2
(C) 12 +
(C)
7√3
4
√3+√15
4
(D) 3 +
(D)
7√3
4
√3+√15
2
Ans.
(C)
7.
Let a, b, c be real numbers greater than 1. If s = log a bc + log b ca + log c ab, then
(A) s ≤ 9
(B) s ≥ 9
(C) s ≤ 6
Ans.
(D)
8.
If cos2x = 4, then the value of (sin4 x + cos 4 x) is equal to
(D) (s ≥ 6
3
11
(A) 32
9
(B) 16
25
(C) 32
23
(D) 32
Ans.
(C)
9.
Number of solutions of the equation |x − 1| + |x − 2| + |x − 3| = 6 is/are
Ans.
(A) 1
(B) 2
(C) 3
(D) Infinitely many
(B)
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10.
APNI KAKSHA STAR BATCH
Number of solution(s) of the equation log √x (x + |x − 3|) = log x (2x + 5 + 2|x − 3|) is
(A) 0
(B) 1
(C) 2
(D) Infinitely many
Ans.
(B)
11.
Sum of solutions of the equation log1/3 (x − 1) + log1/3 (2x − 1) = −1 is
3
(A) 2
(C) −
(B) 2
1
(D) None of these
2
Ans.
(B)
12.
Which of the following is true
(A) tan1 > sin1 > cos1
(B) tan1∘ < sin1∘ < cos1∘
(C) tan1 > cos1 > sin1
(D) sin1∘ < cos1∘ < tan1∘
Ans.
(A)
13.
If 0 < θ < 2 ; and f(θ) = (1 + cotθ − cosecθ)(1 + tanθ + secθ), then f (12) is equal to
π
(A) 0
π
(B) 1
(C) √2
Ans.
(D)
14.
The value of cos12∘ + cos84∘ + cos156∘ + cos132∘ is
1
1
(A) − 2
(B) 8
1
(C) 2
(D) 2
(D) 1
Ans.
(A)
15.
Let product of the solutions of the equation |log10 (x) + 2| = 3 is M. Then |log10 M| =
(B) 3
(A) 2
(C) 4
(D) 1
Ans.
(C)
16.
For a positive number m, Characteristic of log10 m − Mantissa of log10 m = 1.5. Then
log100 (
Ans.
m
)=
√10
(A) 1
(B) -1
(C) 0
(D) Insufficient data
(A)
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17.
APNI KAKSHA STAR BATCH
Given log10 2 = 0.3010, log10 3 = 0.4771. Using this information or otherwise, choose the
correct statement:
(A) 8150 > 9100
(B) 1675 > 8150
(C) log 2/3 3 > 0
(D) None of these is correct
Ans.
(A)
18.
log 72 (√8 + √32) =
1
(A) 1
(B) 2
(C) 2
(D) None of these
Ans.
(B)
19.
Let x1 and x2 be the values of x, which satisfy the equation 9x − 5 ⋅ 3x + 6 = 0. Then
(A) x1 and x2 are both irrational numbers
(B) One of x1 and x2 is rational and the other irrational
(C) x1 and x2 are both rational numbers
(D) None of these
Ans.
(B)
20.
If α and β are solutions of the equation 5log5x + x log5x = 1250; then log β α equals :
Ans.
2
(A) 0
(B) 1
(C) −1
(D) −2
(C)
Section B
•
This Section contain 10 questions (Q.No. 21 to Q.No. 30) whose answer to be filled as
numerical value (Attempt any five)
•
Answer to each question in Section B will be evaluated according to the following marking
scheme:
Full Marks
: +𝟒 for correct answer
Zero Marks
:
0 If the question is unanswered;
Zero Marks
:
𝟎 for incorrect answer
21.
Find the number of integral values of x in [0,4π] such that log (cosx) (sinx) is positive
Ans.
2
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APNI KAKSHA STAR BATCH
22.
The value of (–cosec 6∘ + cosec18∘ + sec24∘ − sec36∘ +cosec42∘ − cosec78∘ ) is equal to
Ans.
−𝟔
23.
If α, β, γ, δ are the smallest positive angles in ascending order which have their sines equal to 9,
7
α
β
γ
δ
then the value of (12 sin 2 + 9 sin 2 + 6 sin 2 + 3 sin 2) is equal to
Ans.
𝟖
24.
The number of integral values of ' a ' such that both roots of the equation
ax 2 + (3a + 1)x − 5 = 0 are integers is
Ans.
1
25.
ABCDEFG is a 7-sided regular polygon such that whose side is 1, then AC + AD =
Ans.
1
26.
If α, β are the roots of equation λ(x 2 − x) + x + 5 = 0 and if λ1 , λ2 are two values of λ that
1
satisfy
α
β
β
4
+ α = 5, then find the value of
λ1
λ22
1
λ
+ λ22
1
Ans.
4048
27.
The value of sec
Ans.
128
28.
Suppose x, y ∈ R and satisfy 7x 2 − 9xy + 7y 2 = 9. If ' a ' is the maximum value and ' b ' is the
π
sec
15
2π
sec
15
3π
sec
15
4π
sec
15
5π
sec
15
6π
sec
15
7π
15
is equal to
minimum value of x 2 + y 2 , then find the number of digits in the number 210a+23b−27.
(log10 2 = 0.301)
Ans.
9
29.
The number of integral values of x for which the expression log (3x−4) (4 + 7x − 2x 2 ) is real, is
Ans.
2
30.
The smallest integral value of x satisfying the inequality log (x2) (2 + x) < 1 is
Ans.
3
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JEE MAIN
CLASS 11th
APNI KAKSHA STAR BATCH
PART – 2 : PHYSICS
•
This question paper contains two sections , section A & B.
•
Section A contains 20 multiple choice questions (MCQs) with four options (A),(B),(C),(D) out
of which only one option is correct.
•
Section B contains 10 Numerical Value Type questions, out of which candidate have to
attempt only 5 questions.
Section A
•
This Section contain 20 questions (Q.No. 1 to Q.No. 20)
•
Answer to each question in Section A will be evaluated according to the following marking
scheme:
Full Marks
: +𝟒 for correct answer
Zero Marks
:
0 If the question is unanswered;
Negative Marks : −𝟏 for incorrect answer
31.
Both the cars are accelerating towards right and their initial velocity is also in rightward
direction as shown in figure. What is maximum separation between car A and B.
(A) 50 m
(B) 150 m
(C) 450 m
(D) 500 m
Ans: B
Sol:
(For maximum separation)
vA = VB
(10 + 2t) = 20 + t
(T = 10)
 dA = 10 × 10 +
1
× 2 × 100 = 200
2
1
dB = 20 × 10 + 2 × 1 × 100 = 250
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32.
APNI KAKSHA STAR BATCH
A particle moves in a straight line with an acceleration of magnitude[6 − K√S]m/s2, where S is
its distance from a fixed-point O on the surface. The acceleration is directed always away from
O. If it starts from rest at O and comes to instantaneous rest again 400 m from O, the value of
the constant K will be:
9
3
(A) 20
2
(B) 5
2
(C) 3
(D) 7
Ans: A
Sol:
Let O is origin
a = 6 − K√S  v
0
dS
= 6 − K√S
400
∫0 vdv = ∫0
33.
dv
(6 − K√S)dS ⇒ 0 = [6S −
2KS3/2
3
400
]
0
9
⇒ K = 20
Two particles P and Q are moving on x-axis such that their v-t graph look like as shown in
diagram. In the graph of P, the initial part up to t = 1 is a parabola of type y = ax 2 , and there
after a straight line. The graph of Q is a straight line throughout and having slope equal to that
of the straight part of P. If the two particles P and Q start from same point, find the separation
between them 2 second after the particle Q starts motion.
(A) 3.5 m
(B) 7 m
(C) 10.5 m
(D) 14 m
Ans: B
Sol:
For P
Parabolic part
V = at 2
Point (1,3) lies on it so
a = 3 ⇒ V = 3t 2
1
S1 = ∫0 3t 2 dt = 1m ⇒ Vt=1 = 3m/s
For Q ⇒ Vt=1 = 0m/s
For motion after 1 sec.
Vrel = 3m/s, arel = 0
Srel = 3 × 2 = 6m
Total separation = 6 + 1 = 7m
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34.
APNI KAKSHA STAR BATCH
Acceleration (a)-time(t) graph for a particle starting from rest at t = 0 is as shown in the figure.
The particle has maximum speed at :
(A) 1 s
(B) 2 s
(C) 3 s
(D) 4 s
Ans: B
Sol:
Between t = 0 and t = 2,
acceleration has same sign which increases speed.
At t = 2 sec.
Direction of acceleration is reversed and speed starts decreasing.
35.
A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up
1
vertically to a height d. Neglecting subsequent motion and air resistance, its velocity v varies
2
with the height h above the ground as
(A)
(B)
(C)
(D)
Ans: A
Sol:
v1 = −√2gd and
d
v2 = +√2g ( )
2
Hence, the correct answer is (A).
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36.
APNI KAKSHA STAR BATCH
The acceleration-velocity graph of a particle moving rectilinearly is as shown in figure. Then
slope of velocity-displacement graph must be
(A) increasing linearly
(B) decreasing linearly
(C) a constant
(D) increasing parabolically
Ans: C
Sol:
a = kv
dv
= kv
dx
dv
dv
{∵ a =
=v }
dt
dx
dv
⇒
=k
dx
v
⇒ Slope of velocity-displacement graph is a constant
Hence, the correct answer is (C).
37.
Is it possible for the motion of an object along a straight line so that its velocity increases while
its acceleration decreases?
(A) No, because if acceleration is decreasing the object will be slowing down
(B) No, because velocity and acceleration must always be in the same direction
(C) Yes, an example would be a falling object near the surface of the moon
(D) Yes, an example would be a falling object in the presence of air resistance
Ans: D
Sol:
ma = mg − 6πηrv
a > 0 and v ↑ v  till mg > 6πηrv
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CLASS 11th
38.
APNI KAKSHA STAR BATCH
For which of the following graphs the average velocity of a body moving along a straight line for
time interval (0, t) must be negative.
(A)
(B)
(C)
(D)
Ans: A
Sol:
s⃗
<v
⃗ >= t =
⃗
Δx
t
(a) +ve disp., +ve velocity
(b) zero disp.
(c) slope is +ve, so velocity is +ve.
(d) slope is –ve, so velocity is –ve.
39.
Body A was dropped from the top of a tall building. At the same instant and from the same
height body B was thrown straight downward. Neglecting the effects of air friction, compare
their accelerations while they were falling.
(A) Their accelerations are equal
(B) Ball A has the greater acceleration
(C) Ball B has the greater acceleration
(D) Initially A has greater acceleration and after some time B has greater acceleration
Ans: A
Sol:
aA = −gm/s
aB = −gm/s
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40.
APNI KAKSHA STAR BATCH
The acceleration due to gravity can be measured by projecting a ball upward and measuring the
time that it takes to pass two given lines in both directions (upward motion and downward
motion). If the time the ball takes to pass a horizontal line A in both direction (from A1 to A 2 )
is TA , and the time to go by a second line B in both directions is (from B1 to B2 ) TB , then
assuming that the acceleration due to gravity to be constant, what is the value of acceleration
due to gravity?
8h
(A) T2 −T2
A
B
8h
(B) T2 +T2
A
(C)
(D)
B
8h
T2A T2B
8hTA TB
T2A T2B
Ans: A
Sol:
Displacement in y direction is zero for A1A 2 and B1B2 in time interval TA and TB
respectively.
1
O = uA TA − gTA2
2
TA g
⇒ uA =
2
1
O = uB TB − gTB2
2
TB g
⇒ uB =
2
u A and u B are velocity components at A1 and B1
u2B = u2A − 2gh
TB2 gL TA2 gL
=
− 2gh
4
4
8h
⇒g= 2
TA − TB2
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41.
APNI KAKSHA STAR BATCH
The displacement time graph of a moving particle with constant acceleration is shown in the
figure. The velocity time graph is best given by :
(A)
(B)
(C)
(D)
Ans: A
Sol:
From figure : (given in problem)
For time interval t = 0 to t = 1 sec
Slope of x-t graph is negative and increasing, so velocity increases in negative direction.
For t = 1 to 2 sec
The slope is +ve and decreasing,
So velocity is decreasing in +ve direction and become zero at t = 2
So, (a) is correct.
42.
The graph shown is a plot of position versus time. For which labeled region is the velocity
positive and the acceleration negative?
(A) a
(B) b
(C) c
(D) d
Ans: D
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Sol:
dx
During ‘d’ slope dt is +ve, so velocity is +ve.
But
d(
d(
dx
)
dt
dt
43.
APNI KAKSHA STAR BATCH
dx
)
dt
dt
=
is −ve because slope decreasing so
d2 x
dt2
is −ve , therefore acceleration is −ve
Starting from rest a particle moves in a stragith line with acceleration a = (2 + |t − 2|)ms −2 .
Velocity (in m/s) of particle at the end of 4 s will be
(A) 20
(B) 10
(C) 25
(D) 12
Ans: D
Sol:
Since, area under a-t graph is equal to the change in velocity, so
1
1
Δv = 2 (2 + 4)(2) + 2 (2 + 4)(2) = 12ms−1
⇒ Δv = v − u = 12ms −1
Since u = 0 ⇒ v = 12ms−1
44.
The graph shows position as a function of time for two trains running on parallel tracks. Which
one of the following statements is true?
(A) At time t B both trains have the same velocity
(B) Both trains have the same velocity at some time
(C) Both trains have the same velocity at some time before t B
(D) Somewhere on the graph, both trains have the same acceleration
Ans: C
Sol:
Slope of curve B and slope of curve A will be same before time TB only.
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45.
APNI KAKSHA STAR BATCH
Each of the three graphs represents acceleration versus time for an object that already has a
positive velocity at time t1 . Which graphs show an object whose speed is increasing for the
entire time interval between t1 and t 2 ?
(A) graph I, only
(B) graphs I and II, only
(C) graphs I and III, only
(D) graphs I, II, and III
Ans: D
Sol:
All the graphs shows that speed is increasing as acceleration is positive in all three graphs for
given time interval. Also we can say that area under a-t graph i.e. change in velocity is positive
so velocity is increasing.
46.
Graph of (1/v) vs x for a particle under motion is as shown, where v is velocity and x is position.
The time taken by particle to move from x = 4 m to x = 12 m is _________ (in sec)
(A) 20
(B) 10
(C) 25
(D) Information is inssuficient
Ans: B
Sol:
1
∫ ydx = ∫ (v) dx = ∫ dt = ΔT
T
∴ T = ∫0 dt = Area under curve from x = 4 to x = 12 m
=
47.
1 1
1 5
( + 2) × (12 − 4) = × × 8 = 10s
2 2
2 2
Velocity of an object depends on displacement as v 3/2 = k8(y)3/4, where v is velocity (in m/s),
y is displacement (in meter) and k is constant, then acceleration (in m/s 2 ) when y = 16 is αk β/γ .
Find α + β + γ ?
(A) 13
(B) 14
(C) 15
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(D) 16
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Ans: C
Sol:
v 3/2 = k8(y)3/4
v = k 2/3 × 4 × (y)1/2
vdv
k 2/3 × 4 × 1
2/3
1/2
a=
= (4 × k × y ) × (
) = 8k 4/3
dx
2 × (y)1/2
48.
Acceleration of a particle is defined as a = (75v 2 − 30v + 3)(m/s 2 ) find constant speed
achieved by the particle:
1
(A) 3 m/s
(B) 5 m/s
(C) 5 m/s
(D) It will never achieve constant speed
Ans: D
Sol:
It will achieve constant speed when ‘a’ becomes 0.
75v 2 − 30v + 3 = 0
Solving the quadratic, we get the value of v.
49.
The drawing shows velocity (v) versus time (t) graphs for two cyclists moving along the same
straight segment of a highway from the same point. The second cyclist starts moving at t = 3
min. At what time do the two cyclists meet?
(A) 4 min
(B) 6 min
(C) 8 min
(D) 12 min
Ans: B
Sol:
Let common velocity at t = 4 min is v 0 .
Then aA =
v0
4
and aB =
v0
1
Let at time t both cyclists meet then s A [in time interval(t − 0)] = s B [in time interval(t − 2)]
1
1
aA t 2 = aB (t − 3)2
2
2
1 v0 2 1 v0
t
(t − 3)2 ⇒ = t − 3 ⇒ t = 6min
⇒
t =
24
21
2
⇒
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50.
APNI KAKSHA STAR BATCH
A particle starts from rest at time t = 0 and moves on a straight line with acceleration as plotted
in figure. The speed of the particle will be maximum at time:
(A) 1 s
(B) 2 s
(C) 3 s
(D) 4 s
Ans: B
Sol:
Area under a-t curve gives change in velocity.
Between 0 to 2 second speed will increase and between 2 to 4 second speed will decrease.
Section B
•
This Section contain 10 questions (Q.No. 21 to Q.No. 30) whose answer to be filled as
numerical value (Attempt any five)
•
Answer to each question in Section B will be evaluated according to the following marking
scheme:
51.
Full Marks
: +𝟒 for correct answer
Zero Marks
:
0 If the question is unanswered;
Zero Marks
:
𝟎 for incorrect answer
Initially car A is 10.5 m ahead of car B. Both cars start moving at time t = 0 in the same direction
along a straight path. The velocity time graph of two cars is shown in figure. The time when the
car B will catch the car A, will be __________ (in secs)
Ans: 21
Sol:
X = 10t ⇒ 10.5 + X =
1
× 1 × t 2 ⇒ t 2 − 20t − 21 = 0 ⇒ t = −1,21sec
2
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52.
APNI KAKSHA STAR BATCH
A car starts from rest with uniform acceleration. Its velocity after 2n second in v0 . The
displacement of the body in last n seconds is
αv0 n
β
. Find  +  .
Ans: 7
Sol:
Let a be the acceleration
at t = 2n
…(i)
v0 = 0 + a × 2n = 2na
at t = n
…(ii)
v = an
Between t = n and t = 2n
v02 − v 2 = 2as
⇒ 4π2 a2 − a2 n2 = 2as
⇒s=
3n2 a
2
v
0
From equation (i) a = (2n)
3n2
So, s =
53.
2
v
3v0 n
2n
4
( 0) =
A car is moving on a straight road with initial velocity u and uniform acceleration f. If the sum of
the distances travelled in t th and (t + 1)th seconds is 100 cm, then its velocity after t seconds,
in cm/s, is :
Ans: 50
Sol:
a=f
v
t
dv
= f ⇒ ∫ dv = ∫ fdt ⇒ v = u + ft
dt
u
0
v=
dx
= u + ft
dt
100
⇒∫
t+1
dx = ∫
0
(u + ft)dt
t−1
⇒ 2u + 2ft = 100
⇒ u + ft = 50
…(i)
⇒ v = u + ft = 50cm/s (by (i))
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54.
APNI KAKSHA STAR BATCH
Two trains, which are moving along different tracks in opposite directions towards each other,
are put on the same track by mistake. Their drivers, on noticing the mistake, start slowing
down the trains when the trains are 300 m apart. Graphs given below show their velocities as
function of time as the trains slow down. The separation between the trains after both the have
stopped, is (in meters)
Ans: 20
Sol:
Final separation = Initial separation - |Relative displacement|
1
1
= 300 − [(2) × 40 × 10 + (2) × 20 × 8] = 20m
55.
A bus, moving with a speed of 50 km h−1 , can be stopped by brakes in minimum 6m distance. If
the same bus is moving at a speed of 100 km h−1 , the minimum stopping distance is (in meters)
Ans: 24
u21
u22
s1
u22
100 2
Sol:
a=
56.
A body starts from rest with constant acceleration a and it is then deaccelerate with a constant
2s1
,a =
2s2
⇒
s2
=
u21
⇒ s2 = 6 × (
50
) = 24m
value b till it is brought to rest. The total time taken between these two rest positions is t. If is
αab
the maximum speed acquired by the body is βa+γb. Find α + β + γ.
Ans: 3
Sol:
v
v
ab
)t
= a; = b ⇒ t1 + t 2 = t ⇒ v = (
t1
t2
a+b
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57.
APNI KAKSHA STAR BATCH
A body is thrown vertically upwards from the ground. It crosses a point at the height of 25 m
twice at an interval of 4 secs. The body was thrown with the velocity (in m/s) is
Ans: 30
Sol:
1
Applying s = uA t + 2 at 2 at height A
0 = uA × 4 −
1
× g × 42
2
1
= uA × 4 − 2 × 10 × 16
⇒ uA = 20m/s
Applying v 2 − u2 = 2as from ground till A
(20)2 − u2 = 2 × (−10) × 25
⇒ u = 30m/s
2
2
2
2
vAy
+ vBx
> vBy
+ vBx
∴ vAy > vBy
R=
H=
T=
58.
2vx vy
g
2v2y
g
2vy
g
∴ RA > RB
∴ HA > HB
∴ TA > TB
A person throws vertical up n balls per second with the same velocity. He throws a ball
g
whenever the previous one is at its highest point. The height to the balls rise is αnβ. Find α + β.
Ans: 4
Sol:
n balls thrown per second
1
Time interval between two balls thrown ⇒ n sec
In this time, it reaches
Highest point ⇒ V = u + at
1
g
g 2
g
0 = V0 − g ( ) ⇒ V0 = V 2 − u2 = 2as ⇒ 02 − ( ) = 2 × (−g)h  h = 2
n
n
n
2n
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59.
APNI KAKSHA STAR BATCH
A lift starts going up from rest. Its acceleration is shown in figure plotted against time. Time
when lift comes to instantaneous rest is (in sec)
Ans: 10
Sol:
Area under a-t curve gives V
1 × 4 = (t − 8) × 2
4 = 2t − 16
t = 10sec.
60.
A bird flies for 4 sec from t = 0 with a velocity of |t − 2|m/s in a straight line, where t = time in
seconds. It covers a distance  meters. Find  .
Ans: 4
Sol:
Here V = |t − 2| = 2 − t for t ≤ 2
⇒ t − 2 for 4 ≥ t > 2
V=
2
dx
dt
x
t
or ∫0 dx = ∫0 Vdt
4
∫0 (2 − t)dt + ∫2 (t − 2)dt = 4meter
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PART – 3 : CHEMISTRY
•
This question paper contains two sections , section A & B.
•
Section A contains 20 multiple choice questions (MCQs) with four options (A),(B),(C),(D) out
of which only one option is correct.
•
Section B contains 10 Numerical Value Type questions, out of which candidate have to
attempt only 5 questions.
Section A
•
This Section contain 20 questions (Q.No. 1 to Q.No. 20)
•
Answer to each question in Section A will be evaluated according to the following marking
scheme:
Full Marks
: +𝟒 for correct answer
Zero Marks
:
0 If the question is unanswered;
Negative Marks : −𝟏 for incorrect answer
61.
Consider the following reaction
2Na + 2NH3 ⎯→ 2NaNH2 + H2(g)
2NaNH2 + C ⎯→ Na2CN2 + 2H2 (g)
Na2CN2 + C ⎯→ 2NaCN
51gm dry ammonia gas is passed over excess heated sodium to form sodamide (NaNH2) which
is further reacted with carbon (excess) to finally form NaCN. Find the total volume of H 2(g)
evolved at 0.5 atm, 273K
(A) 201.6 L
(B) 100.8 L
(C) 403.2 L
Ans.
(A)
Sol.
Since sodium and carbon are present in excess.
(D) 50.4 L
 Other reactants of all 3 reactions get completely converted into products.
 All 3 reactions can be added & converted into a single reaction.
51
Moles of NH3 = 17 = 3
3
9
2
2
Number of moles of H2 produced from NH3 = × 3 = mol.
Using PV = nRT, Vol. =
9
×0.821×273
2
0.5
= 201. 6 
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62.
APNI KAKSHA STAR BATCH
If each orbital were containing three electrons, then the ground state electronic
configuration of iron were:
(A) 1s2 2s2 2p6 3s2 3p6 4s 2 3d6
(B) 1s3 2s3 2p6 3s3 3p6 4s3 3d2
(C) 1s3 2s3 2p9 3s 3 3p8
(D) 1s3 2s3 2p6 3s3 3p6 3d5 .
Ans.
(C)
63.
Bond in acetylene (C2H2) are.
(A) 3 , 2 
Ans.
(B) 2 , 3 
(C) 6 , 2 
(D) 2 , 2 
(A)
Sol.
3 , 2  Bond
64.
34 gm of H2O2 is present in 1135 ml of solution. Volume strength of solution is
(A) 10 V
(B) 20 V
Ans.
(A)
Sol.
Molarity of H2O2 solution =
(C) 30 V
34 / 34
1000
Moles of H 2O 2 1000
=
1000 =
Vol. of solution (in ml)
1135
1135
Volume strength of H2O2 solution = Molarity × 11.35 
65.
(D) 32 V
1000
11.35 = 10
1135
"Electron pairing cannot occur in p, d and f-orbitals until each orbital of a given subshell
contains one electron". This is known as
(A) Aufbau's rule
(B) Pauli's exclusion principle
(C) Hund's rule
(D) (n + ℓ) Rule
Ans.
(C)
66.
Which of the following compound does not contain 3° hydrogen?
(A) (CH3)3 CCH (CH3)2
(B) (CH3)3 CCH2 CH3
(C) (CH3)2 CH CH2 CH2 CH3
(D) [(CH3)2 CH4] C
Ans.
(B)
Sol.
So there is no tertiary (3°) hydrogen in this compound.
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67.
APNI KAKSHA STAR BATCH
1.11 gm of CaCl2 added to water forming 500 ml solution. 20 ml of this solution is
taken and
diluted 10 folds. Find moles of Cl– ions in 2 ml of diluted solution (A) 8 × 10–6
Ans.
(A)
Sol.
Moles of CaCl2 =
(B) 4 × 10–6
(C) 12 × 10–8
(D) 5 × 10–6
1.11
= 0.01 moles
111
0.01 moles are present in 500 ml solution.
Molarity of solution =
0.01
1000 = 0.02
500
Initially 20 ml of the above solution is taken, which is diluted to 10 folds, i.e. solution
finally become 200 ml.
Moles of CaCl2 in 20 ml = Moles of CaCl2 in 200 ml
Molarity1  Initial Vol. = Molarity2  Final Vol.
0.02  20 = M2  200  M2 = 2 × 10–3
Hence No. of moles of CaCl2 in 2 ml solution.
= Molarity  Vol. (in litre) = 2 10–3 × 2 × 10–3 = 4 ×10–6
 No. of moles of Cl– in 2ml = 4 × 10–6 × 2 = 8 × 10–6
Atomic mass of Ca = 40, Cl = 35.5
68.
X − , Y −2 and Z −3 are isotonic and isoelectronic. Thus increasing order of atomic number of X, Y
and Z is
(A) X < Y < Z
(B) Z < Y < X
(C) X = Y = Z
(D) Z < X < Y
Ans.
(B)
69.
For a closed (not rigid) container n = 10 moles of an ideal gas, fitted with movable, frictionless,
weightless piston operating such that pressure of gas remains constant at 0.821 atm, Which
graph represents correct variation of log V vs log T where V is in lit. & T in Kelvin.
(A)
(B)
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(C)
(D)
Ans.
(A)
Sol.
P & n are constant  V & T
V=
nR
T
P
log V = log
nR
+ log T
P
Given:- n = 10, P = 0.821 atm 
log
nR
= log1 = 0
P
Solpe = 1
nR 10  0.0821
=
=1
P
0.821
 log V = log T
 = 45
Option D should be changed
70.
As per the theory which of the following sub-shell is not possible (A) 3 d
(B) 4 f
(C) 5 h
(D) 6 g
Ans.
(C)
71.
A gas cylinder contains 320 gm of O2 at 30 atm and 27°C. What mass (in gram) of O2 would
escape if first the cylinder is heated to 127° C and then valve is held open until the pressure
inside the cylinder becomes 1 atm (the temperature being maintained at 127° C.
(A) 312
Ans.
(A)
Sol.
Initial moles of O 2 =
(B) 315
(C) 340
(D) 320
320
= 10moles
32
Since volume of container remain constant,
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
APNI KAKSHA STAR BATCH
n1T1 n 2 T2
10  300 n 2  400
=

=
 n 2 = 0.25
P1
P2
30
1
Since remaining no of moles = 0.25
 escaped no of moles = 10 − 0.25 = 9.75
Escaped mass of O 2 = (10 − 0.25)  32 = 9.75  32 = 312gm
72.
Which one of the following statements is not correct?
(A) Li shows similarity in properties with Mg
(B) Li, Na and K forms Doberenier's Triads.
(C) In the Lother Meyer's curve alkaline earth metals occupy the ascending positions of the
curve
(D) IUPAC name of the element having atomic number 112 is Uub.
Ans.
(C)
73.
Which of the following combination of gases is most easiest to separate(A) H2 and He
(B) CO2 and N2O
(C) U235F6 and U238F6
(D) C3H8 and C3H6
Ans.
(A)
Sol.
Gases having max. Ratio of higher to lower molecular mass, easiest to separate
(A) H 2 & CH 4
(B) CO 2 & N 2 O
(C) U 235 F6 & U 238 F6
M He 4
= =2
M H2 2
M CO2
=
M N2O
M U 238 F
6
M U 235 F
44
=1
44
=
52
349
6
(D) C3 H 8 & C3 H 6
74.
M C3 H 8
M C3 H 6
=
44
42
Select the correct statement regarding oxides.
(A) As the electronegativity of element increase acidic character of oxide increases.
(B) Down the group the acidic nature of oxide increase.
(C) B2 O3 and Al2 O3 are both acidic oxides.
(D) Nitrogen forms all the three types of oxides (neutral, basic & acidic)
Ans.
(A)
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75.
APNI KAKSHA STAR BATCH
A sample of a gas was heated from 300oC to 600oC at constant pressure. Which of the following
statement(s) is/are true.
(A) Kinetic energy of the gas is doubled
(B) Boyle’s law will apply
(C) Volume of the gas will be doubled
(D) None of the above
Ans.
(D)
Sol.
300ºC to 600ºC
T1 = 300 + 273 = 573k
T2 = 600 + 273 = 873k
K.E.  T, temp. is not doubled
 kinetic energy will not be doubled
Boyle’s law will be applicable at constant temp. & moles
Volume  T , temp. is not doubled,
 volume will not be doubled
76.
Choose the correct statement from the following:
(A) Zeff on ' d ' electron of Sc 2+ is 18
(B) Zeff values on an electron present in 4s and 4p orbital of an atom are identical.
(C) Zeff values on an electron present in 3s and 4s orbital of an atom are identical.
(D) the screening constant value on one electron in H − ion is 0.35.
Ans.
(B)
77.
Average velocity (in m/s) of oxygen gas at 120 Kelvin is [Given R =
25
J/mole-K]
3
(A) 707
(B) 500
(C) 5 10
(D) 316.2
Ans.
(B)
Sol.
Avg. Velocity =
8RT
8  25  120
=
M
3  32  10 −3
= 500m/s
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78.
APNI KAKSHA STAR BATCH
Which of the following is the correct increasing size?
(A) Cl− < Ca2+ < S 2− < Al3+
(B) Mg 2+ < K + < Li+ < Al3+
(C) Mg 2+ < Na+ < F − < O2− < N3−
(D) F − < Na+ < Mg 2+ < O2−
Ans.
(C)
79.
Which of the following changes will double the mean free path of gas in closed container
(A) Increasing temperature two times at constant volume
(B) Increasing temperature four times at constant volume
(C) Increasing temperature two times at constant pressure
(D) Increasing temperature four times at constant pressure
Ans.
(C)
Sol.
=
1
2 N
2
 
*
= =
1 V
2 N
2
=
1  KT
2 2  P
T
T
At constant volume,
is constant
P
P
At constant pressure,   T
 By increasing temp. two times at constant pressure,  will double.
80.
Which of the following statements is correct?
(A) Ionisation energy of A− is greater than A when A is a halogen atom.
(B) Ionisation energy of A+ is greater than that of A2+ when A is the member of alkali metals.
(C) Successive ionisation energy is always increasing for 1st and 2nd period element.
(D) Electron affinity value of ‘ A+ ’ is numerically identical with the ionisation potential of
A− [for any atom].
Ans.
(C)
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APNI KAKSHA STAR BATCH
Section B
•
This Section contain 10 questions (Q.No. 21 to Q.No. 30) whose answer to be filled as
numerical value (Attempt any five)
•
Answer to each question in Section B will be evaluated according to the following marking
scheme:
81.
Full Marks
: +𝟒 for correct answer
Zero Marks
:
0 If the question is unanswered;
Zero Marks
:
𝟎 for incorrect answer
H2 and O2 are kept in mass ratio 1: 8 respectively at 6 atm. If small orifice is made then relative
rate of effusion of H2 with respect to O2 initially is.
Ans.
Sol.
82.
(8)
rH 2
rO2
=
PH 2
PO2
WH 2
32 n H 2
32
1
=
 16 =

 4 =  16  4 = 8
2
rO2
2
WO2
8
If there were 9 periods in the periodic table & each orbital can have maximum 5 electrons,
then how many maximum number of elements will be present in period 9 ?
Ans.
0125
83.
Find double bond equivalent value of given compound.
Ans.
17
Sol.
Double bond equivalent = number of  bond + number of ring = 12 + 5  17
84.
In the barometer shown below, if the height of the liquid column is 68 cm then calculate the
pressure of the gas trapped in inverted tube (in mm of Hg)?
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APNI KAKSHA STAR BATCH
[Take Patm = 760 mm of Hg]
Ans.
10
Sol.
Height of liquid is converted in terms of Height of Hg
h liq d liq = h Hg d Hg = 68  15 = h Hg  13.6  h Hg = 75 cm of Hg
Now, Pgas + h Hg = Patm
Pgas + 75 = 76
Pgas = 1 cm of Hg,
Ans = 10 mm of Hg
85.
The value of σ (screening constant) for 3 s electron of P is 10.5 then what is value of σ for 3p
electron of S.
[If your answer is 3.45 then write 0345]
Ans.
1085
86.
Sum of -electrons present in following compounds.
(A)
(B)
(C)
Ans.
30
Sol.
(A)
number of -electrons = 2  No. of  Bond = 2  5  10
(B)
number of -electrons = 2  5  10
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= number of -electrons = 2  5  10
(C)
Total -electrons = 10+10+10  30
87.
Calculate sum of code numbers of all those orders which are incorrectly written.
S.No.
Parameter
Order
Code No.
1.
Ionisation energy (IE1 )
O− > S −
58
2.
Atomic Size
Mg > Cl > Si
22
3.
Acidic nature of oxides
Na2 O < H2 O < CO2 < SO3
46
4.
Number of unpaired electron
Na+ < Fe2+ < Cr < Mn
113
Ans.
0193
88.
A solution of A (MM = 20) and B (MM = 10), [XB = 0.6] having density = 0.7 gm/mL, then
the
molarity of B in this solution.
Ans.
30
Sol.
If total no. of moles of solution =1
Then no.
of moles of B = 0.6,
No. of moles of A = 0.4
Mass of B = 0.6 × 10 = 6 gm
Mass of A = 0.4 × 20 = 8 gm
Mass of solution = Mass of A + mass of B
Molarity of B =
89.
0.6
No. of moles of B
No. of moles of B
 1000 = 30 M
 1000 =
1000 =
14
Mass of solution
Vol. of solution (in ml)
0.7
density
Among the following compounds, the number of compounds having greater magnitude of
lattice energy than RbBr, are NaCl, KF, RbCl, CsBr, NaBr, KCl, CsI, RbI.
Ans.
0005
90.
Calculate the total number of electron for Fe having n + l + m = 2.
(If your answer is 5 so write is 0005)
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Ans.
APNI KAKSHA STAR BATCH
0004
APNI KAKSHA
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