18-05-2025
1402CJA101021250026 JA
PHYSICS
SECTION-I
1) Water is filled in a fixed container filled upto a height of 13 m from ground. A sniper fires and
makes a small hole in the side at a height of 9m from the ground. The water jet strikes the ground at
X meter from the container. Find the value of X.
(A) 6 m
(B) 9 m
(C) 12 m
(D) 13 m
2) The force acting due to the liquid per unit width on a plane wall inclined at an angle 53° with the
horizontal as shown in figure (density of liquid ρ)
(A)
(B)
(C)
(D)
3) A fluid flowing through a horizontal pipe of varying cross-section, with speed at a point
where the pressure is P. At another point where pressure is P/2 its speed is . If the density of
the fluid is P kg m-3 and the flow is streamline, then is equal to
(A)
(B)
(C)
(D)
4) A block of wood floats in water with of its volume submerged. If the same block just floats
–3
in a liquid, the density of the liquid (in kg m ) is :-
(A) 1250
(B) 600
(C) 400
(D) 800
5) An ideal fluid flows (laminar flow) through a pipe of non - uniform diameter. The maximum and
minimum diameters of the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum and
the maximum velocities of fluid in this pipe is :
(A)
(B)
(C)
(D)
6) Equivalent resistance between A and B will be :
(A) 20 Ω
(B) 80 Ω
(C) 60 Ω
(D) 10 Ω
7) In the shown circuit the resistance R can be varied:
The variation of current through R against R is correctly plotted as:
(A)
(B)
(C)
(D)
8) Consider the circuit shown in the figure. The potential difference between points A and B is:
(A) 6 V
(B) 8 V
(C) 9 V
(D) 12 V
9) A galvanometer of resistance G Ω is converted into an ammeter of range 0 to I A. If the current
through the galvanometer is 0.1% of I A, the resistance of the ammeter is :
(A)
(B)
(C)
(D)
10) Find the current through resistance in the figure shown.
(A) Zero
(B) 0.84 A
(C) 1A
(D) 0.6A
11) A battery of 24 cells, each of emf 1.5 V and internal resistance is to be connected in order to
send the maximum current through a resistor. The arrangement of cells will be
(A) 2 rows of 12 cells connected in parallel
(B) 3 rows of 8 cells connected in parallel
(C) 4 rows of 6 cells connected in parallel
(D) 1 row of 24 cells
12) There is a current of 40 ampere in a wire of area of cross-section. If the number of free
electron per then the drift velocity will be:
(A)
(B)
(C)
(D)
13) Figure shows 3 identical bulbs with switch S closed. The brightness of bulb 1 is B1, bulb 2 is B2
and net brightness of system is BT. If switch is now opened, the brightness of
(A) B1 remains same, B2 increases and BT increases.
(B) B1 remains same, B2 decreases & BT decreases.
(C) B1 decreases, B2 increases & BT decrease
(D) B1 decrease, B2 increase & BT increase
14)
In the circuit shown, the voltmeter reads 30 V when connected across the 400 resistance. The
same voltmeter when connected across 300 will measure:
(A) 20 V
(B) 22.5 V
(C) 25 V
(D) 40 V
15) An ammeter and a voltmeter are joined in series to a cell. Their readings are A and V
respectively. If a resistance is now joined in parallel with the voltmeter,
(A) both A and V will increase
(B) both A and V will decrease
(C) A will decrease, V will increase
(D) A will increase, V will decrease
16) Magnitude of electric field and potential at corner (D) of a rhombus ABCD in case (i) and case (ii)
are respectively E1, V1 and E2, V2 :-
(A) E2 = E1; V2 = V1
(B) E2 = E1; V2 = 0
(C) E2 = ; V2 = 0
(D) E2 = ; V2 = V1
17) A charge –q of mass m is rotating in circle of radius r, around an infinite large uniformly charged
wire as shown. Speed of charge is V then choose the correct option:-
(A)
(B)
(C)
(D)
18) Figure shows, a glass prism ABC (refractive index 1.5) immersed in water (refractive index 4/3).
A ray of light incident normally on face AB. If it is totally reflected at face AC, then :-
(A)
(B)
(C)
sinθ =
(D) < sin θ <
19) An object is moving at some angle with respect to the principal axis of a concave mirror as
shown in figure. Then which of the following diagrams is possible corresponding to the motion of
image :-
(A)
(B)
(C)
(D)
20) A metal wire of uniform mass density having length L and mass M is bent to form a semicircular
arc and a particle of mass m is placed at the centre of the arc. The gravitational force on the particle
by the wire is:
(A)
(B) 0
(C)
(D)
SECTION-II
1) A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected in
combination with a resistance of 10 Ω and a resistance R, to a 100 V mains as shown in figure. For
the heater to operate at 62.5 W, the value of R should be ……………. Ω.
2) 12 identical wire is used in a network as shown. A current i enters into network at A and leaves at
B such that current in wire PQ becomes 1A. Find the value of i in ampere.
3) A cell having emf V and internal resistance 2 Ω is connected with an external resistance R. It
is observed that when R is decreased by 75% power in external circuit remains same. If this power is
P, then P is______
4) Consider a 72 cm long wire AB as shown in the figure. The galvanometer jockey is placed at P on
AB at a distance x cm from A. The galvanometer shows zero deflection. The value of x (in ohm) is :
5) Find the current (in mA) associated with a moving straight wire of linear charge density λ =
2mC/m and of cross-section A = 2 mm2, when the wire is pulled with a speed v = 2 m/s.
CHEMISTRY
SECTION-I
1) Hybridisation of carbon in graphite is :
(A) sp2
(B) sp3
(C) sp3d
(D) sp
2) No. of lone pairs & bond pairs on xenon in XeF4 molecule.
(A) 2, 4
(B) 3, 3
(C) 4, 2
(D) 2, 6
3) Favorable condition for ionic bond formation is
(I) High EA of Non-metal
(II) High Ionisation energy of metal
(III) High lattice energy
(IV) High charge on cation & anion
(A) I, II
(B) I, II only
(C) I, III, IV
(D) II, III
4) Predict the correct order of repulsion among the following.
(A) lone pair - bond pair > bond pair - bond pair > lone pair - lone pair
(B) lone pair - lone pair > lone pair - bond pair > bond pair - bond pair
(C) lone pair - lone pair > bond pair - bond pair > lone pair - bond pair
(D) bond pair - bond pair > lone pair - bond pair > lone pair - lone pair
5) For which of the following molecule significant μ 0.
(a) (b) (c) (d)
(A) Only (c)
(B) (c) and (d)
(C) Only (a)
(D) (a) and (b)
6) In E1 reaction, rate determining step involves the formation of:
(A) free radical
(B) carbanion
(C) carbocation
(D) final product
7)
The major product of the above reaction is obtained by mechanism
(A) SN2
(B) E2
(C) E1cB
(D) SN2, E2 mixed mechanism
8)
Major product obtained when this alkyl halide is subjected to E2 reaction under the treatment of
potassium tert-butoxide will be :
(A)
(B)
(C) Both (A) and (B)
(D) Trans isomer
9) Correct major product formed in given reaction is-
(A)
(B)
(C)
(D)
10) Mole fraction of the toluene in the vapour phase which is in equilibrium with a solution of
benzene (p° = 120 Torr) and toluene (p° = 80 Torr) having 2.0 mol of each is
(A) 0.50
(B) 0.25
(C) 0.60
(D) 0.40
11) The units of rate and rate constant are identical in
(A) fractional order reactions
(B) first order reactions
(C) second order reactions
(D) zero order reactions.
12) The most stable carbanion among the following is
(A)
(B)
(C)
(D)
13) What will be the major product of the following reaction
(A)
(B)
(C)
(D)
14) The compound which would undergo a reaction with ammonia by SN1 mechanism is
(A)
(B)
(C)
(D)
15) The correct order of nucleophilicity of halide ions in polar aprotic solvent is:
(A) F⊝ > Cl⊝ > Br⊝ > I⊝
(B) I⊝ > Br⊝ > Cl⊝ > F⊝
(C) Cl⊝ > F⊝> Br⊝ > I⊝
(D) I⊝ > Cl⊝ > Br⊝ > F⊝
16) A and B are :-
(A) CH3–CN, CH3–NC
(B) CH3–NC, CH3–CN
(C) CH3–CH2CN, CH3CH2NC
(D) CH3–CH2NC, CH3–CH2–CN
17)
Decreasing order towards SN1 reaction for the following compounds is:
(A) a > c > d > b
(B) a > b > c > d
(C) b > d > c > a
(D) d > b > c > a
18)
(A)
(B)
(C)
(D)
19) Product (A) in
(A)
(B)
(C)
(D)
20)
D-exchange is observed in :
(A) E1
(B) E2
(C) E1cB
(D) None of these
SECTION-II
1) How many reactions giving Hoffmann product as major product
(a) (b)
(c) (d)
(e)
2)
How many of the following are less reactive than Me–Cl in an SN2 mechanism reaction
(a) (b) (c) Me – I (d) CH2 = CH – Cl (e) Me3C – Cl
(f) (g) (h) Me – Br
3)
Number of transition states formed during the above reaction is/are :
4) Equal mass of A & B are present in liquid solution then total pressure exerted by vapours is X
torr, then find the value of . [A = 80 gm/mol, B = 120 gm/mol]
5) For a certain first order reaction 32% of the reactant is left after 570 s. The rate constant of this
reaction is _______ × 10–3 s–1. (Round off to the Nearest Integer).
[Given : log102 = 0.301, ln10 = 2.303]
MATHEMATICS
SECTION-I
1) Foot of the normal from the point (4,3) to a circle is (2,1) and a diameter of the circle has the
equation 2x – y = 2 then the equation of the circle is
(A) x2 + y2 – 4y + 2 = 0
(B) x2 + y2 – 4y + 1 = 0
(C) x2 + y2 – 2x – 1 = 0
(D) x2 + y2 – 2x + 1 = 0
2) If a = max{(x + 2)2 + (y – 3)2} and b = min{(x + 2)2 + (y – 3)2}
where x and y satisfying
x2 + y2 + 8x – 10y – 40 = 0, then
(A) a + b = 20
(B) a + b = 172
(C) a – b =
(D) a – b =
3) Let r be radius of the smallest circle which cuts the two circles x2 + y2 = 1 and x2 + y2 + 8x + 8y –
33 = 0 orthogonally then the value of r2 is
(A) 7
(B) 5
(C) 2
(D) 10
4) Locus of midpoint of chords of the circle x2 + y2 = 25, which are tangents to the circle x2 + y2 - 20x
- 20y + 175 = 0 will be :
(A) (10x + 10y + x2 + y2)2 = (x2 + y2)
(B) (10x + 10y - x2 + y2)2 = 100 (x2 + y2)
(C) (10x + 10y - x2 - y2) = 25 (x2 + y2)
(D) (10x + 10y - x2 - y2)2 = 25 (x2 + y2)
5) The circle S ≡ x2+y2+kx +(k+1)y–(k+1)= 0 always passes through two fixed point for every real k.
If the minimum value of the radius of the circle S is then the value of ‘P’ is :
(A) 4
(B) 6
(C) 8
(D) 16
6)
Find the approximate value of (66)1/3 using differentials :
(A) 4.01
(B) 4.040
(C) 4.041
(D) 4.40
7) The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If
initially, the radius of balloon is 3 units and after 5 seconds,, it becomes 7 units, then its radius after
9 seconds is :
(A) 9
(B) 10
(C) 11
(D) 12
8) The area of the triangle formed by the coordinate axes and a tangent to the curve xy = a2 at the
point (x1, y1) on it is
(A)
(B)
(C)
(D)
9) The number of points on the curve y = 54x5 – 135x4 – 70x3 + 180x2 +210x at which the normal
lines are parallel to x + 90y + 2 = 0 is :
(A) 2
(B) 3
(C) 4
(D) 0
10) The point on the straight line 2x + 2y + 7 = 0 which is nearest to curve having parametric
equation
x = t2 – 2t + 3
y = t2 + 2t + 3
(A)
(B) (1, 0)
(C)
(D)
11) If (x – 1)2018(x – 2)2019 is the derivative of ƒ(x) and ƒ(x) = logeg(x) then
(A) at x = 1, g(x) has local maximum
(B) at x = 1, g(x) has a local minimum
(C) at x = 2, g(x) has neither maximum nor minimum
(D) at x = 2, g(x) has minimum
12)
The height of a right circular cone of maximum volume inscribed in a sphere of diameter a is-
(A)
(B)
(C)
(D)
13) If for a function f(x), f '(a) = 0 = f"(a) = ..... = f(n–1)(a) but fn(a) ≠ 0 then at x = a, f(x) is minimum
if-
(A) n is even and fn (a) > 0
(B) n is odd and fn (a) > 0
(C) n is even and fn (a) < 0
(D) n is odd and fn (a) < 0
14) If p(x) be a polynomial of degree three that has a local maximum value 8 at x = 1 and a local
minimum value 4 at x = 2; then p(0) is equal to:
(A) 12
(B) –24
(C) 6
(D) –12
15) A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness the melts
at a rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate (in cm/min.) at which of the
thickness of ice decreases, is :
(A)
(B)
(C)
(D)
16) If , is differentiable everywhere and k = a + b, then value of k is
(A) –1
(B) 2
(C) 0
(D) 1
17) The function
(A) discontinuous at only one point in its domain.
(B) discontinuous at two points in its domain.
(C) discontinuous at three points in its domain.
(D) continuous everywhere in its domain.
18) Let , where a is a constant such that then the
value of a is :
(A) 2
(B) 3
(C) 5
(D) –5
19) If y = at2 + bt + c and t = ax + b, then equals
(A) a3
(B) 2b3
(C) 2a3
(D) a3 + b
20) The range of ƒ(x) = sin–1x + tan–1x is :-
(A)
(B)
(C)
(D) (–π, π)
SECTION-II
1) Let f(x) = 30 – 2x – x3, then find the number of positive integral values of x which satisfies f(f(f(x)))
> f(f(–x))
2) If is monotonic for all for the M – m =
3) If f(x) is continuous and differentiable over [–2, 5] and for all x in (–2, 5), then the
sum of digit of greatest possible value of f(5) – f(–2) is:-
4) The number of real solutions of equation 2017x – 2016x = x is_________
5) For a polynomial g(x) with real coefficient, let mg denote the number of distinct real roots of
g(x). Suppose S is the set of polynomials with real coefficients defined by
S = {(x2 – 1)2 (a0 + a1x + a2x2 + a3x3) : a0, a1, a2, a3 ∈ R}.
For a polynomial f, let f' denote its first and second order derivatives, respectively.
Then the minimum possible value of (mf' + mf"), where f ∈ S, is____
ANSWER KEYS
PHYSICS
SECTION-I
Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A. C C D D D D C B A C A B C B D B C A B D
SECTION-II
Q. 21 22 23 24 25
A. 5 5 8 48 4
CHEMISTRY
SECTION-I
Q. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
A. A A C B B C C B D D D B C A A D C D A C
SECTION-II
Q. 46 47 48 49 50
A. 4 6 6 6 2
MATHEMATICS
SECTION-I
Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
A. C D A D C C A C C D D A A D C D D B C A
SECTION-II
Q. 71 72 73 74 75
A. 2 6 3 2 5
SOLUTIONS
PHYSICS
1) Time taken :
x = ut =
=
2) df = P(wdx)
= (ρgh) (wdx)
3)
4) Weight of the body = Weight of the liquid displaced
In water
...(i)
In liquid,
..(ii)
From equation (i) and (ii), we get
kg m–3
–3
= 800 kg m
7) across R for any value of R.
10)
Using kirchhoff's law and
So current through 2/3 is
11) Suppose m rows are connected in parallel and each row contains n identical cells (each
cell having E = 15 V and r = )
For maximum current in the external resistance R, the necessary condition is
Total cells
On solving equations (i) and (ii) n = 12 and m = 2
i.e. 2 rows of 12 cells are connected in parallel.
12)
=
15)
when a resistance is joined in parallel with voltmeter, effective resistance of circuit will
decrease and current through ammeter increases so, A increases. The sum of A & V is
constant so, V decreases.
16)
AB = BC = CD = DA = BD = a (say)
Figure-1 Figure-2
in Fig. 1
in Fig. 2
V2 = 0
Electric field
So,
17)
Electric field =
= (independent of r.)
18) Since TIR occurs at surface AC, thus
1.5 sin θ = ⇒ sin θ =
Therefore
20) We have
21) Ans. (5)
For heater
V = 25 v
iR = i1 – iH = 5
V = IR
22)
23)
R1 + R2 = r2
24) In Balanced conditions
x = 48 cm
25) Let dq (= λ dℓ) passes through a given vertical plane in time dt.
The, i = = = ℓv = 2 × 10–3 × 2 = 4 mA
CHEMISTRY
40) Nucleophilicity increases down the group in polar aprotic solvent.
42) The rate of SN1 reaction depends upon stability of carbocation which follows the order
∴ Reactivity order
(b) > (d) > (c) > (a)
46)
a,b,c,e
49) Let mass of A & B in solution = y gram then
= 40 × + 30 × = 24 + 12 = 36 torr.
MATHEMATICS
51) C ≡ centre
Find equation of line passing through point (4, 3) and (2, 1) and solve this line with the given
diameter.
52)
C : centre
2 2
53) Let circle be x + y + 2gx + 2fy + c = 0
for orthogonality
2(g.0 + f.0) = c – 1
2(g.4 + f.4) = c – 33
⇒ c = 1 and g + f = –4
54) Equation of chord of the circle x2 + y2 = 25, whose midpoint is P(h, k)
T = S1
xh + yk – 25 = h2 + k2 – 25 ...(1)
⇒ this chord is tangent to the circle
x2 + y2 – 20x – 20y + 175 = 0
Length of perpendicular from centre
= Radius
⇒ (10h + 10k – h2 – k2)2 = 25(h2 + k2)
(10x + 10y – x2 – y2)2 = 25(x2 + y2)
55) x2 + y2 + y – 1 + k(x + y – 1) = 0
centre of required circle lie on AB
⇒
–2k – 1 – 2 = 0
x2 + y2 + y – 1 – (x + y – 1) = 0
⇒ 2x2 + 2y2 + 2y – 2 – 3x – 3y + 3 = 0
⇒ 2x2 + 2y2 – 3x – y + 1 = 0
57)
Let r be the radius of spherical balloon
S = Surface area
S = 4πr2
(constant)
4πr2 = kt + C (C is constant of integration)
For t = 0, r = 3 ⇒ 36π = C
For t = 5 , r = 7 ⇒ K = 32π
4πr2 = 32πt + 36π
r2 = 8t + 9
for t = 9
r2 = 81
r = 9.
58) Since y = ,∴ =
∴ At (x1, y1),
Thus tangent to the curve will be
⇒
This meets the x–axis where y = 0
Point on the
Again, tangent meets the y – axis where x = 0
So point on the y-axis
Required area =
59) Normal of line is parallel to line x + 90y + 2 = 0
Now,
(4) normals
60)
=
⇒t=0
Equation of line through point p
y–x=0
Solve y – x = 0
and 2x + 2y + 7 = 0
61) g(x) = eƒ(x)
⇒ g'(x) = eƒ(x) × ƒ'(x)
sign scheme for g'(x)
minimum at x = 2
64)
Since p(x) has realtive extreme at
x=1&2
so p'(x) = 0 at x = 1 & 2
⇒ p'(x) = A(x – 1) (x – 2)
⇒ p(x) =
p(x) = …(1)
P(1) = 8
From (1)
⇒ …(3)
P(2) = 4
⇒ …(4)
From 3 & 4, C = –12
So P(0) = C =
65) Let thickness of ice be 'h'.
Vol. of ice =
Given and h = 5cm
⇒
66)
67)
At any point ‘t ’ in domain, t \(\) – {0,2,–2}
(t ≠ 0)
⇒ is continuous at all Î Domain
68)
⇒–
⇒a=3
69) y = at2 + bt + c
= a(ax + b)2 + b(ax + b) + c
= a(a2x2 + b2 + 2abx) + abx + b2 + c
= a3x2 + ab2 + 2a2bx + abx + b2 + c
y' = 2a3x + 2a2b + ab
y" = 2a3
70) ƒ(x) = sin–1x + tan–1x, x ∈ [–1, 1]
Range of
71) f ′ (x) = –2 – 3x2 < 0 ⇒ f(x) is decreasing
∵ f(f(f(x))) > f(f(–x))
⇒ f(f(x)) < f(–x) ⇒ f(x) > –x
⇒ 30 – 2x + x3 > –x ⇒ x3 + x – 30 < 0
⇒ (x – 3) (x2 + 3x + 10) < 0 ⇒ x < 3
Positive values of integer x = 1, 2
72)
or
So, D < 0 and 49 – 7 ≠ 0
m = 2, M = 8
M–m=6
73) Apply LMVT
for some x in (–2, 5)
Now,
∴ Greatest possible value of f(5) – f(–2) is 21.
74)
Let f(t) = tx ; x is some real number.
Applying LMVT for f(t) in [2016, 2017]
⇒ x = λ(x-1) (as 2017x – 2016x = x)
⇒ x[λ(x–1) – 1] = 0
⇒ x = 0,1
75)
Now f(1) = f(–1) = 0
⇒ f'(α) = 0, α ∈ (–1, 1) [Rolle's Theorem]
Also f'(1) = f'(–1) = 0 ⇒ f'(x) = 0 has atleast 3 root, –1, α. 1 with –1 < α < 1
⇒ f"(x) = 0 will have at least 2 root, say β, γ such that
–1 < β < α < γ < 1 [Rolle's Theorem]
So, min(mf") = 2
and we find (mf' + mf") = 5.
Thus, Ans. 5.