11-05-2025

7501CJA101001250011 JA

PHYSICS

SECTION-I

1) Refer the block diagram in which four function machines are shown.

Which of the following statements is/are true?

(A)

(B)

(C)

(D)

2) Radius and height of a cone closed at bottom are 3m and 4m respectively at some instant. Radius
is increasing at the rate R m/s and height is increasing at the rate H m/s then :

(A)
Rate of increase in surface area is

(B)
Rate of increase in surface area is
(C) Rate of increase in volume is π(8R + 3H)
(D) Rate of increase in volume is π(3R + 8H)

3) A bird starts from (1, 0, 0) in the direction with a speed 21 m/s for 5 sec, then along

the direction with a speed for 5 sec. Find the final displacement of the bird so
that it reaches the origin?

(A)
(B)
(C)
(D)

4) Sometimes it is convenient to construct a system of units so that all quantities can be expressed in
terms of only one physical quantity. In one such system, dimensions of different quantities are given
in terms of a quantity X as follows : [position] = [Xα] ; [speed] = [Xβ]; [acceleration] = [Xp]; [linear
momentum] = [xq]; [force] = [Xr]. Then,

(A) α + p = 2β
(B) p + q – r = β
(C) p – q + r = α
(D) p + q + r = β

5) Which of the following statements are correct ?

(A)
If y = 2x2 + 8x + 6 then the graph of vs x will be a parabola.

(B)
If S = 1 + x + x2 + x3 + .... + x24 then the value of at x = 1 is 300.

(C)
The maximum value of is 75.
The distance from origin (0,0) to the vertex of the parabola given by the equation (y - 3)2 = 4 (x -
(D)
4) is 5 units.

6) Let us consider a system of units in which mass and angular momentum (angular momentum =m
v r, where v = velocity and r = radius) are dimensionless. If length has dimension of L, which of the
following statement(s) is/are correct ?

(A) The dimension of energy is [L–2]

(B) The dimension of force is [L–3]
(C) The dimension of power is [L–5]
(D) The dimension of linear momentum is [L–1]

SECTION-II

1)

The position vector of point A (7m, 3m, 0) is rotated about y-axis. If the volume (in m3) of the

cone thus formed is 154 × N then find the value of N. (Take : π = )

2) A cylinder of radius R is partially filled with water. There is an inverted cone of cone angle 90°
and base radius R/2 which is falling in it with a constant speed v = 30 cm/s. Find rate of increase of
water level (in cm/s) when half height of cone is immersed in water.

3) Semi vertical angle (θ) of a cone of maximum volume and of given slant height is tan–1 then

the value of x is ?

4)

The following curve represents rate of change of a variable y w.r.t x. The change in the value of y
when x changes from 0 to 11 is:

5) , find θ (in degrees) for which is zero. (value of θ is between 0° and 90°)
6) A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from
each corner and folding up the flaps to form the box. Find the size of the square in cm that should be
cut off so that the volume of the box is maximum possible.

SECTION-III

1) If the unit of mass is made four times, unit of length made double and unit of time doubled the
numerical value of universal gravitational constant (G) becomes n times. The value of n is

2) If f(x) = 2sin x – x and then value of f(x) at local maxima point is . find K.

3) Assume that maximum mass m1 of a boulder swept along by a river, depends on the speed V of the
river, the acceleration due to gravity g, the density d of the boulder. Calculate the percentage
change in maximum mass of the boulder that can be swept by the river, when speed of the river
increases by 1%.

4) If surface area of a cube is changing at a rate of 5 m2/s, find the rate of change of body diagonal at

the moment when side length is 1 m. If your ans is write x in (m/s)

5) Sprit in a bowl evaporates at a rate that is proportional to the surface area of the liquid. Initially,

0 0
the height of liquid in the bowl is H . It becomes in time t . More time needed for the height of

the liquid to becomes will be . Find the value of x.

6) is equal to :

CHEMISTRY

SECTION-I

1) 20 ml of a gaseous compound containing nitrogen and hydrogen only, is passed through a series
of electric spark, by which the volume becomes 40 ml. When the resulting gases are passed through
water, the volume becomes 30 ml. Now, 50 ml oxygen gas is mixed in the resulting gases and fired.
If the final volume becomes 50 ml, the correct information(s) is/are
(All the volumes are measured at the same temperature and pressure). N2 and H2 gases are
completely insoluble in water)

(A) The gaseous compound is NH3

(B) The gaseous compound is N2H4
(C) Only 50% of the gaseous compound was decomposed into N2 and H2
(D) 100 % of the gaseous compound was decomposed into N2 and H2

2) If a definite volume of ‘20 vol’ H2O2 solution is diluted such that the volume of diluted solution
becomes double than that of original volume, then.

(A) The volume strength of diluted solution becomes ‘40 vol’.

(B) The molarity of solution becomes half of its initial molarity.
(C) The molality of solution becomes half of its initial molality.
(D) The maximum amount of O2 gas obtainable from the solution remains the same.

3) Two gases A and B which react according to the equation, aA(g) + bB(g) —→ cC(g) + dD(g) to give two
gases C and D are taken (amount not known) in an Eudiometer tube (operating at a constant
Pressure and temperature) to cause the above. If on causing the reaction there is no volume change
observed then which of the following statement is/are correct.

(A) (a + b) = (c + d)
average molecular mass may increase or decrease if either of A or B is present in limited
(B)
amount.
(C) Vapour Density of the mixture will remain same throughout the course of reaction.
(D) Total moles of all the component of mixture will change.

4) The correct statement(s) regarding 2M MgCl2 aqueous solution is/are

(dsolution = 1.09 gm/ml)

(A) Molality of Cl– is 4.44m

(B) Mole fraction of MgCl2 is exactly 0.035
(C) The conc. Of MgCl2 is 19% w/v
4
(D) The conc. Of MgCl2 is 19×10 ppm

5) The sample(s) containing same number of ‘Na’ atoms as there are ‘Na’ atoms in 5.3 g of Na2CO3,
is/are
(Given: Atomic mass in U Na = 23, H = 1, Cl = 35.5. S = 32, O = 16, P = 31, C = 12)

(A) 4 g of NaOH
(B) 5.85 g of NaCl
(C) 0.25 mole of Na2SO4
(D) 5.467 g of Na3PO4

6) Consider the following reactions

Choose correct option(s). Assuming all other reactant required are in sufficient amount.

(A) Produced mass of NH4Cl is 2.71 gm if 4 gm of NaOH is taken.

(B) 41.25 gm (NH4)2SO4 is required to produce 10.7 gm NH4Cl.
(C) 25 gm NaOH is required to produce 10.7 gm NH4Cl.
Produced moles of NH4Cl (in IInd reaction) are 1.6 times of produced moles of Na2SO4 (in Ist
(D)
reaction), if 4 gm of NaOH is taken.

SECTION-II

1) When 1000 mL of 0.5 M HCl solution was heated, 3.65 gm HCl was lost and volume of solution
became 500 mL. Find the molarity of resulting solution.

2) The density (in mL–1) of a 3.60 M sulphuric acid solution that is 30% H2SO4 (Molar mass = 98 g
mol–1) by mass will be

3) A solution contain substances A and B in H2O (solvent). The mole fraction of 'A' is 0.25 and
molarity of 'B' is 12.5 M. The solution has density 1 gm/ml.

Calculate 'Q', where Q = .

[Molecular weight of A = 32 gm/mol; molecular weight of b = 12 gm/mol]

4) 1L, 1M HCl solution is mixed with 1L, 2M H2SO4 solution. the solution is made 3L by addition of
solvent. The resultant mixture is reacted with 8L solution of NaOH. Find mass (in kg) of NaOH
required for the completion of reaction.

5) A sample of C12H22O11 has 24 g carbon, then let number of hydrogen atoms present in the same
sample be x × 1024. Find x upto 2 decimal places (use avogadro's no = 6 × 1023)

6) A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is
present to the extent of 13.2% by weight. 2.58 g of the mineral on heating lost 1.232 g of CO2.
Calculate the % by weight of the other metal.

SECTION-III

1)

N2O4 dissociates into NO2, if % dissociation of N2O4 is 33.33 %. Calculate average molecular weight
of gaseous mixture formed .

Fill your answer as sum of digits (excluding decimal places) till you get the single digit
answer.

2) Equal mass of KClO3 are reacted as per given two reaction

2KClO3 → 2KCl + 3O2 .........(A)
4KClO3 → 3KClO4 + KCl .........(B)
Ratio of KCl produced from reaction (A) to reaction (B) is.
3) 20 ml of M/X solution of an acid (H3PO4) is required to neutralise 0.8 M, 25 ml NaOH solution.
What is the value of X ?

4) 500 ml of 2M CH3COOH solution is mixed with 600 ml 12% w/v CH3COOH solution, then calculate
the final molarity of solution.

5) How many blood cells of 5ml each having [K+] = 0.1M should burst into 25 ml of blood plasma
having [K+] = 0.02 M so as to give final [K+] = 0.06 M

6) Hg reacts with Cl2 to form Hg2Cl2 and HgCl2. If 1.5 mol of Hg and 1.0 mol of Cl2 are
added.Calculate total moles of HgCl2 and Hg2Cl2 formed ?

MATHEMATICS

SECTION-I

1)

Consider an inequality , then which of the following is/are correct ?

(A) The values of x which satisfying the inequality can be

(B) The values of x which satisfying the inequality can be
(C) Number of integers which does not satisfying the inequality is 9
(D) Number of non negative integers which does not satisfying the inequality is 2

2) If set of values of x which satisfying the inequality , is

then which of the following is/are correct ?

(A) a + b + c + d + e = 0
(B) a + d + e is a prime number
(C)
(D) d + e is a composite number

3) Let S be the universal set, A, B & C are sets such that |(A ∪ B ∪ C)'| = 1 & |S| = 8.
If |A| = |B| = |C|, then which of the following option(s) is\are true ?

(A) Sum of minimum and maximum value of |A ∩ B ∩ C| is 7.

(B) Let X be the set of all possible values of cardinality of A ∩ B ∩ C, then |X| = 7.
(C) Let X be the set of all possible values of cardinality of A ∩ B ∩ C, then |X| = 6.
(D) If |A| = 4, then |A ∩ B ∩ C| ∈ {0, 1, 2}.

4) Consider where . Identify which of the following statement(s)

is(are) correct?

(A) If f (x) = 0 has two distinct real roots for all values of 'b', then sum of values of 'a' is 15
(B) If f (x) = 0 has two distinct real roots for all values of 'b', then sum of values of 'a' is 10
(C) If f (x) = 0 has two real roots for all values of 'b', then sum of values of 'a' is 15
(D) If f (x) = 0 has two real roots for all values of 'b', then sum of values of 'a' is 10

5) If a,b are non-zero real numbers and α,β the roots of x2 + ax + b = 0, then

(A) α2,β2 are the roots of x2 – (2b – a2)x + a2 = 0

(B)
are the roots of bx2 + ax + 1 = 0

(C)
are the roots of bx2 + (2b – a2) x + b = 0
(D) (α – 1), (β – 1) are the roots of the equation x2 + x(a + 2) + 1 + a + b = 0

6) If a and b are the roots of the equation x2 – 6x – 4 = 0, then the value of the expression

is a

(A) real number

(B) composite number
(C) prime number
(D) rational number

SECTION-II

1) The number of values of x satisfying the equation is :

2) The value of the expression

is equal to-

3) The sum of integral solution(s) satisfying the inequality |2x2 – 13x + 5| < |x2 – 3x – 4| + |–x2 + 10x

– 9| is λ, then the value of is equal to:

4) If are the roots of the equation x3 - 5x2 + 3= 0, then the value of is

5) If then the number of integers in the range of the expression is equal to

6) Let α, β be the roots of the equation and α, γ be the roots of the equation

. If , then is equal to (where λ

> 0)

SECTION-III

1)

Number of values of x which satisfying the equation is

2) Sum of all integral values of x satisfying the inequality is

λ, then |λ| is

3) Let f(x) = x7 + x5 + ax3 + bx. The remainder when f(x) is divided by (x + 1) is –6 and the remainder

when it is divided by (x2 – 1) is g(x) then the value of the is equal to :

4) The least value of n such that (n – 2)x2 + 8x + (n + 4) > 0 ∀ x ∈ R where n ∈ N is -

5) If the polynomial P(x) = x5 + x2 + 1 has the roots x1, x2, x3, x4, x5 and g(x) = x2 – 2 then the value of
the expression g(x1) . g(x2) . g(x3) . g(x4) . g(x5) – 30g(x1 . x2 . x3 . x4 . x5) is equal to

6) The number of real values of m for which A ∪ B has exactly three distinct elements, where
A = {x : x2 + (m – 1)x – 2(m + 1) = 0, x ∈ R}
B = {x : (m – 1) x2 + mx + 1 = 0, x ∈ R}
 ANSWER KEYS

PHYSICS

SECTION-I

Q. 1 2 3 4 5 6
A. A,C A,C C A,B B,D A,B,D

SECTION-II

Q. 7 8 9 10 11 12
A. 1.00 2.00 2.00 35.00 30.00 3.00

SECTION-III

Q. 13 14 15 16 17 18
A. 2 3 6 3 2 1

CHEMISTRY

SECTION-I

Q. 19 20 21 22 23 24
A. B,C B,D A,C A,C A,B,D B,C,D

SECTION-II

Q. 25 26 27 28 29 30
A. 0.80 1.17 to 1.18 50.00 0.20 2.20 21.70

SECTION-III

Q. 31 32 33 34 35 36
A. 6 4 3 2 5 1

MATHEMATICS

SECTION-I

Q. 37 38 39 40 41 42
A. A,C A,C,D A,B,D B,C B,C,D A,B,D

SECTION-II

Q. 43 44 45 46 47 48
A. 5.00 3.16 5.20 729.00 1.00 98.00
 SECTION-III

Q. 49 50 51 52 53 54
A. 5 3 6 5 7 7
 SOLUTIONS

PHYSICS

2) Area A = πr2 + πrl

= πr2 + πr

Volume V =

3)

7)

volume =
here r = 7, h = 3

8) Let height of water level is h and depth of cone in water is x

r = x (since cone angle is 90°)
so total volume of water =
V = Vcylinder – Vcone

V = π R2 h – π x3
differentiating with respect to time

0 = π R2 – π x2
 = ...(i)
x+y=h

+ =

=–v

⇒ =v+ ...(ii)
Solving (i) and (ii) we get

=
V = 30 cm/s

put x=

Ans. = 2 cm/s

10)

As

So,

11)

θ = 30°

13) Let unit of mass, length and time by M1, L1, T1 then unit of G is M1–1 L13 T1–2. In new system.
M2 = 4M1, L2 = 2L1, T2 = 2T1 then new unit of

So unit becomes half of old unit of G. So numerical value doubles. Hence n =2.

14) f'(x) = 2cosx – 1

f"(x) = –2sin x

f'(x) = 0 ⇒ ⇒
 At

At

∴ gives maxima

⇒ K=3

17) Let the area of cross section of the bowl at height h be A.

Let the height decrease by dh in further interval dt.
Volume that evaporates = Adh

As per the question

Where k is a positive constant. We have placed a negative sign because h is decreasing with
time.

Let height be at time 't'

Required answer is

18)

= = = tan π/4 – tan 0 = 1

CHEMISTRY

19) NxHy —→ N2 + H2 + NxHy N2 + H2 50 ml

20 ml 40 ml 30ml
Volume of NxHy left undecomposed = 40 – 30 = 10 ml
∴ Volume of NxHy left decomposed = 20 – 10 = 10 ml (50% decomposition)
NxHy —→ N2 + H2
10 ml 5x ml 5y ml
5x + 5y = 30 ⇒ x + y = 6 ......(A)
2H2 + O2 —→ 2H2O

5y ml O

Contraction in volume : ⇒y=4,x=2

∴ Compound = N2H4

20) Strength H2O2 = 20 vol.

Vol. Strength = 11.2 × molarity
20 = 11.2 × molarity

Molarity = = 1.7857 M
Now volume is double. So molarity will be

=
∴ Means molarity of solution become half of initial molarity
Now volume strength = 11.2 × molarity
= 11.2 × 0.8928
= 10 vol. H2O2
H2O2 diluted to doubled its volume but moles remains same so O2 gas remains same after
changing the volume.

21)

aA(g) + bB(g) → c(Cg) + dD(g)

If there is no. volume change the:
(a + b) = (c + d)
Since : (a + b = c + d); hence moles of substances
at any instant of reaction is constant.

V.Dmix = = constant.

22) (A)

molal MgCl2 solution

⊝
Molality of Cl = 2 × molality of MgCl2
= 2 × 2.22
= 4.44 m

(B) Xsolute =

= (Mass of solvent = 900)

(C)
= 19%
(D) The concentration of Mg Cl2 is ppm

23)

Number of Na atoms in Na2CO3 is ; × 2NA = 0.1 NA

(A) Na atoms = × NA = 0.1 NA

(B) Na atoms = × NA = 0.1 NA

(C) Na atoms = 0.25 × 2NA = 0.50 NA

(D) Na atoms = × 3NA = 0.1NA

24)

⇒ = 1.6 times

25)

nHCl = 0.5 × 1000 –

= 500 mmol – 0.1 mol
= 0.5 mol – 0.1 mol
= 0.4 mol
 = 0.8M

26)

Ans. (1.17-1.18)

27)

Let 1000 ml solution

Mass 1000 gm
Mass of B = 12.5 × 12 = 150
Remaining mass = 850 , Let mass of A = x, Mass of water = 850 – x

0.25 =
x = 400
moles of A = 12.5
Moles of A = moles of B
Molarity of A = 12.5
Mole fraction of B = 0.25

28)

HCl + NaOH → NaCl + H2O

1 mole 1 mole
H2SO4 + 2NaOH → Na2SO4 + 2H2O
2 mole 4 mole
Total NaOH = 5 mole
wNaOH = 5 × 40 = 200 gm.

29)

EXPLANATION = The question asks to find the number of hydrogen atoms (expressed as 'x' ×
1024) present in a sample of sucrose (C12H22O11) that contains 24 grams of carbon.
FINAL ANSWER= 2.20.

30)

Let two equimolar carbonates are ACO3 and BCO3.

Wt. of metal A = 2.58 × 0.132 = 0.34 gm
ACO3 + BCO3 → AO + BO + 2CO2
Total two moles provide 2 gm of CO2
or 60 gm provide 44 gm of CO2
Mass of required to produced 1.232 g of CO2

= 1.232 = 1.68 gm
 so total wt. of cabonates = 1.68 gm
wt. of metal (A + B) = 2.58 – 1.68 = 0.90 gm
wt. of B = 0.90 – 0.34

= 0.56 gm ; % of B = × 100
= 21.70 Ans.

31) Meff =

1 + 1/3 =
Mav = 69

32) 2KClO3 → 2KCl + 3O2

↓
n n
4KClO3 → KCl + 3KClO4
n n/4

33) HnA + nNaOH → nH2O + NanA

0.33 × 20 × n = 0.8 × 25
n=3

34)

12 gm CH3COOH is present in 100 ml of solution

120 gm CH3COOH is present in 1000 ml of solution

M2= , Now we are mixing

500 ml, 2M CH3COOH + 2M, 600 ml CH3COOH
M1V1 + M2V2 = M3V3
500 × 2 + 600 × 2 = M3 × 1100 ,

M3 =

35)

0.06 =
x=5

36) Let a mol & b mol HgCl2 & Hg2Cl2 respectively formed.
Hg + Cl2 → HgCl2 + Hg2Cl2
a + 2b = 1.5 ...(i) (POAC on Hg)
2a + 2b = 2 ...(ii) (POAC on Cl)
On solving a = 0.5 and b = 0.5
 a+b=1

MATHEMATICS

37)

(x - 1)

When then
when squaring both sides

38)

a = -3 , b = -1, c= 0, d=1, e=3

40)

For two distinct real roots D > 0

a = 1, 2, 3, 4
sum = 10
For two real roots D 0

a = 1, 2, 3, 4, 5
sum = 15

41)

x2 + ax + b = 0
α + β = –a, αβ = b
α2 + β2 = (α + β)2 – 2αβ = a2 – 2b
Quadratic with roots α2,β2
⇒ x2 – (α2 + β2)x + (αβ)2 = 0
x2 – (a2 – 2b)x + b2 = 0
Quadratic with roots

bx2 + ax + 1 = 0

Quadratic with roots

bx2 – (a2 – 2b)x + b = 0

Quadratic with roots α – 1, β – 1
x2 – ( –a – 2)x + b –(–a) + 1 = 0
x2 + (a + 2)x + b + a + 1 = 0

42)

a17 – 6a16 – 4a15 = 0

b17 – 6b16 – 4b15 = 0
adding (1) and (2)

43) x = 5 or |x – 5| = 1
⇒ x = 6 or x = 4
x2 + 1 = 2x + 1
⇒ x = 0, x = 2

44)

45) If | a | + | b | > | a – b |
i.e. a2 + b2 + 2 | a | | b | > a2 + b2 – 2ab
i.e. | a | | b | > – ab ⇒ ab > 0
Hence, (x2 – 3x – 4) (–x2 + 10x – 9) > 0
( x – 4) (x + 1) (x – 1) (x – 9) < 0

∴ x ∈ (– 1, 1) ∪ (4, 9)
∴ Number of integral values of x is {0, 5, 6, 7, 8} i.e. 5 values.

46)
 are the roots of the equation

Equation having roots

will be :
Product of roots of above equation

47)

Range

48)

α + β = 4λ

∴ αβ = 5
4λ2 = 8 ⇒

∴ .

49)

x = – 1, –3, –4, 3, 7

50)

λ = -3

51)

ƒ(–1) = –6 ⇒ ƒ(1) = 6
ƒ(x) = (x2 – 1)Q(x) + ax + b ; g(x) = ax + b
ƒ(1) = 6 = a + b
ƒ(–1) = –6 = –a + b
On solving both a = 6, b = 0
g(x) = 6x

g(1) = 6
52)

n-2>0&D<0

53) Let y = x2 – 2 = g(x)

y5 + 10y4 + 40y3 + 79y2 + 74y + 23 = 0

54) A = {x : (x – 2)(x + m + 1) = 0, x ∈ R}
B = {x : (m – 1)x2 + mx + 1 =0, x ∈ R}
Case-I : When m –1 = 0 m = 1
A = {2, –2} and B = {–1}
∴ A ∪ B = {2, –2, –1} ⇒ exactly 3 elements
∴ m=1
Case-II : When m ≠ 1

A = {2, –m – 1}, B =
2 = –m – 1 ⇒ m = –3

A = {2}, B
∴ m = –3

2= ⇒ m=

A= , B = {–1, 2}

∴ m=

–m – 1 = ⇒ m2 – 1 = 1
∴ m=±
∴ –m – 1 = –1
⇒ m=0

–1 =
⇒ m–1=1
∴ m=2
Number of values of m = 7