11-05-2025
2302CJA101001250002 JA
PART-A-PHYSICS
SECTION-I(i)
1)
For the given vectors : , and
Then angle between and is :-
(A)
(B)
(C)
(D) None of these
2) Given and . The component of vector along vector :-
(A)
(B)
(C)
(D)
3) The approximate value of x where x = sin 1° cos 1°, is
(A)
(B) 2
(C)
(D)
4)
Find
(A) 3 sin(3x + 2) + c
(B)
(C)
(D)
5) The angle between and is –
(A) 90°
(B) 0°< θ < 180°
(C) 180°
(D) None of these
6) A car is going from one city to another city by travelling 75 km North, 60 km North-West and
20km East. The approximate distance between the two city is :
(A) 120 km
(B) 137 km
(C) 140 km
(D) 170 km
SECTION-I(ii)
1) Which of the following statements about the sum of the two vectors and , is/are correct
(A)
(B)
(C)
(D)
2) Which of the following is/are true statement?
(A) A vector cannot be divided by another vector.
(B) Dot product of two vectors is a scalar.
(C) Since addition of vectors is commutative therefore vector subtraction is also commutative.
The resultant of two equal forces of magnitude F acting at a point is F if the angle between the
(D)
two forces is 120º.
3) Mark the correct option(s)
(A) The x-component of F1 is 3 N
(B) The x-component of F2 is –2 N
(C) The magnitude of resultant force is N
(D) None of the above
4) A particle is moving in x axis according to relation x = (4t – t2 – 4)m :–
(A) Maximum x coordinate of particle is 4m
(B) Magnitude of average velocity is equal to average speed, for time interval t = 0s to t = 2s
(C) Average acceleration is equal to instantaneous acceleration during interval t = 0 to t = 2s
(D) Distance travelled in interval t = 0 to t = 4s is 8 m.
5) The four pairs of force vectors are given, which pairs of force vectors cannot be added to give a
resultant vector of magnitude 10 N?
(A) 2N, 13 N
(B) 5N, 16 N
(C) 7N, 8N
(D) 100N, 105 N
6) Given then
(A)
(B)
(C)
(D)
SECTION-III
1)
2) If then at x = 1 is . Find K.
3) If y = sin2x find . Express your answer in the form . Write value of a
4) A normal human eye can see an object making an angle of 1.8° at the eye approximate height of
object which can be seen by an eye placed at a distance of 1 m from the object is . Find the
value of x.
5) If and , then is equal to
PART-B-CHEMISTRY
SECTION-I(i)
1) Acc. to Bohr, electron absorb energy when it undergoes transition from........ to ....... energy state.
(A) higher, lower
(B) infinity, lowest
(C) Lower, Higher
(D) Higher, any excited
2) What is the work function of the metal if the light of wavelength 4000 Å generates photoelectrons
of velocity 6 × 105 ms–1 form it ?
(Mass of electron = 9 × 10–31 kg
Velocity of light = 3 × 108 ms–1
Planck's constant = 6.626 × 10–34 Js
Charge of electron = 1.6 × 10–19 JeV–1)
(A) 0.9 eV
(B) 4.0 eV
(C) 2.1 eV
(D) 3.1 eV
3) It is known that atoms contain protons, neutrons and electrons. If the mass of neutron is assumed
to half of its original value whereas that of proton is assumed to be twice of its original value then
the atomic mass of will be
(A) Same
(B) 25 % more
(C) 14.28 % more
(D) 28.5 % less
4) 10 g CaCO3 on heating gives 5.6 g CaO and .........g CO2 :-
(A) 4.4
(B) 5.6
(C) 6.5
(D) 4.2
5) A 100 W bulb emits light of wavelength 400 nm number of photons emitted per second by the
bulb is :
(A) 2.0 × 1020
(B) 4.0 × 1019
(C) 6.0 × 1020
(D) 8.0 × 10–20
6) The force of attraction on electron by the nucleus is directly proportional to
(A)
(B)
(C)
(D)
SECTION-I(ii)
1) An aq. solution contains 28% w/v KOH then which of the following is/are correct if density of
solution is 1.28 g/ml ?
(A) Molarity of KOH soltuion = 5 M
(B) Molality of KOH soltuion = 5 m
(C) Mole of fraction of KOH = 0.12
(D) Strength of KOH soltuion in g/L = 280 g/L
2) In the synthesis of urea (NH2CONH2) as given below :
2NH3(g) + CO2(g) → NH2CONH2(aq) + H2O(l)
637.2 g of NH3 are allowed to react with 1142 g of CO2. In this
(A) NH3 is the limiting reactant and CO2 is in excess
(B) NH3 is in excess and CO2 is the limiting reactant
(C) 18.74 moles of urea is formed
(D) 5.00 moles of CO2 is left unreacted
3) The kinetic energy of photoelectron emitted on irradiating a metal surface with frequency is
related by KE = hv – ϕ. The plots of KE vs. incidented frequency v shows :
(A) A straight line with slope equal to Planck's constant.
A straight line with intercept on x-axis equal to the product of threshold frequency and Plank's
(B)
constant.
(C) A straight line with extrapolated intercept on y-axis equal to threshold energy.
(D) A straight line with intercept on x-axis equal to threshold frequency.
4) 1 g atom of nitrogen represents
23
(A) 6.022 × 10 N2 molecules
(B) 22.4 L of N2 at 1 atm and 273 K
(C) 11.2 L of N2 at 1 atm and 273 K
(D) 14 g of N2
5)
Same mass of glucose (C6H12O6) and acetic acid (CH3COOH) contains :
(A) Same number of carbon atoms
(B) Same number of Hydrogen atoms
(C) Same number of oxygen atoms
(D) All of above
6) Which of the following statement(s) are incorrect ?
Photons having energy 400 kJ will break 4 mole bonds of a molecule A2 where A-A bond
(A)
dissociation energy is 100 kJ/mol.
Two bulbs are emitting light having wavelength 2000 Å and 3000 Å respectively. If the bulbs A
and B are 40 watt and 30 watt respectively then the ratio of no. of photons emitted by A and B
(B)
per day is .
(C) When an electron make transition from lower to higher orbit, photon is emitted.
(D) 4eV is sufficient to excite and e– from ground state of H-atom.
SECTION-III
1) If all the oxygen atoms from a mixture of 2 moles H2P2O7, 54 gms H2O and 98 gms H2SO4 are taken
and converted into Ozone, the the number of moles of Ozone gas formed will be:
2) The number of moles of iron in a mixture of 144 gms FeO and 320 gms of Fe2O3 will be :
3) A certain hydrate M.nH2O has 19% by weight of H2O in it. The molar mass of M = 230g mol-1.
Calculate the value of n is ___.
4) The ratio of specific charge (e/m) of a proton and that of an α-particle is
5) Ratio of radii of second and first Bohr orbits of H atom :
PART-C-MATHEMATICS
SECTION-I(i)
1) The value of is:
(A) 3
(B) 6
(C) 5
(D) 4
2) Number of integers satisfying the inequality, is:
(A) 2
(B) 4
(C) 6
(D) 8
3) Value of is:
(A) 2
(B) 1
(C) 0
(D) None
4) Sum of all the real values of x satisfying the equation |x + 1| – |x| = x + 2 is:
(A) –3
(B) 3
(C) – 4
(D) – 2
5) The number of roots of the equation is:
(A) One
(B) Two
(C) Infinite
(D) None
6) Solution set of inequality is:
(A)
x ∈ (–2, –1) ∪
(B)
x ∈ (–3, –1) ∪
(C)
x ∈ (–4, –3) ∪
(D)
x ∈ (–2, –1) ∪
SECTION-I(ii)
1) Which of the following intervals lie in the solution set of > 6 – 3x ?
(A) (1, 4)
(B) (2, 3)
(C) (2, 6)
(D) (2, 4)
2) If the complete solution set of the inequality (log10x)2 > log10 x + 2 is (0, a] [b, ), then:
(A) ab = 10
(B) ab = 1
(C) a2b = 1
(D) a2b = 10
3) The number of value of x satisfying |x – 1| + |x – 3| = 2 is/are :
(A) More than 1
(B) More than 2
(C) More than 3
(D) None of these
4) If , then :
(A) largest positive integer satisfy is 6
(B) smallest integer which satisfy is –2
(C) number of positive integral solution is 5
(D) number of positive integral solution is 7
5) Let , then f(x) > 0 for x belongs to
(A) (–5, –1)
(B) (2, 3)
(C) (8, 9)
(D) (3, 5)
6) If , then:
(A)
maximum value of x is
(B)
x lies between and
(C)
minimum value of x is
(D)
minimum value of x is
SECTION-III
1) If , then possible value of x is ______.
2) The value of is ______.
3) If log(x–1)9 + log9(x – 1) = 2, then the value of log10x, is _________.
4) Let ƒ(x) = x7 + x5 + ax3 + bx. The remainder when ƒ(x) is divided by (x + 1) is –6 and the remainder
when it is divided by (x2 – 1) is g(x) then determine |g(1)| __________.
5) If (x – 3)(x + 2)(x + 3)(x + 8) + 56 = (x2 + qx + r)(x + b)(x + d), then determine |r + 5b + 5d| and
q, r, b, d ∈ Q __________.
ANSWER KEYS
PART-A-PHYSICS
SECTION-I(i)
Q. 1 2 3 4 5 6
A. C D C D B A
SECTION-I(ii)
Q. 7 8 9 10 11 12
A. A,D A,B,D A,B,C B,C,D A,B A,B,C
SECTION-III
Q. 13 14 15 16 17
A. 1 7 2 1 1
PART-B-CHEMISTRY
SECTION-I(i)
Q. 18 19 20 21 22 23
A. C C C A A B
SECTION-I(ii)
Q. 24 25 26 27 28 29
A. A,B,D A,C A,C,D C,D A,B,C,D A,B,C,D
SECTION-III
Q. 30 31 32 33 34
A. 7 6 3 2 4
PART-C-MATHEMATICS
SECTION-I(i)
Q. 35 36 37 38 39 40
A. C D B A D A
SECTION-I(ii)
Q. 41 42 43 44 45 46
A. A,B,D A,C A,B,C A,B,C A,B,C A,B,D
SECTION-III
Q. 47 48 49 50 51
A. 3 1 1 6 3
SOLUTIONS
PART-A-PHYSICS
1)
2)
component of along
3)
5) The angle between two vectors always lie between 0° and 180°
7)
8) 1. vector cannot be divided by vector 2. Ang. displacement is scalar
3. If two for cws 'f' acting at 120° tree resultant
resultant
9)
10) x = (4t – t2 – 4)m
v= = 4 – 2t Velocity is in positive direction initially
x will be maximum when = 0 so t = 2 sec Þ x (t = 2) = 0
In t = 0 to t = 2 sec. direction does not change. Hence average velocity = average speed (B)
Also, acceleration is uniform. So Average acceleration = Instantaneous acceleration (C).
In t = 0 to t = 2 sec. distance travelled is 4m and at t = 4 sec. x = – 4.
So total distance travelled = 8m
12)
13) =
14)
at x = 1
16)
PART-B-CHEMISTRY
19)
hν = ϕ + hn°
ϕ = 3.35 × 10–19 J ⇒ ϕ 2.1 eV
20)
22) Given : Power of bulb (P) = 100 W and wavelength (λ) = 400 nm = 400 × 10–9 m. We know
that energy of one photon
(E) =
= 5 × 10–19 J.
Therefore no. of photons emitted per second (n) = = 2.0 × 1020
(where c=velocity of light equal to 3×10 ms and h=Planck's constant equal to 6.63×10–34 Js)
8 –1
27)
1 g-atom nitrogen = 1 mole nitrogen atom
= mole N2
28)
Glucose molecular mass = 180
CH3COOH molecular mass = 60
∴ Same mass ⇒ a moles of CH3COOH & moles of glucose.
Hence each atom will be same.
32)
n=3
33)
PART-C-MATHEMATICS
35)
= 51
36) We have
Integers satisfying above are
So, there are eight integers.
39) If , multiplying each term by , the given equation reduces
to or or x =1, which is not possible as considering . Thus given equation has
no roots.
40) – >0 ⇒ >0⇒ >0
<0
Ans. : x ∈ (–2, –1) ∪
41) > 6 – 3x
(a) 8 + 2x – x2 ≥ 0 ⇒ x ∈ [–2, 4] .... (i)
Case - I
when (i) 6 – 3x ≥ 0 ⇒ x ≤ 2 ... (ii)
so 8 + 2x – x2 > 36 + 9x2 – 36 x
⇒ 10x2 – 38x + 28 < 0
2
⇒ 5x – 19x + 14 < 0
⇒ (5x – 14) (x – 1)< 0
x∈ .... (iii)
by (1) and (2) and (3)
x ∈ (1, 2]
Case-II
6 – 3x < 0 ⇒ x > 2
+ ve > –ve
so x > 2 .... (iv)
by (1) and (4)
x ∈ (2, 4]
so by case (1) and (2) x ∈ (1, 4]
44) using wavy curve method
45)
47)
48)
= (log62 + log63)3 –3.log62.log63(log62 + log63) + log68.log63
= 1 – log68.log63 + log68.log63 = 1
49) logx–19 = t ⇒ ⇒ t=1
⇒ logx–19 = 1 ⇒ 9 = x – 1
⇒ x = 10 ⇒ log10x = 1
50)
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
51)
(x – 3)(x + 2)(x + 3)(x + 8) + 56
(x2 + 5x – 24)(x2 + 5x + 6) + 56
put x2 + 5x + 6 = t
t(t – 30) + 56 = (t – 2)(t – 28)
= (x2 + 5x + 4)(x2 + 5x – 22)
= (x + 1)(x + 4)(x2 + 5x – 22)
= (x + 1)(x + 4)(x2 + 5x – 22)
b, d = 1, 4 r = –22
