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JEE Exam  >  JEE Tests  >  BITSAT Mock Tests Series & Past Year Papers 2025  >  BITSAT Mock Test - 2 - JEE MCQ

BITSAT Mock Test - 2 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - BITSAT Mock Test - 2

BITSAT Mock Test - 2 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers 2025 preparation. The BITSAT Mock Test - 2 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Mock Test - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Mock Test - 2 below.
Solutions of BITSAT Mock Test - 2 questions in English are available as part of our BITSAT Mock Tests Series & Past Year Papers 2025 for JEE & BITSAT Mock Test - 2 solutions in Hindi for BITSAT Mock Tests Series & Past Year Papers 2025 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt BITSAT Mock Test - 2 | 130 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study BITSAT Mock Tests Series & Past Year Papers 2025 for JEE Exam | Download free PDF with solutions
BITSAT Mock Test - 2 - Question 1
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For a given velocity, a projectile has the same range R for two angles of projection. If t1 and t2 are the times of flight in the two cases, then:

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BITSAT Mock Test - 2 - Question 2
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The number of turns and the radius of cross-section of the coil of a tangent galvanometer are doubled. The value of the reduction factor k will be

Detailed Solution for BITSAT Mock Test - 2 - Question 2

i =
Since, k =
r' = 2r, n' = 2n
Then k' = = k
Then, the value of reduction factor remains the same.

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BITSAT Mock Test - 2 - Question 3
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A square wire of each side L carries a current i. What is the magnetic field at the mid-point of the square?

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BITSAT Mock Test - 2 - Question 4
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The length of a wire of a potentiometer is 100 cm and the emf of its stand and cell is E volt. It is employed to measure the emf of a battery whose internal resistance is 0.5 W. If the balance point is obtained at l = 30 cm from the positive end, then the emf of the battery is
Detailed Solution for BITSAT Mock Test - 2 - Question 4

Potential gradient of the potentiometer wire is
V/cm
Emf of the cell is
, l = 30 cm

BITSAT Mock Test - 2 - Question 5
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The expression up to correct significant figures is equal to
Detailed Solution for BITSAT Mock Test - 2 - Question 5


In the expression, we have minimum three significant figures, hence answer should be up to three significant figures.
Rounding off the result 1.612979, we have

BITSAT Mock Test - 2 - Question 6
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If the internal resistance of the battery is negligible and the capacitance C = 0.5, then what is the steady state current in the 2 resistor shown in the figure below?
Detailed Solution for BITSAT Mock Test - 2 - Question 6

The total resistance of the parallel combination of 2 and 3 resistors is R = 1.2. This resistance is in series with the 2.8 resistor, giving a total effective resistance = 1.2 + 2.8 = 4.0 as in the steady state, no current flows through the capacitor C and hence no current passes through the 5 resistor.
Hence, current through the circuit = 6/4 = 1.5 A
Therefore, potential drop across AB = 1.5 1.2 = 1.8 V
Hence, current through the 2 resistor = 1.8/2 = 0.9 A, which is choice (3).

BITSAT Mock Test - 2 - Question 7
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Directions: The question below is based on the following passage:
Two light springs of force constants k1 = 1.8 Nm-1 and k2 = 3.2 Nm-1 and a block of mass m = 200 g are arranged on a horizontal frictionless table as shown in the figure. One end of each spring is fixed on rigid supports and the other end is free. The distance CD between the free ends of the springs is 60 cm and the block moves with a velocity 'v' = 120 cms-1 between the springs.

When the block moves towards the spring k2, the time taken by it to move from D up to the maximum compression of spring k2 is (in seconds)
Detailed Solution for BITSAT Mock Test - 2 - Question 7

Time taken by block to move from D upto the maximum compression of spring k2 = half of the time period of oscillation of the block if it were attached to the free end of k2, i.e.,

The correct choice is (4).

BITSAT Mock Test - 2 - Question 8
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A coil having N turns is wound tightly in the form of a spiral, with inner and outer radii a and b, respectively. When a current I passes through the coil, the magnetic field at the centre is
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BITSAT Mock Test - 2 - Question 9
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When water flows at a rate Q through a capillary tube of radius r that is placed horizontally, a pressure difference p develops across the ends of the tube. If the radius of the tube is doubled and the rate of flow halved, the pressure difference becomes
Detailed Solution for BITSAT Mock Test - 2 - Question 9
If the length of capillary tube is l, the pressure difference is given by

where η is the coefficient of viscosity of water. If r becomes 2r and Q becomes , we have

Hence the correct choice is (1).
BITSAT Mock Test - 2 - Question 10
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A non-planar loop of conducting wire carrying a current I is placed as shown in the figure. Each of the straight sections of the loop is of length 2a. The magnetic field due to this loop at the point P (a, 0, a) points in the direction
Detailed Solution for BITSAT Mock Test - 2 - Question 10

The magnetic field at P(a, 0, a) due to the loop is equal to the vector sum of the magnetic fields produced by loops ABCDA and AFEBA as shown in the figure.

Magnetic field due to loop ABCDA will be along and due to loop AFEBA, along .
Magnitude of magnetic field due to both the loops will be equal.
Therefore, direction of resultant magnetic field at P will be

BITSAT Mock Test - 2 - Question 11
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The de-Broglie wavelength of the electron in the ground state of hydrogen atom is
(KE = 13.6 eV and 1 eV = 1.602 x 10-19 J)
Detailed Solution for BITSAT Mock Test - 2 - Question 11

According to de-Broglie relation,

h = 6.62 x 10-34 kgm2s-1
KE = 13.6 x 1.602 x 10-19 J
KE =
v =
= 2.18824 x 106 m/s

= 0.3328 x 10-9 m = 0.3328 nm

BITSAT Mock Test - 2 - Question 12
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The following equilibria are given:
a. N2 + 3H2 2NH3; K1
b. N2 + O2 2NO; K2
c. H2 + O2 H2O; K3
The equilibrium constant for the reaction,
2NH3 + O2 2NO + 3H2O in terms of K1, K2 and K3 will be
Detailed Solution for BITSAT Mock Test - 2 - Question 12

From the given equations, we have
K1 =
K2 =
K3 =
For the given reaction,
K =

K =

BITSAT Mock Test - 2 - Question 13
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'A' and 'C' in the following conversion are
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BITSAT Mock Test - 2 - Question 14
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What is the product in the following reaction?
Detailed Solution for BITSAT Mock Test - 2 - Question 14

Yes, it is the correct answer.

BITSAT Mock Test - 2 - Question 15
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The halogen compound which undergoes nucleophilic substitution most readily is
Detailed Solution for BITSAT Mock Test - 2 - Question 15
3-chloro prop-1-ene (allyl halide) undergoes SN1 reactions through the formation of an intermediate carbocation, which is resonance stabilised as follows:

Therefore, it will undergo nucleophilic substitution most readily.
BITSAT Mock Test - 2 - Question 16
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Directions: A phrase is followed by four alternatives. Pick the one which best describes the phrase.
Capable of being approached
Detailed Solution for BITSAT Mock Test - 2 - Question 16

'Accessible' means 'approachable'.

BITSAT Mock Test - 2 - Question 17
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Which of the following images (a), (b), (c) and (d) forms the mirror image of figure (X)? (Mirror is placed on the right side of the image)
Detailed Solution for BITSAT Mock Test - 2 - Question 17

By observing all the figures, we find that figure (d) forms the mirror image of the given figure (X).

BITSAT Mock Test - 2 - Question 18
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Directions: There is some relationship between the two terms to the left of the sign (: :). The same relationship exists between the two terms to its right, out of which one is missing. Find the missing one from the given alternatives.
acE : bdF : : fhJ : ?
Detailed Solution for BITSAT Mock Test - 2 - Question 18

Each letter of the first group is replaced by its successor in the English alphabet to get the second group. Hence, fhJ will become giK to replace the question mark (?).

BITSAT Mock Test - 2 - Question 19
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Directions: The question presents a sentence, all or part of which is underlined. Beneath the sentence you will find four ways of phrasing the underlined part. The first of these repeats the original; the other three are different. If you think the original is best, choose the first answer; otherwise choose one of the others.
The findings of the enquiry committee, as revealed by the chairman, is very encouraging to people.
Detailed Solution for BITSAT Mock Test - 2 - Question 19

(2) is the correct option. 'Are very encouraging' is the correct usage because the subject (findings) is plural and hence, needs a plural verb. So, only (2) satisfies the condition.

BITSAT Mock Test - 2 - Question 20
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The value of is
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=
=
=
BITSAT Mock Test - 2 - Question 21
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The differential equation of all non-vertical lines in a plane is
Detailed Solution for BITSAT Mock Test - 2 - Question 21

The general equation of all non-vertical lines in a plane is: ax + by = 1, where b 0.
a + b = 0
b = 0
= 0
[ b 0]
This is the required differential equation.

BITSAT Mock Test - 2 - Question 22
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The triangle formed by the tangent to the curve f(x) = x2 + bx - b at the point (1, 1) and the co-ordinate axis, lies in the first quadrant. If its area is 2, then the value of b is
Detailed Solution for BITSAT Mock Test - 2 - Question 22
The slope of the tangent at the point (1, 1) is y' = 2 + b.
Hence, the equation of the tangent is:
y = (2 + b)x + c
As the line satisfies the point (1, 1), the value of c = -(1 + b)
Equation of tangent is y = (2 + b)x - (1 + b)
As the triangle formed will be a right-angled triangle made up of intercepts of the tangent,
hence the intercepts are
y = -(1 + b)
x =
Area of triangle = ½ x-intercept y-intercept = 2
=> ½ -(1 + b) = 2
=> b = -3
BITSAT Mock Test - 2 - Question 23
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The smallest positive integer n for which
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BITSAT Mock Test - 2 - Question 24
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A GP consists of an even number of terms (all terms being different). If the sum of all the terms is five times the sum of the terms occupying odd places, then the common ratio will be
Detailed Solution for BITSAT Mock Test - 2 - Question 24

Let 2n be the total number of terms
∴ S2n = 5 [T1 + T3 + ... + T2n - 1]
where T1, T3 ... T2n - 1 are terms occupying odd places
⇒ = 5[a + ar2 + ... + ar2n - 2]
[Sum of G.P., where a is first term and r is common ratio]
⇒ = 5a[1 + r2 + ... + r2n - 2]
⇒
⇒ 1 =
⇒ 1 + r = 5
∴ r = 4

BITSAT Mock Test - 2 - Question 25
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The equation of the circle which touches the coordinate axes and the line and whose centre lies in the first quadrant is x2 + y2 − 2cx − 2cy + c2 = 0, where c is
Detailed Solution for BITSAT Mock Test - 2 - Question 25

The equation of the circle touching the co-ordinate axes is
x2 + y2 - 2cx - 2cy + c2 = 0
i.e. (x - c)2 + (y - c)2 = c2 ...(1)
This touches the line
...(2)
i.e.
i.e. 4x + 3y - 12 = 0 ...(3)
Perpendicular distance between centre (c, c) and line is

⇒
⇒ 7c - 12 = 5c or - 5c
⇒ 7c - 12 = 5c or 5c
⇒ c = 6 or c = 1
Hence, c = 1, 6

BITSAT Mock Test - 2 - Question 26
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Let A = {1, 2, 3, 4} and B = {2, 3, 4, 5, 6}, then A B is equal to
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BITSAT Mock Test - 2 - Question 27
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The equation of smallest degree with real coefficients having 2 + 3i as one of the roots is
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BITSAT Mock Test - 2 - Question 28
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The value of cos 12° + cos 84° + cos 156° + cos 132° is
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BITSAT Mock Test - 2 - Question 29
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The total number of injective mappings from a set with m elements to a set with n elements (m ≤ n) is equal to
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BITSAT Mock Test - 2 - Question 30
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is equal to
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