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JEE Exam  >  JEE Test  >  BITSAT Mock Tests Series & Past Year Papers 2026  >  BITSAT Mock Test - 7 - JEE MCQ

BITSAT Mock Test - 7 Free Online Test 2026


Full Mock Test & Solutions: BITSAT Mock Test - 7 (130 Questions)

You can boost your JEE 2026 exam preparation with this BITSAT Mock Test - 7 (available with detailed solutions).. This mock test has been designed with the analysis of important topics, recent trends of the exam, and previous year questions of the last 3-years. All the questions have been designed to mirror the official pattern of JEE 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 180 minutes
  • - Total Questions: 130
  • - Analysis: Detailed Solutions & Performance Insights
  • - Sections covered: Physics, Chemistry, English Proficiency & Logical Reasoning, Engineering Entrance (Mathematics)

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BITSAT Mock Test - 7 - Question 1
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A thin uniform metallic triangular sheet of mass M has sides AB = BC = L as shown in the figure. What is its moment of inertia about axis AC lying in the plane of the sheet?

Detailed Solution: Question 1

It is clear from the figure that the moment of inertia of triangular sheet ABC = 1/2 x moment of inertia of a square sheet ABCD about its diagonal AC or
Now, mass of square sheet = M + M = 2M.

BITSAT Mock Test - 7 - Question 2
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A charge Q is placed at the mouth of a conical flask. The flux of the electric field through the flask is

Detailed Solution: Question 2

Charge associated with the conical flask is Q/2, Hence, flux through the conical flask is .Q/2∈0

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BITSAT Mock Test - 7 - Question 3
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Inside a uniform sphere of mass M (M is mass of completer sphere) and radius R, a cavity of radius R/3 is made in the sphere as shown.

Which of the following statements is true?

Detailed Solution: Question 3

Gravitational field inside the cavity is , where is mass density and is separation between centre of sphere and centre of cavity.
Applying law of conservation of energy, we get
= 0
Solving, Vesc =

BITSAT Mock Test - 7 - Question 4
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Directions: The following figure shows an experimental arrangement given by Fresnel. It consists of two plane mirrors inclined at an angle of 12 minutes of arc. The distances from the mirrors' intersection to the side 'S' and the screen are 10 cm and 130 cm, respectively. The wavelength of light in Fresnel double mirrors can be looked upon as modified YDSE. Here, the two coherent sources are virtual and mirror images of S. Interference is obtained due to reflected light from the two mirrors. Interference pattern consists of alternate bright and dark fringes similar to YDSE, but with a difference that interference pattern is limited to overlapping region. Fringe width and position of maxima and minima is given by similar expression as in YDSE.

The number of possible fringes formed on the screen is

Detailed Solution: Question 4

The fringes are hyperbolic as the distance difference of a bright fringe from two sources is a constant.
Distance between the two sources is 's' and if the angle between the two mirrors is θ, then
d = 2rθ
Fringe width

Range of fringe formation = 130 x 20 = 130 x 24/60 x π/180 = x
x = 0.907 cm or 9.07 mm
x/w = 8.24 so we can say 8 fringes are observed

BITSAT Mock Test - 7 - Question 5
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Two identical metal plates are given positive charges Q1 and Q2 (< Q1), respectively. If they are brought close together to form a parallel plate capacitor with capacitance C, then the potential difference between them is

Detailed Solution: Question 5

Within the plates electric fields due to charges Q1 and Q2 are
E1 = and E2 =
As these fields are in opposite directions and Q1 > Q2, the net electric field within the plates is
E = E1 - E2 = (Q1 - Q2)
Hence V = Ed = (Q1 - Q2) = which is choice (4).

BITSAT Mock Test - 7 - Question 6
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A shell of mass 2 m, fired with a speed u at an angle θ to the horizontal, explodes at the highest point of its trajectory into two fragments of mass m each. If one fragment falls vertically, the distance at which the other fragment falls from the gun is given by

Detailed Solution: Question 6

At the highest point of trajectory, the projectile has only a horizontal velocity which is u cos θ. After explosion, the fragment falling downwards has no horizontal velocity. If u' is the horizontal velocity of the other fragment, the law of conservation of momentum gives
(2m) u cos θ = m × 0 + mu'
which gives u' = 2u cos θ
Now, the time taken to reach the highest point (as well as the time taken to fall down from this point) is (υ sin θ) / g.
Therefore, the horizontal distance travelled by the other fragment is
u cos θ × (υ sin θ) / g + 2 u cos θ × (υ sin θ) / g
=
Hence the correct choice is (2).

BITSAT Mock Test - 7 - Question 7
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An organ pipe P1, closed at one end vibrating in its first harmonic, and another pipe P2, open at both ends vibrating in its third harmonic, are in resonance with a given tuning fork. The ratio of the lengths of P1 and P2 is

Detailed Solution: Question 7

For the pipe closed at one end vibrating in the fundamental mode,
The frequency is f1 = v/4L1, L1 is the length of the pipe.
For the pipe can open at both end vibrating in 3rd mode of vibration
The frequency is f2 = 3 x v/4L1, L2 is the length of the open pipe.
As per quetion
f1 = f2 ⇒ v/4L1 = 3v/2L2
L1/L2 = 1/6

BITSAT Mock Test - 7 - Question 8
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A right-angled prism is to be made by selecting a proper material and angles A and B (B ≤ A), as shown in the figure. It is desired that a ray of light incident on face AB emerges parallel to the incident direction after two internal reflections. What should be the minimum refractive index (n) for this to be possible?

Detailed Solution: Question 8

Referring to Fig. the path of the ray is PQRS suffering internal reflections at Q and R.
It is clear from the figure that angles α and β should be greater than the critical angle given by
sin ic = 1/n

Also angle A = α and angle B = β. Since A ≥ B; β ≤ α, the minimum value of n is given by 1/n ≤ sin β
∴ nmin = 1/sinβ = 1/sinB
So the correct choice is (2).

BITSAT Mock Test - 7 - Question 9
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The figure shows a series LCR circuit connected to a variable frequency 200 V source. If L = 5 H, C = 80 F and R = 40 , then what is the impedance of the circuit at resonance?

Detailed Solution: Question 9

BITSAT Mock Test - 7 - Question 10
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Two carts of masses 200 kg and 300 kg, standing on horizontal straight rails, are pushed apart by an explosion of a device kept in the connecting mechanism of carts. The coefficient of friction between carts and rails is identical. If the 200 kg cart travels a distance 36 m and stops, then what is the distance covered by the cart weighing 300 kg? 

Detailed Solution: Question 10

Apply conservation of momentum

E1 and E2 are the kinetic energies after explosion and W1 and W2 are the work done in stopping the carts by friction force (= mg).

BITSAT Mock Test - 7 - Question 11
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A ray of light enters a rectangular glass slab of refractive index μ = √3 at an angle of incidence i = 60°. It travels a distance d = 6.0 cm inside the slab before emerging out of it. The lateral displacement of the incident ray is

Detailed Solution: Question 11

BITSAT Mock Test - 7 - Question 12
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In a resonance tube experiment, the first and the second resonance with a given tuning fork were observed at 16.7 cm and 51.7 cm, respectively. The wavelength as deduced from the data is:

Detailed Solution: Question 12

BITSAT Mock Test - 7 - Question 13
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A body 'x' with a momentum 'p' collides with another identical stationary body 'y', one-dimensionally. During the collision, 'y' gives an impulse 'J' to the body 'x'. Then, the coefficient of restitution is

Detailed Solution: Question 13

BITSAT Mock Test - 7 - Question 14
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If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio of 1 : 1 : 0.5 and ΔHf for the formation of XY is -200 kJ mol-1, then the bond dissociation energy of X2 will be

Detailed Solution: Question 14

Formation of XY is shown as follows.
X2 + Y2 → 2XY
ΔH = ∑ Bond energy of reactant - ∑ Bond energy of products
ΔH = (BE)x - x + (BE)y - y - 2(BE)x - y
If the ratio of bond dissociation energies of X-Y, X2 and Y2 is 1 : 1 : 0.5,
then (BE) of X - Y = a
Therefore, (BE) of (X - X) = a and (BE) of (Y - Y) = a/2
ΔHf (X - Y) = -200 kJ
∴ - 400 (for 2 moles XY) = a + a/2 - 2a
- 400 = -a/2
a = + 800 kJ
The bond dissociation energy of X2 is 800 kJ mol-1.

BITSAT Mock Test - 7 - Question 15
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The emf of Daniell cell Zn | ZnSO4 (0.01 M) || CuSO4 (1.0 M) | Cu at 298 K is E1.
When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, then the emf changes to E2. What is the relation between E1 and E2?

Detailed Solution: Question 15

In 1st case:
[Zn2+] = 0.01 M
[Cu2+] = 1.0 M
n = 2
E1 = - log
E1 =
= + 0.0591
In 2nd case:
[Zn2+] = 1.0 M
[Cu2+] = 0.01 M
n = 2
E2 = - log
= log
= - 0.0591
Thus, E1 - E2 = 2 × 0.059 = 0.118 V

BITSAT Mock Test - 7 - Question 16
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An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO3 and the total volume was made to be 100 mL. A silver electrode was dipped in the solution and the emf of the cell set up Pt(s), H2(g) | H+(1M) || Ag+ (aq) | Ag(s) was 0.62 V. If Eocell = 0.80 V, then what is the percentage of Ag in the alloy?
[At 25oC, RT/F = 0.06]

Detailed Solution: Question 16

BITSAT Mock Test - 7 - Question 17
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x moles of potassium dichromate oxidise 1 mole of ferrous oxalate in the acidic medium. Here, x is

Detailed Solution: Question 17

The reaction of oxidation of ferrous oxalate by potassium dichromate in an acidic medium is written as
2FeC2O4 + + 14H+ → 2Fe3+ + 2Cr3+ + 4CO2 + 7H2O
2 moles of FeC2O4 are oxidised by 1 mole of .
1 mole of FeC2O4 will be oxidised by 0.5 mole of .

BITSAT Mock Test - 7 - Question 18
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Directions: Select the Answer figure in which the given Question figure is hidden.

Detailed Solution: Question 18

BITSAT Mock Test - 7 - Question 19
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Directions: Select one figure from the answer figures which will continue the same series as given in the problem figures.

Detailed Solution: Question 19

Problem figure (5) is the same as figure (1). So, the answer figure will be the same as figure (2). Hence, the correct option is (2).

BITSAT Mock Test - 7 - Question 20
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Which of the Answer figures will complete the pattern in the Question figure?

Detailed Solution: Question 20

Answer figure (C) completes the pattern in the Question figure.

BITSAT Mock Test - 7 - Question 21
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Directions: In the following question, choose the word which best fills the blank from the four options given.
Hardly would anyone ___________ that this straw could start a revolution.

Detailed Solution: Question 21

Straw could start a revolution' is hard to believe.

BITSAT Mock Test - 7 - Question 22
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Directions: A piece of paper is folded and punched as shown below.
How will this paper look, when it is unfolded completely?

Detailed Solution: Question 22


Hence, option 3 is correct.

BITSAT Mock Test - 7 - Question 23
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Directions: This question consists of five unmarked figures followed by four figures numbered (1), (2), (3) and (4). Select a figure from the marked figures which will continue the series established by the unmarked figures.

Detailed Solution: Question 23


BITSAT Mock Test - 7 - Question 24
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Evaluate .

Detailed Solution: Question 24

BITSAT Mock Test - 7 - Question 25
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In a moderately skewed distribution, the values of mean and median are 5 and 6, respectively. The value of mode in such a situation is approximately equal to

Detailed Solution: Question 25

BITSAT Mock Test - 7 - Question 26
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The solution of the differential equation = sin x is

Detailed Solution: Question 26

BITSAT Mock Test - 7 - Question 27
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is equal to

Detailed Solution: Question 27

Given series can be rewritten as

Now,
= tan-1 (r + 1) - tan-1(r)

= tan-1 (n + 1) - tan-1 (1)

BITSAT Mock Test - 7 - Question 28
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The value of the determinant is

Detailed Solution: Question 28

BITSAT Mock Test - 7 - Question 29
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What will be the value of 'k', if = ?

Detailed Solution: Question 29

= 3(1)3 - 1 = 3
= x = 4k3 ÷ = 4k3 3k2
= = k
=
3 = k
= k
k =

BITSAT Mock Test - 7 - Question 30
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The intercept on the line y = x by the circle x2 + y2 – 2x = 0 is AB. The equation of the circle with AB as a diameter is:

Detailed Solution: Question 30

y = x -------(1)
x2 + y2 – 2x = 0 -------(2)
Put y = x in equation (2)
x2 + x2 - 2x = 0
x = 0 and 1
when x = 0, y = 0
and x = 1 , y = 0
So, end points of the diameter are (0, 0) and (1 ,1).

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About this JEE Mock Test

This mock test contains 130 questions designed to replicate the pattern, difficulty level and time pressure of the actual JEE exam. It is part of BITSAT Mock Tests Series & Past Year Papers 2026 on EduRev, covering all the topics that are being asked as per the current JEE syllabus. Attempting this test under timed conditions will help you assess your exam readiness, identify subject-wise weak areas, and improve your speed and accuracy before the actual exam.

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