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JEE Main Practice Test- 12 - JEE MCQ


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30 Questions MCQ Test - JEE Main Practice Test- 12

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JEE Main Practice Test- 12 - Question 1

Let p, q and r be three logical statements with truth values F, T and F respectively, then the truth value of (~ p ⇒ ~ q) v r is

Detailed Solution for JEE Main Practice Test- 12 - Question 1

JEE Main Practice Test- 12 - Question 2

A house of height 100 m subtends a right angle at the window of an opposite house. If the height of the window is 64 m, then the distance between the two houses is

Detailed Solution for JEE Main Practice Test- 12 - Question 2

In ΔDAB, tan θ = 64/d ⇒ d = 64 cot θ .....(1)
In ΔCDE, tan (90° - θ) = 36/d  .....(2)

∴ (1) × (2)
⇒ 64 × 36 = d2
⇒ d = 48. Ans.]

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JEE Main Practice Test- 12 - Question 3

Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12}, then the relation R is

Detailed Solution for JEE Main Practice Test- 12 - Question 3

Satisfied condition of reflexive and transitive

JEE Main Practice Test- 12 - Question 4

The parabola y = x2 – 9andy = kx2 intersect each other at the points A and B. If the length AB is equal to 10 units then the value of k is equal to

Detailed Solution for JEE Main Practice Test- 12 - Question 4

solving x2 – 9 = kx2
x2(k – 1) + 0.x + 9 = 0


100 = 36/1-k
100 – 100k = 36 ⇒ k = 64/100 = 16/25

JEE Main Practice Test- 12 - Question 5

An ellipse in the first quadrant is tangent to co-ordinate axes. If one focus is F1 (3, 7), and the other focus is F2(d, 7), then the value of  is equal to

Detailed Solution for JEE Main Practice Test- 12 - Question 5

We have q1q2 = 3d = b2 and p1p2 = 49 = b2
Hence 3d = 49 ⇒ 

JEE Main Practice Test- 12 - Question 6

If A is a square matrix such that A2 + A + 2I = O, then which of the following is INCORRECT?
(Where I is unit matrix of order 2 and O is null matrix of order 2)

Detailed Solution for JEE Main Practice Test- 12 - Question 6

We have A (A + I) = – 2I
⇒ |A (A + I) | = | – 2I |
⇒ |A| |A+ I| = 4 ≠ 0
Thus , | A | ≠ 0
⇒ A is non singular
⇒ A is correct


⇒ D is correct
Also A = 0 does not satisfy the given eqaution ⇒ A ≠ 0

(A2 – B2) + (A – B) = 0
(A – B)(A + B + I) = 0
⇒ A – B = 0 or A + B + I = 0

JEE Main Practice Test- 12 - Question 7

The system of equations : 
2x cosθ + y sin 2θ – 2 sin θ = 0
x sin 2θ + 2y sin2 θ = –2cos θ
x sin θ – y cos θ = 0 , for all values of θ, can

Detailed Solution for JEE Main Practice Test- 12 - Question 7

slope of (1) and (2) is cot θ ⇒ (1) and (2) are parallel and slope of (3) is tanθ ⇒ no solution.
Using R2 → R2 – (2 cosθ) R3 and R1 → R1 + (2 sinθ)R3 , the value of determinat is 4]

JEE Main Practice Test- 12 - Question 8

One circle has a radius of 5 and its center at (0, 5). A second circle has a radius of 12 and its centre at (12, 0). The length of a radius of a third circle which passes through the center of the second circle and both points of intersection of the first 2 circles, is equal to

Detailed Solution for JEE Main Practice Test- 12 - Question 8

JEE Main Practice Test- 12 - Question 9

Locus of a point P(x, y) satisfying the equation

Detailed Solution for JEE Main Practice Test- 12 - Question 9

The given equation denotes that PA + PB = 13


∴ Point P lies on line segment AB

JEE Main Practice Test- 12 - Question 10

P and Q are two points on the base AB of a triangle ABC whose vertices are (–2, 3), (4, – 6) and (1, 1) respectively. If the join of CP and CQ divides the triangle ABC into three triangles of equal areas, then the equation of line pair through the origin and parallel to CP and CQ is equal to

Detailed Solution for JEE Main Practice Test- 12 - Question 10

As length of all the 3 triangles is same line
AP = PQ = QB
Hence equation of CP is y – x = 0

slope of CQ = 4/-1
equation of line through origin and parallel to QC, is y – 0 = – 4(x – 0)
y + 4x = 0
Equation of the line pair (y – x)(y + 4x) = 0 ⇒ y2 + 3xy – 4x2 = 0 

JEE Main Practice Test- 12 - Question 11

The line 4x - 7y + 10 = 0 intersects the parabola, y2 = 4x at the points A and B. The sum of the co-ordinates of the point of intersection of the tangents drawn at the points A and B is

Detailed Solution for JEE Main Practice Test- 12 - Question 11

C.O.C. of P(x1,y1) w.r.t. y2 = 4ax is
yy1 = 2(x + x1)   ....(1)
compare with

4x – 7y + 10 = 0 ....(2)

JEE Main Practice Test- 12 - Question 12

The tangent at a point whose eccentric angle 60° on the ellipse  meet the auxiliary circle at L and M. If LM subtends a right angle at the centre, then eccentricity of the ellipseis

Detailed Solution for JEE Main Practice Test- 12 - Question 12

The equation of the tangent is

Auxiliary circle is x2 + y2 = a2 .........(ii)
C is the centre.
Combined equation of CL, CM is obtained by homgenising (ii) with (i), i.e.,


⇒ 7b2 = 3a2 ⇒ 7a2 (1–e2) = 3a2
Hence e = 2/√7 

JEE Main Practice Test- 12 - Question 13

If the curves  and x2 = cy touch each other at the point (2, √2, 4) then the value of (a2 + b2 + c) is equal to

Detailed Solution for JEE Main Practice Test- 12 - Question 13


JEE Main Practice Test- 12 - Question 14

If the lines x + y + 1 = 0 ; 4x +3y + 4 = 0 and x + αy + β = 0, where α2 + β2 = 2, are concurrent then

Detailed Solution for JEE Main Practice Test- 12 - Question 14

Lines are x + y + 1 = 0 ; 4x + 3y + 4 = 0 and x + αy + β = 0 , where α2 + β2 = 2

1 (3β – 4α) -1 (4β - 4) + 1 (4α - 3)
= 3β – 4α – 4β + 4 + 4α – 3
= – β + 1 = 0 ⇒ β = 1
∴ α = ± 1 

JEE Main Practice Test- 12 - Question 15

The distance from the centre of the circle x2 + y2 = 2x to the straight line passing through the points of intersection of the two circles x2 + y2 + 5x - 8y + 1= 0, x2 + y2 - 3x + 7y - 25 = 0 is :

Detailed Solution for JEE Main Practice Test- 12 - Question 15

Compute perpendicular distance from (1, 0) to the Radical axis of two circles

JEE Main Practice Test- 12 - Question 16

If α + β + γ = π, then the value of

Detailed Solution for JEE Main Practice Test- 12 - Question 16

as α + β + γ = π so

open through R1 = – sinβ(cosγ tanα) + cosγ (sinβ tanα) = 0

JEE Main Practice Test- 12 - Question 17

Number of skew-symmetric matrices of order 3 whose elements are 0, 0, 0, 1, –1, 2, –2, 3, –3 is

Detailed Solution for JEE Main Practice Test- 12 - Question 17


Number of skew symmetric matrices = 3! × 8 = 48. Ans.
[As,diagonal element must be 0and conjugate pair elements are additive inverse of each other in skew-symmetric matrix.
Aliter: 1 can be put by 6 ways
– 1 can be put by 1 way
2 can be put by 4 ways
– 2 can be put by 1 way
3 can be put by 2 ways
– 3 can be put by 1 way
∴ Number of skew symmetric matrices = 6 × 1 × 4 × 1 × 2 × 1 = 48. Ans.]

JEE Main Practice Test- 12 - Question 18

In a class of 100 students few are in science stream and rest are in arts stream. 30 students took physics, 20 took chemistry, 25 took maths, 2 students took physics, chemistry and maths, 5 in physics and Chemistry, 6 in Physics and Maths and 3 in chemistry and Maths. How many took Arts if no students is common in Science & Arts

Detailed Solution for JEE Main Practice Test- 12 - Question 18


n (A ∪ B ∪ C)
∴ (A ∪ B ∪ C)C = 37 

JEE Main Practice Test- 12 - Question 19

If the mean of the numbers 27 + x, 31 + x, 89 + x, 107 + x, 156 + x is 82, then the mean of 130 + x, 126 + x, 68 + x, 50 + x, 1+ x is

Detailed Solution for JEE Main Practice Test- 12 - Question 19

Given,

JEE Main Practice Test- 12 - Question 20

Let p and q be two statements. Then, (~ p v q ) ∧ (~ p∧ ~ q) is a

Detailed Solution for JEE Main Practice Test- 12 - Question 20


∵ neither tautology nor contradiction.

*Answer can only contain numeric values
JEE Main Practice Test- 12 - Question 21

(Instruction to attempt numerical value (integer) type question: If your answer is 100 write 100 only. Do not write 100.0)

Two consecutive numbers from 1, 2, 3, …., n are removed, then arithmetic mean of the remaining numbers is 105/4, then n must be equal to


Detailed Solution for JEE Main Practice Test- 12 - Question 21

Let p and p + 1 be removed numbers from 1, 2, …., n. arithmetic mean of the remaining numbers is

*Answer can only contain numeric values
JEE Main Practice Test- 12 - Question 22

If Sn = 1 + 1/2 + 1/22 + .... + 1/2n-1 and 2 - Sn < 1/100, then the least value of n must be:


Detailed Solution for JEE Main Practice Test- 12 - Question 22

*Answer can only contain numeric values
JEE Main Practice Test- 12 - Question 23

Number of solutions of the equation 


Detailed Solution for JEE Main Practice Test- 12 - Question 23

*Answer can only contain numeric values
JEE Main Practice Test- 12 - Question 24

If α is the common positive root of the equation
x2 - ax + 12 = 0, x2 - bx + 15 = 0, x2 - (a + b) x + 36 = 0
and cosx + cos2x + cos3x = α, then sinx + sin2x + sin3x = ...


Detailed Solution for JEE Main Practice Test- 12 - Question 24

*Answer can only contain numeric values
JEE Main Practice Test- 12 - Question 25

The least period of the function sin π[x]/12 + cos (πx/4) + tan π[x]/3 is λ, then the value of 201λ must be (where [.] denotes the greatest integer function)


Detailed Solution for JEE Main Practice Test- 12 - Question 25

JEE Main Practice Test- 12 - Question 26

The plates of a parallel plate capacitor are charged upto 100 volt. A 2 mm thick plate is inserted between the plates. To maintain the same potential difference, the distance between the capacitor plates is increased by 1.6 mm. The dielectric constant of the plate is

Detailed Solution for JEE Main Practice Test- 12 - Question 26

Potential difference between plates remains same. Decrease in potential difference is counteracted by potential difference due to the extra distance.


E is original electric field, k dielectric constant of plate, t thickness of plate & d extra distance

JEE Main Practice Test- 12 - Question 27

A conducting ring of radius R is placed in uniform inward magnetic field  as shown. If ring is moving with velocity  in its plane, the induced emf across arc PQ will be

Detailed Solution for JEE Main Practice Test- 12 - Question 27


emf = vBl
ℓ is length of component perpendicular to velocity
l = R - R cos 45º

JEE Main Practice Test- 12 - Question 28

An electron beam passes between two parallel plate electrodes as shown in the diagram. The bottom plate is kept at zero potential, while a slowly varying positive voltage is applied to the upper plate, as shown in the graph. After passing between the plates, the beam hits a screen and makes spot. Ignoring gravity, as the potential varies the spot is

Detailed Solution for JEE Main Practice Test- 12 - Question 28


electrostatic force on electrons is opposite to direction of electric field

JEE Main Practice Test- 12 - Question 29

In the circuit shown below, the cell is ideal, with emf = 15 V. Each resistance is of 3W. The potential difference across the capacitor is

Detailed Solution for JEE Main Practice Test- 12 - Question 29

In steady state, capacitor acts as an open circuit.

I1 = 1A;
VA – I1R – IR = VB
⇒ VA – VB = 12

JEE Main Practice Test- 12 - Question 30

Two particles of the same mass carry charges +3Q and –2Q respectively. They are shot into a region that contains a uniform electric field one after the other as shown. The particles have the same initial velocities in the positive x direction. The lines, numbered 1 through 4, indicate possible paths for the particles. If the electric field points in the negative y direction, what will be the resulting paths for these particles?

Detailed Solution for JEE Main Practice Test- 12 - Question 30

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