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JEE Main Part Test - 2 Free Online Test 2026


MCQ Practice Test & Solutions: JEE Main Part Test - 2 (75 Questions)

You can prepare effectively for JEE Mock Tests for JEE Main and Advanced 2026 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "JEE Main Part Test - 2". These 75 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 180 minutes
  • - Number of Questions: 75

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JEE Main Part Test - 2 - Question 1

A particle is projected upward from the surface of earth (radius = R) with a speed equal to the orbital speed of a satellite near the earth’s surface. The height to which it would rise is

Detailed Solution: Question 1

  (orbital speed υ0 of a satellite)

Near the earth’s surface is equal to 1/√2
times the escape velocity of a particle on earth’s surface) Now from conservation of mechanical energy: Decrease in kinetic energy = increase in potential energy

JEE Main Part Test - 2 - Question 2

A spherical hole is made in a solid sphere of radius R. The mass of the original sphere was M.The gravitational field at the centre of the hole due to the remaining mass is

Detailed Solution: Question 2

By the principle of superposition of fields

Here  = net field at the centre of hole due to entire mass

 = field due to remaining mass

 = field due to mass in hole = 0

JEE Main Part Test - 2 - Question 3

Three particles P, Q and R placed as per given figure. Masses of P, Q and R are √3 m, √3 m and m respectively. The gravitational force on a fourth particle ‘S’ of mass m is equal to

Detailed Solution: Question 3

in horizontal direction  

in vertical direction

JEE Main Part Test - 2 - Question 4

A cylindrical tank has a hole of diameter 2r in its bottom. The hole is covered wooden cylindrical block of diameter 4r, height h and density ρ/3.

Situation I : Initially, the tank is filled with water of density ρ to a height such that the height of water above the top of the block is h1 (measured from the top of the block).

Situation II : The water is removed from the tank to a height h2 (measured from the bottom of the block), as shown in the figure.
The height h2 is smaller than h (height of the block) and thus the block is exposed to the atmosphere.

Q. Find the minimum value of height h1 (in situation 1), for which the block just starts to move up?    

Detailed Solution: Question 4

Consider the equilibrium of wooden block.
Forces acting in the downward direction are

(a) Weight of wooden cylinder

(b) Force due to pressure (P1) created by liquid of height h1 above the wooden block is
= P1 × π (2r)2 = [P0 + h1rg] × π (2r)2
Force acting on the upward direction due to pressure P2 exerted from below the wooden block and atmospheric pressure is

At the verge of rising

JEE Main Part Test - 2 - Question 5

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container.
As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere.

Q. If the piston is pushed at a speed of 5 mms–1, the air comes out of the nozzle with a speed of

Detailed Solution: Question 5

From principle of continuity,

JEE Main Part Test - 2 - Question 6

If the elastic limit of copper is 1.5 × 108 N/ m2, determine the minimum diameter a copper wire can have under a load of 10.0 kg if its elastic limit is not to be exceeded.

Detailed Solution: Question 6

Elastic limit (allowable stress) of copper, σel =1.5 ×108 N m−2
Load m =10.0 kg ⇒ F = mg =10.0 × 9.8 = 98 N

So, Option B is the correct answer.

JEE Main Part Test - 2 - Question 7

A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire? Young's modulus for steel: E = 20 × 1010 Pa .

 

 

Detailed Solution: Question 7

JEE Main Part Test - 2 - Question 8

Water is flowing continuously from a tap having an internal diameter 8 × 10-3m. The water velocity as it leaves the tap is 0.4 ms-1. The diameter of the water stream at a distance 2 × 10-1 m below the tap is close to

[AIEEE 2011]

Detailed Solution: Question 8

JEE Main Part Test - 2 - Question 9

There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is h. As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to  

 

Detailed Solution: Question 9

Let ρ = density of the liquid.

 Let α = area of a cross-section of each hole.

 Volume of liquid discharged per second at a hole = αv.

 Mass of liquid discharged per second = αvρ.

 Momentum of liquid discharged per second = αv2ρ.

∴ the force exerted at the upper hole (to the right) = αρv22

 and  the force exerted at the lower hole (to the left) = αρv12

Net force on the tank = 2αρgh. 

*Multiple options can be correct
JEE Main Part Test - 2 - Question 10

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively, are T2 and T1 (T2 >T1) . The rate of heat transfer through the slab, in a steady state is with f equal to 

Detailed Solution: Question 10

The thermal resistance

JEE Main Part Test - 2 - Question 11

In which case change of oxidation number of V is maximum? 

Detailed Solution: Question 11



*Multiple options can be correct
JEE Main Part Test - 2 - Question 12

The complex [Fe(H2O)5NO]2+ is formed in the ring-test for nitrate ion when freshly prepared FeSO4 solution is added to aqueous solution of followed by the addition of conc. H2SO4. NO exists as NO(nitrosyl).

Q. Magnetic moment  of Fe in the ring is 

Detailed Solution: Question 12

Fe2+ is added asFeSO4
Fe+ is formed by charge transfer from NO to Fe2+ 



Fe+ has three unpaired electrons (N).

JEE Main Part Test - 2 - Question 13

The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be -  

[AIEEE-2008]

Detailed Solution: Question 13

JEE Main Part Test - 2 - Question 14

In aqueous solution the ionization constants for carbonic acid are K1 = 4.2 x 10-7 and K2 = 4.8 x 10-11 Selection the correct statement for a saturated 0.034 M solution of the carbonic acid. 

 [AIEEE-2010]

Detailed Solution: Question 14

Carbonic acid is a weak acid it dissociates as follows 

Since both acids are weak cannot be 0.034 M as there is no complete dissociation. 

Total hydrogen ion concentration is approximately equal to concentration of hydrogen ion in first reaction as K1 is larger. This is approximately equal to concentration of 

The concentration of  depends on  K2 and that of H+ on K1. But . So The concentration of H+ is not double that of 

JEE Main Part Test - 2 - Question 15

How many litres of water must be added to litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2? 

 [AIEEE-2013]

Detailed Solution: Question 15

Volume of the original solution = 1 L

pH of the new solution = 2

Volume of the new solution = ?
As per volumetric principle

That means the volume of water added to the original solution of 1 L = 10 −1 = 9 L

JEE Main Part Test - 2 - Question 16

For the following electrochemical cell reaction at 298 K:

Zn(s) + Cu²⁺(aq) ⇌ Cu(s) + Zn²⁺(aq),

Given that the standard cell potential (E° cell) is +1.10 V.

Detailed Solution: Question 16

= 37.22

JEE Main Part Test - 2 - Question 17

A 0.10 M solution of a weak acid, HX, is 0.059% ionized. Evaluate Ka for the acid.

Detailed Solution: Question 17

Since the acid is only 0.059% ionized, therefore the concentration of ions in solution = 0.1 x 0.059 / 100 = 0.000059

Ka = [H+] [X] / [HX] = (0.000059)2 / 0.1 = 3.5 x 10-8

JEE Main Part Test - 2 - Question 18

AgCI(s)is sparingly soluble salt,

AgCl (s)  Ag+(aq) + Cl-(aq)

There is

Detailed Solution: Question 18

When ammonia is added, solubility of AgCl increases due to formation of complex salt which decreases the concentration of radicals in the product side and thus drives the reaction in forward direction.
When we add KCl common ion effect is applied in presence of common ion solubility decreases and reaction goes in backward direction.

JEE Main Part Test - 2 - Question 19

 Which one of the following is the approximate pH of 0.01 M solution of NaOH at 298 k?

Detailed Solution: Question 19

First off, since NaOH is a strong base, it will dissociate completely into Na+ and OH-. Thus, we know that we have 0.01 M OH-.

However, we do not know anything about the concentration of H+. Fortunately, we do not need to, as pH + pOH = 14. So, if we find pOH, we can solve for pH. p is a mathematical function equivalent to -log. So, pH actually means -log[H+] (Note that brackets indicate concentration of).

pOH = -log 0.01M OH-

pOH = 2

pH + 2 = 14

pH = 12

This result makes sense, since a solution of strong base should have a high pH.

JEE Main Part Test - 2 - Question 20

In the expansion of (1+x)60, the sum of coefficients of odd powers of x is

Detailed Solution: Question 20

 (1+x)n= nC0+ nC1x+ nC2x2+...+ nC0xn ....(1)
(1−x)n= nC0 − nC1x+ nC2x2−...+ nC0 xn....(2)
⇒(1+x)n−(1−x)n = nC0+ nC1x+ nC2x2+...+ nC0xn− nC0+ nC1x− nC2x2+...+ nC0xn
 =2(nC1x+nC3x3+....+xn)

Given: n=60 and put x=1
⇒(1+1)60−(1−1)60
 =260
⇒260=2(nC1x+nC3x3+....+xn)

∴Sum of odd powers of x in the expansion is = nC1x+nC3x3+....+xn =260−1 =259

JEE Main Part Test - 2 - Question 21

The middle term in the expansion of (1+x)2n is

Detailed Solution: Question 21

Middle term in the expansion of (1+x)2n; is 
=tn+1 = 2nCn.1(2n−n).xn
= {(2n)!.xn}/(2n−n)!n!
= {{2n(2n−1)(2n−2)(2n−3)....4×3×2×1}/n! n!}/xn
= {{2n[n(n−1)(n−2)....×2×1][(2n−1)(2n−3)....3×1]}/n! n!}xn
= [[(2n−1)(2n−3)....3×1]/n!] 2nxn

JEE Main Part Test - 2 - Question 22

Detailed Solution: Question 22

JEE Main Part Test - 2 - Question 23

The mid points of the sides of a triangle are (5, 0), (5, 12) and (0, 12), then orthocentre of this triangle is

Detailed Solution: Question 23

JEE Main Part Test - 2 - Question 24

Area of a triangle whose vertices are (a cos q, b sinq), (–a sin q, b cos q) and (–a cos q, –b sin q) is

Detailed Solution: Question 24


JEE Main Part Test - 2 - Question 25

If A(cosa, sina), B(sina, – cosa), C(1, 2) are the vertices of a ΔABC, then as a varies, the locus of its centroid is

Detailed Solution: Question 25

JEE Main Part Test - 2 - Question 26

Two perpendicular tangents to the circle x2+y2 = r2 meet at P. The locus of P is

Detailed Solution: Question 26

JEE Main Part Test - 2 - Question 27

The locus of a variable point whose distance from the point (2, 0) is 2/3 times its distance from the line x = 9/2 is

Detailed Solution: Question 27

JEE Main Part Test - 2 - Question 28

The locus of a point such that two tangents drawn from it to the parabola y2 = 4ax are such that the slope of one is double the other is

Detailed Solution: Question 28

Let the point be (h, k)

Now equation of tangent to the parabola y= 4ax whose slope is m is

as it passes through (h, k)]

JEE Main Part Test - 2 - Question 29

The focus of the parabola x2 - 8x + 2y + 7 = 0 is

Detailed Solution: Question 29

Option D is correct; the focus is (4, 4).

Write the x-terms by completing the square: x2 - 8x = (x-4)2 - 16.

Substitute into the equation to get (x-4)2 - 16 + 2y + 7 = 0, which simplifies to (x-4)2 + 2y - 9 = 0.

Rearrange to the standard form: (x-4)2 = -2(y - 9/2).

Compare with the standard parabola form (x - h)2 = -4a (y - k). Equating coefficients gives -4a = -2, so a = 1/2.

From the form, the vertex is (h, k) = (4, 9/2). For a downward-opening parabola the focus is at (h, k - a).

Compute k - a = 9/2 - 1/2 = 4, so the focus is (4, 4). Hence option D is correct.

JEE Main Part Test - 2 - Question 30

is equal to

Detailed Solution: Question 30

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