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JEE Main Part Test - 2 - JEE MCQ


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30 Questions MCQ Test - JEE Main Part Test - 2

JEE Main Part Test - 2 for JEE 2024 is part of JEE preparation. The JEE Main Part Test - 2 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Part Test - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Part Test - 2 below.
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JEE Main Part Test - 2 - Question 1

A planet of mass M is revolving around sun in an elliptical orbit. If dA is the area swept in a time dt, angular momentum can be expressed as

Detailed Solution for JEE Main Part Test - 2 - Question 1

JEE Main Part Test - 2 - Question 2

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 for JEE Main Part Test - 2 - Question 2

  (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

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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 for JEE Main Part Test - 2 - Question 3

in horizontal direction  

in vertical direction

JEE Main Part Test - 2 - Question 4

A uniform ring of mass M and radius R is placed directly above uniform sphere of mass 8M and of same radius R. The centre of the ring is at a distance of d = √3R from the centre of sphere. The gravitational attraction between the sphere and the ring is

Detailed Solution for JEE Main Part Test - 2 - Question 4

Gravitational field due to the ring  at a distance d = √3R on its axis is:

force on sphere

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 for JEE Main Part Test - 2 - 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 for JEE Main Part Test - 2 - Question 6

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? ( Given Young's Modulus for steel is 20 × 1010 Pa.)

Detailed Solution for JEE Main Part Test - 2 - 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 for JEE Main Part Test - 2 - 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 for JEE Main Part Test - 2 - 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. 

JEE Main Part Test - 2 - Question 10

Two soap bubbles with radii   come in contact. Their common surface has a radius of curvature r. Then

Detailed Solution for JEE Main Part Test - 2 - Question 10

Let p0 = atmospheric pressure,

p1 and p2 = pressures inside the two bubbles

Let r = radius of curvature of the common surface

*Multiple options can be correct
JEE Main Part Test - 2 - Question 11

At what temperature is the r.m.s velocity of a hydrogen molecule equal to that of an oxygen molecule at 47°C?

Detailed Solution for JEE Main Part Test - 2 - Question 11

JEE Main Part Test - 2 - Question 12

The oxidation number of P in Ba(H2PO2)2, Ba(H2PO3)2 and Ba(H2PO4)2 are respectively

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



JEE Main Part Test - 2 - Question 13

How much heat will be required at constant pressure to form 1.28 kg of CaC2 from CaO(s) & C(s) ?

Given :

ΔfH°(CaO, s) = -152 kcal/mol

ΔfH°(CaC2, s) = -14 kcal/mol

ΔfH°(CO, g) = -26 kcal/mol

Detailed Solution for JEE Main Part Test - 2 - Question 13

JEE Main Part Test - 2 - Question 14

From the following data, the heat of formation of Ca(OH)2(s) at 18°C is ………..kcal:

Detailed Solution for JEE Main Part Test - 2 - Question 14

The correct answer is Option B.

Ca(s) + O2(g) + H2(g) → Ca(OH)2 , ΔHf = ?

Desired equation = eq (iii) + eq(i) - eq (ii)

ΔHf = (−151.80)+(−15.26)−(−68.37)
ΔHf = (-151.80)+(-15.26)-(-68.37)
ΔHf = −235.43KCalmol−1

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 for JEE Main Part Test - 2 - 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,
 
E°cell = 1.10 V

Detailed Solution for JEE Main Part Test - 2 - Question 16

Zn(s) + Cu2+(aq) ⇌ Cu(s) + Zn2+(aq),

E= +1.10V

∴ Eo = 0.0591/n log10Keq

because at equilibrium, 

Ecell = 0

(n = number of electrons exchanged = 2)

1.10 = 0.0591/2 log10Keq
2.20/0.0591 = log10Keq

Keq = antilog37.225

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 for JEE Main Part Test - 2 - 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 for JEE Main Part Test - 2 - 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 tus 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

If the coefficients of (r +1)th term and (r + 3)th term in the expansion of (1+x)2n be equal then

Detailed Solution for JEE Main Part Test - 2 - Question 19

Tr+1 = 2nCr(x)r (1)2n-r
Tr+3 = 2nCr+2 (x)r+2 (1)2n-r-2
Tr+1 : Tr+3 
= 2nCr = 2nCr+2
=> 2n!/(2n-r)!r! = 2n!/(r+2)!(2n-r-2)!
=> 1/(2n-r)(2n-r-1) = 1/(r+2)(r+1)
=> (r+2)(r+1) = (2n-r)(2n-r-1)
=> r2 + 3r + 2 = 4n2 - 2nr - 2n - 2nr + r^2 + r
=> 0 = 4n2 - 4nr - 2n + r - 3r - 2
=> 0 = 4n2 - 4nr - 2n + r - 3r - 2 - 2 + 2
=> 0 = 4n2 - 4nr - 4 - [2n + 2r - 2]
=> 4n(n-r-1) -2(n-r-1)
Therefore n - r - 1 = 0
=> n = r+1

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 for JEE Main Part Test - 2 - 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 for JEE Main Part Test - 2 - 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

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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 for JEE Main Part Test - 2 - 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 for JEE Main Part Test - 2 - 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 for JEE Main Part Test - 2 - Question 25

JEE Main Part Test - 2 - Question 26

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 for JEE Main Part Test - 2 - Question 26

JEE Main Part Test - 2 - Question 27

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 for JEE Main Part Test - 2 - Question 27

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 28

T is a point on the tangent to a parabola y2 = 4ax at its point P. TL and TN are the perpendiculars on the focal radius SP and the directrix of the parabola respectively. Then

Detailed Solution for JEE Main Part Test - 2 - Question 28


JEE Main Part Test - 2 - Question 29

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

Detailed Solution for JEE Main Part Test - 2 - Question 29

 Parabola is x2 – 8x + 2y + 7 = 0 
∴   (x – 4)2 = – 2y – 7 + 16 
∴   (x – 4)2 = – 2[y – (9/2)]
 ∴   x2 = – 4ay  
 ⇒ x = x – 4, y = y – (9/2) and 2 = 4a
 i.e. a = (1/2) 
 Its focus is given by x = 0 and y = 0 i.e. x – 4 = 0   and     y – (9/3) = 0 
∴    x = 4    and y = (9/2) 
∴ focus [4, (9/2)].

*Answer can only contain numeric values
JEE Main Part Test - 2 - Question 30

The average age of a husband and his wife was 23 years at the time of their marriage. After five years they have a one-year old child. If the average age of the family now is λ, then the number of divisors of λ are


Detailed Solution for JEE Main Part Test - 2 - Question 30

Sum of the present ages of husband, wife and child

= (23 × 2 + 5 × 2) + 1 = 57 years.

∴ Required average = (57/3) = 19 years .

Average age is 19 years which is a prime number, so it has only two divisors {1, 19}.

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