95+ Most Important Questions for JEE Advanced 2019

In this article, we are providing you 95+ most important questions along with their solutions which will boost your preparation for JEE Advanced 2019. The solutions are well explained by experienced Subject Experts.

Rahul Tomar
May 8, 2019, 10:24 IST
Most Important questions for JEE Advanced 2019
Most Important questions for JEE Advanced 2019

In this article, we are providing you 95+ most important questions with answers and explanations from all the three subjects, i.e., Physics, Chemistry and Mathematics. These questions will help you to score good marks and a good rank in JEE Advanced 2019. This year, the exam will be conducted by IIT Roorkee on May 27.

Mathematics for JEE (Advanced)

If you are preparing for engineering entrance exam, then this Mathematics book will help to crack it without any difficulty. It includes step-by-step solution to every equation and has important formulas to solve complex problems within seconds.

Buy this Mathematics Book now @ just Rs. 434

Top 6 tricks to crack IIT JEE Advanced 2019

Few sample questions are given below:

Question:

The Common roots of the equations z3 + 2z2 + 2z + 1 = 0 and z1985 + z100 + 1 = 0 are

(a) ω2, ω3        

(b) ω, ω3

(c) ω, ω2         

(d) None of these

Solution:

The given equation z3 + 2z2 + 2z + 1 = 0 can be rewritten as (z + 1) (z2 + z + 1) = 0. Its roots are

 −1,  ω and ω2.

Let   f(z) = z1985 + z100 + 1

Putting z = −1, ω and ω2 respectively, we get

f(−1) = (−1)1985 + (−1)100 + 1 0

Therefore, −1 is not a root of the equation

f(z) = 0.

Again, f(w) = ω1985 + ω100+ 1

= (ω3)661 ω2 + (ω3)33 ω+1

= ω2 +ω + 1 = 0

Therefore, ω is a root of the equation f(z) = 0.

Similarly, f2) = 0

Hence, ω and ω2 are the common roots.

Hence, the correct option is (c).

Question:

The angle between the pair of tangents drawn to the ellipse 3x2 + 2y2 = 5 from the point (1, 2) is

JEE Advanced Important Questions

Solution:

Here, S = 3x2 + 2y2 -5 = 0

S1 = 3(1)2 + 2(2)2 -5 = 3 + 8 -5 = 6       [∵Point (1, 2)]

T = 3x.x1 + 2y.y1 -5

= 3x(l) + 2y(2) -5

= 3x + 4y -5

Now, use, SS1 = T2

(3x2 + 2y2 -5).6 = (3x + 4y -5)2

18x2 + 12y2 -30 = 9x2 + 16y2 + 24xy + 25 -30x -40y

⇒           9x2 -4y2 -24xy + 30x + 40y -55 = 0

Here, a = 9, b = -4 and h = -12

JEE Advanced Important Questions

Question:

A circular turn table of radius 0.5 m has a smooth groove as shown in figure. A ball of mass 90 g is placed inside the groove along with a spring of spring constant l02 N/cm. The ball is at a distance of 0.1 m from the centre when the turn table is at rest. On rotating the turn .able with a constant angular frequency of 102 sec-1, the ball moves away from the centre by a distance nearly equal to

JEE Advanced Important Questions

(a) 10-1 m                              

(b) 10-2 m

(c) 10-3 m                               

(d) 2 × 10-1 m

Solution:

When the turn table rotates with angular speed a), the particle of mass m describes a circle of radius r. The centrifugal force experienced is mrw2 = (9 × 10−2) (10−1) (102)2 = 9 × 10N

As a result of this force the spring is compressed by a distance x. The restoring force of the spring = kx = 104x

In equilibrium, 9 × 10 = 104x

⇒  x ≈ 10−2m

Hence, the correct option is (b).

Question:

A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two mass less spring toy-guns, each carrying a steel ball of mass 0.05 kg are attached to the platform at a distance 0.25 m from the centre on its either sides along its diameter (see figure). Each  gun  simultaneously  fires  the  balls  horizontally  and  perpendicular  to  the  diameter  in opposite directions. After leaving the platform, the balls have horizontal speed of 9 m/s with respect to the ground. The rotational speed of the platform in rad/s after the balls leave the platform is

JEE Advanced Important Questions

JEE Advanced Important Questions

Question:

Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator, the number of moles of Mohr’s salt required per mole of dichromate is                                                                                                                                       

(a) 3

(b) 4

(c) 5

(d) 6

Solution:

n-factor of dichromate is 6.

Also, n-factor of Mohr’s salt is 1 as:

JEE Advanced Important Questions


∴1 mole of dichromate = 6 equivalent of dichromate

∵6 equivalent of Mohr’s salt is 1, 6 equivalent of it would also be equal to 6 moles.

Hence, 1 mole of dichromate will oxidise 6 moles of Mohr’s salt.

Hence, the correct option is (d).

JEE Advanced Important Questions

JEE Advanced Important Questions

D = -4 + 10 m

If m is negative, then D will also be negative.

So, the roots of z will be purely imaginary.

Hence, the correct option is (a).

 

Question:

In a galvanic cell, the salt-bridge              

(a) does not participate chemically in the cell reaction

(b) slops the diffusion of ions from one electrode to another

(c) is necessary for the occurrence of the cell reaction

(d) ensures mixing of the two electrolytic solutions

Solution:

Functions of salt-bridge are

(i) It connects the two half-cells and completes the cell circuit.

(ii) It keeps the solutions of two half-cells and complete the cell circuit but does not participate chemically in the cell reaction.

(iii) It maintains the diffusion of ions from one electrode to another electrode.

Hence, options (a) and (b) are correct and option (d) is incorrect.

Salt-bridge is not necessary for the occurrence of the cell reaction. Sometimes, both the electrodes dip in the same electrolyte solution. So, option (c) is not correct.

JEE Advanced Important Questions

JEE Advanced Important Questions

JEE Advanced Important Questions

JEE Advanced Important Questions

Question:

The given graph/data I, II, III and IV represent general trends observed for different

physisorption and chemisorptions processes under mild conditions of temperature and pressure. Which of the following choice(s) about I, II, III and IV is (are) correct?   

(a) 1 is physisorption and II is chemisorption

(b) 1 is physisorption and III is chemisorption

(c) IV is chemisorption and II is chemisorption

(d) IV is chemisorption and III is chemisorptions

JEE Advanced Important Questions

Solution:

Graph-I: It represents physical adsorption as in physical adsorption, absorbents are bonded to adsorbate through weak van der Waals' force.

Increasing temperature increases kinetic energy of adsorbed particles increasing the rate of desorption, hence amount of adsorption decreases.

Graph-II: It represents chemical adsorption as it is simple activation energy diagram of a chemical reaction.

Graph-III: It represents physical adsorption as extent of adsorption increasing with pressure.

Graph-IV: It represents chemical adsorption as it represents potential energy diagram for the formation of typical covalent bond.

Hence, the correct options are (a) and (c).

Question:

A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle.  If the initial speed (in m/s) of the particle is zero, the speed (in m/s) after 5 s is 

Solution:

JEE Advanced Important Questions

Download the complete file

Mathematics for JEE (Advanced)

If you are preparing for engineering entrance exam, then this Mathematics book will help to crack it without any difficulty. It includes step-by-step solution to every equation and has important formulas to solve complex problems within seconds.

Buy this Mathematics Book now @ just Rs. 434

 

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