Approximately, 1043739 lakh candidates appeared for JEE Main offline exam for admission to undergraduate courses in NITs, IIITs and other Centrally Funded Technical Institutions. Students who opted for online mode while submitting application form, will have to appear for JEE Main online exam on April 15 and 16. JEE Main is the eligibility test for JEE Advanced.

Such students who think they can easily secure the rank among top 224000 students after checking the answer key released by famous IIT JEE coaching institutes, would have already started their preparation for upcoming JEE Advanced exam which will be conducted on May 20 by IIT Kanpur.

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 2018.

Every day we will update new questions with explanations so that the number of questions keeps on increasing.

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**Few sample questions are given below:**

**Question:**

The Common roots of the equations *z*^{3}* *+* *2*z*^{2}* + *2*z + *1 = 0 and *z*^{1985} + *z*^{100}* *+ 1 = 0 are

(a) ω^{2}, ω^{3}

(b) ω, ω^{3}

(c) ω, ω^{2}

(d) None of these

**Solution:**

The given equation *z*^{3}* + *2*z*^{2} *+ *2*z *+ 1 = 0 can be rewritten as (*z* + 1) (*z*^{2} + *z *+ 1) = 0. Its roots are

−1, ω and ω^{2}.

Let *f*(*z*) = *z*^{1985} + *z*^{100} + 1

Putting z = −1, ω and ω^{2} 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) = ω^{1985} + ω^{100}+ 1

= (ω^{3})^{661 }ω^{2} + (ω^{3})^{33 }ω+1

= ω^{2} +ω + 1 = 0

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

Similarly, *f*(ω^{2}) = 0

Hence, ω and ω^{2} are the common roots.

Hence, the correct option is (c).

**Question:**

The angle between the pair of tangents drawn to the ellipse 3*x*^{2} + 2*y*^{2} = 5 from the point (1, 2) is

**Solution:**

Here, *S* = 3*x*^{2} + 2*y*^{2} -5 = 0

*S*_{1} = 3(1)^{2} + 2(2)^{2} -5 = 3 + 8 -5 = 6 [∵Point (1, 2)]

*T* = 3*x*.*x*_{1} + 2*y*.*y*_{1} -5

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

= 3*x* + 4*y* -5

Now, use, *SS*_{1} =* T*^{2}

(3*x*^{2} + 2*y*^{2} -5).6 = (3*x* + 4*y* -5)^{2}

18*x*^{2} + 12*y*^{2} -30 = 9*x*^{2} + 16*y*^{2} + 24*xy* + 25 -30*x* -40*y*

⇒ 9*x*^{2} -4*y*^{2} -24*xy* + 30*x* + 40*y* -55 = 0

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

**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 l0^{2} *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 10^{2} sec^{-1}, the ball moves away from the centre by a distance nearly equal to

(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 *mr*w^{2} = (9 × 10^{−2}) (10^{−1}) (10^{2})^{2} = 9 × 10*N*

As a result of this force the spring is compressed by a distance *x. *The restoring force of the spring = *kx *= 10^{4}*x*

In equilibrium, 9 × 10 = 10^{4}*x*

⇒ *x* ≈ 10^{−2}m

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

**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:

∴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).

*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.

**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

**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:**

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