# JEE Main 2019: 5 Solved questions on Solid State

In this article, we will provide you 5 solved questions from different topics of all subjects i.e., Physics, Chemistry and Mathematics on daily basis. Find 5 solved questions from the topic Limits. All are important from the exam point of view.

*Solved questions for JEE Main 2019*

The National Testing Agency (NTA) has already released the schedule for JEE Main 2019. According to the schedule, the agency will conduct JEE Main 2019 exam in the month of January and April in fully computer based test (CBT) mode. The eligible students can apply the January exam from Sep 1 to Sep 30, 2018. After submitting application, all the aspirants need to keep a track of their preparedness by practicing more and more questions based on each important topic. It becomes very important to score a good rank in JEE Main 2019 and for this you should have equal grip over all the three subjects. To fulfill the same you will have to practice questions from all the subjects on a daily basis. Here, we will provide you 5 solved questions from different topics of all subjects i.e., Physics, Chemistry and Mathematics on daily basis. Find 5 solved questions from the topic Solid State. All are important from the exam point of view.

Students must go through all the questions listed below to revise the important topics by practicing the questions on a daily basis. These questions will help you to score good marks and a good rank in the engineering entrance exams especially in JEE Main 2019.

**Every day you will get the entirely FRESH 5 Questions for JEE Main 2019.**

**Question:**

Find out the total number of voids in 0.5 mole of a compound forming hexagonal closed packed structure.

(a) 9.034 × 10^{23}

(b) 6.023 × 10^{23}

(c) 18.069 × 10^{23}

(d) 3.011 × 10^{23}

**Solution: (a) **

In hexagonal closed packed structure, there are 6 atoms per unit cell.

Number of octahedral voids = 6

Number of tetrahedral voids = 2 × 6 = 12

**Question:**

CaO and NaCl have the same crystal structure and approximately the same ionic radii. If U is the lattice energy of NaCl, the approximate lattice energy of CaO is

(a) U/2

(b) U

(c) 2U

(d) 4U

**Solution: (d) **

Since, interionic distances in CaO and NaC1 are similar so, r is almost the same. Therefore, lattice energy depends only on charge.

Since, the magnitude of charge on Na^{+} and Cl^{-} ions is same i.e., unity and that on Ca^{2+} and O^{2-} ions is 2 each, therefore, the lattice energy of CaO is four times the lattice energy of NaCl, i.e., 4U.

**Question:**

A cubic solid is made of two elements A and B. Atoms of A are at the corners of the cube and B at the body-centre. What is the formula of the compound?

Options:

(a) AB

(b) A_{2}B_{3}

(c) AB_{2}

(d) A_{2}B

**Solution: (a) **

Atoms of A are present at the corners of the cube.

Therefore, the formula of the compound is AB.

**Question:**

The total number of octahedral voids in the face centered unit cell is ………. .

(a) 4

(b) 8

(c) 10

(d) 12

**Solution: (a) **

There are a total of 4 octahedral voids in the face centered unit cell.

**Question:**

Structure of a mixed oxide is cubic closed packed (*ccp*). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal P and the octahedral voids are occupied by a monovalent metal Q. The formula of the oxide is

(a) PQO_{2}

(b) P_{2}QO_{2}

(c) P_{2}Q_{3}O_{4}

(d) PQ_{2}O_{2}

**Solution: (d)**

Number of atoms in *ccp *= 4 = O^{2−}

Number of tetrahedral voids = 2 × N = 2 × 4

Number of octahedral voids = Number of Q^{+} ions = N = 4

Ratio , O^{2−} : P^{2+} : Q^{+} = 4 : 2 : 4 or 2 : 1 : 2

Therefore, the formula of oxide is PQ_{2}O_{2}.

**Solved questions on Limit**

**Question:**

**Solution: (b)**

**Question:**

**Solution: (c)**

**Question:**

**Solution: (d)**

**Question:**

**Solution: (b)**

**Question:**

**Solution: (b)**

**Solved questions on Law of Motion**

**Question:**

A particle suspended from the ceiling by inextensible light string is moving along a horizontal circle of radius 1.5 m as shown in figure. The string traces a cone of height 2 m. The string breaks and the particle finally hits the floor (which is xy plane 5.76 m below the circle) at point P. The distance OP is

(a) 3

(b) 4

(c) 3.9

(d) 4.1

**Solution: (c)**

Let the string breaks when the particle is 1.5 m right of point O and direction of its velocity *v *is along y-axis.

**JEE Main 2019: Normalisation of Marks**

**Question:**

An explosive of mass 6 kg is projected at 35 m s^{-1} at an angle of 60° with the horizontal. At the top of its flight it explodes, breaking into two parts, one of which has twice the mass of the other. The two fragments land simultaneously. The lighter fragment lands back at the launch point. Where does the other fragment land?

(a) 162

(b) 163

(c) 160

(d) 161

**Question:**

A cylindrical pipe of diameter 1 m is kept on a truck as shown in figure. If the truck now starts moving with a constant acceleration of 1 m s^{-2}, the pipe rolls backward without slipping on the floor of the truck and finally falls on the road. If the pipe moves a total length of 4 m on the floor of the truck, what is the velocity of the pipe in (m s^{-1}) relative to the truck and relative to ground at the instant it leaves contact with the truck? (g = 10 m s^{-2}).

**Question:**

A box of mass 8 kg is placed on a rough inclined plane of inclination. Its downward motion can be prevented by applying an upward pull *F* and it can be made to slide upwards by applying a force 2 *F*. The coefficient of friction between the box and the inclined plane is

(a) (tan θ)/3

(b) 3tan θ

(c) (tan θ)/2

(d) 3tan θ

**Question:**

A ball is released from the top of an inclined plane of inclination. It reaches the bottom with velocity *v*. If the angle of inclination is doubled, then the velocity of ball on reaching the ground is

(a) *v *

(b) 2*v *

**Mathematics all chapters complete notes for JEE 2019**

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