Jharkhand (JSSC) PRT/TGT Exam  >  Jharkhand (JSSC) PRT/TGT Tests  >  Jharkhand TGT (JSSC) Mock Test - 5 - Jharkhand (JSSC) PRT/TGT MCQ

Jharkhand TGT (JSSC) Mock Test - 5 - Jharkhand (JSSC) PRT/TGT MCQ


Test Description

30 Questions MCQ Test - Jharkhand TGT (JSSC) Mock Test - 5

Jharkhand TGT (JSSC) Mock Test - 5 for Jharkhand (JSSC) PRT/TGT 2024 is part of Jharkhand (JSSC) PRT/TGT preparation. The Jharkhand TGT (JSSC) Mock Test - 5 questions and answers have been prepared according to the Jharkhand (JSSC) PRT/TGT exam syllabus.The Jharkhand TGT (JSSC) Mock Test - 5 MCQs are made for Jharkhand (JSSC) PRT/TGT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Jharkhand TGT (JSSC) Mock Test - 5 below.
Solutions of Jharkhand TGT (JSSC) Mock Test - 5 questions in English are available as part of our course for Jharkhand (JSSC) PRT/TGT & Jharkhand TGT (JSSC) Mock Test - 5 solutions in Hindi for Jharkhand (JSSC) PRT/TGT course. Download more important topics, notes, lectures and mock test series for Jharkhand (JSSC) PRT/TGT Exam by signing up for free. Attempt Jharkhand TGT (JSSC) Mock Test - 5 | 180 questions in 120 minutes | Mock test for Jharkhand (JSSC) PRT/TGT preparation | Free important questions MCQ to study for Jharkhand (JSSC) PRT/TGT Exam | Download free PDF with solutions
Jharkhand TGT (JSSC) Mock Test - 5 - Question 1

f : R → and g : [0, ∞) → R is defined by f(x) = x2 and g(x) = √x. Which one of the following is not true?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 1

Here, f(x) = x2 and g(x) = √x

fog(x) = f(g(x))

= [g(x)]2

= [√x]2

= x

⇒ fog(x) = [√x]2

So, fog(2) = (√ 2)2 = 2 and fog(-4) = (√-4)2

But, f(x) belongs to Real number, so fog (-4) = 4 is not true

Jharkhand TGT (JSSC) Mock Test - 5 - Question 2

Lines are drawn parallel to the line , at a distance from the origin. Then which one of the following points lies on any of these lines?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 2

Line parallel to is given as

Distance from origin is

Required lines are

Substituting the given options in the above equations for getting the correct answer.

Clearly satisfies , so there is no need to check other options.

1 Crore+ students have signed up on EduRev. Have you? Download the App
Jharkhand TGT (JSSC) Mock Test - 5 - Question 3

The value of is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 3

Given:

As we know,

Considering the given equation,

The value of is .

Jharkhand TGT (JSSC) Mock Test - 5 - Question 4

What are the direction ratios of the line of intersection of the planes and ?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 4

Given:

and

and

and

If two planes intersect each other, the intersection will always be a line.

Now, the direction ratios of the line are .

Jharkhand TGT (JSSC) Mock Test - 5 - Question 5
If then in which quadrant does lie?
Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 5

Given,

We have,

By eq. eq. (2)

By eq. (1) - eq. (2)

From & both sec and tan are positive only in the first quadrant.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 6

If f(x) = |cos x – sin x|, then f' is equal to-

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 6

Here,

Here,

Jharkhand TGT (JSSC) Mock Test - 5 - Question 7

The order and degree of the differential equation is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 7

We know that,

The order of a differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Given the differential equation is,

Taking cube on both sides, we get,

Taking square on both sides, we get,

Highest derivate is

So, the order of the given differential equation

The power of the highest derivate

So, the degree of the given differential equation

Jharkhand TGT (JSSC) Mock Test - 5 - Question 8

If and , then what is the value of ?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 8

Given,

As we know,

∴ The value of given identities is 8.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 9

Find the value of .

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 9

Let

Jharkhand TGT (JSSC) Mock Test - 5 - Question 10

Consider the following:

Which of the above is/are identity/identities?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 10


So, Statement 1 is true.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 11

What is equal to:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 11


Jharkhand TGT (JSSC) Mock Test - 5 - Question 12
What is the value of
Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 12

Consider,

Jharkhand TGT (JSSC) Mock Test - 5 - Question 13

What is equal to?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 13

Consider

Taking log both sides,

Differentiating both sides,

Putting the value from (2) to (1)

Jharkhand TGT (JSSC) Mock Test - 5 - Question 14

The equation x2 - y2 = 0 represents:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 14

Comparing x2 - y2 = 0 with the general equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we can say that a = 1, b = -1 and all other coefficients are 0.

Δ = abc + 2fgh - af2 - bg2 - cf2 = 0.

∴ The equation represents a pair of straight lines.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 15

Which among the following statement is correct regarding an ideal conductor in a static electric field?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 15

Static field intensity at the surface of a conductor is directly normal to the surface this statement is correct regarding an ideal conductor in a static electric field.

Under the static condition, the electric field inside the solid perfect conductor is zero in electrostatic equilibrium even if it is isolated, charged, and present in the external electrostatic field.

Now, it is known that a perfect solid conductor is an equipotential body. The potential inside the conductor is the same as the potential at the surface. Since the surface of the solid conductor is an equipotential surface hence the electric field will be perpendicular or in other words normal to the perfectly solid conductor surface.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 16

Four graphs between and given for spherical mirrors. Which one of them suitable represents a convex mirror, as per the new Cartesian sign convention?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 16

Since, is always +ve for convex mirror

By the mirror formula

So,

Now,

(negative sign shows negative y-axis)

Jharkhand TGT (JSSC) Mock Test - 5 - Question 17

___________ are those which gets strongly magnetised when placed in an external magnetic field.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 17

Ferromagnetic substances are those which gets strongly magnetised when placed in an external magnetic field. They have strong tendency to move from a region of weak magnetic field to strong magnetic field, i.e., they get strongly attracted to a magnet.

The individual atoms (or ions or molecules) in a ferromagnetic material possess a dipole moment as in a paramagnetic material. However, they interact with one another in such a way that they spontaneously align themselves in a common direction over a macroscopic volume called domain.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 18

Air is an example of ______________.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 18

Air is an example of paramagnetic material since under an external magnetic field it forms an electric dipole.

Paramagnetic substances are those which develop feeble magnetization in the direction of the magnetizing field. Such substances are feebly attracted by magnets and tend to move from weaker to stronger parts of a magnetic field. Magnetic susceptibility is small and positive i.e. x>0.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 19

Which of the following material has the highest relative permittivity?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 19

The relative permittivity of the water is highest among the other materials given.

Permittivity describes the amount of charge needed to generate one unit of electric flux in a particular medium. Accordingly, a charge will yield more electric flux in a medium with low permittivity than in a medium with high permittivity. Thus, permittivity is the measure of a material's ability to resist an electric field.

Relative permittivity is the ratio of its absolute permittivity ' to free space (empty of matter) permittivity ' i.e.,

Jharkhand TGT (JSSC) Mock Test - 5 - Question 20

A ray of light passes from glass to water The value of critical angle will be:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 20

For critical angel,

Jharkhand TGT (JSSC) Mock Test - 5 - Question 21

In class B amplifier, the output current flows for:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 21

In class B amplifier, the output current flows for half input cycle.

Class-B amplifiers use two or more transistors biased in such a way so that each transistor only conducts during one half cycle of the input waveform. In these amplifiers one transistor only amplifies one half or 180° of the input waveform cycle while the other transistor amplifies the other half or remaining 180° of the input waveform cycle. It is also known as a push-pull amplifier configuration.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 22

A metallic rod when placed in a strong magnetic field, aligns itself at right angles to the magnetic field. The nature of the material is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 22

A metallic rod when placed in a strong magnetic field, aligns itself at right angles to the magnetic field. The nature of the material is diamagnetic.

Diamagnetic substances are those which develop feeble magnetization in the opposite direction of the magnetizing field.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 23

The electronegativity of Cesium is 0.7 and that of Fluorine is 4.0. The bond formed between the two is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 23

Such a large difference in electronegativity shows that Cesium is electropositive while Fluorine is electronegative in nature. So, Cesium can donate an electron to fluorine. So they form ionic bond or electrovalent bond.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 24

Deuterium differs from hydrogen:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 24

Deuterium and hydrogen both have same atomic number but different mass number so they have different physical properties but similar chemical properties since they have same electronic configuration.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 25

Which of the following carbohydrates would be the most abundant in the diet of strict vegetarians?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 25

Cellulose would be the most abundant in the diet of strict vegetarians. Cellulose, a fibrous carbohydrate found in all plants, is the structural component of plant cell walls. Because the earth is covered with vegetation, cellulose is the most abundant of all carbohydrates, accounting for over 50% of all the carbon found in the vegetable kingdom.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 26

Starch consists of:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 26

Starch consists of unbranched amylose and branched amylopectin. Starch is a soft, white, tasteless powder that is insoluble in cold water, alcohol, or other solvents. Starch is a polysaccharide comprising glucose monomers joined in α 1, 4 linkages. The simplest form of starch is the linear polymer amylose; amylopectin is the branched form.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 27

Which of the following has largest radius?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 27

Atomic radii decrease across a period in the periodic table due to an increase in the effective nuclear charge.

The element with electronic configuration 1s2, 2s2, 2p6, 3s2 is an s-block element placed at the extreme left of the periodic table. Whereas, the others are p-block elements placed at the extreme right.

Therefore, the atomic radii of the element with electronic configuration 1s2 2s2 2p6 3s2 are largest.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 28

Fibres of _______ resemble that of silk and hence, it is popularly known as ‘artificial silk’.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 28

Rayon was developed as an alternative for silk. It is prepared by the chemical processing of wood pulp or cellulose. It resembles the texture and look of natural silk and can be woven like silk fibres. Hence, it is also called artificial silk.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 29

Structural polysaccharides include:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 29

Structural polysaccharides include cellulose, hemicellulose and chitin.

Cellulose: It is the most abundant organic compound. It is a highly ductile fibrous homopolysaccharide that forms the structural element of the cell wall of all plants, some fungi and protists. Cotton fiber contains about 90% cellulose while wood contains 25-50% cellulose. Other substances in the cell include lignin, hemicellulose, pectin, wax, etc.

Hemicellulose: This refers to a component of the plant cell wall that has a simpler structure than cellulose. It occupies 20 - 30% of the dry weight of the wood.

Chitin: It is the second most common polysaccharide. It is a complex heteropolysaccharide. Found as structural components of fungal wall and external skeleton of arthropods. It is a polymer of N-acetyl glucosamine. It provides both strength and elasticity. These harden due to the impregnation of are linked by monomers.

Jharkhand TGT (JSSC) Mock Test - 5 - Question 30

Which scientist arranged the Periodic Table of Elements in the order of increasing atomic masses?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 5 - Question 30

Dmitri Mendeleev arranged the Periodic Table of Elements in the order of increasing atomic masses.

Mendeleev’s Periodic Table (1869) is based upon Mendeleev’s periodic law, which states, “Properties of the elements are the periodic function of their atomic masses.”

View more questions
Information about Jharkhand TGT (JSSC) Mock Test - 5 Page
In this test you can find the Exam questions for Jharkhand TGT (JSSC) Mock Test - 5 solved & explained in the simplest way possible. Besides giving Questions and answers for Jharkhand TGT (JSSC) Mock Test - 5, EduRev gives you an ample number of Online tests for practice

Top Courses for Jharkhand (JSSC) PRT/TGT

Download as PDF

Top Courses for Jharkhand (JSSC) PRT/TGT