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Jharkhand TGT (JSSC) Mock Test - 4 - Jharkhand (JSSC) PRT/TGT MCQ


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30 Questions MCQ Test Jharkhand (JSSC) TGT Exam Mock Test Series 2024 - Jharkhand TGT (JSSC) Mock Test - 4

Jharkhand TGT (JSSC) Mock Test - 4 for Jharkhand (JSSC) PRT/TGT 2024 is part of Jharkhand (JSSC) TGT Exam Mock Test Series 2024 preparation. The Jharkhand TGT (JSSC) Mock Test - 4 questions and answers have been prepared according to the Jharkhand (JSSC) PRT/TGT exam syllabus.The Jharkhand TGT (JSSC) Mock Test - 4 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 - 4 below.
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Jharkhand TGT (JSSC) Mock Test - 4 - Question 1

Find the equation of the hyperbola whose vertices are and the eccentricity is ?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 1

By comparing the vertices with we get

As we know that,

As we know,

The equation of the hyperbola is given by,

So, the equation of the hyperbola is .

Jharkhand TGT (JSSC) Mock Test - 4 - Question 2

If the plane has the distances and units from the planes and , respectively, then the maximum value of is equal to:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 2

Let the planes,

Given, distance between and is .

And distance between and is .

Thus, the maximum value of .

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Jharkhand TGT (JSSC) Mock Test - 4 - Question 3

The equation of the normal at the point to the hyperbola  is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 3

Given equation of hyperbola .

or Differentiating both sides with respect to ''.

We know that Equation of line at and having slope is:

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

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

Given,

As we know,

Differentiating with respect to , we get

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

If A and B are two sets then A ∩ (B ∪ A)c is equal to-

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

It is given that A and B are two sets.

We have to find A ∩ (B ∪ A)c

As we know, (A ∪ B)c = Ac ∩ Bc

Therefore, (B ∪ A)c = Bc ∩ Ac

Now, A ∩ (B ∪ A)c = A ∩ (Bc ∩ Ac)

= (A ∩ Bc) ∩ (A ∩ Ac) (Using distributive law)

= (A ∩ Bc) ∩ ϕ (∵ x ∩ ϕ = ϕ)

= ϕ

Jharkhand TGT (JSSC) Mock Test - 4 - Question 6

The area enclosed with the curve is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 6

The graph of the given curve is as follows.

Four possible equations arise from the given equation.

x+y=1 when x>0 and y>0

-x+y=1 when x<0 and y>0

-x-y=1 when both x and y are <0

x-y=1 when x>0 and y<0

If you plot them you will get co-ordinates as

(1,0) (0,1) (-1,0) (0,-1)

Join them you will get a square of sides equal to √2 units.

Thus area enclosed = Area of square = (√2)^2

= 2 sq.units

Or

From the, figure ABCD is a square. Whose diagonal AC and BD are of the length of 2 units.

So, required area sq.units

Jharkhand TGT (JSSC) Mock Test - 4 - Question 7

The graph of a linear equation in two variables is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 7

The graph of an equation of degree one in two variables is a straight line. That is the reason it is called a linear equation. Thus, every solution of the linear equation can be represented by a unique point on the graph of the equation. A linear equation in two variable has infinitely many solutions.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 8

Find the area of the region bounded by y2=4x, x=1, x=4, and x-axis in the first quadrant.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 8

Given:

y2 = 4x, x = 1, x = 4

∴ Required area = Area of the shaded portion

sq. units

Jharkhand TGT (JSSC) Mock Test - 4 - Question 9

then the value of is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 9

Given:

The value of is .

Jharkhand TGT (JSSC) Mock Test - 4 - Question 10

The order of the differential equation:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 10

Given the differential equation is:

Highest derivate is .

Here, The given equation is not a polynomial equation.

∴ The order of the differential equation is not defined.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 11

A window in a building is at a height of meters from the ground. The angle of depression of a point  on the ground from the window is . The angle of elevation of the top of the building from the point is . What is the height of the building?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 11

Given:

The window is at a height of meters

Concept used:

perpendicular or height

base

Let the window be .

Height of the building be

Let the point at a distance of meter.

In

In

So, the height of building

Jharkhand TGT (JSSC) Mock Test - 4 - Question 12

According to Dalton's law of partial pressures, (where pb = Barometric pressure, pa = Partial pressure of dry air, and pv = Partial pressure of water vapour).

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 12

According to Dalton's law of partial pressures, the total pressure by a mixture of gases is equal to the sum of the partial pressures of each of the constituent gases. The partial pressure is defined as the pressure each gas would exert if it alone occupied the volume of the mixture at the same temperature -According to Dalton's law of partial pressures, Pb = pa + pv (where pb = Barometric pressure, pa = Partial pressure of dry air, and pv = Partial pressure of water vapour).

Jharkhand TGT (JSSC) Mock Test - 4 - Question 13

The number of images of an object placed between two parallel mirrors is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 13

Number of images formed in two parallel mirror are infinite because an image formed by one mirror can act as an object for the second mirror. And every image formed by the second mirror can act as an object for the first mirror. This image formation will continue for both the mirrors.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 14

A full length image of a distant tall building can definitely be seen by using:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 14

A full length image of a distant tall building can definitely be seen by using a convex mirror.

The convex mirror gives virtual, erect, and diminished images of the objects. It gives a wider view of the image formed in it. It can form full images of a large buildings, cars, etc which are very far behind the mirror.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 15

In Huygens principle, the relation between wavefront and the ray of light is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 15

A wavefront is a locus of all the points in a wave that have the same phase.

The wavefront forms consecutive concentric circles around the source of light. The rays show the direction of propagation of the wave.

Therefore, the rays of light are always normal/perpendicular to the wavefront.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 16

The Peak Inverse Voltage rating of a crystal diode is ________________ that of equivalent vacuum diode.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 16

The maximum value of the reverse voltage that a PN junction or diode can withstand without damaging itself is known as its Peak Inverse Voltage. If the voltage coming across the junction at reverse biased condition increases beyond this specified value, the junction gets damaged. PIV rating of a diode is temperature-dependent. It increases with an increase in temperature and decreases with a decrease in temperature. The Peak Inverse Voltage rating of a crystal diode is lower than that of equivalent vacuum diode.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 17

A wire of length is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were , its new resistance would be:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 17

Let the new length be , keeping volume constant,

Now,

Jharkhand TGT (JSSC) Mock Test - 4 - Question 18

The heat given to an ideal gas in isothermal conditions is used to:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 18

The heat given to an ideal gas in isothermal conditions is used to do external work.

When a thermodynamic system undergoes a physical change in such a way that its temperature remains constant, then the change is known as an isothermal process.

As we know that, the internal energy of the system is a function of temperature alone, so in the isothermal process, the change in internal energy is zero.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 19

The ion with the strongest polarizing capacity is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 19

Polarizing capacity/polarizing power: The ability of a cation to distort an anion is called the polarization power of that cation.

Fajan's rules of polarization: According to this rule, the more the size of an anion and more easily will it be polarized and hence the compound will be more non-polar and Less the size of cation more will be its polarizing power.

The size of Li+ is the smallest in size among all the given options. So according to Fajan's rule, Li+ will have the strongest polarizing capacity.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 20

The volume of hexagonal ice lattice is given by:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 20

Hexagonal ice lattice belongs to the hexagonal crystal system. It has hexag symmetry.
In a hexagonal crystal system, and
where, are the axial distances and are the axial angles.
The base area of the unit cell of HCP:
It is equal to the area of six equilateral triangles each with side and altitude

Base area
Height of unit cell of
The volume of a unit cell of HCP Base area Height .
On substituting the values from equation (i) to equation (ii), we get
The volume of the unit cell ,
Substituting the value of in equation (iii), we get
Volume of the unit cell =6 \sqrt{3} \times\left(\frac{a}{2}\right)^2 \times c \ =6 \sqrt{3} \times \frac{a^2}{4} \times c=\frac{\sqrt{3}}{2} a^2 \cdot c

Jharkhand TGT (JSSC) Mock Test - 4 - Question 21

Which of the following is a thermosetting polymer?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 21

Phenolic resins is a thermosetting polymer. Phenolic resin is notable for being a type of thermoset polymer, meaning that it cures into an altered form than when it is uncured; however, unlike other varieties of plastic polymers, it cannot be re-melted and re-molded.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 22

When beryllium is bombarded with α -particles, extremely penetrating radiations that cannot be deflected by an electrical or magnetic field are given out. These are

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 22

As we know,

Be

(Atom release)

Neutron is a chargeless particle, so it does not get deflected by an electric or magnetic field.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 23

In polysaccharides, monosaccharaides are joined by:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 23

In polysaccharides, monosaccharaides are joined by glycosidic bond. A polysaccharide is a polymer made up of chains of a monosaccharide. It is also known as glycans. A polysaccharide consists of more than ten monosaccharide units, whereas an oligosaccharide consists of three to ten linked monosaccharides. Polysaccharides are a type of carbohydrate. It is a polymer composed of several sugar subunits, called monosaccharides.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 24

Modern periodic table is based on ________ .

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 24

The elements in the modern periodic table are arranged in the order of their increasing atomic numbers.

Atomic number = No. of protons

Mass number = No. of protons + No. of neutrons

There are 7 rows and 18 columns in the modern periodic table.

Rows are known as periods and columns are known as groups.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 25

The compound formed when ethyl bromide is heated with dry silver oxide is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 25

The main product obtained by heating ethyl bromide in presence of dry silver oxide is Diethyl ether, when ethyl bromide heat in the presence of dry silver oxide then it gives diethyl ether and silver bromide as products.


If we take moist then alcohol is formed

Jharkhand TGT (JSSC) Mock Test - 4 - Question 26

Oxymercuration-demercuration of produces:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 26

In oxymercuration-demercuration, more stable carbocation is formed and water attacks on more substituted carbon. The reaction is as follows:

The oxymercuration reaction is an electrophilic addition organic reaction that transforms an alkene into neutral alcohol. The reaction follows Markovnikov's rule (the hydroxy group will always be added to the more substituted carbon) and it is an anti addition (the two groups will be trans to each other).

Jharkhand TGT (JSSC) Mock Test - 4 - Question 27

Which of the following is not a condensation polymer?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 27

Polyacrylonitrile is a synthetic polymer with the linear formula . Although it is thermoplastic it does not melt under normal conditions, since it degrades before melting at above . Almost all Polyacrylonitrile resins are copolymers made from mixtures of monomers, with acrylonitrile as the main component. As it is a non condensation polymer.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 28

Which one among the following most correctly determines the atomic number of an element?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 28

The atomic number or proton number (symbol Z) of a chemical element is the number of protons found in the nucleus of every atom of that element. The atomic number uniquely identifies a chemical element. It is identical to the charge number of the nucleus. In an uncharged atom, the atomic number is also equal to the number of electrons.

Jharkhand TGT (JSSC) Mock Test - 4 - Question 29

Choose the correct statement:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 29

Alpha rays (α): Alpha ray, also called alpha particles alpha ray, contains two protons and two neutrons bound together into a particle just like a helium-4 nucleus.

Beta rays (β): Beta rays, also called beta particles or beta radiation, may be high-energy, high-speed electron or positron emitted by the radioactive decay of an atomic nucleus during the process of beta decay.

Gamma rays (γ): A gamma-ray, or gamma radiation, is penetrating electromagnetic radiation arising from the radioactive decay of atomic nuclei.

The ability of radiation to damage molecules is known as ionizing power.

The ability of each type of radiation to pass through matter is expressed in terms of penetration power.

We know radiation is caused because of radioactivity and the table below represents penetration power and ionization power of different radiation.

Penetrating powers γ – rays > β – rays > α -raysPenetrating powers γ – rays > β – rays > α -rays
Jharkhand TGT (JSSC) Mock Test - 4 - Question 30

PA = (235 y -125 xy)mm of Hg where PA is partial pressure of A

x is mole fraction of B in liquid phase in the mixture of two liquids A and B and y is mole fraction of A in vapour phase, then in mm of Hg is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 4 - Question 30

Given:

Partial pressure of A,

PA = (235 y- 125 xy) mm of Hg......(i)

x = Mole fraction of B in liquid phase

y = Mole fraction of A in vapour phase

We have to find in mm of Hg

Now, according to Dalton's law of partial pressure :

PA = Mole fraction of component A in vapour phase × Total vapour pressure.......(ii)

Equation (i) can be written as:

PA =(235 -125x) y.......(iii)

From equations (i) and (iii),

Total pressure, p = 235 - 125x ...(iv)

Now, total pressure is given by:

p=......(v)

where = Vapour pressure of pure A

=Vapour pressure of pure B

XA = Mole fraction of A

XB = Mole fraction of B

Also, the sum of mole fractions of all the components in solution is always equal to one, i.e.

XA + XB = 1

XA -1 = -XB

=1 - x (As , mole fraction of component B is x)

Substituting values in equation (v), we get

p=pA0(1−x)+pB0x
p=pA0−pA0x+pB0x
p=pA0−(pA0−pB0)x ......(vi)
Comparing equations (iv) and (vi), we get
pA0=235
And,
pA0−pB0=125
pA0 On substituting the value of pA0

We get,

235−pB0=125
pB0=235−125
=110mm of Hg

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