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

Jharkhand TGT (JSSC) Mock Test - 2 for Jharkhand (JSSC) PRT/TGT 2024 is part of Jharkhand (JSSC) TGT Exam Mock Test Series 2024 preparation. The Jharkhand TGT (JSSC) Mock Test - 2 questions and answers have been prepared according to the Jharkhand (JSSC) PRT/TGT exam syllabus.The Jharkhand TGT (JSSC) Mock Test - 2 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 - 2 below.
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Jharkhand TGT (JSSC) Mock Test - 2 - Question 1

The value of is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 1


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

Solve the differential equation

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

It is form of

The solution of the linear equation is given by:

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Jharkhand TGT (JSSC) Mock Test - 2 - Question 3

The greatest integer by which is divisible is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 3

Given:

= 31!

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

The expression , is equal to:

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

Given:

As we know:

Value putting method to solve the question,

Here we can put but we cannot put

Put

Now,

In if we put it is giving which is satisfying

The required answer is .

Jharkhand TGT (JSSC) Mock Test - 2 - Question 5
What is the degree of the differential equation
Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 5

Given:

Now,

For the given differential equation the highest order derivative is .

Now, the power of the highest order derivative is .

We know that the degree of a differential equation is the power of the highest derivative So, the degree of the differential equation is .

Jharkhand TGT (JSSC) Mock Test - 2 - Question 6

What is value of ?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 6

Let f (x) = x5 sin4 x

f (-x) = (-x)5 sin4 (-x) = -x5 (-sin x )4 = -x5 sin4 x

⇒ f(-x) = - f(x)

So, f(x) is an odd function.

We knoiw that, when is odd

So,

Jharkhand TGT (JSSC) Mock Test - 2 - Question 7

What is the equation of the line which bisects the obtuse angle between the lines x - 2y + 4 = 0 and 4x - 3y + 2 = 0?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 7

We know that,

The general equation of angle bisector between two lines:

and is

If and are of the same sign, then evaluate .

If , then taking positive gives acute angle bisector.

And If , then positive gives obtuse angle bisector.

Given, two lines and

Here, and are of the same sign,

And

Taking positive in the formula gives, obtuse angle bisector of the two lines

The equation of the line which bisects the obtuse angle between the two lines is,

Jharkhand TGT (JSSC) Mock Test - 2 - Question 8

In , right angled at , if , then the value of is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 8

Given,

is a right angle at .

As we know,

Now, Put the value of all identities

Jharkhand TGT (JSSC) Mock Test - 2 - Question 9

The point of intersection of diagonals of a square is at the origin and one of its vertices is at . What is the equation of the diagonal

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 9

Given: The point of intersection of diagonals of a square is at the origin and one of its vertices is at .

So, the diagonal passes through the origin

As we know that, the slope of the line joining the points and is: The slope of line is given by

In square , the diagonals and are perpendicular to each other.

Slope of slope of .

So, the slope of is .

As we know that, the equation of a line passing through the point and having the slope ' ' is given as:

The equation of whose slope is and passes through origin is given by:

Jharkhand TGT (JSSC) Mock Test - 2 - Question 10

If and , then what is the value of

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 10

We know that,

It is given that,

P (n, r) = 2520 and C (n, r) = 21

As we know that,

As we know that,

Jharkhand TGT (JSSC) Mock Test - 2 - Question 11

If , then

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 11

Given:

According to the question, we have,

and

We know that,

So,

Now,

Jharkhand TGT (JSSC) Mock Test - 2 - Question 12

If has 17 zeros, then what is the value of ?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 12

It is given that,

n! has 17 zeros

As we know that, the no. of zeros present in a n! = the highest power of 5 in

If then

Hence,

If then

Hence,

If then

So,

Jharkhand TGT (JSSC) Mock Test - 2 - Question 13

What is the value of ?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 13

Let,

By comparing the equation y2 - 2y - 1 = 0 with the standard quadratic equation ax2 + bx + c = 0. We get, a = 1, b = -2 and c = - 1

So,

Jharkhand TGT (JSSC) Mock Test - 2 - Question 14

Total heat is the heat required to _________.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 14

Total heat is the heat required to Convert water into steam and superheat it.

Total Heat is the thermal equivalent of the energy required to convert unit mass of a liquid at one temperature (as the melting point of the substance) into saturated vapor at any other given temperature.

Jharkhand TGT (JSSC) Mock Test - 2 - Question 15

The angle of polarization of glass is 58° and that for water is 53° .The angle of polarization for glass in water is

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 15

Given:

The angle of polarization for glass, ig = 58°

The angle of polarization for water, iw = 53°

According to Brewster's law, the angle of polarization is related to the refractive index of the transparent medium:

.....(i)
and when the medium is water:
.......(ii)
Now, when the glass is in water, the relative refractive index of the system is


Let the angle of polarization be ip when a glass is in water. Then we get

or,

Jharkhand TGT (JSSC) Mock Test - 2 - Question 16
At the centre of a cubical box charge is placed. The value of total
flux that is coming out of each face is:
Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 16

Given:

Charge on cubical box

The Gaussian law is given as,

Where is the total electric flux, is the charge enclosed by the surface and is the permittivity of the free space.

It is given that the charge of is placed at the centre of the box. The Gauss's law provides the information about the distribution of the electric charges for the closed surface. In the closed surface, the electric flux is directly proportional to the electric charges enclosed in the surface.

The cubical surface encloses the charge . Since the cube is the closed surface with the six faces, the gauss law can be applied to find the electric flux of it.

By using the formula of the Gaussian law,

Since the cube contains the six faces, the permittivity of the free space is multiplied by six

to find the flux from each face.

So, the electric flux that is coming out of each flux is obtained as

Jharkhand TGT (JSSC) Mock Test - 2 - Question 17

Magnetism at the center of a bar magnet is ______.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 17

The magnetic field at the center of a bar magnet is zero in an ideal case.

  • The field lines near the center is parallel to the bar magnet.
  • The density of the field lines is maximum at the poles of the bar magnet and zero at the center which means that the magnitude of the magnetic field is small at the center.
Jharkhand TGT (JSSC) Mock Test - 2 - Question 18

In an experiment to trace the path of a ray of light through a glass prism for different values of angle of incidence a student would find that the emergent ray:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 18

In an experiment to trace the path of a ray of light through a glass prism for different values of angle of incidence a student would find that the emergent ray bends at an angle to the direction of incident ray. This happens because the light ray gets refracted two times at different angles.

Jharkhand TGT (JSSC) Mock Test - 2 - Question 19

Which among the following statements is true about Huygen's principle?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 19

Huygen’s principle states that every point on the wavefront may be considered as a source of secondary spherical wavelets that spread out in the forward direction at the speed of light.

  • The new wavefront is the tangential surface of all these secondary wavelets.
  • Secondary sources start making their own wavelets, these waves are similar to that of the primary source
  • Huygens's principle states that each point on a wavefront is a source of wavelets, which spread forward with the same speed.
Jharkhand TGT (JSSC) Mock Test - 2 - Question 20

One proton enters in a magnetic field of of N / Amp- intensity with velocity in parallel of field. The force exerted on proton will be:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 20

N / Amp-

m/s

q = Charge on proton

Culomb

The proton enters parallel to the magnetic field.

Jharkhand TGT (JSSC) Mock Test - 2 - Question 21

The gravitational force between two masses kept at a certain distance is 'P' Newton. The same two masses are now kept in water and the distance between them are same. The gravitational force between these two masses in water is 'Q' Newton then:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 21

According to Newton's law of gravitation, the gravitational force between two masses and is proportional to the product of their masses and inversely proportional to the square of the distance between them.

In this way gravity does not depend on the medium

So,

Jharkhand TGT (JSSC) Mock Test - 2 - Question 22

At room temperature the substances which retain their ferromagnetic property for a long period of time are called?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 22
  • Substances, which at room temperature, retain their ferromagnetic property for a long period of time are called permanent magnets.
  • Ferromagnetic materials are those which get strongly magnetized when placed in an external magnetic field.
  • They get strongly attracted to a magnet.
  • Iron, Nickel, Cobalt, and transition metals are some of the examples of ferromagnetic materials.
Jharkhand TGT (JSSC) Mock Test - 2 - Question 23

The electric field at a point is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 23

The electric field at a point is discontinuous if there is a charge at that point.

The electric field due to any charge will be continuous, if there is no other charge in the medium. It will be discontinuous if there is a charge at the point under consideration.

Jharkhand TGT (JSSC) Mock Test - 2 - Question 24

A potentiometer wire of length 20 m has a resistance of 50 ohms. It is connected in series with a resistance box and a 5 V storage cell. If the potential gradient along the wire is 0.5 mV/cm, what is the resistance unplugged in the box?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 24
Potential gradient along the potentiometer wire,
.



So,


Therefore, the resistance unplugged in the box is 450 ohms.
Jharkhand TGT (JSSC) Mock Test - 2 - Question 25

In the p-n junction, the barrier voltage _________________.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 25

In the p-n junction, the barrier voltage decreases with increases in temperature.

The total charge formed at the p-n junction is called barrier voltage, barrier potential or junction barrier. The size of the barrier voltage at the p-n junction is depends on, the amount of doping, junction temperature and type of material used. It decreases with the increase in temperature.

The transfer of electrons from the n-side of the junction to holes annihilated on the p-side of the junction produces a barrier voltage. This is 0.6 to 0.7 V in silicon and varies with other semiconductors.

Jharkhand TGT (JSSC) Mock Test - 2 - Question 26

A piston-cylinder contains air at , and a volume of . A constant pressure process gives of work out. Find the final volume of the air.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 26

Jharkhand TGT (JSSC) Mock Test - 2 - Question 27
Four capacitors, and are connected as shown in figure below. Calculate equivalent capacitance of the circuit between points and .

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 27
The given arrangement of capacitors can be analysed as below. The combination of capacitors and of capacity and are in series, their combined capacity is given by:

The capacitor is in parallel with , their combined capacity is given by,

Now, and are in series.
Net capacity of the combination is given by,

Jharkhand TGT (JSSC) Mock Test - 2 - Question 28

Which of the following instrument measures arterial blood pressure?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 28

Sphygmomanometer measures arterial blood pressure.

A sphygmomanometer, also known as a blood pressure monitor or blood pressure gauge, is a device used to measure blood pressure, which releases the artery below the cuff in a controlled manner and in a controlled manner. Leaves mercury or aeroid manometer. Measure the pressure.

Jharkhand TGT (JSSC) Mock Test - 2 - Question 29

If we observe that the air is moving from place A to place B, then:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 29

If we observe that the air is moving from place A to place B, then Place A is at high pressure and place B is at low pressure.

From the mean sea level to the top of the atmosphere, the weight of a column of air contained in a unit area is called the atmospheric pressure. It is expressed in units of millibars and measured with the help of a mercury barometer or the aneroid barometer. The average atmospheric pressure at sea level is 1,013.2 millibars. The air near the surface is denser, due to gravity, and so has higher pressure. The pressure decreases with height and its variation is the primary cause of air motion, hence wind moves from high-pressure areas to low-pressure areas.

Hence, the correct option is (B)

Jharkhand TGT (JSSC) Mock Test - 2 - Question 30

In a transistor amplifier, which one of the following capacitors are generally used?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 2 - Question 30

In a transistor amplifier, electrolytic capacitor are generally used because it gives better DC stability.

Electrolytic capacitors are used to prevent interference of a transistor's bias voltage by AC signals. In most amplifier circuits, this is achieved by driving the signal to the base terminal of a transistor through a coupling capacitor. Aluminium electrolytic capacitors are widely used for coupling applications in power amplifiers.

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