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Jharkhand TGT (JSSC) Mock Test - 3 - Jharkhand (JSSC) PRT/TGT MCQ


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30 Questions MCQ Test - Jharkhand TGT (JSSC) Mock Test - 3

Jharkhand TGT (JSSC) Mock Test - 3 for Jharkhand (JSSC) PRT/TGT 2024 is part of Jharkhand (JSSC) PRT/TGT preparation. The Jharkhand TGT (JSSC) Mock Test - 3 questions and answers have been prepared according to the Jharkhand (JSSC) PRT/TGT exam syllabus.The Jharkhand TGT (JSSC) Mock Test - 3 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 - 3 below.
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Jharkhand TGT (JSSC) Mock Test - 3 - Question 1

What is equal to?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 1

As we know,

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

If a line makes the angles with the axes, then the value of is equal to:

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

Given,

The value of is .

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

A person walking along a straight road observes that at two points 1 km apart, the angle of elevation of a pole in front of the person is 30° and 75°. The height of the pole is:

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

Let's first draw the diagram of the given information:

Given:

AB = 1 km, ∠ MAN = 30°, ∠ MBN = 75°

In Δ MAN, tan 30° =

⇒ MN = AM tan 30° ...(1)

And in Δ MBN, tan 75° =

⇒ MN = BM tan 75° ...(2)

Using (1) and (2), AM tan 30° = BM tan 75°

And, MN = BM tan 75°

km = 250 m.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 4
What is the value of
Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 4

We know that .

Let's say that and

and

The given expression can be written as:

Jharkhand TGT (JSSC) Mock Test - 3 - Question 5

A cylindrical vessel of a radius of cm contains water. A solid sphere of radius cm is dipped into the water until it is completely immersed. The water level in the vessel will rise by?

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

The volume of cylinder The volume of the sphere

Jharkhand TGT (JSSC) Mock Test - 3 - Question 6

What is the LCM of and ?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 6

Let

and

LCM of

Jharkhand TGT (JSSC) Mock Test - 3 - Question 7

On increasing the length of wire by four times and increasing the radius of wire by 'X' times the resistance of wire remains unchanged then calculate the value of 'X'.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 7

Given that the resistance remains the same,

Let us assume that initially, the resistance of wire be ,

If the length of wire is increased 4 times and the radius is increased times then,

It is given that resistance remains unchanged,

Radius becomes 2 times.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 8

In a Wien-bridge oscillator, if the resistances in the positive feedback circuit are decreased, the frequency:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 8

In a Wien-bridge oscillator, if the resistances in the positive feedback circuit are decreased, the frequency increases.

The Wien Bridge oscillator is a two-stage RC coupled amplifier circuit that has good stability at its resonant frequency, low distortion. The Wien Bridge Oscillator uses a feedback circuit consisting of a series RC circuit connected with a parallel RC of the same component values producing a phase delay or phase advance circuit depending upon the frequency. If if the resistances in the positive feedback circuit are decreased, the frequency increases.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 9

Stefan Boltzmann law is applicable for heat transfer by __________.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 9

Stefan-Boltzmann law is applicable for radiation it states that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature. The law applies only to black bodies, theoretical surfaces that absorb all incident heat radiation.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 10

________ is the thermodynamics process in which no heat flows between the system and surrounding.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 10

Adiabatic process is the thermodynamics process in which no heat flows between the system and surrounding.

The adiabatic process comes under the thermodynamics process in which there is no exchange of heat.

We can say that ΔQ = 0 (ΔQ is heat exchange between the system and the surroundings)

Jharkhand TGT (JSSC) Mock Test - 3 - Question 11

A convex lens of focal length cm and a concave lens of focal length cm are placed in contact with each other. Then the equivalent focal length of the combination will be:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 11

Given:

cm (convex lens)

cm (concave lens)

We know that:

The focal length of the combination:

cm

Jharkhand TGT (JSSC) Mock Test - 3 - Question 12

Identical cells are joined in series with two cells and with reversed polarities. e.m.f. of each cell is and the internal resistance is . Potential difference across the cell and is :

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 12

Since cells A and B are connected in opposite polarities, they cancel e.m.f. of cells of opposite polarity.

The cells are connected in series, hence their e.m.f. adds up.

Now, let us find, the net e.m.f. in the circuit:

Net e.m.f.

Where, corresponds to emf of and having opposite polarity.

Now, let us calculate the current flowing through the circuit.

Since cells are connected each having internal resistance, total resistance in the circuit, is:

Total internal resistances

Now, applying ohm’s law, we know:

Potential difference across two ends of a conductor is directly proportional to the current through the conductor, resistance being the constant of proportionality.

Where,

Potential Difference

Current

Resistance

Therefore,

Net e.m.f.

Putting the values, we get

Where, is the current flowing through the circuit.

Now, potential difference across and =

Let Potential difference across and

So,

Putting the value of , we get

Thus, on solving the equation, we get

So, finally we arrive at,

Jharkhand TGT (JSSC) Mock Test - 3 - Question 13

Which of the following statements about electric field lines associated with electric charges is false?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 13

Electric field lines form closed loops this statement about electric field lines associated with electric charges is false.

The imaginary lines which are used to represent the electric field are called electric field lines. The field lines emerge from a positive charge and terminate at a negative charge. They originate and end at right angles to the surface of the charge. Electric field lines do not make a loop. The magnitude of the electric field will be maximum where the number of field lines is maximum.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 14

A short bar magnet placed with its axis at with an external field of G experiences a torque of Nm. What is the magnetic moment of the magnet ?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 14

Given,

Magnetic field, G

Torque, Nm

As we know,

Magnetic moment of the magnet

Jharkhand TGT (JSSC) Mock Test - 3 - Question 15

Kirchoff's current law is based on conservation of ______ while Kirchoff's voltage law is based on conservation of _________.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 15

Kirchhoff's current law is based on the law of conservation of charge while Kirchhoff's voltage law is based on the law of conservation of energy.

Kirchhoff’s Current Law states that “The algebraic sum of all currents entering and exiting a node must equal zero.” This law describes how a charge enters and leaves a wire junction point or node on a wire.

Kirchhoff’s Voltage Law states that the sum of voltages around a loop is zero.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 16

The ratio of magnetic field and magnetic moment at the centre of a current carrying circular loop is X. When both the current and radius is doubled the ratio will be

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 16
Given:
The ratio of the magnetic field and magnetic moment at the center of a current-carrying circular loop = X
We have to find the ratio when both the current and radius is doubled.
Magnetic field at the centre of current carrying loop is given by
....(i)
where: =current
r=radius
Let M be the magnetic moment of the current carrying loop, so
.....(ii)
Dividing eq. (i) by eq. (ii), we get


Thus, the ratio is independent of current and inversely proportional to the cube of radius. When the radius is doubled i.e. then the new ratio is


Jharkhand TGT (JSSC) Mock Test - 3 - Question 17

A positive charge q is placed at the center of a spherical conducting shell. What is the electric field inside the shell?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 17

According to Gauss's law, the electric field inside a conductor is always zero. Therefore, the electric field inside the spherical conducting shell in this case will also be zero.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 18

The cooking method by which Idli is cooked is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 18

Cooking with steam is referred to as steaming. Steaming is the cooking method by which Idli is cooked. This is frequently done with a food steamer, which is a kitchen equipment designed expressly to cook food with steam.

Using steam to cook a variety of items is seen to be a healthy cooking method. Because steaming requires less energy to produce and has superior thermodynamic heat transfer.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 19

Which of the following are plasticisers ?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 19

Plasticizers (UK: plasticisers) or dispersants are additives that increase the plasticity or fluidity of a material. The dominant applications are for plastics, especially polyvinyl chloride (PVC).

The properties of other materials are also improved when blended with plasticizers including concrete, clays and related products.

DOP (Dioctyl pthalate) used in flooring materials, carpets, notebook covers and high explosives, such as Semtex.

DBP (Di-n-butyl pthalate) used for cellulose plastics, food wraps, adhesives, perfumes and cosmetics.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 20

The electrons, identified by quantum numbers and (i) , (ii) (iii) and (iv) can be placed in order of increasing energy, from the lowest to highest, as:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 20

The greater is the value of , the greater is the energy of orbitals.
(i) orbital
(ii) s orbital
(iii) orbital
(iv) orbital
According to the Aufbau principle, energies of above-mentioned orbitals are in the order of-
The increasing order of energy (iv)(ii)(iii)(i)

Jharkhand TGT (JSSC) Mock Test - 3 - Question 21

Direction: Choose the correct answer among the alternatives given.

One mole of ethyl acetate on treatment with an excess of in dry ether and subsequent acidification produces.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 21

Ethyl Acetate on treatment with gives 2 moles of Ethyl Alcohol.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 22

The equilibrium constant of the reaction:

at is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 22

For a cell reaction in equilibrium at ,

cell

where KC = equilibrium constant, n = number of electrons involved in the electrochemical cell reaction.

Given,

or,

or,

Jharkhand TGT (JSSC) Mock Test - 3 - Question 23

Rutherford's α-particle experiment showed that the atoms have:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 23

In Rutherford's alpha particle experiment, the scattered alpha particle has large deflection angles. This experiment showed that the positive matter of atom was concentrated in a very small volume and it gave the idea of nucleus of an atom.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 24

Which of the following is a disaccharide.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 24

Trehalose, is a disachharides that is made up of -D-glucopyranosyl-( - -D-glucopyranoside units. While inulin is polysaccharides, raffinose is trisaccharide and cellulose is polysaccharides.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 25

Among the following, correct statement is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 25

Brownian movement is the random movement of particles in a fluid due to collisions with molecules of the surrounding medium. It is more pronounced for smaller particles than for bigger particles. This is because smaller particles are subject to more collisions with the surrounding molecules, which leads to greater movement.
Sols of metal sulphides are lyophilic, which means they have a strong affinity for the solvent molecules. This results in stabilization of the sol and prevents coagulation.
Hardy Schulze law states that the greater the size of the ions, the greater is their coagulating power. This is because larger ions have a greater charge density, which makes them more effective at neutralizing the charge on the particles in the sol, leading to coagulation.
Charcoal is a highly porous material that is capable of adsorbing a wide range of substances. However, the extent of adsorption depends on the nature of the substance being adsorbed. Chlorine is a highly reactive element and is not easily adsorbed by charcoal. On the other hand, hydrogen sulphide is a relatively weakly reactive gas and can be adsorbed by charcoal.
Therefore, the correct statement among the given options is that Brownian movement is more pronounced for smaller particles than for bigger particles.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 26

Three complexes,

absorb light in the visible region. The correct order of the wavelength of light absorbed by them is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 26
A complex having a strong field ligand has the tendency to absorb light of the highest energy. Among the three complexes.
has the weakest field and therefore will absorb light of least energy and highest wavelength.
has a field weaker than the above compound and therefore absorb radiation of lesser energy and more wavelength.
will absorb radiation of highest energy and least wavelength.
Strength of ligand .
Jharkhand TGT (JSSC) Mock Test - 3 - Question 27

Carbon tetrachloride has zero dipole moment because of ________.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 27

Carbon tetrachloride molecule has zero dipole moment even though C and Cl have different electronegativities and each of the C - Cl bond is polar and has some dipole moment. This is because the individual dipole moments cancel out because of the symmetrical tetrahedral shape of the molecule.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 28

The ion which is not tetrahedral in shape is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 28

The ion contains 4 bond pairs of electrons and 2 lone pairs of electrons. The central iodine atom undergoes hybridization. The expected geometry is octahedral and the actual geometry is square planar. and are tetrahedral in shape.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 29

In the Mendeleev periodic table, gaps were left for the undiscovered elements. Which of the following elements found a place in the periodic table later?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 29

Germanium (Ge) was discovered later which fit into the empty spaces left by Mendeleev PT and matched to the most properties

Clemens Winkler discovers Germanium.

Germanium belongs to Group 14 and Period 4.

It is a P-Block element, and the atomic number is 32.

Jharkhand TGT (JSSC) Mock Test - 3 - Question 30

Chitin consists of:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 3 - Question 30

Chitin consists of N-acetyl glucosamine. Chitin is the second most abundant biodegradable polymer produced in nature after cellulose. It is an acetylated polysaccharide composed of N-acetyl-d-glucosamine groups linked by β (1→4) linkages.

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