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Jharkhand TGT (JSSC) Mock Test - 1 - 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 - 1

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

What is the sum of the squares of direction cosines of the line joining the points and ?

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

It is given that, the coordinates on which the line joins are (1, 2, -3) and (-2, 3, 1)
Let the direction cosines be l, m, n
Let  be two point which joins the line.

Now, we know that,

Putting the values according to the question, in equation (1), we get,

Putting the values according to the question, in equation (2), we get,

Putting the values according to the question, in equation (3), we get,

Jharkhand TGT (JSSC) Mock Test - 1 - Question 2

Consider the two curves then:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 2
Solving the two equations, we get,


and
So the two curves meet at two points (1,2) and (1,-2).
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Jharkhand TGT (JSSC) Mock Test - 1 - Question 3

The solution of the differential equation is:

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

The given equation is,

Integrating both side, we get,

We know that,

and

We also know that,

If then

Let

Jharkhand TGT (JSSC) Mock Test - 1 - Question 4

What is the image of the point (1, -2, 3) in the plane?

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

Equation of normal to plane and passing through (1, -2, 3)

Let,

∴ (x - 1) = 2r ⇒ x = 2r +1

And, y + 2 = 3r ⇒ y =3r - 2

Also, z - 3 = -r ⇒ z = -r + 3

i.e., (2r +1, 3r - 2, -r + 3)

Consider Q be the the image of given point

∴ Q = (2r +1, 3r - 2, -r + 3)

Let, R be the midpoint

∴ R =

⇒ R = (r + 1, - 2, + 3)

Since R lies on plane 2x + 3y – z = 7

2 (r + 1) + 3 ( - 2) - ( + 3) = 7

⇒ 2r + 2 + -6 + - 3 = 7

⇒ 4r + 4 + 9r -12 +r - 6 = 14

⇒ 14r - 14 = 14

⇒ r = 2

So, image Q = (2(2) + 1, 3(2) - 2, -2+3)

⇒ (5, 4, 1)

Jharkhand TGT (JSSC) Mock Test - 1 - Question 5

If lies in first quadrant, then the value of is:

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

Given:

As we know,

Now,

By Pythagoras theorem,

The value of ,

The required value is .

Jharkhand TGT (JSSC) Mock Test - 1 - Question 6
The tangents to the curve at the points are parallel to
Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 6
To find tangents to the curve at given points is parallel to which line we just need to find slope of the tangent, Now Slope of the curve


Slope
Now at

Tangent at this points will be parallel to
Jharkhand TGT (JSSC) Mock Test - 1 - Question 7

If and are in , then the straight line will always pass through a fixed point. The fixed point is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 7

Given are in .

So we can write

(i)

Given equation of straight line is (ii)

Comparing (i) and (ii)

So lies on the line.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 8

Cut-off Region of a Junction Field Effect Transistor is also known as:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 8

Cut-off Region of a Junction Field Effect Transistor is also known as pinch off region.

Cut-off Region is also known as the pinch-off region where the gate voltage, VGS is enough to cause the JFET to act as an open circuit as the channel resistance is at maximum. The Pinch-Off value of the JFET refers to the voltage applied between Drain and Source (with the Gate voltage at zero volts) at which maximum current flows. Operating with the Drain/Source voltage below this value is classed is the "Ohmic Region" as the JFET will act rather like a resistor.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 9

Which of the following part of the Carnot engine is a heat reservoir at lower temperature ?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 9

Sink is a heat reservoir at a lower temperature T1 from which the engine rejects heat. The sink has an infinite thermal heat capacity. In the sink, any amount of heat can be added to it without changing the temperature.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 10

The magnetic moment is a:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 10

The magnetic moment of a magnet is a quantity that determines the torque it will experience in an external magnetic field.

It is considered to be a vector having a magnitude and direction. The direction of the magnetic moment points from the South Pole to the North Pole of the magnet.

The magnetic field produced by the magnet is proportional to its magnetic moment.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 11

Study the given ray diagrams and select the correct statement from the following.


Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 11

In the first case, the ray of light is converging, when passing from lens , so it is a convex lens. In the second case, the ray of light is diverging, when getting reflected from the mirror , so it is a concave mirror.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 12

The surface of a spherical shell is uniformly charged. Then what is the electric field inside the spherical shell?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 12
From Gauss's law



Since the surface of the spherical shell is uniformly charged, so the charge inside a spherical shell is zero, the Gaussian Surface encloses no charge.
The Gauss's theorem gives
for
Jharkhand TGT (JSSC) Mock Test - 1 - Question 13

The rate of heat transfer in a conducting rod will increase by-

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 13

The rate of heat transfer in a conducting rod will increase by decreasing the length of the rod

As the heat transfer rate is directly proportional to cross-sectional area and temperature difference of the rod. So by decreasing the area and temperature difference, the heat transfer rate will also decrease. So option (A) and (B) are wrong.

As the heat transfer rate is inversely proportional to length of the rod. By decreasing the length of the rod, the heat transfer rate will increase.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 14

A transistor can be made to operate as a switch by operating it in which of the following regions?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 14

A transistor can be made to operate as a switch by operating it in saturation region and the cut-off region.

The transistor is a three-terminal device. If it were to be used as an amplifier or a switch, it is supposed to be a two-port device. For a two-port device, four terminals are required. One port is referred to as an input port where the signals are fed into the network and the other port is the output port, where the response of the network is available. When both the junctions are forward biased, the transistor is said to be in the saturation region. When both the junctions are reverse biased, the transistor is in cut off region.

If the circuit uses the Bipolar Transistor as a Switch, then the biasing of the transistor, either NPN or PNP is arranged to operate the transistor at both sides of the I-V characteristics curves. The areas of operation for a transistor switch are known as the Saturation Region and the Cut-off Region.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 15

Which among the following statements are true with respect to semiconductor breakdown?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 15

The Zener breakdown occurs in the junctions which are heavily doped and the avalanche breakdown occurs in the junctions, which are lightly doped.

The avalanche breakdown is a phenomenon in which there is an increase in the number of free electrons beyond the rated capacity of the diode; This results in the flow of heavy current through the diode in reverse biased condition. Avalanche breakdown occurs in lightly doped diode

Zener breakdown mainly occurs because of a high electric field; When the high electric field is applied across the PN junction diode, then the electrons start flowing across the PN-junction. Consequently, it develops little current in the reverse bias. The Zener breakdown occurs in heavily doped diodes.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 16

If the current in the wire is doubled then the heat produce will become:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 16

As we know,

and

When the current is ,

....(1)

When the current is doubled,

.....(2)

By equation (1) and equation (2),

So, if the current in the wire is doubled then the heat produce will become four times.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 17

Modern non-stick cookware and the flat end of an electric iron has a coating of a polymer. Identify the name of the polymer.

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 17

A non-stick surface can reduce the ability of other materials to stick on it. These surfaces are coated with a synthetic polymer called teflon. Non-stick cookware and the flat end of an electric iron are coated with teflon.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 18
The equivalent mass of metal is and vapour density of its chloride is . The atomic mass of metal is:
Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 18
Given ;
The equivalent weight of metal
Vapour density of metal chloride
Molecular weight of metal chloride.



Therefore, atomic weight of the metal equivalent weight valency

Jharkhand TGT (JSSC) Mock Test - 1 - Question 19

Li shows diagonal relationship with:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 19

A diagonal relationship is said to exist between certain pairs of diagonally adjacent elements in the second and third periods (first 20 elements) of the periodic table.

These pairs are (lithium (Li) and magnesium (Mg), beryllium (Be) and aluminium (Al), boron (B) and silicon (Si).

Magnesium and Lithium are almost the same sizes and electronegativity values are nearly the same.

Therefore, Magnesium and Lithiumthe showing similar properties to each other.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 20

Which of the following ore is concentrated using group 1 cyanide salt?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 20

Sphalerite ore : ZnS

Calamine ore : ZnCO3

Siderite ore : FeCO3

Malachite ore : Cu(OH)2.CuCO3

It is possible to separate two sulphide ores by adjusting proportion of oil to water or by using' depressants'. In case of an ore containing ZnS and PbS, the depressant used is NaCN.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 21

Which of the following fibres is used for making parachutes?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 21

Nylon threads are strong, elastic and lightweight. A nylon rope is actually stronger than a steel wire. Hence, it is preferred for making parachutes. Harness straps, suspension lines, tents, sleeping bags, sails, rope, tennis strings, fishing poles and lines, etc. are also made from nylon fibers because of their strength.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 22
Boiling point of water at is . How much sucrose is to be added to of water such that it boils at Molal elevation constant for water is .
Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 22
Here, elevation of boiling point

Mass of water,
Molar mass of sucrose

Molal elevation constant,
We know that:



(approximately)
so, of sucrose is to be added.
Jharkhand TGT (JSSC) Mock Test - 1 - Question 23

When a primary amine reacts with chloroform in alcoholic KOH. the product is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 23

When primary amine reacts with chloroform in ethanoic KOH, it follows carbylamine reaction and the product formed is an isocyanide. Isocyanide is also known as carbylamine. Isocyanide gives off a very offensive odor, so this reaction is also used as a test for primary amines.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 24

Consider the elements Mg, Al, S, P and Si, the correct increasing order of their first ionization enthalpy is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 24

In general from left to right in a period, ionistion enthalpy increases due to effective nuclear charge increases.

But due to extra stability of half filled and full filled electronic configuration, required ionisation enthalpy is more from neighbouring elements. i.e., first ionisation enthalpy order is:

Al < Mg < Si < S < P

Jharkhand TGT (JSSC) Mock Test - 1 - Question 25

Electronic configuration of Cr is:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 25

Electronic configuration of Chromium (Cr).

Cr- 1s2 2s2 2p63s23p64s13d5

Jharkhand TGT (JSSC) Mock Test - 1 - Question 26

The bond length between hybridized carbon atom and other carbon atom is minimum in:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 26

Since propyne has a triple bond, therefore it has minimum bond length.

Jharkhand TGT (JSSC) Mock Test - 1 - Question 27

The first order rate constant for the decomposition of ethyl iodide by the reaction

at is . Its energy of activation is . Calculate the rate constant of the reaction at .

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 27

We know that

Jharkhand TGT (JSSC) Mock Test - 1 - Question 28

If equal volumes of and solutions are used to oxidise in acidic medium, then will be oxidised by:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 28

The redox half cell reactions are written below:

Jharkhand TGT (JSSC) Mock Test - 1 - Question 29

The electrode potential of M / M of 3d-series elements shows positive value of:

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 29

Only copper shows positive value for electrodepotential of of 3d-series elements. The rest of the metals are good oxidizing agents with

Jharkhand TGT (JSSC) Mock Test - 1 - Question 30

Which of the following compounds will be suitable for Kjeldahl's method for nitrogen estimation?

Detailed Solution for Jharkhand TGT (JSSC) Mock Test - 1 - Question 30

The Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in rings (e.g. pyridine, quinoline, isoquinoline) as nitrogen of these compounds does not convert to ammonium sulphate under the conditions of this method.

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