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JEE Main Practice Test- 17 - JEE MCQ


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30 Questions MCQ Test - JEE Main Practice Test- 17

JEE Main Practice Test- 17 for JEE 2024 is part of JEE preparation. The JEE Main Practice Test- 17 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Practice Test- 17 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Practice Test- 17 below.
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JEE Main Practice Test- 17 - Question 1

In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to

Detailed Solution for JEE Main Practice Test- 17 - Question 1

JEE Main Practice Test- 17 - Question 2

A transformer is employed to reduce 220 V to 11 V. The primary draws a current of 5 A and the secondary 90 A. The efficiency of the transformer is

Detailed Solution for JEE Main Practice Test- 17 - Question 2

Primaray  power = 220 *5 = 1100 w 
Secondary power = 11*90 = 990 w
So , efficiency = 990w/1100w = 90 %

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JEE Main Practice Test- 17 - Question 3

Moderator used in nuclear reactor is

Detailed Solution for JEE Main Practice Test- 17 - Question 3

Heavy water is used in certain types of nuclear reactors, where it acts as a neutron moderator  to slow down neutrons so that they are more likely to react with the fissile uranium-235 than with uranium-238, which captures neutrons without fissioning. 

JEE Main Practice Test- 17 - Question 4

The unit of intensity of electric field is

Detailed Solution for JEE Main Practice Test- 17 - Question 4

The SI unit of electric field strength is newtons per coulomb (N/C) or volts per meter (V/m). The force experienced by a very small test charge q placed in a field E in a vacuum is given by E = F/q, where F is the force experienced.

JEE Main Practice Test- 17 - Question 5

In a hydrogen atom, when an electron jumps from second orbit to first orbit, the wavelength of spectral line emitted by hydrogen atom is

JEE Main Practice Test- 17 - Question 6

“If a particle of mass m is projected from the lowest point of smooth vertical circle of radius 'A' with velocity √(5ag) the force at its lowest point will be”

Detailed Solution for JEE Main Practice Test- 17 - Question 6

At lowest point centrifugal force is equal to mv2/a in downward direction

=> total downward force = mg + mv2/a
= mg + m 5ag/a
= mg + 5 mg
= 6mg

JEE Main Practice Test- 17 - Question 7

A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is

Detailed Solution for JEE Main Practice Test- 17 - Question 7

By conservation of energy, the increase in kinetic energy will be the decrease in potential energy.

Thus, the kinetic energy obtained is simply: qEy

JEE Main Practice Test- 17 - Question 8

An electron having charge e and mass m is moving in a uniform electric field E. Its acceleration will be

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JEE Main Practice Test- 17 - Question 9

25 N force is required to raise 75 kg mass from a pulley. If rope is pulled 12 m, then the load is lifted to 3 m, the efficiency of pulley system will be

Detailed Solution for JEE Main Practice Test- 17 - Question 9

Efficiency = output/input = (75x3) / (25x12)
                                       =75/100 =75%

JEE Main Practice Test- 17 - Question 10

Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the magnetic moments is

Detailed Solution for JEE Main Practice Test- 17 - Question 10


Let magnetic moment of square wire be μ1  = I1A1
Let magnetic moment of circle wire be μ2= I2A2 
If the wires carry same current then the ratio of their magnetic moments will be
 

JEE Main Practice Test- 17 - Question 11

In a capillary tube, water rises by 1.2 mm. The height of water that will rise in another capillary tube having half the radius of the first, is

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JEE Main Practice Test- 17 - Question 12

Three forces start acting simultaneously on a particle moving with velocity v → . These forces are represented in magnitude and direction by the three sides of a triangle ABC (as shown). The particle will now move with velocity

Detailed Solution for JEE Main Practice Test- 17 - Question 12


According to triangle law of vector addition, if three vectors are represented by three sides of a triangle taken in same order, then their resultant will be zero. Here, three forces acting on the particle are represented by the three sides of a triangle taken in one order, thus the resultant is zero. So, in presence of zero external force, the particle's velocity remains unchanged. 

JEE Main Practice Test- 17 - Question 13

Four bodies P, Q, R and S are projected with equal velocities having angles of projection 15º, 30º, 45º and 60º with the horizontal respectively. The body having shortest range is

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When the angle of projection is very far from 45° then range will be minimum.

JEE Main Practice Test- 17 - Question 14

 An electric current of 5 A is same as

Detailed Solution for JEE Main Practice Test- 17 - Question 14

Hence 5A of current is same as 5 C/sec

JEE Main Practice Test- 17 - Question 15

The nucleus of 56Ba141 contains ---- neutrons

Detailed Solution for JEE Main Practice Test- 17 - Question 15

Given , 

=> atomic mass=141

=>atomic number=56

So, neutrons=atomic mass-atomic number

=141-56 

=85

JEE Main Practice Test- 17 - Question 16

Assertion : The work done during a round trip is always zero.
Reason : No force is required to move a body in its round trip.

Detailed Solution for JEE Main Practice Test- 17 - Question 16

In a round trip work done is zero only when the force is conservative in nature.  Force is always required to move a body in a conservative or non-conservative field

JEE Main Practice Test- 17 - Question 17

In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer-
Assertion(A): Study of Fraunhofer lines in the spectrum of sun led to the discovery of hydrogen.
Reason(R):Fraunhofer lines are dark lines in the continuous solar spectrum.

Detailed Solution for JEE Main Practice Test- 17 - Question 17

In 1868 Lockyer observed a line among the Fraunhofer lines which could not be traced in the spectrum of any element on earth. So it was assumed that there was some new element in the sun's chromsphere. It was named helium. In 1895 Ramsay found this gas on earth. Thus, the study of Fraunhofer lines led to the discovery of new elements.

JEE Main Practice Test- 17 - Question 18

A body A of mass M while falling vertically downwards under gravity breaks into two parts, a part B of mass M/3 and body C of mass 2M/3. The centre of mass of the bodies B and C taken together

Detailed Solution for JEE Main Practice Test- 17 - Question 18

No horizontal external force is acting
∴ acm = 0
since vcm = 0
∴ Δxcm = 0

JEE Main Practice Test- 17 - Question 19

A Carnot engine used first an ideal monoatomic gas and then an ideal diatomic gas. If the source and sink temperature are 411ºC and 69ºC respectively and the engine extracts 1000 J of heat in each cycle, then area enclosed by the P-V diagram is

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JEE Main Practice Test- 17 - Question 20

The Martians use force (F), acceleration (A) and time (T) as their fundamental physical quantities. The dimensions of length on Martians system are

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Acceleration

JEE Main Practice Test- 17 - Question 21

Wave optics is based on wave theory of light put forward by Huygen and modified later by

Detailed Solution for JEE Main Practice Test- 17 - Question 21

The Huygens–Fresnel principle (named after Dutch physicist Christiaan Huygens and French physicist Augustin-Jean Fresnel) is a method of analysis applied to problems of wave propagation both in the far-field limit and in near-field diffraction.

 

It states that every point on a wavefront is itself the source of spherical wavelets. The sum of these spherical wavelets forms the wavefront.

JEE Main Practice Test- 17 - Question 22

Doppler's effects in sound and light are

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An apparent change in the frequency of waves, as of sound or light, occurring when the source and observer are in motion relative to each other, with the frequency increasing when the source and observer approach each other anddecreasing when they move apart.

JEE Main Practice Test- 17 - Question 23

The ray of light

Detailed Solution for JEE Main Practice Test- 17 - Question 23

A perpendicular drawn to the surface of a wavefront at any point in the direction of propagation of light is called as wave normal. The direction of propagation of a wave front is given by a ray of light. Therefore, a wave normal represents a ray of light.

JEE Main Practice Test- 17 - Question 24

The phase difference between two points separated by 0.8 m in a wave of frequency 120 Hz is 90º. Then the velocity of wave will be

Detailed Solution for JEE Main Practice Test- 17 - Question 24

Here, path difference Δx=0.8m Frequency n =120 Hz Phase difference is ϕ=0.5π The phase difference is given by  

 Hence, the wave velocity is given by v = nλ (where λ is the wavelength of wave) =120 × 2.3 = 384m/s

JEE Main Practice Test- 17 - Question 25

Which of the following is not a statement?

JEE Main Practice Test- 17 - Question 26

When the electron in a hydrogen atom of mass M undergoes transition from an orbit of higher quantum number n2 to an orbit of lower quantum number n1, the recoil velocity acquired by the atom is (Rydberg's constant = R, Planck's constant = h)

Detailed Solution for JEE Main Practice Test- 17 - Question 26

The wave number of the photon emitted because of the electron transition is

ν' = 1/λ = R(1/(n1)2 – 1/(n2)2) where λ is the wave length of the photon and R is Rydberg’s constant.

The momentum of the photon is p = h/λ = hR(1/(n1)2 – 1/(n2)2) where h is Planck’s constant.

When the photon is emitted with this momentum, the atom recoils (like a gun firing a bullet) with an equal and opposite momentum. Therefore, the recoil velocity of the atom is given by

v = p/M = (hR/M)(1/(n1)2 – 1/(n2)2).

JEE Main Practice Test- 17 - Question 27

A 5.0 μ F capacitor is charged to a potential difference of 800 V and discharged through a conductor. The energy given to the conductor during the discharge is

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JEE Main Practice Test- 17 - Question 28

A stone of mass 1 kg tied to a light inextensible string of length L = 10/3 m is whirling in a circular path of radius L in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if g is taken to be 10 m/sec2, the speed of the stone at the highest point of the circle is

Detailed Solution for JEE Main Practice Test- 17 - Question 28

The ratio of maximum (T2) to the minimum tension is

Now, the difference between the two tensions should be 6 mg.

Hence, we have

Now, the tension at the top of the circle is 


JEE Main Practice Test- 17 - Question 29

If six identical cells each having an e.m.f of 6V are connected in parallel, the E.M.F of the combination is

Detailed Solution for JEE Main Practice Test- 17 - Question 29

 

 In parallel combination  Eeq=E=6V

JEE Main Practice Test- 17 - Question 30

There are 20 divisions in 4 cm of the main scale. The vernier scale has 10 divisions. The least count of the instrument is

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