JEE Main Practice Test- 17 - JEE MCQ

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

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. 

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.

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

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

Thus, the kinetic energy obtained is simply: qEy

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

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
 

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. 

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

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

Given , 

=> atomic mass=141

=>atomic number=56

So, neutrons=atomic mass-atomic number

=141-56 

=85

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

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.

No horizontal external force is acting
∴ a_cm = 0
since v_cm = 0
∴ Δx_cm = 0

Acceleration

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.

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.

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.

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

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).

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 

 In parallel combination  E_eq=E=6V