Physics: CUET Mock Test - 4 - JEE MCQ

Given: Output power (P_o) = 100 W; I_p = 0.5 A; V_p = 250 V
Input power (P_i) = V_p × I_p
P_i = 250 × 0.5 = 125 W
Efficiency = (Output power/Input power) × 100%
Efficiency = (100/125) x 100
Efficiency = 80
Therefore, the efficiency of the transformer is 80%.

When the refractive index of the liquid is the same as the lens material, no light will be reflected by the lens and hence it will not be visible. So, the lens immersed in a transparent liquid will not be visible.

We know that, P_av = V_rms I_rms cos⁡Φ ……1 and cosΦ;= R/Z ....2
In LCR, cosΦ = V_R/V = IR/IZ and I_rms = V_rms/Z ....3
Substituting 2 and 3 in 1


Given: V_rms = 24 V; Resistance (R) = 70 Ω; Inductance (I) = 30 mH = 20 × 10^-3 H; Capacitance (C) = 300 μF
X_L = 2πυL = 2π (60)(30 × 10^-3)
X_L= 11.304 Ω
X_C = (1/(2πvC)) = (1/(2π(60)(300×10⁻⁶)))
X_C = 8.846 Ω
For series LCR circuit → Z = 

Z = 70.04 ≈ 70 Ω
So, the energy used in 1000 seconds is → P_avt =

Therefore, the energy dissipated in the circuit at 1000 seconds is 8.22 × 10³J.

Given: N_s = 20; N_p = 100; V_p = 120 V
For a step-down transformer → N_s/N_p = V_s/V_p

V_s = 24 V
Therefore, the voltage output from the secondary coil is 24 V.

The blue color of the ocean is due to the preferential scattering of light by water molecules. This is the underlying reason for many other phenomena such as why the sky is blue or the sky is reddish at the time of sunrise or sunset.

The portion of a refracting medium, whose curved surface forms the part of a sphere, is known as a spherical refracting surface. Spherical refracting surfaces are of two types, namely, convex refracting spherical surface and concave refracting spherical surface.

As the focal length is negative, the lens used is concave. When a person is prescribed a concave lens, then the person is considered to be suffering from myopic. Therefore, the boy is suffering from myopia.

μ = c/v_t
μ = 3 × (10⁸/2.4) x 10⁸
μ = 1.25

Fringe width is given as:
β = Dλ/d
As the screen is moved closer to the two slits, the distance denoted by D decreases and so fringe width β increase.

According to Bohr’s second postulate, an electron revolves around the nucleus of an atom in orbits and, therefore, the angular momentum of the revolution is an integral multiple of (h/2p), where h is the Planck’s constant. Thus, the expression for angular momentum is given as: L = nh/2p

Lens is a portion of a transparent refracting medium bound by two spherical surfaces or one spherical surface and the other plane surface. Lenses are divided into two classes, namely, convex lens or converging lens, and concave lens or diverging lens.

On doubling the diameter, the area of the objective increases four times. Its light-gathering power increases four times. The brightness of the image also increases four times. Therefore, the intensity of the image increases by four.

λ = 589 nm, μ = 1.33.
Wavelength in water, λ_W = λ/μ
λ_W = 589/1.33
λ_W = 443 nm.

Fringe width is expressed as:
β = Dλ/d.
Since the wavelength are related as shown → λ_G < λ_Y < λ_O, we can say that → β_G < β_Y < β_O.

de – Broglie hypothesis did modify Bohr’s second postulate. This postulate of Bohr regarding the quantization of the angular momentum of an electron was further explained by Louis de Broglie. According to de – Broglie, a moving electron in its circular orbit behaves like a particle-wave.

When the lens is thicker in the middle than at the edges, then it is convex lens. When the lens is thicker at the edges than in the middle, then it is concave lens. Therefore, X = convex lens, and Y = concave lens.

The fly will not be seen in the final image. But the intensity of the final image gets reduced because light will not enter the telescope from that part of the object where the fly is sitting. Therefore, a reduction effect will be observed in the final image of the fly.

When a light wave enters from the air medium into the glass medium, the energy of the wave decreases because a part of the light wave is reflected in the air. Therefore, in this way, energy will be affected.

Fringe width is given as:
β = Dλ/d.
So β = (1×500×10⁻⁹)/(10⁻³)
β = 0.5 mm
Therefore, the fringe separation when blue-green light of wavelength 500 nm is used is 0.5 mm.

A string is used to represent the behavior of a particle analogously to the waves traveling on it. Particle waves can lead to standing waves held under resonant conditions. When a stationary string is plucked, it causes several wavelengths to be excited. But, we know that only the ones which have nodes at the ends will survive.

The resistivities of semiconductors lie between 10-6 Ωm and 104 Ωm. A semiconductor material has an electrical conductivity value falling between that of a conductor, such as metallic aluminum, and an insulator, such as wood. Silicon and germanium are typical semiconductors.

In a telescope, the objective should have a large focal length than the eyepiece. So the lens of 50 cm focal length will be used as the objective. So, the smaller one of the two, i.e. the 5 cm focal length lens will be used as the eyepiece.

When a wave undergoes refraction, the phase change will be zero. The phase change is what guides the process. Therefore, no phase change occurs to a wave during refraction.

Resultant intensity at the point of destructive interference will be as follows:
I = I₀ + I₀ + 2√I₀ I₀ cos 180^o
I = 0
Therefore, the value of the resultant intensity at a point of destructive interference is zero.

Given: number of turns (n) = 200 turns per meter; current (I) = 2 A; Relative permeability (µ_r) = 200
Magnetic intensity is given by: H = nI = 200 × 2 = 400 A/m
μ_r = 1 + γ → γ is the magnetic susceptibility of the material
γ = μ_r – 1
Magnetization → M = γ x H
M = (μ_r – 1) × H
M = (200 – 1) × 200
M = 199 × 200
M = 79600
M = 7.96 × 10⁴ A/m ≈ 8 × 10⁴ A/m

The light-gathering power of the objective will increase and even faint objects will become visible. This is done by increasing the diameter of the objective. So, by increasing the diameter of the objective, the range of the telescope can be increased.

When a wave is reflected in the rarer medium from the surface of a denser medium, it undergoes a phase change of π radian. Therefore, when a wave undergoes reflection at a denser medium, the phase change will be π radian.

The positions of bright and dark fringes will change rapidly. Such rapid changes cannot be detected by our eyes. Uniform illumination is seen on the screen i.e., the interference pattern disappears.

de – Broglie concluded his modification of Bohr’s second postulate by stating that the wavelengths of matter waves can be quantized. This implies that the electrons can exist in those orbits which had a complete set of several wavelengths.

Thin lens formula is given as:
1/f = 1/v - 1/u
Where u = distance of the object from the optical center of the lens; v = distance of the image from the optical center of the lens; f = focal length of a lens. According to the thin lens formula, f is positive for converging or convex lens and f is negative for a diverging or concave lens.