UP PGT Physics Mock Test - 5 - UPTET MCQ

Lines in the spectrum were due to transitions in which an electron moved from a higher-energy orbit with a larger radius to a lower-energy orbit with smaller radius. The orbit closest to the nucleus represented the ground state of the atom and was most stable; orbits farther away were higher-energy excited states.

The forces acting on an electron in uniform electric and magnetic fields pointing in the same direction are explained below:

• Magnetic Force: Since the electron is moving parallel to the fields, the magnetic force (v × B) is zero.
• Electric Force: The electric field exerts a force (F_e = -E) on the electron, acting opposite to its direction of motion.

As a result:

• The electron experiences a backward force due to the electric field.
• This force causes the electron to decelerate, reducing its velocity.

While in the pike position the body decreases in radius as each segment moves closer to the axis of rotation, resulting in angular velocity increasing and a decrease of moment of inertia. Thus, during a dive, angular momentum is constant meaning that moment of inertia is inversely proportional to angular velocity.

For the equilibrium,  m_Bg = μm_Ag

Given X is greater than 2Ω when the bridge is balanced

When the resistances are interchanged the jockey shifts 20 cm. Therefore

Electric Field Intensity E = F\q

Concept:

• As per Indian standard for domestic power supply we use A.C current, which is also known as alternating current since they change its polarity at a certain interval of time.
• Now the two principal properties of the A.C electric power supply are Voltage and frequency.
• And we use 220 V and 50 Hz power supply for our common domestic use
• Where 220 V is the potential difference and 50 Hz is its frequency.

Explanation: 

The voltage and frequency of domestic power supply used in India are 220 volt and 50 Hz respectively

The chemical behavior of atoms is determined by the number and the configuration of the electrons in the atom. The number of protons in the nucleus determines the number of electrons in the atom.

Here, are perpendicular to each other and the velocity does not change; therefore

Also,

According to the question, there can be two cases in which the point mass is at a distance a from the surface of the spherical shell. 

1. The point mass is at the center of the shell

2. The point mass is outside the shell at a distance 2a from  the center of the shell.

Superposition principle dictates that the potential and field at any point will be the sum of the potentials and fields respectively at that point due to the individual masses.

It can be seen that in both the cases, neither field nor potential is zero inside the shell. This is due to the contribution of the point mass.

We know that,
 ω=(1/2)CV²
After putting the values,
=(1/2)x9.2x22.5x22.5
=2328.75J
Hence option B is the answer.

g' = acceleration due to gravity at latitude
g = acceleration due to gravity at poles
ω = angular velocity of earth
(θ) = latitude angle
Now,
g' = g - Rω² cos²(θ)
As, g' = 0 (weightless):
g = Rω² cos²(θ)
g = Rω² cos²(30)
g = Rω² (3/4)
ω² = 4g/3R
ω = √(4g/3R)

Acceleration 

e = NBAω = NBA²πn = 8 x 0.5 x 0.09 x 2 x 3.14 x 60 = 136V

Davisson and Germer Experiment, for the first time, proved the wave nature of electrons and verified the de Broglie equation. de Broglie argued the dual nature of matter back in 1924, but it was only later that Davisson and Germer experiment verified the results. The results established the first experimental proof of quantum mechanics. In this experiment, we will study the scattering of electrons by a Ni crystal.

Given    P=50W, V_rms​=220V,
by     P=V_rms^2​​/R
or      R=V_rms²/P​​= 220²/50=968Ω
The peak voltage of the source is,
E_rms=E₀/√2
=>E₀=Erms x √2=220√2=311.13V
Now, 
I_rms​=V_rms​​/R
=220/968​
=0.227A
Hence option B is correct.

Force due to electric field will make the charged particle released from rest to move in the straight line (that of electric field). Since the force due to magnetic field is zero, therefore, the charged particle will move in a straight line.

Unit of velocity = m/s
1m = 10⁶  Micro meter
And 1 sec = 10⁶  Micro sec
Thus 1 m/s = 10⁶ 10⁶  micro meter /micro second

X-rays are absorbed by the atmosphere and therefore the source of X-rays must lie outside the atmosphere to carry out X-ray astronomy and therefore satellites orbiting the earth are necessary but radio waves and visible light can penetrate through the atmosphere and therefore optical and radio telescopes can be built on the ground.

Solution:

To calculate the work done in moving a charge in an electric field, we use the formula:

• Work = Charge × Voltage Difference

Here, we have:

• Charge = -2 x 10^-9 C
• Initial position P = (0, 3 cm, 0)
• Final position Q = (0, 4 cm, 0)

The voltage (potential) at a distance r from a point charge is given by:

• V = k × Q / r
• Where k = 9 x 10⁹ Nm²/C² and Q = 6 mC = 6 x 10^-3 C

Calculating the voltage at points P and Q:

• At P (0, 3 cm): r = 0.03 m
• V_P = (9 x 10⁹) × (6 x 10^-3) / 0.03 = 1.8 x 10⁶ V
• At Q (0, 4 cm): r = 0.04 m
• V_Q = (9 x 10⁹) × (6 x 10^-3) / 0.04 = 1.35 x 10⁶ V

Now, the voltage difference:

• ΔV = V_Q - V_P = 1.35 x 10⁶ - 1.8 x 10⁶ = -0.45 x 10⁶ V

Finally, calculating the work done:

• Work = (-2 x 10^-9) × (-0.45 x 10⁶) = 0.0009 J = 0.9 J

Thus, the work done is 0.9 J.

The magnitude of induced emf is independent of the resistance of the coil.
∴e=−dϕ/dt=−N (d/dt) [(BA)/dt]
Thus, e depends upon N, B and A.
But induced current depends upon R.

A freely falling body observes a constant downwards force of gravity which hence applies a constant downward acceleration upon the body.

Horizontal component is a component of Earth’s magnetic field along the horizontal direction in the magnetic meridian. It is denoted by B_H. If B is the intensity of Earth’s total magnetic field, then the horizontal component of Earth’s magnetic field is given by:
B_H=Bcosδ

When we use white light, the central bright will have light from all wavelengths as none of them cancel out. Hence the central bright fringe appears white.
For other bright fringes, depending on the wavelength of light constructive interference will not take place for certain wavelengths. Hence they will not be white, rather will be coloured, hence differentiated from central fringe.

The earth's magnetic field is Be​ and its horizontal and vertical components are H_e​ and H_v​
cosθ= H_e​​/B_e​
∴cos60^o= (​0.26×10^−4​/ B_e )T
⇒B_e​=(​0.26×10⁻⁴)/ (½)​=0.52×10⁻⁴T=0.52G

NOTE : Electr o mag n et sh ould be amen abl e to magnetisation &  demagnetization.

∴ retentivity should be low & coercivity should be low

∴ T ∝ √L , as time period decreases when the length of pendulum decreases, the time period of shorter pendulum (T_s) is smaller than that of longer pendulum (T_l). That means shorter pendulum performs more oscillations in a given time.

  It is given that after n oscillations of shorter pendulum, both are again in phase. So, by this time longer pendulum must have made (n – 1) oscillations. 

So, the two pendulum shall be in the same phase for the first time when the shorter pendulum has 
completed 5/4  oscillation  

When rubber ball completely filled with water its centre of gravity will be at its centre, as water will fall through hole its COG will shift towards lower side leading to increase in length of pendulum and thus T, when very small amount of water will be left in rubber ball its COG will again shifts upward causing decrease in length and thus T, and finally when rubber ball becomes empty its COG will be at its centre and T will remains same as earlier.