UP PGT Physics Mock Test - 5 - UPTET MCQ

Explanation:

for monoatomic  gas C_V = 1.5R, C_P =  2.5R

At const volume, 

Q = 500 J

Q=nCVΔT

500=1×1.5×8.31(T₁−300)

T₁=340K

At const pressure Q = 500 J

Q=nCPΔT

500=1×2.5×8.31(340−T₂)

T₂=316K

In balanced condition torque of the both sides must be equal so Fa × Da = Fb × Db
50 × 20 = Fb × 40
Fb = 25 N

M=k√L₁L₂

here k is cofficient of coupling. Its maximum value is 1 for tight coupling.

• The no load current is about 2-5% of the full load current and it accounts for the losses in a transformer.
• These no-load losses include core(iron/fixed) losses, which contains eddy current losses & hysteresis losses and the copper (I²R) losses due to the no Load current.

We know that the continuity theorem says that if the cross sectional area of the water flow decreases, the speed must increase to maintain the volume of water flown. And according to Bernoulli's principle if the speed of water flow increases , then the pressure must decrease.

λ=hc/E​=1240evnm /15×103 ​=0.083 nm
Since the order of wavelength is between 10 nm and .001 nm, it belongs to X-ray. part of the spectrum.

Let’s keep this simple and to the point. We know that:
1.in a series circuit the same current flows through each component
2.the voltage across an ideal inductor L is 90˚ ahead of an AC current through it
3.the voltage across an ideal capacitor C is 90˚ behind an AC current through it
So putting these facts together we can conclude that given an AC series current the voltages across any L and C must have a phase difference of 180˚

Using Hooke ‘s Law we get

Stress directly proportional to stress = Load/Area=F/pie*r*r

And rp:rq=2:1

When both the wires are under the same stress,strain produced will be the same.
 

When both the wires are under the same stress,strain produced will be the same.

2.when both the wires are loaded by same weight then

Strain p/strain q=(rq)²/(rp)²=¼

The newton second (also newton-second, symbol N s or N.s) is the derived SI unit of impulse. It is dimensionally equivalent to the momentum unit kilogram metre per second (kg.m/s). One newton second corresponds to a one-newton force applied for one second.

Dipole is the combination of two equal but opposite charges, kept at a certain distance. If it is placed in a uniform electric field, both the charges will suffer the same but opposite forces on them. As a result, the net force on the dipole becomes zero, but due to equal and opposite forces acting on two different points, there is a net torque acting on the dipole.

Electric Field Intensity E = F\q

Magnetic field due to current carrying loop = Magnetic field due to straight wires

B = B₁cos θ + B₂cos θ = 2 B₁cos θ

The current is from P to Q in wire 1 and from R to S in wire 2.

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 - Rω^2 cos^2(θ)
As, g' = 0 (weightless):
g = Rω^2 cos^2(θ)
g = Rω^2 cos^2(30)
g = Rω^2 (3/4)
ω^2 = 4g/3R
ω = √(4g/3R)

Acceleration 

The rate of loss of heat by a body is directly proportional to its excess temperature over that of the surroundings provided that this excess is small.

Natural Frequency of an LC Oscillator

- The natural frequency of an LC oscillator is determined by the values of the inductance (L) and capacitance (C) in the circuit.

- The LC oscillator is a type of oscillator circuit that uses a combination of inductors and capacitors to generate oscillations at a specific frequency.

- The natural frequency of an LC oscillator can be calculated using the formula:

f = 1 / (2 * π * √(LC))
 

Time taken by first drop to cover 5 cm (u = 0), is: 
h = 1/2 gt²
5 = 1/2 x 10 x t²
t = 1 sec
Hence, the interval is 0.5 sec for each drop.
Now distance fallen by second drop in 0.5 sec
h₁ = 1/2 gt²
= 1/2 x 10 x (0.5)²         
= 5 x 0.25 
= 1.25 m
Height above the ground ( of 2^nd drop) = 5 - 1.25 = 3.75 m

Explanation for the Gravitational Force Between the Earth and the Moon

• According to Newton's Third Law of Motion, every action has an equal and opposite reaction. This law applies to the gravitational forces between two bodies as well.
• The gravitational force with which the Earth attracts the Moon is equal to the gravitational force with which the Moon attracts the Earth. This is because the forces are action and reaction pairs, as described by Newton's Third Law.
• Therefore, if the Earth attracts the Moon with a gravitational force of 10²⁰ N, then the Moon attracts the Earth with the same gravitational force of 10²⁰ N.

Hence, the correct answer is b. 10²⁰ N.

q=6x10^-10 c
Q=-2x10^-9c
r₁=3x10^-2m
r₂=4x10^-2m
△v_Q=w/q
(1/4πε_o)-2x10^-9/10^-2[(1/4)-(1/3)]=W/6x10^-3
9x10⁹x2x10^-9x6x10^-3/12x10^-2=W
W=9x10^-3x10²
W=0.9J

f​₁=7.5×10⁶ Hz
f₂​=12×10⁶ Hz
λ_1​=c/f₁​=40 m
λ₂​=c/f₂​=25 m
 
So, the range is 40m to 25m

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

T_Initial  = t K
Work, W = 9R
Ratio of specific heats, γ = C_p / C_v = 4/3
In an adiabatic process, we have
W = R(T_Final – T_initial) / (1-γ)
9R = R (T_Final – t) / (1 – 4/3)
T_Final – t = 9 (-1/3) = -3
T_Final  = (t-3) K

∴ 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  

Length of the wire, l =15 m
Area of cross-section of the wire, a = 6.0 × 10⁻⁷ m²
Resistance of the material of the wire, R = 5.0 Ω
Resistivity of the material of the wire = ρ
Resistance is related with the resistivity as,
ρ=RA/l
=5.0×6×10⁻⁷/15
=0.2×10⁻⁶Ωm

The negative sign signifies that the induced emf always opposes any change in the magnetic flux.

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.

Electric Field Intensity due to a short dipole is given by

E = K2P/r^3

where K is the proportionality constant

P is the electric dipole moment

So electric field intensity is inversely proportional to r^3