UP PGT Physics Mock Test - 2 - UPTET MCQ

Weight becomes zero in a freefall or when the total force on your body is zero. For example, in a freely falling lift the contact force on your body is zero, so the weighing​ machine would show zero reading. So you are weightless.

Definition based

Suppose a be the downward acceleration of the 4 kg mass, therefore, 2 a    is the upward acceleration of the I kg mass. Hence, equations of motion are:

Adding, after multiplying the equation (i) by 2.

Thus, the acceleration of the mass
a 1 kg is g/2 upwards.

Compare the given eqn. with the standard from I=I0sinωt
I₀=8, Irms=I₀/√2=8/√2=5.656A

E_mech. = 8 J
(A) At x > x₄, U = constant = 6 J
K = E_mech. - U = 2 J = constant
(B) At x < x₁, U = constant = 8 J
K = E_mech - U = 8 - 8 = 0 J
The particle is at rest.
(C) At x = x₂, U = 0 ⇒ E_mech. = K = 8 J
K.E. is greatest and the particle is moving at fastest speed.
(D) At x = x₃, U = 4 J
U + K = 8 J
K = 4 J

In a convex lens,  (1/v)−(1/u)=(1/f)​
Or (1/v)​−(1/f)​=(1/u)​ where u is always negative and f is always positive.
(1/v)​=(1/f)​−(1/u)​
So, as u is increased 1/u will decrease and  (1/f)​−(1/u)​  will increase, so, 1/v​ increases or v decreases.
So, its graph will be a hyperbola

When a light travels from one medium to another then deviation in the path of light known as refraction. And in the prism surrounding medium is air and material medium is glass due to the change of medium refraction will occur in prism.

Two sources are said to be coherent if there always exists a constant phase difference between the waves emitted by these sources. But when the sources are coherent, then the resultant intensity of light at a point will remain constant and so interference fringes will remain stationary.

The relative error in a physical quantity raised to the power k is the k times the relative error in the individual quantity.

Suppose Z = A²,
Then,
ΔZ/Z = (ΔA/A) + (ΔA/A) = 2 (ΔA/A).
Hence, the relative error in A² is two times the error in A.

A photon is massless, has no electric charge, and is a stable particle. In vacuum, a photon has two possible polarization states. The photon is the gauge boson for electromagnetism, and therefore all other quantum numbers of the photon (such as lepton number, baryon number, and flavour quantum numbers) are zero.
 

As point O is along the length of segments L and M, so the field at O due to these segments will be zero. 
Also, point O is near one end of a long wire.
The resultant field at O, B_R​=B_P​+B_Q​, this field will be into the plane paper.
⇒B_R​= μ₀I/4π​​RO​+​​ μ₀I/4πSO​
But, RO=SO=0.02m
Hence, BR​=2× (μ₀/4π) ​​×(10/0.02)=2×10⁻⁷ (10/0.02) ​=10⁻⁴Wb m⁻²

Let R be the normal contact force by wall on the block.

R = 10 N
f_L = w and f = μR
∴ μR = w
or w = 0.2 x 10 = 2N

In an RLC circuit, the voltage is always used as a reference and according to the phase of the voltage, the phase of the other parameters is decided.

The magnetic field at a point on the axis of a circular loop at a distance x from centre is,

CONCEPT:

• Angular velocity (ω): How fast the angular position or orientation of an object changes with time or How fast an object rotates or revolves relative to another point is known as angular velocity.
  • Angular velocity in terms of RPM (revolutions per minute) 

ω = 2 π f / 60

where ω is in terms of radian per second and f is in revolution per minute.

• Angular acceleration (α): The rate of change of angular velocity is known as angular acceleration.

The relation between α and changing ω is given by: 

ω₁ = ω₀ + α t

where ω₁ is the final angular velocity, ω₀ is the initial angular velocity α is the angular acceleration.

CALCULATION:

Given that f₀ = 2100 rpm; f₁ = 0 rpm; t = 2 sec.

ω₀ = 2 π f₀ / 60 = 2 π 2100 / 60 = 70π 

ω₁ = 0

ω₁ = ω₀ + α t

0 = 70π + α 2

α = -35π rad/sec²

So the correct answer is option 1.

As the balls collide head-on and no other external force is acting, momentum of the system realins conserve and so will be the energy of the system (i.e. KE)
Thus we get for the mass of other ball be m and initial velocity of mass 1 be v
1 x v = 1 x -v/3 + m x V         -1
For some V that is velocity of m after collision.
Also  ½ 1 x v² =  ½ 1 x v² /9 +  ½ m x V2                -2
Now from eq 1 we get mV = 4v/3
And from eq 2 we get  mV²  = 8v² /9
Thus we get V =  (8v² /9) / (4v/3)
=  2v/3
Thus we get m x 2v/3 = 4v/3
So m = 2kg

In case I :


In Case II :


From equations (i) and (ii),

The Measurement of a given quantity is essentially an act or result of comparison between a quantity whose magnitude (amount) is unknown, with a similar quantity whose magnitude (amount) is known, the latter quantity being called a Standard.

When frequency decreases, capacitive reactance XC​=1/2πνC​ increases and hence impedance in the circuit Z=√(R2+XC2​)​ increases and so current I=V/Z​ decreases. As a result, the brightness of the bulb is reduced.
 

Explanation : The coil of galvanometer is wound on a light metal frame. When the coil and frame rotate in the field of the permanent magnet, the eddy current set up in the frame oppose the motion so that the coil returns to zero quickly.

KEY CONCEPT : The heat produced is given by

Thus heat (H) is doubled if both length (l) and radius (r) are doubled.

Φ=BAN
=5x10^-3x2x10^-3x500
=5x10^-3x1000x10^-3
=5x10^-3

Angular velocity is defined as the rate of change of angular displacement with respect to time. It is usually expressed by a Greek letter ω (omega).
Mathematically, angular velocity,
ω =dθ/dt
If a body is rotating at the rate of N r.p.m. (revolutions per minute), then its angular velocity,
ω = 2πΝ / 60 rad/s

Radiation alludes to the mechanism in which heat is transmitted without any physical contact between objects. It uses electromagnetic waves for transfer of heat.

Change n in magnetic flux Δϕ=0.1−1=−0.9 Weber
Time taken  Δt=0.1 sec
So, emf induced in the coil  E=− Δϕ​/Δt
⟹ E=− (−0.9)​/ 0.1=9 Volts
Hence option C is the correct answer.

By the principle of superposition of fields

Here  = net field at the centre of hole due to entire mass

 = field due to remaining mass

 = field due to mass in hole = 0

The Huygens-Fresnel principle states that every point on a wavefront is a source of wavelets. These wavelets spread out in the forward direction, at the same speed as the source wave. The new wavefront is a line tangent to all of the wavelets.
The correct answer is option B.

Pressure = Force/Area = ML¹T^-2/ L² = [M L^-1 T^-2]
Stress = Force/Area   = ML¹T^-2/ L² = [ML^-1T^-2]
Coefficient of elasticity = Stress/Strain = [ML^-1T^-2]

Hence, the correct answer is C.