UP PGT Physics Mock Test - 4 - UPTET MCQ

Force on EF,
F_mag​​=i∫(dl×B)=iLBsin(90−θ)=iLBcosθ
F_mag​​ up the inclined plane.
Component of weight of EF down the inclined plane: mgsinθ
For EF to be in equilibrium, iLBcosθ=mgsinθ
∴iLB=mgtanθ
For F_mag​ to be up the inclined plane, it needs to flow from E to F.

Since R_AB = 2R_CD therefore, current in AB will be half as compared to current in CD.

Here P₄ = Power dissipation in 4Ω

P₅ = Power dissipation in 5Ω

Y=F x l/A x Δ l
Δ l=0.5cm=0.5x10^-2m, l=10M, F=940N
Y=20x10¹⁰pa
20x10¹⁰=940x10/πr²x0.5x10^-10
πr²=94x100/5x10^-3x2x10¹¹=94x10²/10x10⁸
r²=94/π x 10^-7 =2.99 x 10^-6
r2 ≅3x10^-6
r=1.13x10^-10 m
diameter=2r=3.6mm

As, U=−Gm₁m₂/r, 
it can be used to describe the potential energy in a system of point charges (or radially symmetric spheres) as a function of their separation distance r, then the maximum value is zero at infinite separation.
Hence, its maximum value is zero at infinity.
 

The principle of homogeneity of dimensions states that “For an equation to be dimensionally correct, the dimensions of each term on LHS must be equal to the dimensions of each term on RHS.”

Since the charge –q is moving in elliptical orbit so to make its motion stable the total angular momentum of the charge is constant since it experience a centripetal force from the charge +Q so it follow the motion as the motion of earth around sun.

No, this statement is false. The Earth always has a vertical component except at the equator. Similarly, the earth always has a horizontal component except at the poles.

Option B is the correct answer.

In equilibrium position, mg = k x₀

When the bullet gets embedded, the new equilibrium position changes.  Let it is a distance x from initial equilibrium position. 

Applying conservation of linear momentum, 

So, at t = 0, mass 3m/2   is at a distance mg/2k  from it mean position and moving up with velocity u/3

Long-sighted people who have lost their spectacles can still read a book by looking through a small (3-4 mm) hole in a sheet of paper because by doing so due to the diffraction at the hole the focal length of the eye lens is effectively decreased. Hence the correct option is C.
 

Explanation:Avogadro's law states that, "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules"For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.

As the power of 10 is irrelevant to the determination of significant figures hence, the numbers of significant figures are 3 (1, 8 and 4).

When a resistor is added parallel to other resistor in circuit it results in the reduction of overall resistance, since there are multiple pathways by which charge can flow. adding another resistor in a separate branch provides another pathway by which to direct charge through the main area of resistance within the circuit. This decreased resistance resulting from increasing the number of branches will have the effect of increasing the rate at which charge flows i.e. the current.

Field intensity is proportional to the inverse of the square of the distance separating the point charges:

F = k*q1*q2/d^2

Since k, q1, and q2 are assumed to be constant, their product can be combined to form a single constant:

K = k*q1*q2, so

F = K/d^2 (and by algebraic manipulation we have K=F*d^2).

Therefore we know that

F(at d=4) = K/4^2, and

F(at d=2) = K/2^2.

Since K is a constant, we can equate K=F*d^2 for each case:

K=F(at d=4)*4^2 = F(at d=2)*2^2

so

F(at d=2) = (F(at d=4)*4^2)/2^2

F(at d=2) = 200N/C * (4^2)/(2^2) = 200N/C *16/4 = 800N/C.

• As per rule, zero between two non zero digits are significant.
• The number of a significant figure in the length of the side is 4.
• So the calculated volume should be rounded off to 4 significant figures. Multiplying numbers do not increase significant figures.

Therefore, the correct answer is 4. 

From the first law of thermodynamics,
we know, dU = dQ - dW ; (work done BY the system is considered +ve)
For an adiabatic process, dQ = 0, and hence, dU = -dW
For an ideal gas expansion, we see that work done
BY the system is +ve (recall the sign convention for work done), i.e., dW > 0.
Therefore, dU is less than 0, and thus, the internal energy decreases.

Clearly, the force is time-dependent.  

In the direction of electric field the electric potential decreases. This is because electric potential is the work done against the direction of the electric field.

 absence of wall W2, the time period of block would have been  But due to the presence of W₂, it alters. But for the left part of oscillation (from the mean position shown), the period will be 

For the right side part, d = A sinω t [from x = A sinω t ] 

This is the time taken by the block to reach W₂ from mean position. Collision with W₂ is perfectly elastic (given). 

Time taken for right side part of oscillation will be 

∴ Total time period is 

According to the second law the orbital radius and angular velocity of the planet in the elliptical orbit will vary. The planet travels faster when closer to the Sun, then slower when farther from the Sun. Hence we can say that the transverse acceleration is zero while radial and longitudinal accelerations are not zero.

Lenz’s law is used for determining the direction of induced current.
Lenz’s law of electromagnetic induction states that the direction of induced current in a given magnetic field is such that it opposes the induced change by changing the magnetic field.
Following is the formula of Lenz’s law:
ϵ=−N (∂ϕ_B/∂t)
Where,

• ε is the induced emf
• ∂ΦB is the change in magnetic flux
• N is the number of turns in the coil
   

Lenz’s law finds application in electromagnetic braking and in electric generators

The voltage developed by the cells in parallel connection cannot be increased by increasing the number of cells present in the circuit. It is because they do not have the same circular path. In parallel connection the connection provides power based on one cell.
The voltage will remain the same.

The central fringe will be white because all the component color bands will superimpose there. The option (a) is true.
Since the fringes of different wavelengths of light have different fringe-widths, there will not be a completely dark fringe. Option (b) is true.
Since the wavelength of the violet light is nearly half of the red light, the fringe-width of the violet light is half of the red light. Since the central bright violet fringe combines with other fringes to form the white fringe, the next to it will be a minima for violet light. So violet will not be next to the central white fringe. Next will be a very thin band of white light minus violet light which will give a shade of off white. In this way, the components of indigo, blue, green, yellow and orange colors will be withdrawn respectively in the very thin consecutive bands. These will effect in different off white shades which cannot be clearly differentiated from the white band. The first color away from the center will be recognized as red when in these very thin bands yellow + orange + red, orange + red or red remains. Hence the option (d) is true.

Young's Double Slit Experiment with White light

Given,
We are given the current I=1.5Am where the wire will break.
We are asked to determine the root mean square of the current Irms. The maximum current here represents the current that just after it the wire will break. The maximum value of the current is the amplitude of the current wave and it should be larger than the root mean square of the current. Using equation, we can get the I_rms in the form,
I_rms = I_max/√2
The term, 1/√2, times any factor represents the root mean square of this factor, Now, plug the value for Imax into equation 1 and get I_rms
I_rms=I_max/√2
     =1.5A/√2
     =1.06A.

The refraction takes place at both the air-glass interface and glass-air interface of a rectangular glass slab. When the light ray incident on the air-glass interface (DC) obliquely, it bends towards the normal.

Explanation:By using the phenomenon of mutual induction, transformers allow us to easily change voltage of AC. This is necessary to cut down poer losses while supplying electricity to our homes

To find the characteristic temperature using Wien's displacement law, we start with the formula λ_m T = 0.29 cmK. We need to convert the wavelength λ_m from meters to centimeters: λ_m = 5×10^-7 m = 5×10^-5 cm. Substitute this into the formula: T = 0.29 cmK / 5×10^-5 cm, which gives T = 5800 K. Due to rounding and context of black body radiation, this temperature is closer to 6000 K. Thus, the characteristic temperature range is approximately 6000 K.