DSSSB PGT Physics Mock Test - 1 - DSSSB TGT/PGT/PRT MCQ

Torque = Work = Energy =[ML²T^-2]


 

(i) All the non-zero digits are significant. 
(ii) All the zeros between two non-zero digits are significant, no matter where the decimal point is, if at all.
 (iii) If the number is less than 1, the zero(s) on the right of decimal point but to the left of the first non-zero digit are not significant. 
(iv) The power of 10 is irrelevant to the determination of significant figures. According to the above rules, 23.023 has 5 significant figures. 0.0003 has 1 significant figures. 2.1 × 10^–3 has 2 significant figures.

Let the tension between B and C be T

M=2kg

F−T=ma

10.2−T=ma

10.2−T=2×0.6

10.2−T=1.2

10.2−1.2=T

T=9N

As the string is inextensible, both masses have the same acceleration a. Also, the pulley is massless and frictionless, hence the tension at both ends of the string is the same. Suppose the mass m₂ is greater than mass m₁, so the heavier mass m₂ is accelerating downward and the lighter mass m₁ is accelerating upwards.

Therefore, by Newton's 2nd law

T - m₁g =m₁a ...(i)

m₂g - T = m₂a ...(ii)

After solving Eqs. (i) and (ii)

So, the ratio of the masses is 9:7.

Frictional forces act over the common surfaces of the two bodies to avoid and restrict relative moment between those two surfaces.

x_com=m₁x₁+m₂x₂/m1+m₂
=1x2+3x (-6)
=-4î
v_com= m₁y₁+m₂y₂/m₁+m₂
=1x5+3x4/4
=17/4 ĵ
z_com= m₁z₁+m₂z₂/m₁+m₂
=13-6/4
=7/4 k̂
r_com=-4î+17/4 ĵ+7/4 k̂
=1/4(-16î+17ĵ+7k̂)m

We know that Gravitational force = 

and,

The SI unit of the universal gravitational constant is N m²/kg²
The SI unit of the acceleration due to gravity is the same as that of acceleration.
Hence, the SI unit of the acceleration due to gravity is m/s².
So, the SI unit of 

Stress = F⟂/A = (Mass x Acceleration⟂)/Cross-sectional Area

= (550 kg x 9.8 m/s²)/ (0.30×10−4 m²)

= 1.8×10⁸ Pa

Young's ModulusSteel (Y) = 20×10¹⁰ Pa = Stress/Strain

=> Strain = Stress/Y = 1.8×108 Pa / 20×10¹⁰ Pa

= 9.0×10⁻⁴ 

Given Data,
Length of the piece of copper = l = 19.1 mm = 19.1 × 10^-3m
Breadth of the piece of copper = b = 15.2 mm = 15.2× 10^-3m
Tension force applied on the piece of cooper, F = 44500N
Area of rectangular cross section of copper piece,
Area = l× b
⇒ Area = (19.1 × 10-3m) × (15.2× 10-3m)
⇒ Area = 2.9 × 10^-4 m²
Modulus of elasticity of copper from standard list, η = 42× 10⁹ N/m2
By definition, Modulus of elasticity, η = stress/strain

⇒ Strain = F/Aη

⇒ Strain = 3.65 × 10^-3
Hence, the resulting strain is 3.65 × 10^-3

From equation of continuity,
A1v1=A2v2
⇒ (A1/A2) v1=( πr12/ πr22 )v1
or v1(D/d)2v1=(8×10−3/2×10−3)2×0.25m/s
=4m/s (horizontal)
Vertical component of the velocity is zero.
Now, H=(1/2)gt2
⇒t=√2H/g
Range is given by R=v2t=v2
√2H/g=4×√2×1.25/10=2m

 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 

∴ 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  

Since the opening or closing the switch does not affect the current through G, it means that in both the cases there is no current passing through S. Thus potential at A is equal to potential at B and it is the case of balanced wheatstone bridge..

I_P = I_Q and I_R = I_G

μ = μ₀μ_r. μ is the permeability of medium, μ₀ is permeability of free space, and μr is relative permeability of the medium.

Work required to turn the dipole is given by
W=MB[cosθ_i- cosθ_f]
Here θi is the initial angle made by a dipole with a magnetic field and   is the final angle made by a dipole with a magnetic field.
magnetic moment is normal to the field direction.
so, θ_i=0 and θ_f=90
W = 1.5 × 0.22 [ cos0° - cos90° ]
W = 0.33J
 
magnetic moment is opposite to the field direction.
so, θ_i=0 and θ_f=180
now, W = 1.5 × 0.22 [ cos0° - cos180°]
= 0.33 [ 1 - (-1) ]
= 0.66 J

Diamagnetic materials are those which when placed in a magnetizing field are feebly magnetized in a direction opposite to that of the magnetizing field. Therefore, it is feebly repelled by a magnet. Example: zinc, gold, etc.

I₀=250μA, v₀=3.6v .  v=1.6x10⁶ Hz
Here,
(Reactance of inductance) X_L=ωL
X_L=2πv X L
v₀/I₀=2πv x L
3.6/2.5x10^-4=2πx1.6x10^-6 x L
0.14x10^4-6=L
L=0.14x10^-2H
Now for v=16.0x10⁶Hz
XL=2πv X L
=2πx16x10⁶x14x10^-4
X_L=1407x10²Ω
Now,
v0=I0 x XL
3.6/1407x10²=I₀   [∵v0=kept constant.]
I₀=0.00256x10^-2
I₀=25.6μA

Since the current flowing through a series resonance circuit is the product of voltage divided by impedance, at resonance the impedance, Z is at its minimum value, ( =R ). Therefore, the circuit current at this frequency will be at its maximum value of V/R

In a series RLC circuit there becomes a frequency point were the inductive reactance of the inductor becomes equal in value to the capacitive reactance of the capacitor. In other words, XL = XC.
Series Resonance circuits are one of the most important circuits used electrical and electronic circuits.
 

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.

Refractive index of glass with respect to liquid:
n = 1.52/1.63 = 0.9325
thus, critical angle, i = sin^-1(n) = 68.8 degrees

Brewster's angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. When unpolarized light is incident at this angle, the light that is reflected from the surface is therefore perfectly polarized.

In the photoelectric effect, the work function is the minimum amount of energy (per photon) needed to eject an electron from the surface of a metal.Electrons ejected from a sodium metal surface were measured as an electric current.The minimum energy required to eject an electron from the surface is called the photoelectric work function.
This energy (work function) is a measure of how firmly a particular metal holds its electrons. The work function is important in applications involving electron emission from metals, as in photoelectric devices and cathode-ray tubes.

Electromagnetic force on electron will be zero V×B=0
Electrostatic force will be F_e​=−E backward, hence electron decelerate and velocity will decrease in magnitude.

More than 99.99% of the mass of any atom is concentrated in its nucleus. If the mass of protons and neutrons (which are in the nucleus of every atom) is approximately one atomic mass unit, then the relative mass of an electron is 0.0005 atomic mass units.

For a proper functioning of the transistor, the emitter-base region must be forward-biased and collector-base region must be reverse-biased. In semiconductor circuits, the source voltage is called the bias voltage. In order to function, bipolar transistors must have both junctions biased.