DSSSB PGT Physics Mock Test - 3 - DSSSB TGT/PGT/PRT MCQ

Explanation:Equal vectors are vectors that have the same magnitude and the same direction. Equal vectors may start at different positions.

When friction is absent

a₁ = gsinθ

When friction is present

a₂ = gsin θ - μ_k gcos θ

From Eqs. (i) and (ii)

In the frame of wedge, the force diagram of block is shown in figure. From free body diagram of wedge.

For block to remain stationary

ma cos α = mg sin α

or         a = g tan α

It is advised to drive along the slippery road slowly, because friction is reduced and if the driver is driving at a larger speed there is a risk of skidding.

CONCEPT:

• Angular displacement (θ): The angular displacement of a body is the angle in degrees or radians or revolutions through which a point revolves around a point or axis.
• 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.
  • It is the rate of change of angular displacement.​
• Angular acceleration (α): The rate of change of angular velocity is known as angular acceleration.
• Angular momentum (L): The moment of momentum is known as angular momentum.

CALCULATION:

• The rate of change of angular velocity is known as angular acceleration. 

Mathematically,

α = dω/dt

• So the correct answer is option 3.

The forces acting on the body are the force of gravity and the air friction. By work energy theorem, the total work done on the body is
W=1/2m(1.2√gH)2−0=0.72mgH
The work done by the force of gravity is mgH.
. Hence, the work done by the air friction is,
=0.72mgH-mgH
=00.28mgH.
 

Average force = change in momentum/time
Change in momentum = m(v1 - v2)
=0.01[ 5 - (-5)]
=0.1
Average force = 0.1/0.01
= 10 N

An earth satellite is moving in a circular motion due to change in the direction of velocity. Its angular momentum about centre is conserved as there is no external force acting on the system.

A communications satellite is in a GEO (geostationary equatorial orbit) with a period of 24 hours. The spin rate ω_s about its axis of symmetry is 1 revolution per minute, and the moment of inertia about the spin axis is 550 kg · m².

Explanation:External force permanently distubed the equilibrium position of the interatomic ( or intermolecular ) forces between the particles of solid bodies.

The property by virtue of which the free surface of a liquid at rest behaves like an elastic stretched membrane tending to contract so as to occupy minimum surface area is known as surface tension. By the definition of surface tension.

Correct Answer : C

Explanation : The adiabatic condition is given by the relation between pressure volume and temperature volume as:

(PV)γ = constant 

where, γ = Cp/Cv is ratio of the specific heats

These relations suggest that an decrease in volume is associated with increase in temperature

ATQ 

P₁(V₁)γ = (P₂V₂)γ

=> P₁(1)² = P₂(4)²

P₁/P₂ = 16

When the angular displacement of the disc is θ the deformation (elongation and compression) in each  spring is x = Rθ .

We know that the electric field and area both are vector quantity and the electric flux is expressed as E.s (taking dot product). But if the angle between the two vectors is θ then the formula becomes E^-.s^-.cosθ. Cosθ will be maximum if theta is zero, in all other cases, the value of Cosθ is less than 1. Therefore flux will be the maximum if the angle is 0 degrees.

On equatorial line, the direction of electric field is reversed to that of angle of axial line. Therefore, the angle between dipole moment and electric field is 180°.

Potential difference between two points is given by

V_a - V_b = W/q₀

Work, W = 2 J

Charge, q₀ = 20 C

Potential difference = 2/20 = 0.1 V

The correct option is C.

Definition based

C=(Kε0A/d)∝K/d
Hence,C₁/C₂=d2/d1=(d/3)d=3
C₂=3C₁

Capacitance of an isolated spherical conductor =4π€R , where €=permittivity of vacuum
C=2×10^-6F
C=4π€R
4×3.14×8.85×10^-12×R=2×10^-6
R=2×10^-6/4×3.14×8.85×10^-12
=1.8×10⁴ m

Total external resistance will be the total resistance of whole length of box. It should be connected between A and D.

Maximum voltage

e = Rate of change of magnetic field x area

=0.02×120×10⁻⁴=0.24mV

 So the range is 88 - 198 pF

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.

3/2=v/u…...1
2/3=(v-16)/(u+16)......2
from these equation
we get v=48, u=32
so now f=(vxu)/(v+u)=96/5=19.2
The correct answer is option B.

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.