DSSSB PGT Physics Mock Test - 4 - DSSSB TGT/PGT/PRT MCQ

Unit is the reference used as the standard measurement of a physical quantity.

• The unit in which the fundamental quantities are measured are called fundamental unit.
• Units used to measure derived quantities are called derived units.

The average acceleration is 1/√2 ms^-2 towards the northwest.

Explanation:

The car is in uniform circular motion because it’s moving at a constant speed along a curve that is a segment of a circle. Hence we know 

This is the minimum turning radius because arad is the maximum centripetal acceleration.

Limiting friction, f = μmg = 0.6 × 1 × 9.8 = 5.88 N

Applied force = F = ma = 1 × 5 = 5 N

As F < f, so force of friction = 5 N

As the plane is frictionless, thus there will be only slipping and no rolling. So, for all of them acceleration will be ‘gsinθ’

In free space, neither acceleration due to gravity for external torque act on the rotating solid sphere. Therefore, taking the same mass of sphere if the radius is increased then a moment of inertia, rotational kinetic energy and angular velocity will change but according to the law of conservation of momentum, angular momentum will not change.
In free space, neither acceleration due to gravity for external torque act on the rotating solid sphere. Therefore, taking the same mass of sphere if the radius is increased then a moment of inertia, rotational kinetic energy and angular velocity will change but according to the law of conservation of momentum, angular momentum will not change.

A uniform circular disc of radius 50 cm at rest is free to turn about an axis having perpendicular to its plane and passes through its centre. This situation can be shown by the figure given below:

Therefore,

Angular acceleration = α = 2 rad/s²

Angular speed, ω = αt = 4 rad/s

Centripetal acceleration, a_{c = ω}²r

4² x 0.5

=16 x 0.5

= 8m/s²

Linear acceleration at the end of 2 s is,

a_t = αt = 2 x 0.5 = 1 m/s²​

Therefore, the net acceleration at the end of 2.0 sec is given by 

When two forces of equal magnitude opposite in direction and acting along parallel straight lines, then they are said to form a couple. The perpendicular distance between the two force forming a couple is called the arm of the couple.

Let the density be d for both the planets. Given that R_A ​= 2 R_B​
Now, mass of A, M_A​ = 4 d π R_A​³/ 3 ​= 32 dπ R_B​³ ​/ 3
similarly, M_B​ = 4 dπ R_B​³ / 3
Escape velocity for a planet is given by V = √2 GM ​​/ R
So, V_A ​= ​√2 G M_A / 3 R_A ​​​= √​64 G dπ R_B​³ / 6R_B ​​=√32 G dπ R_B​² / 3​​
 
Similarly, V_B​ = 8 G dπ R_B​² / 3​​
 
Taking the ratio, ​V_A / V_B ​​= 32 G dπ R_B​² ​/ 3 ​× ​√3 / 8 G dπ R_B^​2​ ​= 2

Gravitational Potential V = -GM/R for hollow spherical shell at the centre. If we replace R by R/2 then we get V = -2GM/R. Therefore it decreases.

Escape velocity is the minimum velocity with which the body has to be projected vertically upwards from the surface of the earth so that it crosses the gravitational field of earth and never returns back on its own.

Stress has its own SI unit called the Pascal. 1 Pascal (Pa) is equal to 1 N/m². In imperial units stress is measured in pound force per square inch which is often shortened to "psi". The dimension of stress is same as that of pressure.

Hence 1.6 mm upper, 1.0 mm lower is correct.

Bernoulli’s theorem, in fluid dynamics, relation among the pressure, velocity, and elevation in a moving fluid (liquid or gas), the compressibility and viscosity (internal friction) of which are negligible and the flow of which is steady, or laminar. First derived (1738) by the Swiss mathematician Daniel Bernoulli, the theorem states, in effect, that the total mechanical energy of the flowing fluid, comprising the energy associated with fluid pressure, the gravitational potential energy of elevation, and the kinetic energy of fluid motion, remains constant. Bernoulli’s theorem is the principle of energy conservation for ideal fluids in steady, or streamline, flow and is the basis for many engineering applications.

Gravity does not play any part in radiation and conduction because in both these processes heat is transferred without any motion of the medium particles.

The time period of a physical pendulum is 

where, I = MI of the system about point of sduspension 

M = mass of the system and

l = distance between COM of the system and point suspension

 Now, M = m + m = 2m

in time   first function completes one oscillation, second function two oscillations and the third three

so for the combination T = 2π/ω

Given X is greater than 2Ω when the bridge is balanced

When the resistances are interchanged the jockey shifts 20 cm. Therefore

We know that P=V²/R

For a given potential difference at a particular temperature

It is given that the powers of the bulbs are in the order 

100W > 60 W > 40W

We know that R=ρ l/a

Where l is the length of the conductor through which the current flows and a is the area of cross section.

Here l = L and a = L × t

∴ R is independent of L

The resistance is R=200ohm the inductor is L=0.4 H, the capacitance is C=5  µF and the amplitude voltage is V=CV
The frequency depends on the inductance and the capacitance and it is given by,
f₀=1/ (2π√(LC))  
So, plug the values of L and C into equation 1 to get f₀
f₀=1/ (2π√(LC))=2/(2π√[0.4H)(5x10^-6)]=113Hz
for the current, we use ohm’s law to get the current,
I=V/R
Now, plug the values for V and R to get I
I=V/R=3V/200Ω=0.015A=15mA
 

P(t)=V_mI_mSin²ωt
P(t)=0.5V_mI_m(2sin2ωt)
P(t)= 0.5V_mI_m(1-cos2ωt)
Therefore, frequency is doubled for the instantaneous power so, frequency of instantaneous power is 100Hz.
 

Impedance(X_C)=1/ω_C
=1/ 2πfC
=1/2πx50x10x10^-6
=10⁶/1000π
X_C=10³/π=(1000/π) Ω

The series resonance circuit is known as an acceptor circuit. The impedance of the acceptor circuit is minimum but the voltage can be magnified.
The Acceptor circuit provides the maximum response to currents at its resonant frequency.
Series resonance circuit is known as acceptor circuit because the impedance at the resonance is at its minimum so as to accept the current easily such that the frequency of the accepted current is equal to the resonant frequency.
 

Wavelength of light beam, λ

Distance of the screen from the slit, D=1m

For first minima, n=1

Distance between the slits is d

Distance of the first minimum from the centre of the screen can be obtained as, x = 2.5mm = 2.5×10⁻³

Now, nλ = xd/D

⇒ d= nλD/x = 0.2mm

Therefore, the width of the slits is 0.2 mm.

Substance left undecayed, N₀​−(3/4)​N₀​=(1/4)​N₀​
N/N0​ ​=1/4​=(1/2​)ⁿ
 
∴ Number of atoms left undecided, n=2 i.e, in two half-lives
∴ t=nT=2×4=8 months

Idea The normal reaction between the blocks will be non-zero only if the initial acceleration of block m₂ is more than block m₁.
Then, only the block m₂ will push block m₁ and normal reaction will develop and it is possible only when μ₂< μ₁ The normal reaction between the blocks will not depend on the masses m₁ and m₂.


So, there is no relative motion between the blocks, so normal reaction between them is zero.
TEST Edge Both are identical planes, if m does not slip then M will?

M will also not slip because in this case motion of block will depend only on (l not on mass of the block.