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

A projectile can have the same range if the angle of projection are complementary ie, θ and (90⁰ - θ). Thus, in both cases:

The minimum force required to start pushing a body up a rough inclined plane is 
F1=mgsinθ+ μgcosθ            ......(i)
Minimum force needed to prevent the body from sliding down the inclined plane is
F2=mgsinθ− μgcosθ           ...........(ii)
Divide (i) by (ii), we get
F1/F2
=sinθ+μcosθ/ sinθ−μcosθ 
= tanθ+μ/tanθ−μ
= 2μ+μ/2μ−μ 
=3              (tanθ=2μ (given))

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.

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

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.

Hence 1.6 mm upper, 1.0 mm lower is correct.

No, heat is not transferred as the flask is insulated from the surroundings ∴dQ=0

Explanation:Avogadro's law states that, "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules".

Explanation:

for monoatomic  gas C_V = 1.5R, C_P =  2.5R

At const volume, 

Q = 500 J

Q=nCVΔT

500=1×1.5×8.31(T₁−300)

T₁=340K

At const pressure Q = 500 J

Q=nCPΔT

500=1×2.5×8.31(340−T₂)

T₂=316K

Restoring torque is τ = mg ( O ' P ) = − I₀ α

In the figure, θ is very small 

This is a direct result from NCERT. In Young's double slit experiment, the distance of the nth dark fringe from the centre is 

► Electric field (E) = 26i - k
► A = 20k
Electric flux = electric field.area
= E.A
= (26i - k)*(20k)
= (26*0 -1)*20
= -20 units

We know that 
P=We know that 
P=V²/R
For constant value of potential difference (V) we have
For constant value of potential difference (V) we have

This is a case of balanced Wheatstone bridge R1 = 1Ω Case (ii)

Clearly the equivalent resistance (R2) will be less than 1Ω.
Case (iii)

Since, R₂ < R₁ < R₃

∴ P₂ > P₁ > P₃

In case of a meter bridge

For finding the value of R

For finding the value of ΔR

Therefore, R = (60 ± 0.25)Ω

The magnetic moment of a current carrying loop is given by

the direction is towards positive z-axis.

No, aluminum is a paramagnetic material. Paramagnetic materials are those, which when placed in a magnetic field are feebly magnetized in the direction of the magnetizing field, i.e. it is feebly attracted by a magnet.

• When a conductive material is subjected to a time-varying magnetic flux, eddy currents are generated in the conductor.
• These eddy currents circulate inside the conductor generating a magnetic field of opposite polarity as the applied magnetic field. The interaction of the two magnetic fields causes a force that resists the change in magnetic flux.
• However, due to the internal resistance of the conductive material, the eddy currents will be dissipated into heat and the force will die out. As the eddy currents are dissipated, energy is removed from the system, thus producing a damping effect.

At resonance, the potential drop across the capacitance is equal and opposite to that of inductance i.e. 110V.

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.
 

α (Radian)=10/2000=1/200rad
α=h/F=h/1.2
h= α x 1.2
h=(1/200) x1.2 cm
=6mm

A particle representing a quantum of light or other electromagnetic radiation. A photon carries energy proportional to the radiation frequency but has zero rest mass.

The number of electrons emitted per second is observed to be directly proportional to the intensity of light. “Ok, so light is a wave and has energy. It takes electrons out of a metal, what is so special about that!” First of all, when the intensity of light is increased, we should see an increase in the photocurrent (number of photoelectrons emitted). Right?
As we see, this only happens above a specific value of frequency, known as the threshold frequency. Below this threshold frequency, the intensity of light has no effect on the photocurrent! In fact, there is no photocurrent at all, however high the intensity of light is.

The graph between the photoelectric current and the intensity of light is a straight line when the frequency of light used is above a specific minimum threshold value.

In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67×10−27) is less than the mass of incident α−particles (6.64×10−27). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α−particles would not bounce back if solid hydrogen is used in the α−particle scattering experiment.

Half-life of radioactive sample, i.e., the time in which the number of undecayed nuclei becomes half (T) is 10 s.
Mean life, τ=T/loge​2=10s/0.693=1.443×10=14.43s ≈ 15s

The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.

When a transistor is used in the common emitter mode as an amplifier, the base-emitter junction acts as input junction and collector-emitter junction acts as output junction. The input junction is made forward biased by applying forward voltage and output junction is made reverse biased by applying reverse voltage. The input signal is connected in series with the voltage applied to forward bias the base-emitter junction.

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.