DSSSB PGT Physics Mock Test - 5 - DSSSB TGT/PGT/PRT MCQ

Difference in masses = 20.17 – 20.15 = 0.02 g
In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

Explanation

To determine the number of significant digits in the number 8.1000, we can follow a systematic approach based on the established rules of significant figures

1Identify non-zero digits. In the number 8.1000, the digits '8' and '1' are non-zero. According to the rules of significant figures, all non-zero digits are considered significant. Therefore, we have two significant digits from these numbers

2Examine the zeros. The zeros in the number 8.1000 need to be evaluated based on their position. The first zero appears after the '1', and since it is sandwiched between non-zero digits (1 and 0), it is significant

3Analyze trailing zeros. The last three zeros in 8.1000 come after the decimal point and follow the last non-zero digit (which is '1'). According to the rules, trailing zeros in a decimal number are significant. Therefore, these three zeros also count as significant digits

4Count the total significant digits. Adding the significant digits identified: 2 (from '8' and '1') + 1 (from the zero between '1' and '0') + 3 (trailing zeros) gives us a total of 5 significant digits

Explanation:Although vectors and scalars represent different types of physical quantities, it is sometimes necessary for them to interact. While adding a scalar to a vector is impossible because of their different dimensions in space, it is possible to multiply a vector by a scalar.

Explanation:Mechanical energy is the sum of kinetic and potential energy in an object that is used to do work. In other words, it is energy in an object due to its motion or position, or both. In case of conservative forces total mechanical energy remains conserved because potential energy applicable only for conservative forces.

In case of a projectile, the angular momentum is minimum at the starting point.
 

As the relative velocity of the satellite with respect to the earth is zero, it appears stationary from the Earth surface and therefore it is called is geostationary satellite or geosynchronous satellite.

Let the horizontal displacement of the plank, at an instant during its oscillation is x and its velocity is v.

During its oscillations, at a particular instant let the coordinates of the particle are (x, y)  The total energy of the particle is 

As oscillations are very small, we can ignore the middle term. 

The bob of a pendulum moves in such a way that it repeats its positions several time but alternately the time gap between these positions is always equal which proves that the motion of a pendulum is oscillatory.

B because of constructive interference.

Let m be mass of each ball and q be charge on each ball. Force of repulsion,


In equilibrium
Tcosq = mg ...(i)
Tsinq = F ...(ii)
Divide (ii) by (i), we get,

From figure (a),



Divide (iv) by (iii), we get

The needle will point along W-E if the result of the earth’s magnetic field and magnetic field due to the coil have a resultant in the W-E direction. This happens if B cos 45^o=Earth's field
⟹ (μ0​nI​/2r) cos45^o=(4π×10−7×30×0.35​)/(2×0.12) ×(1/√2)​​=0.39G=Earth's magnetic field.

The formula for magnetic flux is Φ = B × A × cosθ, where θ is the angle between the magnetic field (B) and the normal to the surface (area vector A).

Since the plane of the strip is at 60° to the field, the normal is at 30° to B.

Given: Φ = 1 mWb = 1 × 10⁻³ Wb, A = 0.02 m², θ = 30°.

So, B = Φ / (A × cos30°)
cos30° = √3/2 ≈ 0.866
B = (1 × 10⁻³) / (0.02 × 0.866) = (1 × 10⁻³) / 0.01732 ≈ 0.058 T

Therefore, the correct answer is 0.058 T.

Φ=BAN
=5x10^-3x2x10^-3x500
=5x10^-3x1000x10^-3
=5x10^-3

ε=− dϕ​/dt
=−β(dA/dt)

​⇒∣ε∣=βπd(R_0​+t)²
=2πβ(R_0​+t)
Since inward flux is increasing the EMF will be induced counter clockwise.
 

In an inductor, current lags behind the input voltage by a phase difference of π/2.
Current and voltage are in the same phase in the resistor whereas current leads the voltage by π/2 in a capacitor.
So, the circuit must contain an inductor only.

The images of clouds and trees in water are always less bright than in reality because only a portion of the incident light is reflected and quite a large portion goes mid water.

The final image of the moon should be formed at a distance of 100 cm. Using a concave lens of focal length 100 cm will enable him  to see the moon distinctly.

Focal length of lens =-100 cm

The ratio of angular dispersion to the angle of deviation for the mean wavelength is called Dispersive Power. Represented by 

where δ_v − δ_r is the angular dispersion for violet and red lights and δ_r the deviation suffered by the mean light.

 Suppose the angle between the transmission axes of the analyser and the polarizer is θ. The completely plane polarized light form the polarizer is incident on the analyser. If E₀ is the amplitude of the electric vector transmitted by the polarizer, then intensity I₀ of the light incident on the analyser is
I ∞ E₀²
 
The electric field vector E₀ can be resolved into two rectangular components i.e E₀ cosθ and E₀ sinθ. The analyzer will transmit only the component ( i.e E₀ cosθ ) which is parallel to its transmission axis. However, the component E₀sinθ will be absorbed by the analyser. Therefore, the intensity I of light transmitted by the analyzer is,
 
I ∞ ( E₀ x cosθ )²
 
I / I₀ = ( E₀ x cosθ )² / E₀² = cos2θ
 
I = I₀ x cos²θ
when light passes from polarizer it's intensity becomes half and when passed through analyser it becomes,
I = I₀ x cos²θ/2
 

Lines in the spectrum were due to transitions in which an electron moved from a higher-energy orbit with a larger radius to a lower-energy orbit with smaller radius. The orbit closest to the nucleus represented the ground state of the atom and was most stable; orbits farther away were higher-energy excited states.

Given that 90% is left un-decayed after time 't'.
Hence, 10% decays in time 't'.
Initially assume that the amount of substance is 'x'
After time 't' 10% is decayed.
i.e. Amount of substance left =0.9x
After further time 't' another 10% is decayed.
i.e. 0.1×0.9x is decayed 
Leaving behind 0.81x.
Hence after time 2t we see that 0.19x has decayed, which is 19%.
 

When a p−n junction is formed, some of the free electrons in the n region diffuse across the junction and combine with holes to form negative ions. In doing so they leave behind positive ions at the donor impurity sites. Similarly, holes from p side diffuse to the n side and thus form a layer called diffusion layer at the junction.
So the depletion region is caused by the diffusion of charge carriers.
 

In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms.