VITEEE: Subject Wise and Full Length VITEEE Physics Test - 2 Free Online

 Answer :- d

Solution :- power =1/T ∫v(t)i(t)dt 

In sinusoidal wave, V and I differ by 90 ° 

Pa=ω2π∫Aωsin(ωt)cos(ωt)dt 

where A is a constant depending on the value of the inductance L or capacitance C of the component.

P=Aω²/2π  [ ∫2π/sin(2ωt)dt]

period of sin(2ωt) is T′=π/ω 

interval of integration of P is exactly two periods of sin(2ωt)

so finally :

P=Aω²/2π∫sin(2ωt)dt=0

The frequency of characteristic X-ray spectral line is directly proportional to the square of atomic number of target element.
⇒ v ∝ Z² i . e . √ v ∝ Z
or √ v = a (z − b)    
Here a is constant of proportionality b is screening constant.

ln balance Wheatstone bridge, the galvanometer arm can be neglected so equivalent resistance = R.

In the photoelectric effect, electrons are emitted from a metal surface when it is illuminated by light of a certain frequency (or wavelength). However, this emission depends on the energy of the incident photons.

The energy of a photon is given by the equation: E = hc/λ where hhh is Planck's constant, c is the speed of light, and λ is the wavelength of the light.

For the photoelectric effect to occur, the energy of the photons must be greater than or equal to the work function of the metal, which is the minimum energy required to eject an electron from the surface. This implies that the wavelength of the incident light must be sufficiently short (or the frequency must be sufficiently high) to provide the necessary energy.

Hence, the correct condition for the emission of electrons is when the incident light has sufficiently low wavelength (or equivalently, high frequency), which corresponds to answer D.

"When a sheet of aluminium is inserted in the air gap of a parallel plate capacitor without touching any of the two plates of the capacitor. Then the capacitance of the capacitor is

where t is the thickness of the aluminium and k for aluminium is infinity.
Therefore,
So, the capacitance of the capacitor is invariant for all positions of the sheet."

When an electric current is passed through water that is acidified with dilute sulfuric acid, electrolysis of water occurs. The electrolysis of water can be represented by the following chemical reactions:

At the cathode (reduction reaction):

At the anode (oxidation reaction):

For every two molecules of hydrogen gas produced at the cathode, one molecule of oxygen gas is produced at the anode. Therefore, the volume ratio of hydrogen to oxygen produced during the electrolysis of water is 2 : 1. This is known as the "volume ratio" in electrolysis, where hydrogen (at the cathode) is produced in twice the volume of oxygen (at the anode).
Hence, the correct answer is A.

Option C. Final value 3 Ω

Use the relation R(t) = R(0)[1 + α t], where R(0) denotes the resistance at 0^oC.

Rearrange to get R(0) = R(t) / [1 + α t].

Substitute the values R(100) = 4.2 Ω, α = 0.004^oC^-1 and t = 100.

R(0) = 4.2 / (1 + 0.004 × 100) = 4.2 / (1 + 0.4) = 4.2 / 1.4 = 3 Ω

In this case, the cell has an emf of 3 V and an internal resistance of 0.1 Ω. The external resistance connected to the cell is 2.9 Ω.

To calculate the voltage across the cell, we first need to determine the total resistance in the circuit. The total resistance is the sum of the internal resistance of the cell and the external resistance:
R_total = R_internal + R_external = 0.1 Ω + 2.9 Ω = 3.0 Ω
Next, we calculate the total current in the circuit using Ohm’s law:
I = V_emf / R_total = 3 V / 3.0 Ω = 1 A
Now, the voltage across the external resistance is given by:
V_external = I × R_external = 1 A × 2.9 Ω = 2.9 V
Since the external resistance has a voltage drop of 2.9 V, the voltage across the cell will be 2.9 V, which matches the emf of the cell.

Therefore, the correct answer is C: 2.9 V.

Therefore, Option (a) 2 Mhz is the correct answer.

 Answer :- a

Solution :- The current always flows in the same direction as the electric field. The electrons flow in the opposite direction, because they are negatively charged.

The particle moves in a circular path with radius d if it is just to miss the wall.

The magnetic force must provide the necessary centripetal force for circular motion. Therefore, we have the equation mv = Bqd, where B is the magnetic field, q is the charge, and m is the mass of the particle.

Rearranging this equation, we get B = v/(qd).

Since the specific charge s is defined as q/m, we substitute s into the equation to get B = v/(sd). Thus, the minimum magnetic field required is v/sd.

The electromagnetic spectrum consists of various types of waves, each having a different wavelength. The wavelengths increase from gamma rays to radiowaves.

Here’s the order from the shortest to the longest wavelength:

• X-rays have very short wavelengths, typically between 10^-12 m and 10^-8 m.
• Ultraviolet (UV) rays have wavelengths ranging from 10^-8 m to 10^-12 m.
• Infrared (IR) rays have wavelengths between 10^-3 m and 10^-6 m.
• Radiowaves have the longest wavelengths in the electromagnetic spectrum, ranging from several meters to thousands of kilometers (up to 10³ m).

Therefore, radiowaves have the maximum wavelength among the options, and the correct answer is D: Radiowaves.

When equal charges are given to two conducting spheres of different radii, the potential on a sphere is given by the formula:

where V is the potential, Q is the charge, ϵ_0​ is the permittivity of free space, and R is the radius of the sphere.

From the formula, we can see that the potential is inversely proportional to the radius of the sphere. This means that for the same charge Q, the smaller sphere (with a smaller radius) will have a higher potential compared to the larger sphere (with a larger radius).

Therefore, the potential will be more on the smaller sphere, and the correct answer is A.

First, let's find the initial energy stored in the capacitor, which is given by the formula:

E_initial = (1/2) * C * V^2

where E_initial is the initial energy, C is the capacitance, and V is the voltage.

E_initial = (1/2) * 2 * 10^-6 F * (500 V)^2
E_initial = 1 * 10^-6 F * (250000 V^2)
E_initial = 0.25 J

Now, when the plates are joined through a resistance, the capacitor will discharge and its energy will be transferred into heat in the resistor. The final energy stored in the capacitor will be zero as it is completely discharged. So, the heat produced in the resistor will be equal to the initial energy stored in the capacitor.

Heat produced (in Joules) = Initial energy stored in the capacitor (in Joules)
Heat produced = 0.25 J

Co-ordinates of the point = (x, y).
Electric potential (V) = −Kxy
Distance of the point from origin 

The magnetic field due to a current carrying straight wire at distance a is given by

The current is same in both the wires, hence magnetic field strength will also be the same.

The potential energy U of an electric dipole in a uniform electric field E is given by the equation:

Where:

•  is the dipole moment,
•  is the electric field,
• θ is the angle between  and .

To find when the potential energy is maximum, we need to maximize the value of U. Since cos(θ) ranges from -1 to 1, the potential energy is maximized when cos(θ) = 1, which happens when θ = 180^o (i.e., when the dipole is aligned opposite to the electric field).

Therefore, the potential energy of the dipole becomes maximum when θ = 180^o, and the correct answer is D.

• Energy is continuously created in the sun due to nuclear fusion.
• Nuclear fusion is a nuclear reaction in which two or more atomic nuclei join together or fuse to form a single heavier nucleus.
• During fusion, a large amount of energy is released.
• The main composition of the sun is Hydrogen and Helium.
• Two hydrogen nuclei fused together to form single helium with the release of a large amount of energy.
• Sun is the primary source of energy in our solar system.

The photoelectric effect occurs when the energy of the incident light (photon) is greater than or equal to the work function of the metal, causing the emission of electrons from the metal surface.

The energy of a photon is given by the formula:
E = hc/λ

where:

• h is Planck's constant (6.626 × 10^-34 J·s),
• c is the speed of light (3 × 10⁸ m/s),
• λ is the wavelength of the incident light.

The threshold wavelength (λ_threshold) corresponds to the minimum wavelength that has sufficient energy to overcome the work function of the metal. For this metal, the threshold wavelength is 30 Å.

Now, for the incident light with a wavelength of 2000 Å:

1. The energy of the photon with wavelength 2000 Å is calculated as:
   
2. The energy corresponding to the threshold wavelength of 30 Å is:
   

Since the energy of the incident photon (9.939 × 10^-19 J) is much lower than the threshold energy required (6.626 × 10^-18 J), the photon does not have enough energy to overcome the work function of the metal and therefore cannot emit electrons.

Thus, the correct answer is D: Electrons will not be emitted.

f = - 15 cm concave  ​​​​​

Let's analyze the comparison between a half-wave rectifier and a full-wave rectifier in terms of efficiency, average current, and average output voltage:

Option A: Efficiency: The efficiency of a half-wave rectifier is around 40.6%, while the efficiency of a full-wave rectifier is around 81.2%. Therefore, a full-wave rectifier is more efficient than a half-wave rectifier.
Option B: Average Current:: The average current provided by a full-wave rectifier is higher than that of a half-wave rectifier because it utilizes both halves of the AC input cycle.
Option C: Average Output Voltage: The average output voltage of a full-wave rectifier is also higher than that of a half-wave rectifier because it is using both halves of the AC cycle.

Thus, none of the given options (efficiency, average current, average output voltage) are lower for the full-wave rectifier in comparison to the half-wave rectifier. Therefore, the correct answer is D: none of these.

In full wave rectifier, two diodes are used to rectify the input A.C. voltage into D.C. voltage. Hence, for each cycle of input signal pulsating, unidirectional output will obtain. Thus, the fundamental frequency in the ripple will be doubled, i.e., 100 Hz.

Solution :- Current gain of transistor is

α = ΔIC/ΔiE

 Where ΔiC is change in collector current, and ΔiE is change in emitter current.

Given, α=0.96,ΔiE=7.2,

ΔiC=0.96×7.2=6.91mA

Also ΔiE=ΔiC+ΔiB

∴ ΔiB = ΔiE + ΔiC

= 7.2−6.91 = 0.29mA

To obtain electrons as majority charge carriers in a semiconductor, the impurity added should be a pentavalent element.

When a pentavalent impurity (such as phosphorus or arsenic) is added to a semiconductor like silicon, it donates an extra electron because pentavalent elements have five valence electrons, and only four are needed to bond with the silicon atoms. This results in an excess electron in the material, which becomes the majority charge carrier. This process is called n-type doping.

Thus, the correct answer is D: Pentavalent.