Physics Test - 36

Given: V₂ = 10 V; V₁ = 20 V; C = 5 μF; Inductance (I) = 0.5 mH

Initial charge on the capacitor (q₁) = C × V₁ = 5 × 10^-6 × 20 ……………….1

q₁ = 10^-4 C …………..B

The instantaneous charge on the capacitor as the capacitor discharges through the inductor ⇒ q₂

q₂ = q₁cos (ωt) ⇒ q₂/q₁ = cos (ωt) ………………..A

Also, q₂ = C × V₂ = 5 × 10^-6 × 10 …………………….2

q₂ = 0.5 × 10^-4 C

From 1 and 2 ⇒ q₂/q₁ = V₂/ V₁ ⇒ q₂/q₁ = 0.5 = 1/2

From equation A, we can equate as follows ⇒ cos (ωt) = 1/2

ωt = π/3 rad ………………..3

ω=20000rad/s …………………….4

The current through the circuit is given as:

Current (I)=-dq/dt

Current: I = -dq/dt = q₁ωsin(ωt) = 10⁻⁴ × 20,000 × sin(π/3) = 2 × 0.866 = 1.732 A

Charge decreases with respect to time, so, dq/dt obtained will be negative and this is why we add a negative sign to make a curren

Calculation:

Let Voltage V = V_mSin(wt)

V = 200Sin(wt)

When connected to "X",

The current flowing lags "V" by π/2

I = 5Sin(wt - π/2)

Since the current lags "V" by π/2, the element "X" would be the inductor.

Then reactance "X" would be = 200/5 = 40 ohm

The reactance "X" = 40j.

When connected to "Y" , the current 5 A is in phase with the voltage.

So, the "Y" will be a resistor.

Resistance "Y" = 200/5 = 40 ohm.

The "X" and "Y" are connected in series to the same applied voltage.

The net impedance would be = 40 + 40j

The current I' flowing, when connected in series, will be,

Concept:

• For a plano-convex lens, one of the surfaces is flat (Plano) and the other is curved (convex).
• The radius of curvature of the curved surface is given as 20 cm, and the refractive index of the material of the lens is 1.5.
• The focal length of the lens can be calculated using the lens maker's formula:

1/f = (n - 1) × (1/R1 - 1/R2)

where

f is the focal length of the lens,

n is the refractive index of the lens material,

R1 is the radius of curvature of the first surface (the flat surface in this case, which is infinite), and

R2 is the radius of curvature of the second surface (the curved surface in this case).

Since the first surface is flat, the radius of curvature R1 is infinite, and the term (1/R1) becomes zero. Therefore, the lens maker's formula simplifies to:

1/f = (n - 1) × (1/R2)------(1)

Calculation:

Substituting the values in equation (1), we get:

1/f = (1.5 - 1) × (1/20)

1/f = 0.025

f = 1/0.025

f = 40 cm

Therefore, the focal length of the plano-convex lens is 40 cm.

The correct answer is option (4)

Concept:

The height of the image formed by a convex mirror can be calculated using the mirror formula:

1/f = 1/v + 1/u

where

f is the focal length of the mirror,

v is the distance of the image from the mirror, and

u is the distance of the object from the mirror.

Calculation:

In this case, the object distance u is equal to the focal length f, so we can simplify the formula to:

1/f = 1/v + 1/f ----(1)

From sign convention u = -f

Putting u = -f in equation (1), we get,

v = f/2

This means that the image is formed at a distance of half the focal length from the mirror.

Now, to find the height of the image, we can use the magnification formula:

m = -v/u

where m is the magnification of the image. Since the image formed by a convex mirror is always virtual and upright, the magnification is negative.

Substituting the values of v and u, we get:

m = -f/2f = -1/2

This means that the image is half the size of the object, and since the object is 1 meter tall, the height of the image will be:

Height of image = m * height of object = (-1/2) * 1m = -0.5m

Therefore, the height of the image is 0.5 meters or 50 centimeters.

The correct answer is option (4)

CONCEPT:

Dual nature of matter:

• According to de Broglie, the matter has a dual nature of wave-particle.
• The wave associated with each moving particle is called matter waves.
• ​​de Broglie wavelength associated with the particle

Where, h = Planck's constant, m = mass of a particle and v = velocity of a particle

Work-Energy Theorem:

• The work-energy theorem states that the net work done by the forces on an object is equal to the change in its kinetic energy.

​⇒ W = ΔKE

Where W = work done and ΔKE = change in kinetic energy

CALCULATION:

Given V = 2 volt, charge of an electron e = 1.6×10^-19 C and m = 9.1×10^-31 kg

• We know that if a charge of e coulombs is moved through a potential difference of V volts then the work done is given as,

⇒ W = eV -----(1)

By work-energy theorem, the work done in accelerating an electron through the electric field will be equal to the kinetic energy of the electron,

Concept:

Uniform Magnetic Field:

• A uniform magnetic field is a magnetic field that has the same magnitude and direction throughout the region under consideration, thus the field lines need to be both parallel and spaced out evenly.
• It is symbolled as B.
• It is a vector quantity.

Electric field lines show the direction of the electric field at the point. If the electric field lines were tangential, parallel, or opposite to the equipotential surface, a tangential field will exist on the surface and work done in moving a charge on the surface is not zero.

Therefore electric field lines are always perpendicular to the equipotential surface.

Aluminium electrolytic capacitors have the Aluminium foil anode (positive terminal) which is attached and covered with a layer of Aluminium Oxide which acts as a dielectric. The whole assembly is covered using a paper separator soaked in electrolyte such as, Borax or Glycol and covered by Aluminium foil which acts as cathode ( negative electrode)

Given, the maximum kinetic energy: K_max​=4eV

If V0​ be the stopping potential, then K_max​=eV₀​

⇒eV_0​=4eV 

⇒V₀​=4V