Physics Test 231

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Physics Test 231

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Question 1
4 / -1

Directions For Questions

Consider the following statements:

1. SI unit of magnetisation is Am^-3

2. SI unit of permeability TmA^-1

3. SI unit of magnetic field Tesla

4. SI unit of susceptibility is Am^-1

...view full instructions

Which of the above statements are correct

A
1) and 3) only

B
2) and 3) only

C
2) and 4) only

D
1), 2), 3) and 4)

Solution

Hence option 2) is correct.
Question 2
4 / -1

Directions For Questions

Consider the following the statements

1. Meter bridge works on the principal of the Wheatstone bridge.

2. A Meter bridge is used to find the unknown resistance.

3. A meter bridge is used to find the internal resistance of the battery.

...view full instructions

Which of the following statement is INCORRECT?

A
Only 1

B
Only 2

C
Only 3

D
All 1, 2 and 3
Solution
EXPLANATION:
- From above it clear that the meter bridge works on the principle of Wheatstone bridge. Therefore, option 1 is correct.
- The Meter Bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance. Therefore option 2 is correct.
- The Meter bridge can't measure the internal resistance of a battery.
- A potentiometer is a device used to measure the internal resistance of the battery. Therefore option 3 is incorrect.
Question 3
4 / -1

For the given circuit diagram, if the current flow in the galvanometer G is zero, find the relation between resistances P, Q, R, and S.

A

B

C
Both 1 and 2 are correct

D
Neither 1 nor 2 is correct

Solution
Question 4
4 / -1

In the given circuit, the capacitors C₁ = 4 μF, C₂ = 8 μF, C₃ = 3 μF, C₄ = 6 μF are connected to a voltage source 60V. Then the ratio of charge across the capacitors of two arms AB and DC is

A
4 : 3

B
2 : 3

C
1 : 1

D
2 : 5
Solution
The correct answer is option 1) i.e. 4 : 3

CONCEPT:
- Capacitor: A capacitor is an electrical component with two terminals used to store charge in the form of an electrostatic field in it.
  
  It consists of two parallel plates each possessing equal and opposite charges, separated by a dielectric constant.
  
  Capacitance is the ability of a capacitor to store charge in it. The capacitance C is related to the charge Q and voltage V across them as:
Question 5
4 / -1

Electric field intensity due to thin infinite parallel sheets of charge in region 1 is

( a field pointing from left to right is taken as positive and the one pointing from right to left is taken as negative. )

A

B
0

C

D
Solution
Where ϵ_o = Absolute electrical permittivity of free space, E = Electric field, and σ = surface charge density.

EXPLANATION:

Let σ₁ = Uniform surface density of charge on A, σ₂ = Uniform surface density of charge on B, E₁, E₂ = Electric field intensities at a point due to charged sheet A and B respectively.
- The arrangement shows three regions I, II, and III.
- We apply the superposition principle to calculate the net field intensity in the three regions. As a matter of convention, a field pointing from left to right is taken as positive and the one pointing from right to left is taken as negative.
Question 6
4 / -1

An inductance L, a capacitance C and a resistance R may be connected to an AC source of angular frequency ω, in three different combinations of RC, RL and LC in series. Assume that

The power drawn by the three combinations are P₁, P₂, P₃, respectively. Then

A
P₁ > P₂ > P₃

B
P₁ = P₂ < P₃

C
P₁ = P₂ > P₃

D
P₁ = P₂ = P₃
Solution
Concept:

The power dissipated in the RC circuit depends on both the resistor and the capacitor.

The impedance for the RC circuit is given by:

Z_RC = [(R₂ + (1/ωC)²)]^1/2

The power factor for an RC circuit is: cos(φRC) = R / Z_RC

Explanation:
- The power dissipated in the RC circuit is:
  
  P1 = (V²_rmsR) / Z²RC
  
  The power dissipated in the RL circuit depends on both the resistor and the inductor.
- The impedance for the RL circuit is given by:
  
  Z_RL = [(R₂ + (ωL)²)]^1/2
  
  The power factor for an RL circuit is:
  
  cos(φRL) = R / Z_RL
- The power dissipated in the RL circuit is:
  
  P2 = (V²_rms R) / Z²_RL
- In an LC circuit, the inductive reactance and capacitive reactance cancel each other out when
  
  ωL = 1/ωC.
- This results in a purely reactive circuit with no real power dissipation. Hence, the power in the LC circuit is:
  
  P3 = 0
Thus, Both the RC and RL circuits dissipate real power, and because both involve resistive elements, they will dissipate the same amount of power.
The LC circuit, being purely reactive, dissipates no real power.
P1 = P2 > P3

∴ The correct option is 3
Question 7
4 / -1

Which of the following processes represent a gamma decay ?

A
^AX_Z + λ → ^AX_Z–1 + a + b

B
^AX_Z +¹n₀ → ^A–3X_Z–2 + c

C
^AX_Z → ^AX_Z + f

D
^AX_Z + e – 1 → ^AX_Z–1 + g

Solution
Question 8
4 / -1

In an L-R circuit, an inductance of 0.1 H and a resistance of 1 Ω are connected in series with an ac source of voltage V = 5 sin 10t. The phase difference between the current and applied voltage will be

A
π/2

B
π/4

C
π/6

D
0

Solution

Calculation:

The given voltage is V(t) = 5 sin(10t),

where the angular frequency ω is 10 rad/s.

The inductive reactance XL is given by,

⇒ XL = ωL

⇒ XL = 10 × 0.1 = 1 Ω

The total impedance Z of the L-R circuit is given by

⇒ Z = √(R² + XL²)

⇒ Z = √(1² + 1²) = √2 Ω

The phase difference φ between the current and the voltage is given by

⇒ (φ) = tan⁻¹(XL/R)

⇒ φ = tan⁻¹(1/1) = tan⁻¹(1) = π/4 radians

The phase difference between the current and the applied voltage in the L-R circuit is π/4 radians, or 45 degrees.

∴ The correct option is 2.
Question 9
4 / -1

A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic field at the center is

A

B

C

D
Solution
Concept:

Magnetic Field Due to a Single Loop

The magnetic field at the center of a circular loop of radius r carrying a current I is given by.

B = (μ₀ I) / (2r)

Where:
- μ₀ = permeability of free space (4π × 10⁻⁷ Tm/A)
- I = current in the loop
- r = radius of the loop
Calculation:

For a tightly wound coil with varying radius from a to b, we integrate the magnetic field contributions from all the turns.

Consider a differential element of the coil at radius r, contributing a small magnetic field dB at the center,

dB = (μ₀ I / 2r) × (dN / N)

Where dN is the number of turns in a small radial segment between r and r + dr.

The number of turns per unit length (radially) is N / (b - a),

dN = (N / (b - a)) dr

Integrating to Find the Total Magnetic Field

The total magnetic field at the center of the spiral coil is given by integrating from a to b,

⇒ B = ∫ (μ₀ I / 2r) × (N / (b - a)) dr

⇒ B = (μ₀ I N / (2(b - a))) ∫ (1 / r) dr

Integrating (1 / r) from a to b gives,

⇒ B = (μ₀ I N / (2(b - a))) [ln(b) - ln(a)]

⇒ B = (μ₀ I N / (2(b - a))) ln(b/a)

⇒ B = (μ₀ I N / (2(b - a))) ln(b/a)

∴ The correct answer is 3
Question 10
4 / -1

A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam incident normally on the face AB is totally reflected to reach the face BC, if

A
sin θ ≥ 8/9

B
2/3 < sin θ < 8/9

C
sin θ < 2/3

D
None of these

Solution

Concept:

Total internal reflection occurs when the angle of incidence inside the glass prism exceeds the critical angle, θc, for the interface between the glass and the water.

The critical angle θc is given by: sin(θc) = n₂ / n₁

Where:

n₂ is the refractive index of the medium outside (water),
n₁ is the refractive index of the medium inside (glass).

Calculation:

Given values: sin(θc) = (4/3) / 1.5 = (4/3) × (2/3) = 8/9

Now, calculate θc,

θc = sin-1(8/9)

Given that sin(θc) ≈ 0.8889,

θc ≈ 62.73°

The light beam will be totally internally reflected at the face BC if the angle of incidence at that face exceeds approximately 62.73°.

∴ The correct option is 1.