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CONCEPT:

Two-Body Spring-Mass System

• When two masses $m_1$ and $m_2$ are connected by a spring and oscillate, the system can be treated as a single body with an effective mass called the reduced mass ($\mu$).
• The reduced mass ($\mu$) is calculated as:

$\mu = \frac{m_1 \cdot m_2}{m_1 + m_2}$

• The time period ($T$) of the oscillations for such a system is given by the formula:

$T = 2\pi \sqrt{\frac{\mu}{K}}$

where $K$ is the spring constant.

EXPLANATION:

• Given parameters:
• Mass of the first body ($m_1$) = 1 kg
• Mass of the second body ($m_2$) = 4 kg
• Spring constant ($K$) = 5 N/m
• Step 1: Calculate the reduced mass ($\mu$):
• $\mu = \frac{1 \times 4}{1 + 4}$
• $\mu = \frac{4}{5} \text{ kg}$
• Step 2: Calculate the time period ($T$):
• Using the formula $T = 2\pi \sqrt{\frac{\mu}{K}}$:
• $T = 2\pi \sqrt{\frac{4/5}{5}}$
• $T = 2\pi \sqrt{\frac{4}{25}}$
• $T = 2\pi \times \frac{2}{5}$
• $T = \frac{4\pi}{5} \text{ s}$

Therefore, the time period of the system is $\frac{4\pi}{5} \text{ s}$.

CONCEPT:

Mutual Induction and EMF

• Mutual Induction is the property of two coils such that a change in current in one coil induces an electromotive force (EMF) in the neighboring coil.
• The magnitude of the induced EMF ($e$) is directly proportional to the rate of change of current ($dI/dt$) in the first coil.
• The mathematical relationship is expressed as:

$e = M \frac{dI}{dt}$

• Where:
• $e$ is the induced EMF in Volts (V).
• $M$ is the coefficient of mutual inductance in Henrys (H).
• $dI$ is the change in current in Amperes (A).
• $dt$ is the time interval in seconds (s).

EXPLANATION:

• From the problem, we have the following given values:
• Mutual Inductance ($M$) = $2\text{ H}$
• Change in current ($dI$) = $10\text{ A} - 0\text{ A} = 10\text{ A}$
• Time interval ($dt$) = $0.5\text{ s}$
• To find the induced EMF, we substitute these values into the mutual induction formula:
• $e = M \times \frac{dI}{dt}$
• $e = 2 \times \frac{10}{0.5}$
• Calculating the rate of change of current:
• $\frac{10}{0.5} = 20\text{ A/s}$
• Multiplying by the mutual inductance:
• $e = 2 \times 20$
• $e = 40\text{ V}$

Therefore, the magnitude of the induced EMF is 40 V.

CONCEPT:

Fringe Width in Young's Double Slit Experiment (YDSE)

• Fringe width (β) is defined as the distance between two consecutive bright or dark fringes on the interference pattern.
• The formula to calculate the fringe width is:

β = (λD) / d

• Where:
• λ is the wavelength of the light used.
• D is the distance between the slits and the screen.
• d is the distance between the two slits (slit separation).

EXPLANATION:

• Given parameters:
• Distance between slits (d) = 2 mm = 2 × 10^-3 m
• Distance between slits and screen (D) = 1.6 m
• Wavelength of light (λ) = 500 nm = 500 × 10^-9 m
• Substituting the values into the fringe width formula:
• β = (500 × 10^-9 m × 1.6 m) / (2 × 10^-3 m)
• β = (800 × 10^-9) / (2 × 10^-3)
• β = 400 × 10^-6 m
• Converting the result into millimeters (mm):
• β = 0.4 × 10^-3 m
• β = 0.4 mm

Therefore, the fringe width is 0.4 mm.

CONCEPT:

Radius of a Charged Particle in a Magnetic Field

• When a charged particle moves perpendicular to a uniform magnetic field, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path.
• The radius (r) of this circular trajectory is given by the formula:

r = (m × V) / (q × B)

• Where:
• m is the mass of the particle.
• V is the velocity of the particle.
• q is the charge of the particle.
• B is the magnetic field strength.

EXPLANATION:

• Given parameters for the proton:
• Velocity (V) = 4 × 10⁵ m/s
• Magnetic Field (B) = 0.01 T
• Mass of a proton (m) ≈ 1.67 × 10^-27 kg
• Charge of a proton (q) ≈ 1.6 × 10^-19 C
• Substituting the values into the radius formula:
• r = (1.67 × 10^-27 kg × 4 × 10⁵ m/s) / (1.6 × 10^-19 C × 0.01 T)
• r = (6.68 × 10^-22) / (1.6 × 10^-21)
• r = 4.175 × 10^-1 m
• r ≈ 0.4175 m
• The calculated radius is approximately 0.4 m.

Therefore, the radius of the trajectory of the proton is 0.4 m.

CONCEPT:

Telescope Length and Lens Power

• The power (P) of a lens is the reciprocal of its focal length (f). When power is expressed in Diopters (D), the focal length is in meters (m):

f = 1 / P

• For an astronomical telescope in normal adjustment (where the final image is formed at infinity), the length of the telescope (L) is defined as the distance between the objective lens and the eye lens.
• The formula for the length of a telescope in normal adjustment is:

L = f_o + f_e

where f_o is the focal length of the objective lens and f_e is the focal length of the eye lens.

EXPLANATION:

• Given values:
• Power of objective lens (P_o) = 2 D
• Power of eye lens (P_e) = 20 D
• Step 1: Calculate the focal length of the objective lens (f_o):
• f_o = 1 / P_o = 1 / 2 m
• f_o = 0.5 m = 50 cm
• Step 2: Calculate the focal length of the eye lens (f_e):
• f_e = 1 / P_e = 1 / 20 m
• f_e = 0.05 m = 5 cm
• Step 3: Calculate the length of the telescope (L):
• L = f_o + f_e
• L = 50 cm + 5 cm
• L = 55 cm

Therefore, the length of the telescope is 55 cm.

CONCEPT:

Moment of Inertia (M.O.I) of a Thin Rod

• The moment of inertia (I) of a uniform thin rod of mass (M) and length (L) about an axis passing through its center of mass (COM) and perpendicular to its length is defined by the formula:

$I = \frac{1}{12} ML^{2}$

• It is important to ensure all units are in the SI system (Mass in kg and Length in m) before calculation.

EXPLANATION:

• Given data:
• Mass (M) = 100 g = 0.1 kg (since 1000 g = 1 kg)
• Length (L) = 0.3 m
• Substituting the given values into the M.O.I formula for a rod about its COM:
• $I = \frac{1}{12} \times 0.1 \times (0.3)^{2}$
• $I = \frac{1}{12} \times 0.1 \times 0.09$
• $I = \frac{0.009}{12}$
• $I = 0.00075 \text{ kg.m}^{2}$
• Converting the result into scientific notation:
• $I = 7.5 \times 10^{-4} \text{ kg.m}^{2}$

Therefore, the Moment of Inertia of the thin rod about its center of mass is 7.5 × 10^-4 kg.m².

CONCEPT:

Terminal Voltage (V) and Internal Resistance

• Terminal voltage is the potential difference across the terminals of a cell when current is being drawn from it.
• The total current (I) in a circuit with an external resistance (R), a cell of electromotive force (EMF, E), and internal resistance (r) is given by:

I = E / (R + r)

• The terminal voltage (V) can be calculated using either the external circuit parameters or the internal cell parameters:

V = I × R

OR

V = E - Ir

EXPLANATION:

• In the given problem:
• EMF (E) = 12 V
• Internal resistance (r) = 1 Ω
• External resistance (R) = 5 Ω
• Step 1: Calculate the total current (I) flowing through the circuit:
• I = E / (R + r)
• I = 12 / (5 + 1)
• I = 12 / 6
• I = 2 A
• Step 2: Calculate the terminal voltage (V) using the external resistance:
• V = I × R
• V = 2 A × 5 Ω
• V = 10 V
• Alternatively, calculating using the EMF and internal voltage drop:
• V = E - Ir
• V = 12 V - (2 A × 1 Ω)
• V = 12 V - 2 V
• V = 10 V

Therefore, the terminal voltage is 10 V.

CONCEPT:

Mobility (μ)

• Mobility is defined as the magnitude of the drift velocity per unit electric field.
• Mathematically, it is expressed as:

μ=vdEμ=vdE

where vdvd

is the drift velocity and EE

is the electric field.

• The SI unit of mobility is m2V−1s−1m2V−1s−1

or m⋅s−1⋅N−1⋅Cm⋅s−1⋅N−1⋅C

.

EXPLANATION:

• To find the dimensions of mobility (μμ

), we use the formula:

μ=vdEμ=vdE

• Dimensions of Drift Velocity (vdvd

):

[vd]=[LT−1][vd]=[LT−1]

• Dimensions of Electric Field (EE

):

• Electric field is force per unit charge: E=FqE=Fq
• Dimensions of Force (FF

) = [MLT−2][MLT−2]

• Dimensions of Charge (q=I×tq=I×t

) = [AT][AT]

• So, [E]=[MLT−2][AT]=[MLT−3A−1][E]=[MLT−2][AT]=[MLT−3A−1]
• Dimensions of Mobility (μμ

):

• [μ]=[vd][E][μ]=[vd][E]
• [μ]=[LT−1][MLT−3A−1][μ]=[LT−1][MLT−3A−1]
• [μ]=[M−1L1−1T−1−(−3)A1][μ]=[M−1L1−1T−1−(−3)A1]
• [μ]=[M−1L0T2A][μ]=[M−1L0T2A]

Therefore, the dimension of mobility is [M−1L0T2A][M−1L0T2A]

.

CONCEPT:

Relationship between Electric and Magnetic Fields in Electromagnetic Waves

• In an electromagnetic wave traveling through a vacuum or air, the electric field (E) and the magnetic field (B) amplitudes are related by the speed of light (c).
• The relationship is expressed by the formula:

c = E / B

where c is approximately 3 × 10⁸ m/s.

• From this, the magnitude of the magnetic field can be calculated using:

B = E / c

EXPLANATION:

• Identify the given parameters:
• Electric field amplitude (E) = 5.6 V/m
• Frequency (f) = 50 MHz (Note: The frequency is not required for determining the ratio between E and B).
• Speed of light (c) = 3 × 10⁸ m/s
• Substitute the values into the magnetic field formula:
• B = E / c
• B = 5.6 / (3 × 10⁸)
• B = (5.6 / 3) × 10^-8
• B ≈ 1.866... × 10^-8 T
• Rounding the result to two significant figures to match the provided choices:
• B ≈ 1.9 × 10^-8 T

Therefore, the value of B is 1.9 × 10^-8 T.

CONCEPT:

Fundamental Frequency of a Pipe Closed at One End

• In a pipe closed at one end, the fundamental mode of vibration consists of a node at the closed end and an antinode at the open end.
• The length (L) of the pipe is equal to one-fourth of the wavelength (λ), i.e., L = λ/4.
• The fundamental frequency (f) is given by the formula:

f = v / 4L

where:

• v is the speed of sound in the medium.
• L is the length of the pipe.

EXPLANATION:

• Given parameters:
• Velocity of sound (v) = 330 m/s
• Length of the pipe (L) = 55 cm = 0.55 m
• Applying the values to the fundamental frequency formula:
• f = v / (4 × L)
• f = 330 / (4 × 0.55)
• f = 330 / 2.2
• f = 150 Hz

Therefore, the fundamental frequency of the pipe is 150 Hz.