Solution: Oscillating Block on a Spring

STEP 1 - Given Data

Key interpretation: Moving between the two endpoints (extreme to extreme) is half of a full oscillation.

$$\frac{T}{2} = 0.15 \text{ s} \implies T = 0.30 \text{ s}$$

STEP 2 - Solve

(a) Frequency

$$f = \frac{1}{T} = \frac{1}{0.30} = 3.33 \text{ Hz}$$

(b) Amplitude

The endpoints are separated by $2A$:

$$2A = 0.40 \text{ m} \implies A = 0.20 \text{ m}$$

(c) Force Constant

$$T = 2\pi\sqrt{\frac{m}{k}} \implies k = \frac{4\pi^2 m}{T^2}$$

$$k = \frac{4 \times (3.14159)^2 \times 0.25}{(0.30)^2} = \frac{9.8696}{0.09} = 109.66 \text{ N/m}$$

Summary

Digital Circuit: $y = (AB + \bar{C}BA)(B + C)$

Step 1: Extract Given Data

Boolean expression: $$y = (AB + \bar{C}BA)(B + C)$$

Inputs: $A$, $B$, $C$ Gates allowed: OR, AND, NOT

Step 2: Simplify the Expression

The first bracket: $$AB + \bar{C}BA = AB + AB\bar{C} = AB(1 + \bar{C}) = AB$$

(using $1 + \bar{C} = 1$)

So: $$y = AB(B + C)$$

Expand: $$y = AB\cdot B + AB\cdot C = ABB + ABC$$

Since $B\cdot B = B$: $$y = AB + ABC = AB(1 + C) = AB$$

Therefore: $$\boxed{y = AB}$$

Step 3: Circuit Diagram (drawn as per given expression)

                        AB
A ──┬──────────[AND1]──────────┐
    │            │             │
B ──┼──┬─────────┘             ├─[OR1]─┐
    │  │                       │       │
    │  │   ┌─[AND2]────────────┘       │
    │  │   │  ABC̄                      ├─[AND3]── y
C ──┼──┼─[NOT]                          │
    │  │   (C̄)                          │
    │  └───────────────[OR2]────────────┘
    │                   │  (B+C)
C ──┴───────────────────┘

Gate list:

Step 4: Truth Table

Conclusion

The circuit simplifies to $y = AB$. Output $y = 1$ only when both $A = 1$ and $B = 1$, independent of $C$.

Quantity Value ------ Potential difference $V = 100 \text{ V}$ Charge of alpha particle $q = 3.2 \times 10^{-19} \text{ C}$ Mass of alpha particle $m = 6.70 \times 10^{-27} \text{ kg}$ Initial velocity $u = 0$ (released from rest) Note o...

Hall Voltage Across the Width of a Copper Slab

STEP 1 - Given Data

STEP 2 - Solve

Hall Voltage Formula

$$V_H = \frac{BI}{net}$$

where $t$ is the dimension of the slab parallel to the magnetic field $B$.

Since $B$ is perpendicular to the plane of the slab (i.e. through the thickness), the relevant dimension in the formula is the thickness $t = 0.5\times10^{-2}\ \text{m}$, and the Hall voltage develops across the width $w$.

Calculation

Numerator: $$B \cdot I = 1.5 \times 50 = 75$$

Denominator: $$n \cdot e \cdot t = (8.4 \times 10^{28})(1.6 \times 10^{-19})(0.5 \times 10^{-2})$$

$$= 8.4 \times 1.6 \times 0.5 \times 10^{28-19-2} = 6.72 \times 10^{7}$$

Hall Voltage: $$V_H = \frac{75}{6.72 \times 10^{7}} = 1.116 \times 10^{-6}\ \text{V}$$

$$\boxed{V_H \approx 1.12\ \mu\text{V}}$$

Result

The magnitude of the Hall voltage across the width of the copper slab is approximately

$$V_H \approx 1.12\ \mu\text{V}$$

This very small value is characteristic of a metal such as copper, where the high free electron concentration produces a very small drift velocity and hence a tiny Hall voltage.

• Kinetic energy of α particle: $E = 6.8$ MeV - Radius of emitting nucleus: $r = 8 \times 10^{-15}$ m Constants used: - Mass of α particle: $m = 4u = 4 \times 1.66 \times 10^{-27} = 6.64 \times 10^{-27}$ kg - Planck's constant: $h = 6.62...

Solution: Oscillating Block on a Spring

STEP 1 - Given Data

Key interpretation: Moving between the two endpoints (extreme to extreme) is half of a full oscillation.

$$\frac{T}{2} = 0.15 \text{ s} \implies T = 0.30 \text{ s}$$

STEP 2 - Solve

(a) Frequency

$$f = \frac{1}{T} = \frac{1}{0.30} = 3.33 \text{ Hz}$$

(b) Amplitude

The endpoints are separated by $2A$:

$$2A = 0.40 \text{ m} \implies A = 0.20 \text{ m}$$

(c) Force Constant

$$T = 2\pi\sqrt{\frac{m}{k}} \implies k = \frac{4\pi^2 m}{T^2}$$

$$k = \frac{4 \times (3.14159)^2 \times 0.25}{(0.30)^2} = \frac{9.8696}{0.09} = 109.66 \text{ N/m}$$

Summary

Digital Circuit: $y = (AB + \bar{C}BA)(B + C)$

Step 1: Extract Given Data

Boolean expression: $$y = (AB + \bar{C}BA)(B + C)$$

Inputs: $A$, $B$, $C$ Gates allowed: OR, AND, NOT

Step 2: Simplify the Expression

The first bracket: $$AB + \bar{C}BA = AB + AB\bar{C} = AB(1 + \bar{C}) = AB$$

(using $1 + \bar{C} = 1$)

So: $$y = AB(B + C)$$

Expand: $$y = AB\cdot B + AB\cdot C = ABB + ABC$$

Since $B\cdot B = B$: $$y = AB + ABC = AB(1 + C) = AB$$

Therefore: $$\boxed{y = AB}$$

Step 3: Circuit Diagram (drawn as per given expression)

                        AB
A ──┬──────────[AND1]──────────┐
    │            │             │
B ──┼──┬─────────┘             ├─[OR1]─┐
    │  │                       │       │
    │  │   ┌─[AND2]────────────┘       │
    │  │   │  ABC̄                      ├─[AND3]── y
C ──┼──┼─[NOT]                          │
    │  │   (C̄)                          │
    │  └───────────────[OR2]────────────┘
    │                   │  (B+C)
C ──┴───────────────────┘

Gate list:

Step 4: Truth Table

Conclusion

The circuit simplifies to $y = AB$. Output $y = 1$ only when both $A = 1$ and $B = 1$, independent of $C$.

Quantity Value ------ Potential difference $V = 100 \text{ V}$ Charge of alpha particle $q = 3.2 \times 10^{-19} \text{ C}$ Mass of alpha particle $m = 6.70 \times 10^{-27} \text{ kg}$ Initial velocity $u = 0$ (released from rest) Note o...

Hall Voltage Across the Width of a Copper Slab

STEP 1 - Given Data

STEP 2 - Solve

Hall Voltage Formula

$$V_H = \frac{BI}{net}$$

where $t$ is the dimension of the slab parallel to the magnetic field $B$.

Since $B$ is perpendicular to the plane of the slab (i.e. through the thickness), the relevant dimension in the formula is the thickness $t = 0.5\times10^{-2}\ \text{m}$, and the Hall voltage develops across the width $w$.

Calculation

Numerator: $$B \cdot I = 1.5 \times 50 = 75$$

Denominator: $$n \cdot e \cdot t = (8.4 \times 10^{28})(1.6 \times 10^{-19})(0.5 \times 10^{-2})$$

$$= 8.4 \times 1.6 \times 0.5 \times 10^{28-19-2} = 6.72 \times 10^{7}$$

Hall Voltage: $$V_H = \frac{75}{6.72 \times 10^{7}} = 1.116 \times 10^{-6}\ \text{V}$$

$$\boxed{V_H \approx 1.12\ \mu\text{V}}$$

Result

The magnitude of the Hall voltage across the width of the copper slab is approximately

$$V_H \approx 1.12\ \mu\text{V}$$

This very small value is characteristic of a metal such as copper, where the high free electron concentration produces a very small drift velocity and hence a tiny Hall voltage.

• Kinetic energy of α particle: $E = 6.8$ MeV - Radius of emitting nucleus: $r = 8 \times 10^{-15}$ m Constants used: - Mass of α particle: $m = 4u = 4 \times 1.66 \times 10^{-27} = 6.64 \times 10^{-27}$ kg - Planck's constant: $h = 6.62...

Consider a free particle moving along the x-direction. The generalized wavefunction is taken as: $$\psi = A e^{i(kx - \omega t)} \quad \cdots (1)$$ where: - $k$ = wave number (propagation constant) - $\omega$ = angular frequency - $A$ = ...

Stern-Gerlach Experiment: Beam Separation

Given Data

Physics

Force on each spin component: $F = \mu_B \dfrac{dB}{dz}$ (magnitude), directed oppositely for $m_s = \pm\tfrac12$.

Step 1: Force

$$F = (9.274\times 10^{-24})(3\times 10^4) = 2.782\times 10^{-19}\text{ N}$$

Step 2: Acceleration

$$a = \frac{F}{m} = \frac{2.782\times 10^{-19}}{1.67\times 10^{-27}} = 1.666\times 10^{8}\text{ m/s}^2$$

Step 3: Transit time

$$t = \frac{L}{v} = \frac{0.20}{10^4} = 2\times 10^{-5}\text{ s}$$

Step 4: Deflection of one component

$$z = \tfrac12 a t^2 = \tfrac12 (1.666\times 10^{8})(2\times 10^{-5})^2$$ $$= \tfrac12(1.666\times 10^{8})(4\times 10^{-10}) = 3.33\times 10^{-2}\text{ m}$$

Step 5: Total separation

$$\Delta z = 2z = 6.66\times 10^{-2}\text{ m} \approx 6.66\text{ cm}$$

$$\boxed{\Delta z \approx 6.66\text{ cm}}$$

The two components are separated by about 6.66 cm on emerging from the magnet.

--- Oxidation is one of the key steps in IC fabrication. In this process, the silicon wafer is exposed to an oxidizing environment (oxygen or steam) at high temperatures (around 900°C-1200°C) to grow a thin layer of silicon dioxide (SiO₂...

In an extrinsic semiconductor, impurity atoms (donor or acceptor) are added to a pure semiconductor. These impurity atoms have an ionization energy (also called activation energy) $Ei$, which is the energy required to ionize the impurity...

Length of the Unit Cube for Sodium (BCC)

Step 1 - Extract: Given Data

Step 2 - Solve

Atoms per unit cell (BCC): $$n = 8\times\tfrac{1}{8} + 1 = 2$$

Density relation: $$\rho = \frac{nM}{N_a,a^3}\quad\Rightarrow\quad a^3 = \frac{nM}{\rho N_a}$$

Substitute: $$a^3 = \frac{2\times 23}{0.971\times 6.022\times 10^{23}} = \frac{46}{5.847\times 10^{23}}$$

$$a^3 = 7.867\times 10^{-23}\ \text{cm}^3$$

Cube root: $$a = (7.867\times 10^{-23})^{1/3} = (78.67\times 10^{-24})^{1/3}\ \text{cm}$$

$$78.67^{1/3} \approx 4.286,\qquad (10^{-24})^{1/3} = 10^{-8}$$

$$\boxed{a \approx 4.29\times 10^{-8}\ \text{cm} = 4.29\ \text{Å}}$$

Result

The length of the unit cube of sodium is approximately $4.29\ \text{Å};(4.29\times10^{-8}\ \text{cm})$, which agrees well with the accepted experimental value of ~4.28 Å.

Quantity Value ------ Moment of inertia, $I$ $0.5\ \text{kg m}^2$ Initial rotation rate, $n0$ $2\ \text{rev/s}$ Final rotation rate, $n$ $0\ \text{rev/s}$ Friction torque, $T$ $0.4\ \text{N m}$ Newton's second law for rotation: $T = I\al...

Torque on a Current-Carrying Rectangular Loop in a Magnetic Field

Setup and Description

Consider a rectangular loop of wire ABCD of dimensions a × b (width × length), carrying current I, mounted on a pivot rod (axis of rotation), and placed in a uniform external magnetic field B.

Let the plane of the loop make an angle φ with the magnetic field direction, so the normal to the loop makes an angle θ with B, where θ + φ = 90°.

Forces on Each Side of the Loop

Using the force law on a current-carrying conductor:

$$\vec{F} = I(\vec{L} \times \vec{B})$$

or magnitude: $F = BIL\sin\alpha$, where $\alpha$ is the angle between the current direction and B.

Side AB and Side CD (parallel to the pivot axis, length = b)

• The current in AB and CD flows parallel to the pivot rod axis.
• The force on each is:

$$F_{AB} = BIb\sin 90° = BIb$$ $$F_{CD} = BIb\sin 90° = BIb$$

• These two forces are equal in magnitude but opposite in direction, forming a couple (they do not cancel because they act on opposite sides).

Side BC and Side DA (perpendicular to the pivot axis, length = a)

• The forces on BC and DA are equal, opposite, and collinear along the pivot axis.
• They produce no net torque (they cancel each other).

Calculation of Torque

The two forces on AB and CD (each of magnitude $F = BIb$) act at perpendicular distances from the pivot axis.

The perpendicular distance between the two forces depends on the orientation of the loop. If the normal to the loop makes angle θ with B, the moment arm for each force is $\frac{a}{2}\sin\theta$.

Therefore, the net torque is:

$$\tau = F \cdot a\sin\theta$$

$$\tau = BIb \cdot a\sin\theta$$

$$\boxed{\tau = BIA\sin\theta}$$

where $A = ab$ is the area of the loop.

Expression in Terms of Magnetic Dipole Moment

The magnetic dipole moment of the current loop is defined as:

$$\vec{\mu} = I\vec{A} = IA\hat{n}$$

where $\hat{n}$ is the unit normal to the loop (direction given by the right-hand rule), and $A$ is the area of the loop.

The torque can then be written in vector form as:

$$\boxed{\vec{\tau} = \vec{\mu} \times \vec{B}}$$

with magnitude:

$$\tau = \mu B\sin\theta = IAB\sin\theta$$

Special Cases

Potential Energy of the Dipole

The potential energy of the magnetic dipole in the field is:

$$E_p = -\vec{\mu} \cdot \vec{B} = -\mu B\cos\theta$$

• Minimum energy (stable) at θ = 0°: $E_p = -\mu B$
• Maximum energy (unstable) at θ = 180°: $E_p = +\mu B$

Alternative Ways to Increase the Torque on the Coil

From the expression $\tau = NIAB\sin\theta$, the torque can be increased by:

1. Increasing the number of turns N: For a coil of N turns, $\tau = NIAB\sin\theta$. More turns means greater torque for the same current and field.

2. Increasing the current I: A larger current directly increases the magnetic dipole moment $\mu = NIA$, hence increasing torque.

3. Increasing the area A of the loop: A larger loop area (larger $a \times b$) increases the torque proportionally.

4. Increasing the magnetic field strength B: Using a stronger external magnetic field increases the torque directly.

5. Orienting the loop so that θ = 90°: The torque is maximum when the plane of the loop is parallel to B (i.e., the normal to the loop is perpendicular to B), giving $\tau_{max} = NIAB$.

6. Using a ferromagnetic core inside the coil: This concentrates the magnetic field lines and effectively increases the flux, thereby increasing the torque (used in practical galvanometers and motors).

Summary

The torque on a current-carrying rectangular loop in a magnetic field is:

$$\tau = NIAB\sin\theta$$

This principle is the fundamental operating principle of electric motors and galvanometers, where electrical energy is converted to mechanical rotation through the torque produced on a current loop in a magnetic field.

In a crystalline solid (semiconductor), electrons and holes do not move as free particles. They experience the periodic potential of the crystal lattice. To simplify the analysis, we use the concept of effective mass, which accounts for ...

Heisenberg Uncertainty Principle

Step 1 - Given Data

• Mass of small particle: $m = 10^{-6}$ g $= 10^{-6} \times 10^{-3}$ kg $= 10^{-9}$ kg
• Uncertainty in speed: $\Delta v_x = 10^{-6}$ m/s
• Mass of electron: $m_e = 9.11 \times 10^{-31}$ kg
• Planck's constant: $h = 6.626 \times 10^{-34}$ J·s

Step 2 - Solution

The uncertainty principle:

$$\Delta x \cdot \Delta p_x \geq \frac{h}{4\pi}$$

With $\Delta p_x = m,\Delta v_x$, the minimum position uncertainty is:

$$\Delta x = \frac{h}{4\pi, m, \Delta v_x}$$

Part (a): Small particle

$$\Delta x = \frac{6.626 \times 10^{-34}}{4\pi \times 10^{-9} \times 10^{-6}}$$

Denominator:

$$4\pi \times 10^{-15} = 1.2566 \times 10^{-14}$$

$$\Delta x = \frac{6.626 \times 10^{-34}}{1.2566 \times 10^{-14}} = 5.27 \times 10^{-20}\ \text{m}$$

$$\boxed{\Delta x \approx 5.27 \times 10^{-20}\ \text{m}}$$

This is negligibly small for a macroscopic particle.

Part (b): Electron

$$\Delta x = \frac{6.626 \times 10^{-34}}{4\pi \times 9.11 \times 10^{-31} \times 10^{-6}}$$

Denominator:

$$4\pi \times 9.11 \times 10^{-37} = 12.566 \times 9.11 \times 10^{-37} = 1.1448 \times 10^{-35}$$

$$\Delta x = \frac{6.626 \times 10^{-34}}{1.1448 \times 10^{-35}} = 57.9\ \text{m}$$

$$\boxed{\Delta x \approx 57.9\ \text{m}}$$

Conclusion

The uncertainty principle is significant only at the microscopic scale. For the macroscopic particle the position uncertainty is unmeasurably small, while for the electron it is enormous, reflecting the fundamental quantum limitation on simultaneous knowledge of position and momentum.

(Note: If the simpler form $\Delta x,\Delta p \geq h/2\pi$ or $\hbar$ is used, results differ by a constant factor of 2, but the $h/4\pi$ minimum-uncertainty form used here is standard.)

RTL and TTL Logic Gates: Memory and Clock Circuits

1. RTL (Resistor-Transistor Logic) Gates

Definition

RTL (Resistor-Transistor Logic) is one of the earliest logic gate families where resistors and bipolar junction transistors (BJTs) are used to implement logic functions. The resistors are used for input coupling and the transistors act as switching elements.

Basic RTL NOR Gate Circuit

Vcc (+5V)
     |
    [Rc]
     |
     +-------- Output (Y)
     |
    [Q1]    [Q2]
     |        |
    [R1]    [R2]
     |        |
    A         B

• When either input A or B is HIGH (logic 1), the corresponding transistor saturates (turns ON), pulling output LOW.
• When both inputs are LOW, both transistors are OFF, output is pulled HIGH through Rc.
• This implements a NOR gate: Y = (A + B)'

Truth Table (NOR)

Characteristics of RTL

• Simple and low cost
• Slow switching speed
• High power dissipation
• Low noise immunity
• Voltage levels: Logic 0 = 0V, Logic 1 = 3.6V (approx.)

2. TTL (Transistor-Transistor Logic) Gates

Definition

TTL (Transistor-Transistor Logic) is a logic gate family where multiple-emitter BJTs are used for both input and logic functions. It replaced RTL due to higher speed and better noise immunity. TTL operates on a +5V supply.

Basic TTL NAND Gate Circuit

Vcc (+5V)
     |
    [R1]        [R3]
     |            |
     +---[Q2]----+----[Q4]---- Output (Y)
     |    |      |
    [Q1]  |     [Q3]
   /   \  |      |
  A     B [R2]  GND
           |
          GND

Key components:

• Q1: Multi-emitter input transistor (one emitter per input)
• Q2: Phase splitter transistor
• Q3, Q4: Totem-pole output stage
• R1, R2, R3, R4: Biasing resistors

Working of TTL NAND Gate

Case 1: Any input is LOW (0)

• The emitter of Q1 connected to LOW input conducts
• Q1 base-emitter junction forward biased
• Q2 and Q3 are turned OFF
• Q4 turns ON
• Output is HIGH (logic 1)

Case 2: All inputs are HIGH (1)

• All emitters of Q1 are at HIGH voltage
• Q1 operates in reverse active mode
• Q2 and Q3 turn ON (saturate)
• Q4 turns OFF
• Output is LOW (logic 0)

This implements: Y = (A.B)' (NAND function)

Truth Table (NAND)

Characteristics of TTL

3. Memory Circuit Using Logic Gates (SR Latch)

A memory circuit (latch) stores one bit of information. It can be built using NOR gates (RTL) or NAND gates (TTL).

SR Latch Using NOR Gates (RTL-based)

        S ----[NOR]---- Q
               |   \
               |    \
               +-----[NOR]---- Q'
               |   /
        R ----/

Circuit Diagram:

S ---+--[NOR Gate 1]---+--- Q
     |        ^        |
     |        |        |
     +--------+        |
                       |
R ---+--[NOR Gate 2]---+--- Q'
     |        ^        |
     |        |        |
     +--------+

More precisely:

S ----\
       [NOR1] ---- Q ----\
      /                   [NOR2] ---- Q'
Q' --/                   /
                    R ---/

Working of SR Latch

Explanation:

• When S=1, R=0: NOR1 output goes LOW (Q=0 initially), but feedback