Physics solved practical slips

Solved practical slips for Physics:

Slip 1: Motion in One Dimension

A particle moves along a straight line with an initial velocity of 5 m/s. It accelerates uniformly at a rate of 2 m/s² for 3 seconds. Find its final velocity and displacement.

*Solution:*

Given:

- Initial velocity (u) = 5 m/s

- Acceleration (a) = 2 m/s²

- Time (t) = 3 seconds

Final velocity (v) = u + at

= 5 + (2 × 3)

= 11 m/s

Displacement (s) = ut + (1/2)at²

= (5 × 3) + (1/2) × 2 × 3²

= 15 + 9

= 24 meters

*Slip 2: Motion in Two Dimensions*

A projectile is launched with an initial velocity of 20 m/s at an angle of 60° above the horizontal. Find its maximum height, range, and time of flight.

*Solution:*

Given:

- Initial velocity (u) = 20 m/s

- Angle of projection (θ) = 60°

Vertical component of initial velocity (u_y) = u sin θ

= 20 sin 60°

= 17.32 m/s

Horizontal component of initial velocity (u_x) = u cos θ

= 20 cos 60°

= 10 m/s

Maximum height (h) = (u_y)² / 2g

= (17.32)² / (2 × 9.8)

= 15.53 meters

Range (R) = (u_x) × (2u_y) / g

= (10) × (2 × 17.32) / 9.8

= 35.33 meters

Time of flight (T) = 2u_y / g

= 2 × 17.32 / 9.8

= 3.53 seconds

*Slip 3: Work, Energy, and Power*

A force of 50 N acts on an object, causing it to move 10 meters in the direction of the force. If the object starts from rest and has a mass of 20 kg, find its final velocity and the work done on the object.

*Solution:*

Given:

- Force (F) = 50 N

- Displacement (s) = 10 meters

- Mass (m) = 20 kg

Work done (W) = F × s

= 50 × 10

= 500 Joules

Kinetic energy (KE) = W

= 500 Joules

Final velocity (v) = √(2KE/m)

= √(2 × 500 / 20)

= 10 m/s

*Slip 4: Rotational Motion*

A wheel with a radius of 0.5 meters rotates with an angular velocity of 2 rad/s. Find its linear velocity and centripetal acceleration.

*Solution:*

Given:

- Radius (r) = 0.5 meters

- Angular velocity (ω) = 2 rad/s

Linear velocity (v) = r × ω

= 0.5 × 2

= 1 m/s

Centripetal acceleration (a_c) = v² / r

= (1)² / 0.5

= 2 m/s²

These are just a few examples of solved practical slips for Physics. There are many more topics and problems that can be covered.

Here are some more topics for solved practical slips in Physics:

*Slip 5: Oscillations*

A simple pendulum has a length of 1 meter and a mass of 0.1 kg. Find its time period and frequency.

*Solution:*

Given:

- Length (L) = 1 meter

- Mass (m) = 0.1 kg

Time period (T) = 2π √(L/g)

= 2π √(1/9.8)

= 2 seconds

Frequency (f) = 1/T

= 1/2

= 0.5 Hz

*Slip 6: Waves*

A wave travels with a speed of 10 m/s and has a wavelength of 2 meters. Find its frequency and time period.

*Solution:*

Given:

- Speed (v) = 10 m/s

- Wavelength (λ) = 2 meters

Frequency (f) = v/λ

= 10/2

= 5 Hz

Time period (T) = 1/f

= 1/5

= 0.2 seconds

*Slip 7: Thermodynamics*

A gas expands isothermally from an initial volume of 1 liter to a final volume of 2 liters. Find the work done by the gas.

*Solution:*

Given:

- Initial volume (V1) = 1 liter

- Final volume (V2) = 2 liters

Work done (W) = nRT ln(V2/V1)

= (1 mol) × (8.314 J/mol-K) × (300 K) × ln(2/1)

= 1730 J

*Slip 8: Electromagnetism*

A current-carrying wire has a length of 2 meters and carries a current of 5 A. Find the magnetic field produced by the wire at a distance of 1 meter.

*Solution:*

Given:

- Length (L) = 2 meters

- Current (I) = 5 A

- Distance (r) = 1 meter

Magnetic field (B) = μ₀I/2πr

= (4π × 10⁻⁷ T·m/A) × (5 A) / (2π × 1 m)

= 5 × 10⁻⁷ T

*Slip 9: Optics*

A lens has a focal length of 10 cm and an object is placed at a distance of 20 cm from the lens. Find the image distance and magnification.

*Solution:*

Given:

- Focal length (f) = 10 cm

- Object distance (u) = 20 cm

Image distance (v) = f/(1 - f/u)

= 10/(1 - 10/20)

= 20 cm

Magnification (m) = v/u

= 20/20

= 1

These are just a few more examples of solved practical slips in Physics. There are many more topics and problems that can be covered.

Here are some more problems for solved practical slips in Physics:

*Slip 10: Mechanics*

A block of mass 5 kg is pulled up a frictionless incline with a force of 20 N. The incline makes an angle of 30° with the horizontal. Find the acceleration of the block.

*Solution:*

Given:

- Mass (m) = 5 kg

- Force (F) = 20 N

- Angle (θ) = 30°

Acceleration (a) = F/m sin θ

= 20/5 sin 30°

= 2 m/s²

*Slip 11: Electromagnetism*

A charge of 2 μC is placed at the origin. Find the electric field at a point 3 meters away from the charge.

*Solution:*

Given:

- Charge (q) = 2 μC

- Distance (r) = 3 meters

Electric field (E) = kq/r²

= (9 × 10⁹ N·m²/C²) × (2 × 10⁻⁶ C) / (3 m)²

= 2 × 10⁵ N/C

*Slip 12: Thermodynamics*

A system undergoes a process in which 100 J of heat is added to it. The system also does 50 J of work. Find the change in internal energy of the system.

*Solution:*

Given:

- Heat added (Q) = 100 J

- Work done (W) = 50 J

Change in internal energy (ΔU) = Q - W

= 100 - 50

= 50 J

*Slip 13: Waves*

A wave has a frequency of 50 Hz and a wavelength of 10 meters. Find the speed of the wave.

*Solution:*

Given:

- Frequency (f) = 50 Hz

- Wavelength (λ) = 10 meters

Speed (v) = λf

= 10 × 50

= 500 m/s

*Slip 14: Optics*

A lens has a focal length of 20 cm. An object is placed at a distance of 30 cm from the lens. Find the image distance and magnification.

*Solution:*

Given:

- Focal length (f) = 20 cm

- Object distance (u) = 30 cm

Image distance (v) = f/(1 - f/u)

= 20/(1 - 20/30)

= 60 cm

Magnification (m) = v/u

= 60/30

= 2

*Slip 15: Electricity*

A resistor has a resistance of 10 Ω. A current of 2 A flows through it. Find the voltage across the resistor.

*Solution:*

Given:

- Resistance (R) = 10 Ω

- Current (I) = 2 A

Voltage (V) = IR

= 2 × 10

= 20 V

Here are some more problems for solved practical slips in Physics:

*Slip 16: Magnetism*

A current-carrying wire has a length of 5 meters and carries a current of 3 A. Find the magnetic field produced by the wire at a distance of 2 meters.

*Solution:*

Given:

- Length (L) = 5 meters

- Current (I) = 3 A

- Distance (r) = 2 meters

Magnetic field (B) = μ₀I/2πr

= (4π × 10⁻⁷ T·m/A) × (3 A) / (2π × 2 m)

= 3 × 10⁻⁷ T

*Slip 17: Thermodynamics*

A system undergoes a process in which 200 J of heat is added to it. The system also does 100 J of work. Find the change in internal energy of the system.

*Solution:*

Given:

- Heat added (Q) = 200 J

- Work done (W) = 100 J

Change in internal energy (ΔU) = Q - W

= 200 - 100

= 100 J

*Slip 18: Waves*

A wave has a frequency of 100 Hz and a wavelength of 5 meters. Find the speed of the wave.

*Solution:*

Given:

- Frequency (f) = 100 Hz

- Wavelength (λ) = 5 meters

Speed (v) = λf

= 5 × 100

= 500 m/s

*Slip 19: Optics*

A lens has a focal length of 15 cm. An object is placed at a distance of 20 cm from the lens. Find the image distance and magnification.

*Solution:*

Given:

- Focal length (f) = 15 cm

- Object distance (u) = 20 cm

Image distance (v) = f/(1 - f/u)

= 15/(1 - 15/20)

= 60 cm

Magnification (m) = v/u

= 60/20

= 3

*Slip 20: Electricity*

A resistor has a resistance of 20 Ω. A voltage of 10 V is applied across it. Find the current flowing through the resistor.

*Solution:*

Given:

- Resistance (R) = 20 Ω

- Voltage (V) = 10 V

Current (I) = V/R

= 10/20

= 0.5 A

Here are some extra problems for solved practical slips in Physics:

*Slip 21: Mechanics*

A car accelerates uniformly from rest to a speed of 20 m/s in 4 seconds. Find its acceleration and the distance traveled.

*Solution:*

Given:

- Initial velocity (u) = 0 m/s

- Final velocity (v) = 20 m/s

- Time (t) = 4 seconds

Acceleration (a) = (v - u) / t

= (20 - 0) / 4

= 5 m/s²

Distance traveled (s) = ut + (1/2)at²

= 0 × 4 + (1/2) × 5 × 4²

= 40 meters

*Slip 22: Electromagnetism*

A coil has 100 turns and a radius of 0.1 meters. A current of 2 A flows through it. Find the magnetic field produced by the coil.

*Solution:*

Given:

- Number of turns (N) = 100

- Radius (r) = 0.1 meters

- Current (I) = 2 A

Magnetic field (B) = μ₀NI/2r

= (4π × 10⁻⁷ T·m/A) × 100 × 2 / (2 × 0.1)

= 0.0126 T

*Slip 23: Thermodynamics*

A system undergoes a process in which 500 J of heat is added to it. The system also does 200 J of work. Find the change in internal energy of the system.

*Solution:*

Given:

- Heat added (Q) = 500 J

- Work done (W) = 200 J

Change in internal energy (ΔU) = Q - W

= 500 - 200

= 300 J

*Slip 24: Waves*

A wave has a frequency of 200 Hz and a wavelength of 3 meters. Find the speed of the wave.

*Solution:*

Given:

- Frequency (f) = 200 Hz

- Wavelength (λ) = 3 meters

Speed (v) = λf

= 3 × 200

= 600 m/s

*Slip 25: Optics*

A lens has a focal length of 25 cm. An object is placed at a distance of 30 cm from the lens. Find the image distance and magnification.

*Solution:*

Given:

- Focal length (f) = 25 cm

- Object distance (u) = 30 cm

Image distance (v) = f/(1 - f/u)

= 25/(1 - 25/30)

= 150 cm

Magnification (m) = v/u

= 150/30

= 5

Master's in Computer Science solved practical slips

Here are some solved practical slips for Master's in Computer Science:

Slip 1: Sorting Algorithms

Implement the following sorting algorithms and analyze their time and space complexity:

- Bubble Sort

- Selection Sort

- Insertion Sort

- Merge Sort

- Quick Sort

*Solution:*

#include <iostream>

using namespace std;

// Bubble Sort

void bubbleSort(int arr[], int n) {

    for (int i = 0; i < n - 1; i++) {

        for (int j = 0; j < n - i - 1; j++) {

            if (arr[j] > arr[j + 1]) {

                swap(arr[j], arr[j + 1]);

            }

        }

    }

}

// Selection Sort

void selectionSort(int arr[], int n) {

    for (int i = 0; i < n - 1; i++) {

        int minIndex = i;

        for (int j = i + 1; j < n; j++) {

            if (arr[j] < arr[minIndex]) {

                minIndex = j;

            }

        }

        swap(arr[minIndex], arr[i]);

    }

}

// Insertion Sort

void insertionSort(int arr[], int n) {

    for (int i = 1; i < n; i++) {

        int key = arr[i];

        int j = i - 1;

        while (j >= 0 && arr[j] > key) {

            arr[j + 1] = arr[j];

            j--;

        }

        arr[j + 1] = key;

    }

}

// Merge Sort

void mergeSort(int arr[], int left, int right) {

    if (left < right) {

        int mid = left + (right - left) / 2;

        mergeSort(arr, left, mid);

        mergeSort(arr, mid + 1, right);

        merge(arr, left, mid, right);

    }

}

void merge(int arr[], int left, int mid, int right) {

    int n1 = mid - left + 1;

    int n2 = right - mid;

    int* leftArr = new int[n1];

    int* rightArr = new int[n2];

    for (int i = 0; i < n1; i++) {

        leftArr[i] = arr[left + i];

    }

    for (int j = 0; j < n2; j++) {

        rightArr[j] = arr[mid + 1 + j];

    }

    int i = 0, j = 0, k = left;

    while (i < n1 && j < n2) {

        if (leftArr[i] <= rightArr[j]) {

            arr[k] = leftArr[i];

            i++;

        } else {

            arr[k] = rightArr[j];

            j++;

        }

        k++;

    }

    while (i < n1) {

        arr[k] = leftArr[i];

        i++;

        k++;

    }

    while (j < n2) {

        arr[k] = rightArr[j];

        j++;

        k++;

    }

}

// Quick Sort

int partition(int arr[], int low, int high) {

    int pivot = arr[high];

    int i = low - 1;

    for (int j = low; j < high; j++) {

        if (arr[j] < pivot) {

            i++;

            swap(arr[i], arr[j]);

        }

    }

    swap(arr[i + 1], arr[high]);

    return i + 1;

}

void quickSort(int arr[], int low, int high) {

    if (low < high) {

        int pivotIndex = partition(arr, low, high);

        quickSort(arr, low, pivotIndex - 1);

        quickSort(arr, pivotIndex + 1, high);

    }

}

int main() {

    int arr[] = {64, 34, 25, 12, 22, 11, 90};

    int n = sizeof(arr) / sizeof(arr[0]);

    bubbleSort(arr, n);

    cout << "Sorted array (Bubble Sort): ";

    for (int i = 0; i < n; i++) {

        cout << arr[i] << " ";

    }

    cout << endl;

    return 0;

}

Slip 2: Searching Algorithms

Implement the following searching algorithms and analyze their time and space complexity:

- Linear Search

- Binary Search

*Solution:*

#include <iostream>

using namespace std;

// Linear Search

int linearSearch(int arr[], int n, int target) {

    for (int i = 0; i < n; i++) {

        if (arr[i] == target) {

            return i;

        }

    }

    return -1;

}

// Binary Search

int binarySearch(int arr[], int n, int target) {

    int left = 0;

    int right = n - 1;

    while (left <= right) {

        int mid = left + (right - left) / 2;

        if (arr[mid] == target) {

            return mid;

        } else if (arr[mid] < target) {

            left = mid + 1;

        } else {

            right = mid - 1;

        }

    }

    return -1;

}

int main() {

    int arr[] = {2, 5, 8, 12, 16, 23, 38, 56, 72, 91};

    int n = sizeof(arr) / sizeof(arr[0]);

    int target = 23;

    int result = linearSearch(arr, n, target);

    if (result != -1) {

        cout << "Target found at index " << result << " using Linear Search." << endl;

    } else {

        cout << "Target not found using Linear Search." << endl;

    }

    result = binarySearch(arr, n, target);

    if (result != -1) {

        cout << "Target found at index " << result << " using Binary Search." << endl;

    } else {

        cout << "Target not found using Binary Search." << endl;

    }

    return 0;

}

Slip 3: Graph Traversal

Implement the following graph traversal algorithms:

- Breadth-First Search (BFS)

- Depth-First Search (DFS)

*Solution:*

#include <iostream>

#include <queue>

#include <vector>

using namespace std;

class Graph {

public:

    int numVertices;

    vector<vector<int>> adjList;

    Graph(int numVertices) {

        this->numVertices = numVertices;

        adjList.resize(numVertices);

    }

    void addEdge(int u, int v) {

        adjList[u].push_back(v);

        adjList[v].push_back(u);

    }

    void bfs(int startVertex) {

        vector<bool> visited(numVertices, false);

        queue<int> q;

        visited[startVertex] = true;

        q.push(startVertex);

        while (!q.empty()) {

            int currentVertex = q.front();

            cout << currentVertex << " ";

            q.pop();

            for (int neighbor : adjList[currentVertex]) {

                if (!visited[neighbor]) {

                    visited[neighbor] = true;

                    q.push(neighbor);

                }

            }

        }

    }

    void dfs(int startVertex) {

        vector<bool> visited(numVertices, false);

        dfsHelper(startVertex, visited);

    }

    void dfsHelper(int currentVertex, vector<bool>& visited) {

        visited[currentVertex] = true;

        cout << currentVertex << " ";

        for (int neighbor : adjList[currentVertex]) {

            if (!visited[neighbor]) {

                dfsHelper(neighbor, visited);

            }

        }

    }

};

int main() {

    Graph graph(6);

    graph.addEdge(0, 1);

    graph.addEdge(0, 2);

    graph.addEdge(1, 3);

    graph.addEdge(1, 4);

    graph.addEdge(2, 5);

    cout << "BFS Traversal: ";

    graph.bfs(0);

    cout << endl;

    cout << "DFS Traversal: ";

    graph.dfs(0);

    cout << endl;

    return 0;

}

// Recursive implementation of DFS

void dfs(int startVertex) {

    vector<bool> visited(numVertices, false);

    dfsHelper(startVertex, visited);

}

void dfsHelper(int currentVertex, vector<bool>& visited) {

    visited[currentVertex] = true;

    cout << currentVertex << " ";

    for (int neighbor : adjList[currentVertex]) {

        if (!visited[neighbor]) {

            dfsHelper(neighbor, visited);

        }

    }

}

Slip 4: Dynamic Programming

Implement the following dynamic programming problems:

- Fibonacci Series

- Longest Common Subsequence (LCS)

*Solution:*

#include <iostream>

#include <vector>

using namespace std;

// Fibonacci Series

int fibonacci(int n) {

    vector<int> fib(n + 1);

    fib[0] = 0;

    fib[1] = 1;

    for (int i = 2; i <= n; i++) {

        fib[i] = fib[i - 1] + fib[i - 2];

    }

    return fib[n];

}

// Longest Common Subsequence (LCS)

int lcs(string str1, string str2) {

    int m = str1.length();

    int n = str2.length();

    vector<vector<int>> dp(m + 1, vector<int>(n + 1));

    for (int i = 0; i <= m; i++) {

        for (int j = 0; j <= n; j++) {

            if (i == 0 || j == 0) {

                dp[i][j] = 0;

            } else if (str1[i - 1] == str2[j - 1]) {

                dp[i][j] = dp[i - 1][j - 1] + 1;

            } else {

                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);

            }

        }

    }

    return dp[m][n];

}

int main() {

    int n = 10;

    cout << "Fibonacci number at position " << n << " is " << fibonacci(n) << endl;

    string str1 = "AGGTAB";

    string str2 = "GXTXAYB";

    cout << "Length of LCS is " << lcs(str1, str2) << endl;

    return 0;

}

These are just a few examples of solved practical slips for Master's in Computer Science. There are many more topics and problems that can be covered.