class 11 problem solving

Introduction to Problem Solving Class 11 Notes | CBSE Computer Science

Latest Problem Solving Class 11 Notes includes Problem Solving, steps, algorithm and its need, flow chart, pseudo code with lots of examples.

• 1 What is Problem Solving?
• 2 Steps for problem solving
• 3 What is Algorithm?
• 4 Why do we need Algorithm?
• 5.1 Flow chart
• 5.2 Flow Chart Examples
• 5.3 Pseudo code
• 5.4 Pseudo Code Example
• 6.1 Selection
• 6.2 Algorithm, Pseudocode, Flowchart with Selection ( Using if ) Examples
• 6.3 Repetition
• 6.4 Algorithm, Pseudocode, Flowchart with Repetition ( Loop ) Examples
• 7 Decomposition

What is Problem Solving?

Problem solving is the process of identifying a problem, analyze the problem, developing an algorithm for the identified problem and finally implementing the algorithm to develop program.

Steps for problem solving

There are 4 basic steps involved in problem solving

Analyze the problem

• Developing an algorithm
• Testing and debugging

Analyzing the problem is basically understanding a problem very clearly before finding its solution. Analyzing a problem involves

• List the principal components of the problem
• List the core functionality of the problem
• Figure out inputs to be accepted and output to be produced

Developing an Algorithm

• A set of precise and sequential steps written to solve a problem
• The algorithm can be written in natural language
• There can be more than one algorithm for a problem among which we can select the most suitable solution.

Algorithm written in natural language is not understood by computer and hence it has to be converted in machine language. And to do so program based on that algorithm is written using high level programming language for the computer to get the desired solution.

Testing and Debugging

After writing program it has to be tested on various parameters to ensure that program is producing correct output within expected time and meeting the user requirement.

There are many standard software testing methods used in IT industry such as

• Component testing
• Integration testing
• System testing
• Acceptance testing

What is Algorithm?

• A set of precise, finite and sequential set of steps written to solve a problem and get the desired output.
• Algorithm has definite beginning and definite end.
• It lead to desired result in finite amount of time of followed correctly.

Why do we need Algorithm?

• Algorithm helps programmer to visualize the instructions to be written clearly.
• Algorithm enhances the reliability, accuracy and efficiency of obtaining solution.
• Algorithm is the easiest way to describe problem without going into too much details.
• Algorithm lets programmer understand flow of problem concisely.

Characteristics of a good algorithm

• Precision — the steps are precisely stated or defined.
• Uniqueness — results of each step are uniquely defined and only depend on the input and the result of the preceding steps.
• Finiteness — the algorithm always stops after a finite number of steps.
• Input — the algorithm receives some input.
• Output — the algorithm produces some output.

What are the points that should be clearly identified while writing Algorithm?

• The input to be taken from the user
• Processing or computation to be performed to get the desired result
• The output desired by the user

Representation of Algorithm

An algorithm can be represented in two ways:

Pseudo code

• Flow chart is visual representation of an algorithm.
• It’s a diagram made up of boxes, diamonds and other shapes, connected by arrows.
• Each step represents a step of solution process.
• Arrows in the follow chart represents the flow and link among the steps.

Flow Chart Examples

Example 1: Write an algorithm to divide a number by another and display the quotient.

Input: Two Numbers to be divided Process: Divide number1 by number2 to get the quotient Output: Quotient of division

Step 1: Input a two numbers and store them in num1 and num2 Step 2: Compute num1/num2 and store its quotient in num3 Step 3: Print num3

• Pseudo code means ‘not real code’.
• A pseudo code is another way to represent an algorithm.  It is an informal language used by programmer to write algorithms.
• It does not require strict syntax and technological support.
• It is a detailed description of what algorithm would do.
• It is intended for human reading and cannot be executed directly by computer.
• There is no specific standard for writing a pseudo code exists.

Keywords used in writing pseudo code

Pseudo Code Example

Example:  write an algorithm to display the square of a given number.

Input, Process and Output Identification

Input: Number whose square is required Process: Multiply the number by itself to get its square Output: Square of the number

Step 1: Input a number and store it to num. Step 2: Compute num * num and store it in square. Step 3: Print square.

INPUT num COMPUTE  square = num*num PRINT square

Example: Write an algorithm to calculate area and perimeter of a rectangle, using both pseudo code and flowchart.

INPUT L INPUT B COMPUTER Area = L * B PRINT Area COMPUTE Perimeter = 2 * ( L + B ) PRINT Perimeter

Flow of Control

An algorithm is considered as finite set of steps that are executed in a sequence. But sometimes the algorithm may require executing some steps conditionally or repeatedly. In such situations algorithm can be written using

Selection in algorithm refers to Conditionals which means performing operations (sequence of steps) depending on True or False value of given conditions. Conditionals are written in the algorithm as follows:

If <condition> then                 Steps to be taken when condition is true Otherwise                 Steps to be taken when condition is false

Algorithm, Pseudocode, Flowchart with Selection ( Using if ) Examples

Example: write an algorithm, pseudocode and flowchart to display larger between two numbers

INPUT: Two numbers to be compared PROCESS: compare two numbers and depending upon True and False value of comparison display result OUTPUT: display larger no

STEP1: read two numbers in num1, num2 STEP 2: if num1 > num2 then STEP 3: display num1 STEP 4: else STEP 5: display num2

INPUT num1 , num2 IF num1 > num2 THEN                 PRINT “num1 is largest” ELSE                 PRINT “num2 is largest” ENDIF

Example: write pseudocode and flowchart to display largest among three numbers

INPUT: Three numbers to be compared PROCESS: compare three numbers and depending upon True and False value of comparison display result OUTPUT: display largest number

INPUT num1, num2, num3 PRINT “Enter three numbers” IF num1 > num2 THEN                 IF num1 > num3 THEN                                 PRINT “num1 is largest”                 ELSE                                 PRINT “num3 is largest”                 END IF ELSE                 IF num2 > num3 THEN                                 PRINT “num2 is largest”                 ELSE                                 PRINT “num3 is largest”                 END IF END IF

• Repetition in algorithm refers to performing operations (Set of steps) repeatedly for a given number of times (till the given condition is true).
• Repetition is also known as Iteration or Loop

Repetitions are written in algorithm is as follows:

While <condition>, repeat step numbers                 Steps to be taken when condition is true End while

Algorithm, Pseudocode, Flowchart with Repetition ( Loop ) Examples

Example: write an algorithm, pseudocode and flow chart to display “Techtipnow” 10 times

Step1: Set count = 0 Step2: while count is less than 10, repeat step 3,4 Step 3:                  print “techtipnow” Step 4:                  count = count + 1 Step 5: End while

SET count = 0 WHILE count<10                 PRINT “Techtipnow”                 Count = count + 1 END WHILE

Example: Write pseudocode and flow chart to calculate total of 10 numbers

Step 1: SET count = 0, total = 0 Step 2: WHILE count < 10, REPEAT steps 3 to 5 Step 3:                  INPUT a number in var Step 4:                  COMPUTE total = total + var Step 5:                  count = count + 1 Step 6: END WHILE Step 7: PRINT total

Example: Write pseudo code and flow chart to find factorial of a given number

Step 1: SET fact = 1 Step 2: INPUT a number in num Step 3: WHILE num >=1 REPEAT step 4, 5 Step 4:                  fact = fact * num Step 5:                  num = num – 1 Step 6: END WHILE Step 7: PRINT fact

Decomposition

• Decomposition means breaking down a complex problem into smaller sub problems to solve them conveniently and easily.
• Breaking down complex problem into sub problem also means analyzing each sub problem in detail.
• Decomposition also helps in reducing time and effort as different subprograms can be assigned to different experts in solving such problems.
• To get the complete solution, it is necessary to integrate the solution of all the sub problems once done.

Following image depicts the decomposition of a problem

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NCERT Solutions for Class 11 - 2024-25

• NCERT Solutions

Class 11 NCERT Solutions - FREE PDF Download

NCERT Solutions for Class 11 is a comprehensive guide to help students understand and excel in their studies. Based on the latest CBSE syllabus, these solutions cover all subjects, including mathematics, physics, chemistry, biology, business studies, accounting, and economics. Each solution provides detailed explanations and step-by-step answers to the exercises and problems in the NCERT textbooks .

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Prepare early by studying your textbooks thoroughly.

Understand each chapter completely before moving on.

Test your understanding by solving exercises without looking at the answers.

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The Economics syllabus of Class 11 includes two important components. The first important component is Statistics, and the second one is Indian Economic Development. Both the topics are important as the first one shows us how to compute big numbers into easier and simpler values, and the second one looks at the history of economic development in our country. Notes for both of these important components have been included in the NCERT Solutions’ Class 11 PDFs which will help you score exceptionally well in your Economics paper.

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It is indeed very crucial to practice as much as possible since Class 11 is primarily focused on fundamentals which can be challenging and it would be quite difficult for you to complete the syllabus and score well if you do not grasp the concepts. NCERT questions are also well-structured and created with strict respect to the CBSE syllabus for Class 11. You can get the NCERT Solutions for Class 11 free of cost. 

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NCERT books are of utmost importance to properly grasp all of your fundamentals. The different problems are structured properly with different levels of difficulty. These books help you understand the topics in an easy language. The exercises in the NCERT books will give an apt idea of the paper structure. Furthermore, you can also reinforce your arsenal of knowledge with other guide books.

7. How many Chapters are there in the NCERT Solutions Class 11 Economics?

The NCERT Class 11 Indian Economic Development book has ten chapters. Each chapter covers a certain area and its topics, which are adequately expanded and explained, as well as carefully structured and conform strictly to the CBSE syllabus. From the standpoint of the test, it is critical to understand all of the principles, study the NCERT book thoroughly, and practice all of the exercises. Please refer to the link NCERT Solutions for Class 11 on Vedantu website (vedantu.com) for additional information.

8. How to study class 11 easily?

To study class 11 easily, follow these steps:

Organise: Make a timetable to cover all subjects.

Understand Concepts: Focus on understanding rather than rote learning.

Take Notes: Write short notes for quick revisions.

Practice Regularly: Solve problems and practice past papers.

Stay Consistent: Study a little every day.

Ask for Help: Don’t hesitate to ask teachers or friends if you’re stuck.

9. Can NCERT Class 11 Solutions help in competitive exam preparation? 

Yes, NCERT Solutions Class 11 can help in competitive exam preparation. They cover the basic concepts and fundamentals that are essential for exams like JEE, NEET, and others. By studying these solutions, you can build a strong foundation, which is crucial for understanding advanced topics. NCERT solutions provide clear explanations and step-by-step answers, making complex topics easier to grasp. 

10. How should I use NCERT Solutions Class 11 for effective learning? 

To use NCERT Solutions Class 11 for effective learning, follow these steps:

Read the Chapter: Understand the concepts before looking at the solutions.

Attempt Exercises: Try solving questions on your own first.

Check Solutions: Compare your answers with NCERT solutions to identify mistakes.

Understand Steps: Study the step-by-step solutions to grasp the correct method.

Clarify Doubts: Use the solutions to clear any confusion.

Revise Regularly: Go through the solutions during revision to improve learning.

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Chapter 4 : Introduction to Problem Solving | class 11th | revision notes computer science

• Chapter 4 : Introduction to…

Introduction to Problem Solving

• Steps for problem solving ( analysing the problem, developing an algorithm, coding, testing and debugging).
• flow chart and
• pseudo code,

Decomposition

Introduction

Computers is machine that not only use to develop the software. It is also used for solving various day-to-day problems.

Computers cannot solve a problem by themselves. It solve the problem on basic of the step-by-step instructions given by us.

Thus, the success of a computer in solving a problem depends on how correctly and precisely we –

• Identifying (define) the problem
• Designing & developing an algorithm and
• Implementing the algorithm (solution) do develop a program using any programming language.

Thus problem solving is an essential skill that a computer science student should know.

Steps for Problem Solving-

1.   Analysing the problem

Analysing the problems means understand a problem clearly before we begin to find the solution for it. Analysing a problem helps to figure out what are the inputs that our program should accept and the outputs that it should produce.

2. Developing an Algorithm

It is essential to device a solution before writing a program code for a given problem. The solution is represented in natural language and is called an algorithm.

Algorithm:  A set of exact steps which when followed, solve the problem or accomplish the required task.

Coding is the process of converting the algorithm into the program which can be understood by the computer to generate the desired solution.

You can use any high level programming languages for writing a program.

4. Testing and Debugging

The program created should be tested on various parameters.

• The program should meet the requirements of the user.
• It must respond within the expected time.
• It should generate correct output for all possible inputs.
• In the presence of syntactical errors, no output will be obtained.
• In case the output generated is incorrect, then the program should be checked for logical errors, if any.

Software Testing methods are

• unit or component testing,
• integration testing,
• system testing, and
• acceptance testing

Debugging – The errors or defects found in the testing phases are debugged or rectified and the program is again tested. This continues till all the errors are removed from the program.

Algorithm is a set of sequence which followed to solve a problem.

Algorithm for an activity ‘riding a bicycle’: 1) remove the bicycle from the stand, 2) sit on the seat of the bicycle, 3) start peddling, 4) use breaks whenever needed and 5) stop on reaching the destination.

Algorithm for Computing GCD of two numbers:

Step 1: Find the numbers (divisors) which can divide the given numbers.

Step 2: Then find the largest common number from these two lists.

A finite sequence of steps required to get the desired output is called an algorithm. Algorithm has a definite beginning and a definite end, and consists of a finite number of steps.

Characteristics of a good algorithm

• Precision  — the steps are precisely stated or defined.
• Uniqueness  — results of each step are uniquely defined and only depend on the input and the result of the preceding steps.
• Finiteness  — the algorithm always stops after a finite number of steps.
• Input  — the algorithm receives some input.
• Output  — the algorithm produces some output.

While writing an algorithm, it is required to clearly identify the following:

• The input to be taken from the user.
• Processing or computation to be performed to get the desired result.
• The output desired by the user.

Representation of Algorithms

There are two common methods of representing an algorithm —

Flowchart — Visual Representation of Algorithms

A flowchart is a visual representation of an algorithm. A flowchart is a diagram made up of boxes, diamonds and other shapes, connected by arrows. Each shape represents a step of the solution process and the arrow represents the order or link among the steps. There are standardised symbols to draw flowcharts.

Start/End –  Also called “Terminator” symbol. It indicates where the flow starts and ends.

Process  – Also called “Action Symbol,” it represents a process, action, or a single step. Decision  – A decision or branching point, usually a yes/no or true/ false question is asked, and based on the answer, the path gets split into two branches.

Input / Output  – Also called data symbol, this parallelogram shape is used to input or output data.

Arrow  – Connector to show order of flow between shapes.

Question: Write an algorithm to find the square of a number. Algorithm to find square of a number. Step 1: Input a number and store it to num Step 2: Compute num * num and store it in square Step 3: Print square

The algorithm to find square of a number can be represented pictorially using flowchart

Table of Contents

A pseudocode (pronounced Soo-doh-kohd) is another way of representing an algorithm. It is considered as a non-formal language that helps programmers to write algorithm. It is a detailed description of instructions that a computer must follow in a particular order.

• It is intended for human reading and cannot be executed directly by the computer.
• No specific standard for writing a pseudocode exists.
• The word “pseudo” means “not real,” so “pseudocode” means “not real code”.

Keywords are used in pseudocode:

Question : Write an algorithm to calculate area and perimeter of a rectangle, using both pseudocode and flowchart.

Pseudocode for calculating area and perimeter of a rectangle.

INPUT length INPUT breadth COMPUTE Area = length * breadth PRINT Area COMPUTE Perim = 2 * (length + breadth) PRINT Perim The flowchart for this algorithm

Benefits of Pseudocode

• A pseudocode of a program helps in representing the basic functionality of the intended program.
• By writing the code first in a human readable language, the programmer safeguards against leaving out any important step.
• For non-programmers, actual programs are difficult to read and understand, but pseudocode helps them to review the steps to confirm that the proposed implementation is going to achieve the desire output.

Flow of Control :

The flow of control depicts the flow of process as represented in the flow chart. The process can flow in

In a sequence steps of algorithms (i.e. statements) are executed one after the other.

In a selection, steps of algorithm is depend upon the conditions i.e. any one of the alternatives statement is selected based on the outcome of a condition.

Conditionals  are used to check possibilities. The program checks one or more conditions and perform operations (sequence of actions) depending on true or false value of the condition.

Conditionals are written in the algorithm as follows: If is true then steps to be taken when the condition is true/fulfilled otherwise steps to be taken when the condition is false/not fulfilled

Question : Write an algorithm to check whether a number is odd or even. • Input: Any number • Process: Check whether the number is even or not • Output: Message “Even” or “Odd” Pseudocode of the algorithm can be written as follows: PRINT “Enter the Number” INPUT number IF number MOD 2 == 0 THEN PRINT “Number is Even” ELSE PRINT “Number is Odd”

Repetitions are used, when we want to do something repeatedly, for a given number of times.

Question : Write pseudocode and draw flowchart to accept numbers till the user enters 0 and then find their average. Pseudocode is as follows:

Step 1: Set count = 0, sum = 0 Step 2: Input num Step 3: While num is not equal to 0, repeat Steps 4 to 6 Step 4: sum = sum + num Step 5: count = count + 1 Step 6: Input num Step 7: Compute average = sum/count Step 8: Print average The flowchart representation is

Once an algorithm is finalised, it should be coded in a high-level programming language as selected by the programmer. The ordered set of instructions are written in that programming language by following its syntax.

The syntax  is the set of rules or grammar that governs the formulation of the statements in the language, such as spelling, order of words, punctuation, etc.

Source Code:  A program written in a high-level language is called source code.

We need to translate the source code into machine language using a compiler or an interpreter so that it can be understood by the computer.

Decomposition is a process to ‘decompose’ or break down a complex problem into smaller subproblems. It is helpful when we have to solve any big or complex problem.

• Breaking down a complex problem into sub problems also means that each subproblem can be examined in detail.
• Each subproblem can be solved independently and by different persons (or teams).
• Having different teams working on different sub-problems can also be advantageous because specific sub-problems can be assigned to teams who are experts in solving such problems.

Once the individual sub-problems are solved, it is necessary to test them for their correctness and integrate them to get the complete solution.

Author:  noor arora

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Dot & Line Blog

Physics Numericals for Class 11: Mastering Concepts with Practical Problems

• Kanwal Hafeez
• July 21, 2023

Table of Contents

1. introduction to physics numericals.

Physics is a branch of science that deals with the study of matter, energy, and their interactions. Numerical problem-solving plays a crucial role in mastering physics concepts and understanding their practical applications. By solving physics numericals, students can deepen their understanding of theories, equations, and principles while developing problem-solving skills. In this section, we will explore some fundamental concepts and units commonly encountered in physics numericals.

1.1 Basic Concepts and Units

• Displacement: Displacement is the change in position of an object in a particular direction. It is a vector quantity and is measured in units of length, such as meters (m) in the International System of Units (SI).
• Velocity: Velocity is the rate of change of displacement with time. It is also a vector quantity and is expressed in units of distance per time, such as meters per second (m/s).
• Acceleration: Acceleration is the rate of change of velocity with time. It is a vector quantity and is measured in units of distance per time squared, like meters per second squared (m/s²).
• Newton’s Second Law of Motion: This law states that the force acting on an object is directly proportional to the mass of the object and the acceleration produced. Mathematically, F = m * a, where F is the force in Newtons (N), m is the mass in kilograms (kg), and a is the acceleration in meters per second squared (m/s²).

Example Numerical:

A car of mass 1000 kg is accelerating uniformly at 2 m/s². Calculate the force acting on the car. Solution: Given: Mass (m) = 1000 kg, Acceleration (a) = 2 m/s² Using Newton’s Second Law of Motion: F = m * a F = 1000 kg * 2 m/s² = 2000 N

1.2 Measurement and Error Analysis

• Precision and Accuracy: Precision refers to the closeness of measurements to each other, while accuracy is the closeness of a measurement to the true value. High precision means less variability among repeated measurements, while high accuracy implies less systematic error.
• Absolute Error: The absolute error is the difference between the measured value and the true value of a quantity.
• Relative Error: The relative error is the ratio of the absolute error to the true value, usually expressed as a percentage.
• Significant Figures: Significant figures in a measurement are the digits that carry meaningful information. They include all the certain digits plus one uncertain or estimated digit.

A student measures the length of a pencil as 18.2 cm using a ruler with millimeter markings. The true length of the pencil is 18.0 cm. Calculate the absolute and relative error in the measurement. Solution: Given: Measured length = 18.2 cm, True length = 18.0 cm Absolute Error = |Measured length – True length| = |18.2 cm – 18.0 cm| = 0.2 cm Relative Error = (Absolute Error / True length) * 100 Relative Error = (0.2 cm / 18.0 cm) * 100 ≈ 1.11%

By mastering these basic concepts and units, and understanding measurement and error analysis, students can confidently tackle more complex physics numericals and gain a deeper appreciation for the practical aspects of the subject. Regular practice with solved numericals will undoubtedly enhance their problem-solving skills and pave the way for success in physics examinations and beyond.

2. Kinematics Numericals

2.1 displacement, velocity, and acceleration calculations.

In kinematics, we study the motion of objects without considering the forces causing that motion. This section focuses on problems related to displacement, velocity, and acceleration.

Numerical 1:

A car starts from rest and accelerates uniformly at 3 m/s² for 8 seconds. Calculate its final velocity and displacement during this time. Solution: Given: Initial velocity (u) = 0 m/s, Acceleration (a) = 3 m/s², Time (t) = 8 seconds Using the equation v = u + at: Final velocity (v) = 0 + (3 m/s² * 8 s) = 24 m/s Using the equation s = ut + (1/2)at²: Displacement (s) = (0 m/s * 8 s) + (0.5 * 3 m/s² * (8 s)²) = 96 m

Numerical 2:

An object is moving with a constant velocity of 5 m/s. Calculate the displacement of the object after 10 seconds. Solution: Given: Velocity (v) = 5 m/s, Time (t) = 10 seconds Using the equation s = vt: Displacement (s) = 5 m/s * 10 s = 50 m

Numerical 3:

A train decelerates uniformly at 2 m/s² until it comes to rest. If its initial velocity was 20 m/s, calculate the time taken for the train to stop. Solution: Given: Initial velocity (u) = 20 m/s, Acceleration (a) = -2 m/s² (negative as it’s deceleration), Final velocity (v) = 0 m/s Using the equation v = u + at: 0 m/s = 20 m/s + (-2 m/s² * t) Solving for t: t = 20 m/s / 2 m/s² = 10 seconds

Numerical 4:

An athlete runs along a straight track. He covers the first 40 meters in 5 seconds and then stops for 2 seconds. After that, he runs at a constant velocity of 6 m/s for 8 seconds. Calculate the total distance covered by the athlete. Solution: Total distance covered = Distance in the first 5 seconds + Distance during constant velocity Distance in the first 5 seconds = 40 meters (given) Distance during constant velocity = 6 m/s * 8 s = 48 meters Total distance covered = 40 meters + 48 meters = 88 meters

Numerical 5:

A stone is dropped from the top of a tower 100 meters high. Calculate its velocity when it hits the ground. Solution: Given: Initial velocity (u) = 0 m/s (since it’s dropped), Displacement (s) = 100 meters, Acceleration (a) = 9.8 m/s² (acceleration due to gravity, downwards) Using the equation v² = u² + 2as: Final velocity (v) = √(0 m/s)² + 2 * 9.8 m/s² * 100 m ≈ 44.29 m/s (rounded to two decimal places)

2.2 Projectile Motion

Projectile motion refers to the motion of an object thrown or projected into the air, under the influence of gravity. In this section, we will solve numerical problems related to projectile motion.

A ball is thrown horizontally from the top of a cliff 80 meters high with an initial velocity of 20 m/s. Calculate the time it takes for the ball to reach the ground and the horizontal distance traveled. Solution: Given: Initial vertical velocity (uy) = 0 m/s (thrown horizontally), Displacement in the y-direction (sy) = -80 meters (negative as it’s downward), Acceleration in the y-direction (ay) = -9.8 m/s² (acceleration due to gravity, downward), Horizontal velocity (ux) = 20 m/s Time to reach the ground can be calculated using the equation: t = (2 * |sy| / |ay|)^(1/2) t = (2 * 80 m / 9.8 m/s²)^(1/2) ≈ 4.04 seconds (rounded to two decimal places) The horizontal distance (dx) can be calculated using the equation: dx = ux * t dx = 20 m/s * 4.04 s ≈ 80.8 meters (rounded to one decimal place)

A soccer player kicks a ball at an angle of 30 degrees above the horizontal with a velocity of 15 m/s. Determine the maximum height reached by the ball and the total time of flight. Solution: Given: Launch angle (θ) = 30 degrees, Launch velocity (v) = 15 m/s, Acceleration due to gravity (g) = 9.8 m/s² The time taken to reach the maximum height can be calculated using the equation: t = uy / g uy = v * sin(θ) = 15 m/s * sin(30 degrees) ≈ 7.5 m/s t = 7.5 m/s / 9.8 m/s² ≈ 0.77 seconds (rounded to two decimal places) The maximum height (H) can be calculated using the equation: H = (uy²) / (2 * g) H = (7.5 m/s)² / (2 * 9.8 m/s²) ≈ 2.29 meters (rounded to two decimal places) The total time of flight can be calculated using: Total time = 2 * t Total time ≈ 2 * 0.77 s ≈ 1.54 seconds (rounded to two decimal places)

A stone is thrown horizontally from the top of a cliff with a velocity of 12 m/s. Calculate the horizontal distance it travels before hitting the ground. Solution: Given: Initial horizontal velocity (ux) = 12 m/s, Time of flight (t) = ? The horizontal distance (dx) can be calculated using the equation: dx = ux * t We need to find the time of flight first. Since the stone is thrown horizontally, the vertical component of velocity (uy) is 0 m/s. The time of flight (t) can be calculated using: t = 2 * (vertical displacement) / g Vertical displacement = 0 m (since the stone starts and ends at the same height) t = 2 * 0 m / 9.8 m/s² = 0 seconds Now, the horizontal distance can be calculated: dx = 12 m/s * 0 s = 0 meters

A football is kicked from the ground with an initial velocity of 25 m/s at an angle of 45 degrees above the horizontal. Calculate the range of the football. Solution: Given: Launch angle (θ) = 45 degrees, Launch velocity (v) = 25 m/s, Acceleration due to gravity (g) = 9.8 m/s² The horizontal distance (range) can be calculated using the equation: Range (R) = (v² * sin(2θ)) / g Range (R) = (25 m/s)² * sin(2 * 45 degrees) / 9.8 m/s² ≈ 31.88 meters (rounded to two decimal places)

A ball is thrown vertically upward with an initial velocity of 30 m/s. Calculate the maximum height reached by the ball and the time taken to reach it. Solution : Given: Initial velocity (u) = 30 m/s (upward), Acceleration due to gravity (g) = 9.8 m/s² The maximum height (H) can be calculated using the equation: H = (u²) / (2 * g) H = (30 m/s)² / (2 * 9.8 m/s²) ≈ 45.92 meters (rounded to two decimal places) The time taken to reach the maximum height (t) can be calculated using the equation: t = u / g t = 30 m/s / 9.8 m/s² ≈ 3.06 seconds (rounded to two decimal places)

2.3 Circular Motion Problems

Circular motion involves an object moving in a circular path. In this section, we will explore numerical problems related to circular motion.

A car moves around a circular track with a radius of 50 meters at a constant speed of 20 m/s. Calculate its centripetal acceleration. Solution: Given: Radius of the circular track (r) = 50 meters, Speed of the car (v) = 20 m/s Centripetal acceleration (a) can be calculated using the equation: a = v² / r a = (20 m/s)² / 50 m ≈ 8 m/s²

A stone tied to a string is whirled around in a horizontal circle with a constant speed of 6 m/s. If the radius of the circle is 2 meters, calculate the centripetal force acting on the stone. Solution: Given: Speed of the stone (v) = 6 m/s, Radius of the circular path (r) = 2 meters, Mass of the stone (m) = ? Centripetal force (F) can be calculated using the equation: F = m * a Centripetal acceleration (a) = v² / r = (6 m/s)² / 2 m = 18 m/s² Since F = m * a, we need to find the mass (m) of the stone to calculate the centripetal force.

A Ferris wheel has a radius of 20 meters and completes one revolution in 40 seconds. Calculate the linear speed of a passenger at the top and bottom of the wheel. Solution: Given: Radius of the Ferris wheel (r) = 20 meters, Time for one revolution (T) = 40 seconds Linear speed at the top can be calculated using the equation: v_top = 2 * π * r / T v_top = 2 * 3.14 * 20 m / 40 s ≈ 3.14 m/s (rounded to two decimal places) Linear speed at the bottom is the same as at the top because the distance traveled in one revolution is the same.

A car moves around a horizontal circular track with a radius of 100 meters. If the car completes one revolution in 60 seconds, calculate the magnitude of the centripetal acceleration and the net force acting on the car if its mass is 1500 kg. Solution: Given: Radius of the circular track (r) = 100 meters, Time for one revolution (T) = 60 seconds, Mass of the car (m) = 1500 kg Centripetal acceleration (a) can be calculated using the equation: a = 4 * π² * r / T² a = 4 * 3.14² * 100 m / (60 s)² ≈ 2.62 m/s² (rounded to two decimal places) Net force (F) acting on the car can be calculated using the equation: F = m * a F = 1500 kg * 2.62 m/s² ≈ 3930 N (rounded to two decimal places)

A small object is tied to a string and whirled around in a horizontal circle with a radius of 0.5 meters. If the speed of the object is 4 m/s, calculate the period of its motion and the centripetal acceleration. Solution : Given: Radius of the circular path (r) = 0.5 meters, Speed of the object (v) = 4 m/s Period (T) of motion can be calculated using the equation: T = 2 * π * r / v T = 2 * 3.14 * 0.5 m / 4 m/s ≈ 0.785 seconds (rounded to three decimal places) Centripetal acceleration (a) can be calculated using the equation: a = v² / r a = (4 m/s)² / 0.5 m ≈ 32 m/s²

By solving these kinematics and circular motion numericals, students can gain a strong grasp of the fundamental principles of motion and further enhance their problem-solving skills in physics. Regular practice and a solid understanding of these concepts will enable them to tackle more complex physics problems with confidence.

3. Laws of Motion Numericals

3.1 newton’s first law applications.

Newton’s First Law of Motion states that an object at rest will remain at rest, and an object in motion will continue moving with a constant velocity unless acted upon by an external force. This law is also known as the law of inertia. Let’s explore some numerical problems that apply Newton’s First Law.

A book is placed on a table and remains at rest. Explain the forces acting on the book and how Newton’s First Law is demonstrated in this scenario. Solution: In this scenario, the forces acting on the book are gravity (acting downwards) and the normal force (acting upwards) exerted by the table. According to Newton’s First Law, the book remains at rest because the forces acting on it are balanced. The gravitational force pulling the book downwards is balanced by the normal force exerted by the table in the opposite direction. As a result, there is no net force acting on the book, and it stays at rest.

A hockey puck slides on a frictionless ice surface and keeps moving with a constant velocity. Explain how Newton’s First Law is demonstrated in this situation. Solution: In this situation, the only force acting on the hockey puck is its initial forward push. Once the puck starts moving, there is no significant external force acting on it to change its velocity or direction. Newton’s First Law explains that the puck will continue moving with the same velocity (constant speed and direction) as there is no net external force acting on it. The absence of friction on the ice surface allows the puck to maintain its motion without slowing down or changing direction.

A car comes to a stop after hitting the brakes. Explain how Newton’s First Law is involved in this scenario. Solution: When the driver hits the brakes of the car, frictional forces between the tires and the road come into play. The frictional force opposes the forward motion of the car and acts in the opposite direction. As a result, there is a net external force acting on the car in the direction opposite to its initial motion. According to Newton’s First Law, the car will tend to maintain its initial forward motion due to its inertia, but the net external force from braking overcomes this inertia, causing the car to slow down and eventually come to a stop.

3.2 Newton’s Second Law Problems

Newton’s Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The law is mathematically expressed as F = ma, where F is the net force, m is the mass of the object, and a is the acceleration. Let’s solve some numerical problems involving Newton’s Second Law.

A force of 40 N is applied to an object with a mass of 5 kg. Calculate the acceleration of the object. Solution: Given: Force (F) = 40 N, Mass (m) = 5 kg Using Newton’s Second Law: a = F / m a = 40 N / 5 kg = 8 m/s²

A 20 kg box is pushed with a force of 100 N. If the coefficient of friction between the box and the floor is 0.4, calculate the acceleration of the box. Solution: Given: Force applied (F) = 100 N, Mass of the box (m) = 20 kg, Coefficient of friction (μ) = 0.4 Frictional force (F_friction) = μ * (normal force) Normal force (F_normal) = m * g (where g is the acceleration due to gravity, approximately 9.8 m/s²) F_friction = 0.4 * (20 kg * 9.8 m/s²) = 78.4 N Net force (F_net) = F – F_friction = 100 N – 78.4 N = 21.6 N Using Newton’s Second Law: a = F_net / m a = 21.6 N / 20 kg ≈ 1.08 m/s² (rounded to two decimal places)

An object is pushed with a force of 30 N and accelerates at 5 m/s². Calculate the mass of the object. Solution: Given: Force (F) = 30 N, Acceleration (a) = 5 m/s² Using Newton’s Second Law: m = F / a m = 30 N / 5 m/s² = 6 kg

3.3 Newton’s Third Law Numericals

Newton’s Third Law of Motion states that for every action, there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B exerts an equal and opposite force on object A. Let’s solve some numerical problems based on Newton’s Third Law.

A person pushes a wall with a force of 50 N. Explain the reaction force according to Newton’s Third Law.

According to Newton’s Third Law, the wall exerts an equal and opposite force of 50 N on the person. When the person pushes the wall, the wall pushes back on the person with the same magnitude of force but in the opposite direction.

A rocket is launched into space by expelling gases at a high velocity. Explain the forces involved and their directions based on Newton’s Third Law.

The rocket propulsion is based on Newton’s Third Law. The rocket engine expels high-velocity gases downward (action). As a reaction to the downward expulsion of gases, the rocket experiences an equal and opposite force pushing it upward. This upward force propels the rocket into space.

A ball is dropped on the floor, and it bounces back upward. Explain the forces involved and their directions according to Newton’s Third Law.

When the ball hits the floor, it exerts a downward force on the floor (action). According to Newton’s Third Law, the floor exerts an equal and opposite force on the ball (reaction). This upward reaction force causes the ball to bounce back upward.

By practicing these numerical problems related to Newton’s Laws of Motion, students can better comprehend the practical applications of these laws and improve their problem-solving abilities in physics . Understanding these foundational principles is crucial for building a strong foundation in mechanics and other branches of physics.

4. Work, Energy, and Power Numericals

4.1 work-energy theorem applications.

The Work-Energy Theorem states that the work done on an object is equal to the change in its kinetic energy. Work is done on an object when a force is applied to it and it moves in the direction of the force. Let’s explore some numerical problems that apply the Work-Energy Theorem.

A force of 20 N is applied to push a box a distance of 5 meters along a horizontal surface. Calculate the work done on the box. Solution: Given: Force (F) = 20 N, Displacement (s) = 5 meters Work done (W) can be calculated using the equation: W = F * s W = 20 N * 5 m = 100 Joules

A spring with a spring constant of 200 N/m is compressed by 0.1 meters. Calculate the work done to compress the spring. Solution: Given: Spring constant (k) = 200 N/m, Compression (x) = 0.1 meters Work done (W) to compress the spring can be calculated using the equation: W = (1/2) * k * x² W = (0.5) * 200 N/m * (0.1 m)² = 1 Joule

A ball is thrown vertically upward with an initial velocity of 15 m/s. Calculate the work done by gravity on the ball during its ascent until it reaches its highest point. Solution: At the highest point of its trajectory, the ball’s vertical velocity becomes 0 m/s. Therefore, its kinetic energy at the highest point is 0. Work done by gravity (W) can be calculated using the equation: W = ΔKE (change in kinetic energy) Since the kinetic energy at the highest point is 0, the work done by gravity is also 0.

4.2 Conservation of Mechanical Energy Problems

Conservation of Mechanical Energy states that the total mechanical energy of an isolated system remains constant if only conservative forces, like gravity, are acting on it. In these numerical problems, we’ll explore scenarios where mechanical energy is conserved.

A roller coaster car with a mass of 500 kg starts from the top of a hill with an initial velocity of 10 m/s. Calculate its total mechanical energy at the top of the hill and when it reaches the bottom, assuming no frictional forces are acting. Solution: Given: Mass (m) = 500 kg, Initial velocity (v_initial) = 10 m/s, Height of the hill (h) = ? At the top of the hill, the roller coaster’s kinetic energy is 0 since its velocity is 0. Therefore, the total mechanical energy is equal to its potential energy: Potential energy (PE) at the top = m * g * h At the bottom of the hill, the roller coaster’s height is 0, and its potential energy is 0. Therefore, the total mechanical energy is equal to its kinetic energy: Kinetic energy (KE) at the bottom = (1/2) * m * v_final² Since mechanical energy is conserved, the total mechanical energy at the top is equal to the total mechanical energy at the bottom: m * g * h = (1/2) * m * v_final² Solving for v_final: v_final = √(2 * g * h) Substitute the given values to find v_final.

A pendulum swings back and forth between two points. At its highest point (Point A), it has a potential energy of 80 J, and at its lowest point (Point B), it has a potential energy of 20 J. Assuming no energy losses due to friction, calculate its speed at Point B. Solution: At Point A, the kinetic energy is 0, so the total mechanical energy is equal to its potential energy: Total mechanical energy at A = Potential energy (PE) at A = 80 J At Point B, the potential energy is 20 J, and the kinetic energy is: Total mechanical energy at B = Potential energy (PE) at B + Kinetic energy (KE) at B Total mechanical energy at B = 20 J + KE at B Since mechanical energy is conserved, the total mechanical energy at A is equal to the total mechanical energy at B: 80 J = 20 J + KE at B Solving for KE at B: KE at B = 80 J – 20 J = 60 J The kinetic energy at Point B is 60 J. Use the formula for kinetic energy (KE = (1/2) * m * v²) to find the speed (v) at Point B.

A skier starts from rest at the top of a hill and slides down. At the bottom of the hill, the skier has a speed of 15 m/s. Calculate the loss in mechanical energy due to non-conservative forces (like friction). Solution: Given: Initial velocity (v_initial) = 0 m/s, Final velocity (v_final) = 15 m/s The change in kinetic energy (ΔKE) can be calculated using the equation: ΔKE = (1/2) * m * (v_final² – v_initial²) Since the skier starts from rest (v_initial = 0 m/s), the change in kinetic energy simplifies to: ΔKE = (1/2) * m * v_final² Calculate the change in kinetic energy, and this value represents the loss in mechanical energy due to non-conservative forces like friction.

4.3 Power Calculations

Power is the rate at which work is done or the rate at which energy is transferred or converted. It is the time derivative of work or energy. Let’s explore some numerical problems involving power calculations.

A motor lifts a 500 kg crate vertically at a constant speed of 2 m/s. Calculate the power output of the motor. Solution: Given: Mass (m) = 500 kg, Vertical speed (v) = 2 m/s Work done (W) by the motor to lift the crate can be calculated using the equation: W = m * g * h, where h is the height lifted. Since the crate moves at a constant speed, there is no change in height (h) during the lift. Therefore, the work done (W) is 0, as no energy is transferred in the vertical direction. Power (P) can be calculated using the equation: P = W / t, where t is the time taken to lift the crate. Since W = 0, the power output of the motor is also 0.

A machine does 2000 Joules of work in 5 seconds. Calculate the power output of the machine. Solution: Given: Work done (W) = 2000 Joules, Time (t) = 5 seconds Power (P) can be calculated using the equation: P = W / t P = 2000 Joules / 5 seconds = 400 Watts

A pump lifts 100 liters of water from a well to the ground level in 2 minutes. The height of the well is 5 meters. Calculate the power output of the pump. Solution: Given: Volume of water (V) = 100 liters = 100,000 cm³ = 100,000 mL, Height (h) = 5 meters, Time (t) = 2 minutes = 120 seconds Work done (W) by the pump to lift the water can be calculated using the equation: W = m * g * h, where m is the mass of the water. Since the density of water is approximately 1 g/cm³, the mass of 100 liters (100,000 mL) of water is 100,000 grams (100 kg). W = 100 kg * 9.8 m/s² * 5 meters = 4,900 Joules Power (P) can be calculated using the equation: P = W / t P = 4,900 Joules / 120 seconds ≈ 40.83 Watts (rounded to two decimal places)

By solving these work, energy, and power numericals, students can better understand the concepts of energy transformation, conservation, and power calculations in various physical scenarios. These concepts play a fundamental role in understanding how energy works in real-world situations and are essential for solving more complex problems in physics .

FAQ’s

1. what is the purpose of solving physics numericals in class 11.

Solving physics numericals in Class 11 serves multiple purposes. It helps students develop problem-solving skills, gain a deeper understanding of theoretical concepts, and apply mathematical techniques to real-world scenarios. Numericals also reinforce the application of fundamental principles, preparing students for more complex topics in higher classes and future scientific endeavors.

2. How do I approach solving physics numericals effectively?

To effectively solve physics numericals, follow these steps:

• a) Carefully read the problem statement and identify the given data and required quantities.
• b) Choose the appropriate formula or equations that relate to the given scenario.
• c) Substitute the given values into the equations and solve for the unknown quantities.
• d) Pay attention to units and dimensions to ensure correct results.
• e) Check your answers and reevaluate if needed.

3. I find physics numericals challenging. How can I improve my problem-solving skills?

Improving problem-solving skills in physics requires practice and a systematic approach. Start with simple numericals and gradually progress to more complex ones. Seek help from your teachers or peers if you encounter difficulties. Analyze your mistakes and learn from them. Engage in group discussions, participate in online forums, and explore additional resources like textbooks and video tutorials to reinforce your understanding.

4. Can I rely solely on theory and skip practicing physics numericals?

While understanding theoretical concepts is essential, practicing physics numericals is equally crucial. Numericals provide a practical application of theory, helping you grasp the concepts more effectively. They also enhance problem-solving abilities, critical thinking, and mathematical skills, which are valuable in various fields of study and everyday life.

5. Are there any specific strategies to excel in physics numericals during exams?

Yes, there are strategies to excel in physics numericals during exams:

• a) Familiarize yourself with common numerical patterns and formulas beforehand.
• b) Practice solving a variety of numericals to gain confidence and speed.
• c) Create a formula sheet with key equations for quick reference during exams.
• d) Read each question carefully, paying attention to units and significant figures.
• e) If you get stuck, move on to the next question and come back later with a fresh perspective.
• f) Review and revise your solutions to avoid simple mistakes and maximize your score.

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NCERT Solutions for Class 11 (Updated For 2023-2024)

NCERT Solutions for Class 11 are solved by experts of LearnCBSE.in in order to help students to obtain excellent marks in their board examination. All the questions and answers that are present in the CBSE NCERT Books has been included in this page. We have provided all the Class 11 NCERT Solutions with a detailed explanation i.e., we have solved all the questions with step by step solutions in understandable language. So students having great knowledge over NCERT Solutions Class 11 can easily make a grade in their board exams. Read on to find out more about  NCERT Solutions for Class 11 . 

NCERT Solutions for Class 11

In this page, each and every question originate with a step-wise solution. Working on NCERT Solutions for Class 11 will help students to get an idea about how to solve the problems. With the help of these NCERT Solutions for Class 11 Science, Commerce and Humanities   you can easily grasp basic concepts better and faster. Moreover, it is a perfect guide to help you to score good marks in CBSE board examination. Just click on the chapter wise links given below to practice the NCERT Solutions for the respective chapter. 

Class 11 NCERT Solutions All Subjects

Ncert solutions for class 11 maths, ncert solutions for class 11 physics, ncert solutions for class 11 chemistry, ncert solutions for class 11 biology, ncert solutions for class 11 english.

NCERT Solutions for Class 11 English Woven Words Short Stories

NCERT Solutions for Class 11 English Woven Words Essay

NCERT Solutions for Class 11 English Woven Words Poetry

• NCERT Solutions for Class 11 English Snapshots
• NCERT Solutions for Class 11 English Hornbill
• NCERT Solutions for Class 11 Hindi
• NCERT Solutions for Class 11 Hindi Aroh  (आरोह भाग 1)
• NCERT Solutions for Class 11 Hindi Vitan  (वितान भाग 1)
• NCERT Solutions for Class 11 Sanskrit
• NCERT Solutions for Class 11 Business Studies

NCERT Solutions for Class 11 Accountancy

Ncert solutions for class 11 psychology, ncert solutions for class 11 entrepreneurship.

• Class 11 Indian Economic Development

NCERT Solutions for Class 11 Computer Science (Python)

• NCERT Solutions for Class 11 History
• NCERT Solutions for Class 11 Geography
• NCERT Solutions for Class 11 Sociology
• NCERT Solutions for Class 11 Political Science
• Chapter 1 Sets
• Chapter 2 Relations and Functions
• Chapter 3 Trigonometric Functions
• Chapter 4 Principle of Mathematical Induction
• Chapter 5 Complex Numbers and Quadratic Equations
• Chapter 6 Linear Inequalities
• Chapter 7 Permutation and Combinations
• Chapter 8 Binomial Theorem
• Chapter 9 Sequences and Series
• Straight Lines Exercise 10.2
• Straight Lines Exercise 10.3
• Straight Lines Exercise Miscellaneous Questions
• Chapter 11 Conic Sections
• Chapter 12 Introduction to Three Dimensional Geometry
• Chapter 13 Limits and Derivatives
• Chapter 14 Mathematical Reasoning
• Chapter 15 Statistics
• Chapter 16 Probability
• Chapter 1 Physical World
• Chapter 2 Units and Measurements
• Chapter 3 Motion in a Straight Line
• Chapter 4 Motion in a plane
• Chapter 5 Laws of motion
• Chapter 6 Work Energy and power
• Chapter 7 System of particles and Rotational Motion
• Chapter 8 Gravitation
• Chapter 9 Mechanical Properties Of Solids
• Chapter 10 Mechanical Properties Of Fluids
• Chapter 11 Thermal Properties of matter
• Chapter 12 Thermodynamics
• Chapter 13 Kinetic Theory
• Chapter 14 Oscillations
• Chapter 15 Waves
• Chapter 1 Some Basic Concepts of Chemistry
• Chapter 2 Structure of The Atom
• Chapter 3 Classification of Elements and Periodicity in Properties
• Chapter 4 Chemical Bonding and Molecular Structure
• Chapter 5 States of Matter
• Chapter 6 Thermodynamics
• Chapter 7 Equilibrium
• Chapter 8 Redox Reactions
• Chapter 9 Hydrogen
• Chapter 10 The S-Block Elements
• Chapter 11 The P-Block Elements
• Chapter 12 Organic Chemistry: Some Basic Principles and Techniques
• Chapter 13 Hydrocarbons
• Chapter 14 Environmental Chemistry
• Chapter 1 The Living World
• Chapter 2 Biological Classification
• Chapter 3 Plant Kingdom
• Chapter 4 Animal Kingdom
• Chapter 5 Morphology of Flowering Plants
• Chapter 6 Anatomy of Flowering Plants
• Chapter 7 Structural Organisation in Animals
• Chapter 8 Cell The Unit of Life
• Chapter 9 Biomolecules
• Chapter 10 Cell Cycle and Cell Division
• Chapter 11 Transport in Plants
• Chapter 12 Mineral Nutrition
• Chapter 13 Photosynthesis in Higher Plants
• Chapter 14 Respiration in Plants
• Chapter 15 Plant Growth and Development
• Chapter 16 Digestion and Absorption
• Chapter 17 Breathing and Exchange of Gases
• Chapter 18 Body Fluids and Circulation
• Chapter 19 Excretory Products and their Elimination
• Chapter 20 Locomotion and Movement
• Chapter 21 Neural Control and Coordination
• Chapter 22 Chemical Coordination and Integration

Class 11 English Hornbill NCERT Solutions  Prose

• Hornbill Chapter 1 The Portrait of a Lady
• Hornbill Chapter 2 We’re Not Afraid to Die…if We Can All Be Together
• Hornbill Chapter 3 Discovering Tut: the Saga Continues
• Hornbill Chapter 4 Landscape of the Soul
• Hornbill Chapter 5 The Ailing Planet: the Green Movement’s Role
• Hornbill Chapter 6 The Browning Version
• Hornbill Chapter 7 The Adventure
• Hornbill Chapter 8 Silk Road

NCERT Solutions for Class 11 English Hornbill Poem

• Hornbill Chapter 1 A Photograph (Poem)
• Hornbill Chapter 3 Lubrnum Top
• Hornbill Chapter 4 The Voice of the Rain
• Hornbill Chapter 6 Childhood
• Hornbill Chapter 8 Father to Son

Class 11 English Snapshots NCERT Solutions

• Snapshots Chapter 1 The Summer of the Beautiful White Horse
• Snapshots Chapter 2 The Address
• Snapshots Chapter 3 Ranga’s Marriage
• Snapshots Chapter 4 Albert Einstein at School
• Snapshots Chapter 5 Mother’s Day
• Snapshots Chapter 6 The Ghat of the only World
• Snapshots Chapter 7 Birth
• Snapshots Chapter 8 The Tale of Melon City
• Chapter 1 The Lament
• Chapter 2 A Pair of Mustachios
• Chapter 3 The Rocking-horse Winner
• Chapter 4 The Adventure of the Three Garridebs
• Chapter 5 Pappachi’s Moth
• Chapter 6 The Third and Final Continent
• Chapter 7 Glory at Twilight
• Chapter 8 The Luncheon
• Chapter 1 The Third and Final Continent
• Chapter 2 My Watch
• Chapter 3 My Three Passions
• Chapter 4 Patterns of Creativity
• Chapter 5 Tribal Verse
• Chapter 6 What is a Good Book?
• Chapter 7 The Story
• Chapter 8 Bridges
• Chapter 1 The Peacock
• Chapter 2 Let Me Not to the Marriage of True Minds
• Chapter 3 Coming
• Chapter 4 Telephone Conversation
• Chapter 5 The World is too Much With Us
• Chapter 6 Mother Tongue
• Chapter 7 Hawk Roosting
• Chapter 8 For Elkana
• Chapter 9 Refugee Blues
• Chapter 10 Felling of the Banyan Tree
• Chapter 11 Ode to a Nightingale
• Chapter 12 Ajamil and the Tigers

NCERT Solutions for Class 11 English Long Reading Texts/Novels

• The Canterville Ghost
• Up From Slavery
• Value-Based Questions

NCERT Solutions for Class 11 English Reading

• Factual Passages
• Discursive Passages
• Passages for Note-Making and Summarizing

NCERT Solutions for Class 11 English Writing

• Advertisements
• Business/Official Letters
• Letters to The Editor
• Application for a Job
• Letter to School/College Authorities
• Report Writing

NCERT Solutions for Class 11 English Grammar

• Determiners
• Tenses in Conditional Sentences
• Present Tense
• Future Tense
• Active and Passive Voice
• Integrated Exercises

NCERT Solutions for Class 11 Hindi Aroh (आरोह)

पाठ्यपुस्तक एवं पूरक पाठ्यपुस्तक आरोह, भाग-1 (पाठ्यपुस्तक)

(अ) काव्य भाग

• Chapter 1 हम तौ एक एक करि जांनां, संतों देखत जग बौराना
• Chapter 2 मेरे तो गिरधर गोपाल दूसरो न कोई, पग घुँघरू बाधि मीरां नाची
• Chapter 3 पथिक
• Chapter 4 वे आँखें
• Chapter 5 घर की याद
• Chapter 6 चंपा काले-काले अच्छर नहीं चीन्हती
• Chapter 7 गजल
• Chapter 8 हे भूख! मत मचल, हे मेरे जूही के फूल जैसे ईश्वर
• Chapter 9 सबसे खतरनाक
• Chapter 10 आओ, मिलकर बचाएँ

(ब) गद्य भाग

• Chapter 11 नमक का दारोगा
• Chapter 12 मियाँ नसीरुद्दीन
• Chapter 13 अपू के साथ ढाई साल
• Chapter 14 विदाई-संभाषण
• Chapter 15 गलता लोहा
• Chapter 16 स्पीति में बारिश
• Chapter 17 रजनी
• Chapter 18 जामुन का पेड़
• Chapter 19 भारत माता
• Chapter 20 आत्मा का ताप

NCERT Solutions for Class 11 Hindi Vitan (वितान)

वितान, भाग-1 (पूरक पाठ्यपुस्तक)

• Chapter 1 भारतीय गायिकाओं में बेजोड़ : लता मंगेशकर
• Chapter 2 राजस्थान की रजत बूँदें
• Chapter 3 आलो-आँधारि

खंड- क : अपठित-बोध

• अपठित गद्यांश
• अपठित काव्यांश

खंड-ख : कार्यालयी हिंदी और रचनात्मक लेखन

NCERT Solutions for Class 11 Hindi निबंध

NCERT Solutions for Class 11 Hindi कार्यालयीन पत्र

• कार्यालयी पत्र

जनसंचार माध्यम और लेखन

• जनसंचार माध्यम
• पत्रकारिता के विविध आयाम
• फीचर, रिपोर्ट, आलेख लेखन

NCERT Solutions for Class 11 Hindi मौखिक परीक्षा

• मौखिक परीक्षा

NCERT Solutions for Class 11 Business Studies (BST)

• Chapter 1 Nature and Purpose of Business
• Chapter 2 Forms of Business Organisation
• Chapter 3 Private, Public and Global Enterprises
• Chapter 4 Business Services
• Chapter 5 Emerging Modes of Business
• Chapter 6 Social Responsibilities of Business and Business Ethics
• Chapter 7 Formation of a Company
• Chapter 8 Sources of Business Finance
• Chapter 9 Small Business
• Chapter 10 Internal Trade
• Chapter 11 International Business-I
• Chapter 12 International Business-II
• Chapter 1 Introduction to Accounting
• Chapter 2 Theory Base of Accounting
• Chapter 3 Recording of Transactions – I
• Chapter 4 Recording of Transactions – II
• Chapter 5 Bank Reconciliation Statement
• Chapter 6 Trial Balance and Rectification of Errors
• Chapter 7 Depreciation, Provisions and Reserves
• Chapter 8 Bills of Exchange
• Chapter 1 What is Psychology?
• Chapter 2 Methods of Enquiry in Psychology
• Chapter 3 The Bases of Human Behaviour
• Chapter 4 Human Development
• Chapter 5 Sensory, Attentional And Perceptual Processes
• Chapter 6 Learning
• Chapter 7 Human Memory
• Chapter 8 Thinking
• Chapter 9 Motivation And Emotion
• Chapter 1 Concept and Functions
• Chapter 2 An Entrepreneur
• Chapter 3 Entrepreneurial Journey
• Chapter 4 Entrepreneurship as Innovation and Problem Solving
• Chapter 5A Concept of Market: Market, Market, Where are you?
• Chapter 5B Analysing the Market Environment
• Chapter 5C Researching the Market Facts That Matter
• Chapter 5D Expanding Markets
• Chapter 5E Know Thy Business
• Chapter 5F Marketing Mix
• Chapter 6 Business Finance and Arithmetic
• Chapter 7 Resource Mobilization

NCERT Solutions for Class 11 Indian Economic Development

• Chapter 1 Indian Economy on the Eve of Independence
• Chapter 2 Indian Economy 1950-1990
• Chapter 3 Liberalisation, Privatisation and Globalisation -An Appraisal
• Chapter 4 Poverty
• Chapter 5 Human Capital Formation in India
• Chapter 6 Rural Development
• Chapter 7 Employment-Growth, Informalisation and Related Issues
• Chapter 8 Infrastructure
• Chapter 9 Environment Sustainable Development
• Chapter 10 Comparative Development Experience of India with its Neighbours

Unit 1 : Computer Fundamentals

1.  Computer overview and its Basics 2.  Software Concepts 3.  Data Representation 4.  Microprocessor and Memory Concepts

Unit 2 : Programming Methodology

5.  Programming Methodology 6.  Algorithms and Flowcharts

Unit 3 : Introduction to Python

7.  Introduction to Python 8.  Getting Started with Python 9.  Operators in Python 10.  Functions 11.  Conditional and Looping Constructs

Unit 4 : Programming with Python

12.  Strings 13.  Lists, Dictionaries and Tuples

Advantages of Solving NCERT Solutions for Class 11 from LearnCBSE.in

•   All the Class 11 NCERT Textbook Solutions  provided in this page are clear and concise in nature.
• NCERT Solutions for  Class 11 Books  are solved in easily understandable language to help students to grasp everything on the go.
• Accessible to everyone at any time anywhere without any difficulty.
• All the questions are solved strictly based on the NCERT (CBSE) Syllabus and Books. So mastering physics these solutions will definitely help students to score good marks in the examination.
• NCERT Solutions for  Class XI   given in this page are of free of cost.

CBSE Class 11 study materials are provided here for all the subjects in PDF format. These materials can be easily downloaded by individual student, which they can use it offline as well. The materials provided here by LearnCBSE are NCERT solutions, revision notes, syllabus, sample papers, previous year question papers and important questions of CBSE class 11, as per the latest CBSE syllabus 2019-20. All these materials are designed by our experts and experienced teachers in each subject. 

Students can use these materials as reference, while preparing for the final exams to score good marks. In 11th standard students are introduced with elective subjects of Maths and Biology, Physics and Chemistry are common for both the stream. Till class 10th the education level was medium and students can easily grasp the knowledge. But, after reaching to grade 11, the level of education increases to an extent, where students are required with more focus in studies. Hence, to make it easier for them we have prepared these learning materials which will clear almost all the doubts of the students with an ease.  

Student will get here the best solutions for all the questions asked in the NCERT textbooks. Also, we are providing here the notes for reference, to be used as per their convenience. 

We hope the NCERT Solutions for Class 11 provided in this page helps in your board exam preparation.

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Introduction to problem solving Computer Science Class 11 Notes

By Sanjay Parmar

This article – introduction to problem solving Computer Science Class 11 offers comprehensive notes for Chapter 4 of the CBSE Computer Science Class 11 NCERT textbook.

Topics Covered

Introduction to problem solving Computer Science class 11

Computers, mobiles, the internet, etc. becomes our essentials nowadays for our routine life. We are using the to make our tasks easy and faster.

For example, earlier we were going to banks and standing in long queues for any type of transaction like money deposits or withdrawals. Today we can do these tasks from anywhere without visiting banks through internet banking and mobiles.

Basically, this was a complex problem and solved by a computer. The system was made online with the help of computers and the internet and made our task very easy.

This process is termed “Computerisations”. The problem is solved by using software to make a task easy and comfortable. Problem solving is a key term related to computer science.

The question comes to your mind how to solve a complex problem using computers? Let’s begin the article introduction to problem-solving Computer Science 11.

Introduction to problem solving Computer Science Class 11 – Steps for problem solving

“Computer Science is a science of abstraction -creating the right model for a problem and devising the appropriate mechanizable techniques to solve it.”

Solving any complex problem starts with understanding the problem and identifying the problem.

Suppose you are going to school by your bicycle. While riding on it you hear some noise coming from it. So first you will try to find that from where the noise is coming. So if you couldn’t solve the problem, you need to get it repaired.

The bicycle mechanic identifies the problem like a source of noise, causes of noise etc. then understand them and repair it for you.

So there are multiple steps involved in problem-solving. If the problem is simple and easy, we will find the solution easily. But the complex problem needs a few methods or steps to solve.

So complex problem requires some tools, a system or software in order to provide the solution. So it is a step-by-step process. These steps are as follows:

Analysing the problem

Developing an algorithm, testing and debugging.

The first step in the introduction to problem solving Computer Science Class 11 is analyzing the problem.

When you need to find a solution for a problem, you need to understand the problem in detail. You should identify the reasons and causes of the problem as well as what to be solved.

So this step involves a detailed study of the problem and then you need to follow some principles and core functionality of the solution.

In this step input and output, elements should be produced.

The second step for introduction to problem solving Computer Science class 11 is developing an algorithm.

An algorithm is a step-by-step process of a solution to a complex problem. It is written in natural language. An algorithm consists of various steps and begins from start to end. In between input, process and output will be specified. More details we will cover in the next section.

In short, the algorithm provides all the steps required to solve a problem.

For example:

Finding the simple interest, you need to follow the given steps:

• Gather required information and data such as principle amount, rate of interest and duration.
• Apply the formula for computing simple interest i.e. si=prn/100
• Now store the answer in si
• Display the calculated simple interest

In the above example, I have started and completed a task in a finite number of steps. It is completed in 4 finite steps.

Why algorithm is needed?

The algorithm helps developers in many ways. So it is needed for them for the following reasons:

• It prepares a roadmap of the program to be written before writing code.
• It helps to clearly visualise the instructions to be given in the program.
• When the algorithm is developed, a programmer knows the number of steps required to follow for the particular task.
• Algorithm writing is the initial stage (first step) of programming.
• It makes program writing easy and simple.
• It also ensures the accuracy of data and program output.
• It increases the reliability and efficiency of the solution.

Characteristics of a good algorithm

The characteristics of a good algorithm are as follows:

• It starts and ends with a finite number of steps. Therefore the steps are precisely stated or defined.
• In the algorithm, the result of each step is defined uniquely and based on the given input and process.
• After completion of the task, the algorithm will end.
• The algorithm accepts input and produces the output.

While writing the algorithm the following things should be clearly identified:

• The input required for the task
• The computation formula or processing instructions

After writing the algorithm, it is required to represent it. Once the steps are finalised, it is required to be represented logically. This logical representation of the program clearly does the following:

• Clears the logic of the program
• The execution of the program

The algorithm is steps written in the form of text. So it is difficult to read sometimes. So if it is represented in pictorial form it would be better for analysis of the program.

The flowchart is used to represent the algorithm in visual form.

Flowchart – Visual representation of an algorithm

A flowchart is made of some symbols or shapes like rectangles, squares, and diamonds connected by arrows. Every shape represents each step of an algorithm. The arrow basically represents the order or link of the steps.

The symbols used in the flow chart are as follows:

Coding is an essential part of the introduction to problem solving ComputerScience11.

• It is pronounced as soo-doh-kohd
• It is one of the ways of representing algorithms in a systematic way
• The word pseudo means not real, therefore pseudocode means not real code
• It is non-formal language, that helps programmers to write code
• It is written in human understandable language
• It cannot be directly read by computers
• There is no specific standard or way of writing pseudocode is there

When an algorithm is prepared, the next step is writing code. This code will be written in a specific programming language. The code follows certain rules and regulations of the programing language and provides solutions.

When coding is done you need to maintain it with proper documentation as well. The best practices for coding procedures must be followed. Because this code can be reviewed a number of times for further development and upgradation.

Let’s understand this step with a simple example!!

When your mother prepares a cake at your home, she will give peace of cake to someone before serving it to check the taste of the cake, right!!! If anything is needed like sugar or softness or hardness should be improved she will decide and do the improvement.

Similarly after writing code testing and debugging are required to check the software whether is providing the solution in a good manner not.

Have look at this also: Computer Science Class XI

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NCERT solutions for Class 11 Computer Science chapter 4 - Introduction to Problem Solving [Latest edition]

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Solutions for chapter 4: introduction to problem solving.

Below listed, you can find solutions for Chapter 4 of CBSE NCERT for Class 11 Computer Science.

NCERT solutions for Class 11 Computer Science Chapter 4 Introduction to Problem Solving Exercise [Pages 83 - 85]

Write a pseudocode that reads two numbers and divides one by another and displays the quotient.

Two friends decide who gets the last slice of a cake by flipping a coin five times. The first person to win three flips wins the cake. An input of 1 means player 1 wins a flip, and a 2 means player 2 wins a flip. Design an algorithm to determine who takes the cake.

Write the pseudocode to print all multiples of 5 between 10 and 25 (including both 10 and 25).

Give an example of a loop that is to be executed a certain number of times.

Suppose you are collecting money for something. You need ₹ 200 in all. You ask your parents, uncles, and aunts as well as grandparents. Different people may give either ₹ 10, ₹ 20, or even ₹ 50. You will collect till the total becomes 200. Write the algorithm.

Write the pseudocode to print the bill depending on the price and quantity of an item. Also print Bill GST, which is the bill after adding 5% of the tax to the total bill.

Write pseudocode that will perform the following:

• Read the marks of three subjects: Computer Science, Mathematics, and Physics, out of 100.
• Calculate the aggregate marks.
• Calculate the percentage of marks.

Write an algorithm to find the greatest among two different numbers entered by the user.

Write an algorithm that performs the following:

Ask a user to enter a number. If the number is between 5 and 15, write the word GREEN. If the number is between 15 and 25, write the word BLUE. if the number is between 25 and 35, write the word ORANGE. If it is any other number, write that ALL COLOURS ARE BEAUTIFUL.

Write an algorithm that accepts four numbers as input and finds the largest and smallest of them.

Write an algorithm to display the total water bill charges of the month depending upon the number of units consumed by the customer as per the following criteria:

• for the first 100 units @ 5 per unit
• for next 150 units @ 10 per unit
• more than 250 units @ 20 per unit

Also, add meter charges of 75 per month to calculate the total water bill.

What are conditionals? When they are required in a program?

Match the pairs.

Following is an algorithm for going to school or college. Can you suggest improvements in this to include other options?

Reach_School_Algorithm

• Take lunch box
• Take the bus
• Get off the bus
• Reach school or college

Write pseudocode to calculate the factorial of a number (Hint: Factorial of 5, written as 5! = 5 × 4 × 3 × 2 × 1).

Draw a flowchart to check whether a given number is an Armstrong number. An Armstrong number of three digits is an integer such that the sum of the cubes of its digits is equal to the number itself. For example, 371 is an Armstrong number since 3**3 + 7**3 + 1**3 = 371.

Following is an algorithm to classify numbers as “Single Digit”, “Double Digit” or “Big”.

Verify for (5, 9, 47, 99, 100, 200) and correct the algorithm if required.

For some calculations, we want an algorithm that accepts only positive integers up to 100.

• On what values will this algorithm fail?
• Can you improve the algorithm?

NCERT solutions for Class 11 Computer Science chapter 4 - Introduction to Problem Solving

Shaalaa.com has the CBSE Mathematics Class 11 Computer Science CBSE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. NCERT solutions for Mathematics Class 11 Computer Science CBSE 4 (Introduction to Problem Solving) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.

Concepts covered in Class 11 Computer Science chapter 4 Introduction to Problem Solving are Flowchart, Pseudocode, Introduction to Flow of Control, Sequence, Selection, Repetition, Verifying Algorithms, Comparison of Algorithm, Coding, Decomposition, Problem Solving, Steps for Problem Solving, Algorithms, Why Do We Need an Algorithm?, Representation of Algorithms.

Using NCERT Class 11 Computer Science solutions Introduction to Problem Solving exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Solutions are essential questions that can be asked in the final exam. Maximum CBSE Class 11 Computer Science students prefer NCERT Textbook Solutions to score more in exams.

Get the free view of Chapter 4, Introduction to Problem Solving Class 11 Computer Science additional questions for Mathematics Class 11 Computer Science CBSE, and you can use Shaalaa.com to keep it handy for your exam preparation.

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NCERT Solutions for Class 11 Maths

Ncert solutions for class 11 maths – free pdf.

NCERT Solutions Class 11 Maths provided here by Mathongo are made by the best teachers, academicians and experienced professors as per the guidelines prescribed by CBSE (Central Board of Secondary Education). All the questions are solved in a manner that the students will find it easy and at the same time understand the concept immediately. The NCERT maths textbook is usually used by all the CBSE affiliated schools in India.

However, many students from state board schools are also asked to solve questions from CBSE NCERT books as it is the best book which gives conceptual clarity along with practical application of knowledge. Studying from CBSE NCERT Maths Class 11 book will not only help you to achieve this, but also help you to crack exams like IIT JEE, NEET, AIIMS and other government exams like NDA and UPSC. The NCERT solutions of class 11 are available for free download in PDF format. It is easily accessible. It compromises of 16 chapters in total. The solutions enable the students to have an overall better learning of Mathematics along with enhancing their problem-solving skills.

All the difficult questions have solved in the simplest way possible explained in step by step method so that it becomes easy to grasp and help in quick learning. It is also advisable that students also solve the exemplar problems of NCERT along with exercise problems. The Pdf of NCERT Maths class 11 will give you proper guidance and help you in achieving the best scores you can. Click on the links to get the PDF download online.

NCERT Class 11 Maths Chapterwise Solutions

• Chapter 1 Sets
• Chapter 2 Relations and Functions
• Chapter 3 Trigonometric Functions
• Chapter 4 Principles of Mathematical Induction
• Chapter 5 Complex Numbers and Quadratic Equations
• Chapter 6 Linear Inequalities
• Chapter 7 Permutations and Combinations
• Chapter 8 Binomial Theorem
• Chapter 9 Sequences and Series
• Chapter 10 Straight lines
• Chapter 11 Conic Sections
• Chapter 12 Introduction to Three-Dimensional Geometry
• Chapter 13 Limits and Derivatives
• Chapter 14 Mathematical Reasoning
• Chapter 15 Statistics
• Chapter 16 Probability

Why Go for Mathongo’s NCERT Maths Solutions for Class 11

Mathongo provides NCERT class 11 maths solution at free of cost online and it’s easily accessible in the form of PDF. Each PDF constitutes of the chapter in which exercise wise solutions are given. It is very easy for the students to locate any answer from the exercise. All the solutions are made with the help of the best teachers and academicians.

The solutions are plain easy and can be understood by anyone. The approaches and the methods used to solve the questions are also given in step by step method so that students score good in paper. For additional conceptual clarity, diagrams and formulas are also used.

Mathongo has provided the best solutions for class 11 maths Ncert textbooks. Students can refer these solutions which will help them in solving the previous year question papers. These NCERT maths class 11 solution will help the students to study more effectively and also present the answers in the most effective way to get the highest scores.

The back-end exercise questions given in NCERT books are very important for the class 11 finals. These exercises often test the conceptual clarity and the subject knowledge of the student. NCERT Solutions will help you to understand these concepts better. On the other hand, the students who have problem in solving maths and require assistance and guidance can also refer class 11 Ncert Maths Solutions to clarify the complex questions concepts.

Maths is all about practice. The more you practice, the better you will be. Ncert maths class 11 pdf download can help students to solve all their problems and also save their time as everything is at one place, one click away.

Is class 11 maths tough?

Is Maths tough? Well this question is pretty subjective. Maths is one of the subjects which causes phobia among many students. But if you develop interest with constant practice, slowly and slowly you will get it. Transition of maths especially in CBSE is quite difficult as the pattern and type of questions change from class 10 th to class 11 th .

This often leads to fall in internal marks too as students are not used to these new concepts. It takes a bit of time to hang onto new theories and concepts. Only by practice, you can get through this change. Chapters like Trigonometry and Permutation and Combination may seem hard at first but as said earlier with practice it will be A-Okay.

Which book is best for class 11 maths?

Well, no book can take place of NCERT class 11 maths textbook. This is your bible which you must not forget. Almost half of the question paper comes from NCERT and all the competitive exams are based on NCERT.

However, once you are done with Ncert, you can go through Senior Secondary School Mathematics: for Class 11 by RS Agarwal and Mathematics for Class 11 by RD Sharma.

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Introduction to Problem Solving Class 11 Questions and Answers

Teachers and Examiners ( CBSESkillEduction ) collaborated to create the Introduction to Problem Solving Class 11 Questions and Answers . All the important Information are taken from the NCERT Textbook Computer Science (083) class 11 .

1. Write pseudocode that reads two numbers and divide one by another and display the quotient. Answer – Input num1 Input num2 Calculate div = num1 / num2 Print div

2. Two friends decide who gets the last slice of a cake by flipping a coin five times. The first person to win three flips wins the cake. An input of 1 means player 1 wins a flip, and a 2 means player 2 wins a flip. Design an algorithm to determine who takes the cake? Answer – Set p1 = 0 Set p2 = 0 For i in range (5): Input coin If coin = 1 then P1 += 1 Elif coin = then P2 += 1 If p1 > 2 then P1 wins Elif p2 > 2 then P2 wins

3. Write the pseudocode to print all multiples of 5 between 10 and 25 (including both 10 and 25). Answer – FOR num := 10 to 25 DO IF num % 5 = 0 THEN PRINT num END IF END LOOP

4. Give an example of a loop that is to be executed a certain number of times. Answer – SET i: = 1 FOR i: = 1 to 10 do PRINT i END LOOP

5. Suppose you are collecting money for something. You need ` 200 in all. You ask your parents, uncles and aunts as well as grandparents. Different people may give either ` 10, ` 20 or even ` 50. You will collect till the total becomes 200. Write the algorithm. Answer – Step 1 : Start Step 2 : Set money := 0 Step 3 : While Loop (money <200) Input money Step 4 : money = money + money Step 5 : End Loop Step 6 : Stop

6. Write the pseudocode to print the bill depending upon the price and quantity of an item. Also print Bill GST, which is the bill after adding 5% of tax in the total bill. Answer – INPUT Item INPUT price CALCULATE bill := Item * price PRINT bill CALCULATE tax := bill * (5 / 100) CALCULATE GST_Bill := bill + tax PRINT GST_Bill

7. Write pseudocode that will perform the following: a) Read the marks of three subjects: Computer Science, Mathematics and Physics, out of 100 b) Calculate the aggregate marks c) Calculate the percentage of marks Answer – INPUT computer, maths, phy COMPUTE average := (computer + maths + phy) / 3 COMPUTE percentage := (average / 300) * 100 PRINT average PRINT percentage

8. Write an algorithm to find the greatest among two different numbers entered by the user. Answer – INPUT num1, num2 IF num1 > num2 THEN PRINT num1 ELSE IF num2 > num1 THEN PRINT num2 END IF

9. Write an algorithm that performs the following: Ask a user to enter a number. If the number is between 5 and 15, write the word GREEN. If the number is between 15 and 25, write the word BLUE. if the number is between 25 and 35, write the word ORANGE. If it is any other number, write that ALL COLOURS ARE BEAUTIFUL. Answer – INPUT num IF num >=5 AND num < 15 THEN PRINT ‘GREEN’ ELSE IF num >= 15 AND num < 25 THEN PRINT ‘BLUE’ ELSE IF num >= 25 AND num < 35 THEN PRINT ‘ORANGE’ ELSE PRINT ‘ALL COLOURS ARE BEAUTIFUL’ END IF

10. Write an algorithm that accepts four numbers as input and find the largest and smallest of them. Answer – INPUT max SET min := max FOR i: = 1 to 3 do INPUT num IF num<max THEN SET max :=num ELSE SET min := num END LOOP PRINT max PINT min

11. Write an algorithm to display the total water bill charges of the month depending upon the number of units consumed by the customer as per the following criteria: • for the first 100 units @ 5 per unit • for next 150 units @ 10 per unit • more than 250 units @ 20 per unit Also add meter charges of 75 per month to calculate the total water bill . Answer – INPUT units SET bill := 0 IF units > 250 THEN CALCULATE bill := units * 20 ELIF units <= 100 THEN CALCULATE bill := units * 5 ELSE CALCULATE bill := 100 * 5 + (units – 100) * 10 END IF END IF CALCULATE totalBill := bill + 75 PRINT totalBill

12. What are conditionals? When they are required in a program? Answer – Conditionals are programming language elements used in computer science that execute various computations or actions based on whether a boolean condition supplied by the programmer evaluates to true or false.

When a software needs to calculate a result based on a given circumstance, they are necessary (s).

14. Following is an algorithm for going to school or college. Can you suggest improvements in this to include other options? Reach_School_Algorithm a) Wake up b) Get ready c) Take lunch box d) Take bus e) Get off the bus f) Reach school or college Answer – a) Wake up b) Brush your teeth c) Take bath d) Dress up e) Eat breakfast f) Take lunch box g) Take Bus h) Get off the bus i) Reach school or college

15. Write a pseudocode to calculate the factorial of a number (Hint: Factorial of 5, written as 5!=5 4 3 21 ×××× ). Answer – INPUT num SET fact := 1, i := 1 WHILE i <= num DO CALCULATE fact := fact * i INCREASE i by 1 END LOOP PRINT fact

16. Draw a flowchart to check whether a given number is an Armstrong number. An Armstrong number of three digits is an integer such that the sum of the cubes of its digits is equal to the number itself. For example, 371 is an Armstrong number since 3**3 + 7**3 + 1**3 = 371. Answer –

17. Following is an algorithm to classify numbers as “Single Digit”, “Double Digit” or “Big”. Classify_Numbers_Algo INPUT Number IF Number < 9 “Single Digit” Else If Number < 99 “Double Digit” Else “Big” Verify for (5, 9, 47, 99, 100 200) and correct the algorithm if required Answer – INPUT Number IF Number <= 9 “Single Digit” Else If Number <= 99 “Double Digit” Else “Big”

18. For some calculations, we want an algorithm that accepts only positive integers upto 100.

Accept_1to100_Algo INPUT Number IF (0<= Number) AND (Number <= 100) ACCEPT Else REJECT a) On what values will this algorithm fail? b) Can you improve the algorithm?

Answer – INPUT number IF (number>0) AND (number <=100) ACCEPT Else REJECT

Computer Science Class 11 Notes

• Unit 1 : Basic Computer Organisation
• Unit 1 : Encoding Schemes and Number System
• Unit 2 : Introduction to problem solving
• Unit 2 : Getting Started with Python
• Unit 2 : Conditional statement and Iterative statements in Python
• Unit 2 : Function in Python
• Unit 2 : String in Python
• Unit 2 : Lists in Python
• Unit 2 : Tuples in Python
• Unit 2 : Dictionary in Python
• Unit 3 : Society, Law and Ethics

Computer Science Class 11 MCQ

Computer science class 11 ncert solutions.

• Unit 2 : Tuples and Dictionary in Python

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CBSE Class 11 | Problem Solving Methodologies

Problem solving process.

The process of problem-solving is an activity which has its ingredients as the specification of the program and the served dish is a correct program. This activity comprises of four steps : 1. Understanding the problem: To solve any problem it is very crucial to understand the problem first. What is the desired output of the code and how that output can be generated? The obvious and essential need to generate the output is an input. The input may be singular or it may be a set of inputs. A proper relationship between the input and output must be drawn in order to solve the problem efficiently. The input set should be complete and sufficient enough to draw the output. It means all the necessary inputs required to compute the output should be present at the time of computation. However, it should be kept in mind that the programmer should ensure that the minimum number of inputs should be there. Any irrelevant input only increases the size of and memory overhead of the program. Thus Identifying the minimum number of inputs required for output is a crucial element for understanding the problem.

2. Devising the plan: Once a problem has been understood, a proper action plan has to be devised to solve it. This is called devising the plan. This step usually involves computing the result from the given set of inputs. It uses the relationship drawn between inputs and outputs in the previous step. The complexity of this step depends upon the complexity of the problem at hand.

3. Executing the plan: Once the plan has been defined, it should follow the trajectory of action while ensuring the plan’s integrity at various checkpoints. If any inconsistency is found in between, the plan needs to be revised.

4. Evaluation: The final result so obtained must be evaluated and verified to see if the problem has been solved satisfactorily.

Problem Solving Methodology(The solution for the problem)

The methodology to solve a problem is defined as the most efficient solution to the problem. Although, there can be multiple ways to crack a nut, but a methodology is one where the nut is cracked in the shortest time and with minimum effort. Clearly, a sledgehammer can never be used to crack a nut. Under problem-solving methodology, we will see a step by step solution for a problem. These steps closely resemble the software life cycle . A software life cycle involves several stages in a program’s life cycle. These steps can be used by any tyro programmer to solve a problem in the most efficient way ever. The several steps of this cycle are as follows :

Step by step solution for a problem (Software Life Cycle) 1. Problem Definition/Specification: A computer program is basically a machine language solution to a real-life problem. Because programs are generally made to solve the pragmatic problems of the outside world. In order to solve the problem, it is very necessary to define the problem to get its proper understanding. For example, suppose we are asked to write a code for “ Compute the average of three numbers”. In this case, a proper definition of the problem will include questions like : “What exactly does average mean?” “How to calculate the average?”

Once, questions like these are raised, it helps to formulate the solution of the problem in a better way. Once a problem has been defined, the program’s specifications are then listed. Problem specifications describe what the program for the problem must do. It should definitely include :

what is the input set of the program

What is the desired output of the program and in what form the output is desired?

2. Problem Analysis (Breaking down the solution into simple steps): This step of solving the problem follows a modular approach to crack the nut. The problem is divided into subproblems so that designing a solution to these subproblems gets easier. The solutions to all these individual parts are then merged to get the final solution of the original problem. It is like divide and merge approach.

Modular Approach for Programming :

The process of breaking a large problem into subproblems and then treating these individual parts as different functions is called modular programming. Each function behaves independent of another and there is minimal inter-functional communication. There are two methods to implement modular programming :

• Top Down Design : In this method, the original problem is divided into subparts. These subparts are further divided. The chain continues till we get the very fundamental subpart of the problem which can’t be further divided. Then we draw a solution for each of these fundamental parts.
• Bottom Up Design : In this style of programming, an application is written by using the pre-existing primitives of programming language. These primitives are then amalgamated with more complicated features, till the application is written. This style is just the reverse of the top-down design style.

3. Problem Designing: The design of a problem can be represented in either of the two forms :

The ways to execute any program are of three categories:

• Sequence Statements Here, all the instructions are executed in a sequence, that is, one after the another, till the program is executed.
• Selection Statements As it is self-clear from the name, in these type of statements the whole set of instructions is not executed. A selection has to be made. A selected number of instructions are executed based on some condition. If the condition holds true then some part of the instruction set is executed, otherwise, another part of the set is executed. Since this selection out of the instruction set has to be made, thus these type of instructions are called Selection Statements.

Identification of arithmetic and logical operations required for the solution : While writing the algorithm for a problem, the arithmetic and logical operations required for the solution are also usually identified. They help to write the code in an easier manner because the proper ordering of the arithmetic and logical symbols is necessary to determine the correct output. And when all this has been done in the algorithm writing step, it just makes the coding task a smoother one.

• Flow Chart : Flow charts are diagrammatic representation of the algorithm. It uses some symbols to illustrate the starting and ending of a program along with the flow of instructions involved in the program.

4. Coding: Once an algorithm is formed, it can’t be executed on the computer. Thus in this step, this algorithm has to be translated into the syntax of a particular programming language. This process is often termed as ‘coding’. Coding is one of the most important steps of the software life cycle. It is not only challenging to find a solution to a problem but to write optimized code for a solution is far more challenging.

Writing code for optimizing execution time and memory storage : A programmer writes code on his local computer. Now, suppose he writes a code which takes 5 hours to get executed. Now, this 5 hours of time is actually the idle time for the programmer. Not only it takes longer time, but it also uses the resources during that time. One of the most precious computing resources is memory. A large program is expected to utilize more memory. However, memory utilization is not a fault, but if a program is utilizing unnecessary time or memory, then it is a fault of coding. The optimized code can save both time and memory. For example, as has been discussed earlier, by using the minimum number of inputs to compute the output , one can save unnecessary memory utilization. All such techniques are very necessary to be deployed to write optimized code. The pragmatic world gives reverence not only to the solution of the problem but to the optimized solution. This art of writing the optimized code also called ‘competitive programming’.

5. Program Testing and Debugging: Program testing involves running each and every instruction of the code and check the validity of the output by a sample input. By testing a program one can also check if there’s an error in the program. If an error is detected, then program debugging is done. It is a process to locate the instruction which is causing an error in the program and then rectifying it. There are different types of error in a program : (i) Syntax Error Every programming language has its own set of rules and constructs which need to be followed to form a valid program in that particular language. If at any place in the entire code, this set of rule is violated, it results in a syntax error. Take an example in C Language

In the above program, the syntax error is in the first printf statement since the printf statement doesn’t end with a ‘;’. Now, until and unless this error is not rectified, the program will not get executed.

Once the error is rectified, one gets the desired output. Suppose the input is ‘good’ then the output is : Output:

(ii) Logical Error An error caused due to the implementation of a wrong logic in the program is called logical error. They are usually detected during the runtime. Take an example in C Language:

In the above code, the ‘for’ loop won’t get executed since n has been initialized with the value of 11 while ‘for’ loop can only print values smaller than or equal to 10. Such a code will result in incorrect output and thus errors like these are called logical errors. Once the error is rectified, one gets the desired output. Suppose n is initialised with the value ‘5’ then the output is : Output:

(iii) Runtime Error Any error which causes the unusual termination of the program is called runtime error. They are detected at the run time. Some common examples of runtime errors are : Example 1 :

If during the runtime, the user gives the input value for B as 0 then the program terminates abruptly resulting in a runtime error. The output thus appears is : Output:

Example 2 : If while executing a program, one attempts for opening an unexisting file, that is, a file which is not present in the hard disk, it also results in a runtime error.

6. Documentation : The program documentation involves :

• Problem Definition
• Problem Design
• Documentation of test perform
• History of program development

7. Program Maintenance: Once a program has been formed, to ensure its longevity, maintenance is a must. The maintenance of a program has its own costs associated with it, which may also exceed the development cost of the program in some cases. The maintenance of a program involves the following :

• Detection and Elimination of undetected errors in the existing program.
• Modification of current program to enhance its performance and adaptability.
• Enhancement of user interface
• Enriching the program with new capabilities.
• Updation of the documentation.

Control Structure- Conditional control and looping (finite and infinite)

There are codes which usually involve looping statements. Looping statements are statements in which instruction or a set of instructions is executed multiple times until a particular condition is satisfied. The while loop, for loop, do while loop, etc. form the basis of such looping structure. These statements are also called control structure because they determine or control the flow of instructions in a program. These looping structures are of two kinds :

In the above program, the ‘for’ loop gets executed only until the value of i is less than or equal to 10. As soon as the value of i becomes greater than 10, the while loop is terminated. Output:

In the above code, one can easily see that the value of n is not getting incremented. In such a case, the value of n will always remain 1 and hence the while loop will never get executed. Such loop is called an infinite loop. Output:

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Steps for Problem Solving

Last updated at April 16, 2024 by Teachoo

• Analyzing the Problem: Involves identifying the problem , inputs the program should accept and the desired output of the program.
• Developing an Algorithm: The solution to the problem represented in natural language is called Algorithm. For a given problem, more than one algorithm is possible and we have to select the most suitable solution.
• Coding: Different high level languages can be used for writing the code based on the algorithm developed.
• Testing and Debugging: To ensure that the software meets all the business and technical requirements and works as expected . The errors or defects found in the testing phases are debugged or rectified and the program is again tested . This continues till all the errors are removed from the program.  

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NCERT Class 11 Physics Chapter 11 Thermodynamics

NCERT Class 11 Physics Chapter 11 Thermodynamics   Solutions , NCERT Class 11 Physics Chapter 11 Thermodynamics Notes to each chapter is provided in the list so that you can easily browse throughout different chapters  NCERT Class 11 Physics Chapter 11 Thermodynamics Question Answer and select needs one.

Also, you can read the SCERT book online in these sections NCERT Class 11 Physics Chapter 11 Thermodynamics Solutions by Expert Teachers as per SCERT ( CBSE ) Book guidelines. These solutions are part of SCERT  All Subject Solutions . Here we have given NCERT Class 11 Physics Chapter 11 Thermodynamics Solutions for All Subjects, You can practice these here.

Thermodynamics

Chapter: 11

1. A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 10 4 J/g?

Ans: Energy required to heat the water:

Mass of water 

= 3.0 litres/minute × 1000 g/litre

= 3000 g/minute

Heat capacity of water = 4.184  J/g°C

Temperature change = 77°C – 27°C = 50°C

Energy = 3000 g/minute × 4.184 J/g°C × 50°C

= 627600 J/minute

Energy into joules per second (J/s):

627600 J/minute ÷ 60 s/minute = 10460 J/s

Heat of combustion = 4.0 × 10 4 J/g

Rate of fuel consumption:

= 10460 J/s ÷ 4.0 × 10 4 J/g

= 0.2615 g/s  or 15.69 g/minute.

2. What amount of heat must be supplied to 2.0 × 10 –2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular Mass of N2 = 28; R = 8.3 J mol –1 K –1 .)

Q = n × Cp × ΔT

n = mass / molecular mass

= 2.0 × 10 –2 kg / 28 kg/kmol

= 7.14 × 10 –4 k mol

Cp = (7/2) × R

= (7/2) × 8.3 J/mol·K

= 29.05 J/mol·K

Q = 7.14 × 10 –4 k mol × 29.05 J/mol·K × 45 K

3. Explain why:

(a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2 )/2.

Ans: Law of conservation of Energy says that the energy is transferred from one body to another and from one format to another. The final temperature can be the mean temperature only when the thermal capacities of both the bodies are equal.

(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.

Ans: If the specific heat of the coolant is high, it can absorb large amounts of heat without heating itself much. Hence, a coolant should have high specific heat.  

(c) Air pressure in a car tyre increases during driving.

Ans: Temperature of air in the tyre increases due to friction of tyre with road. Therefore, air pressure inside the tyre increases according to Charle’s law. 

(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

Ans: This is because in a harbour town, the relative humidity is more than in a desert town. Hence, the climate of harbour town is without extremes of hot and cold.

4. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the ab pressure of the gas increase if the gas is compressed to half its original volume?

Ans: Here:  

P 1 V 1 γ = P 2 V 2 γ

where γ is the adiabatic index (approximately 1.4 for hydrogen).

V 2 = V 1 / 2

Rearrange the equation to solve for P 2 /P 1 :

P 2 /P 1 = (V 1 /V 2 ) γ

5. In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)

9.35 cal × 4.19 J/cal = 39.15 J

The first law of thermodynamics:

ΔU = Q – W

For the adiabatic process:

ΔU = 0 – (-22.3 J) 

ΔU = 22.3 J

For the non-adiabatic process:

ΔU = 39.15 J – W

Since ΔU is the same for both processes:

22.3 J = 39.15 J – W

W = 16.85 J.

6. Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock suddenly opened. Answer the following:

(a) What is the final pressure of the gas in A and B?  

P 2 V 2 = P 1 V 1

P 1 = 1 atm,

V 1 = V 

V 2 = 2V and P 2 = ?

P 2 = P 1 V 1 /V 2

= 1 × V/ 2V

= 0.5 atm. 

(b) What is the change in internal energy of the gas?

Ans: Since the temperature of the system remains unchanged, and the process is adiabatic, the change in internal energy of the gas is zero.

(c) What is the change in the temperature of the gas?  

Ans: The system being internally insulated, there is no change in temperature. 

(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?

Ans: No, because the process called free expansion is rapid and cannot be controlled. Therefore, the intermediate states are non – equilibrium states and the gas equation is not satisfied in these states. The gas cannot return to an equilibrium state which lie on the P – T- V surface.

7. An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal

energy increasing?

Ans: The first law of thermodynamics states:

dU/dt = dQ/dt – dW/dt

dQ/dt = 100 W 

dW/dt = 75 J/s 

dU/dt = 100 W – 75 J/s

8. A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.13).

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.

Ans: As it is clear from the above figure, Change in pressure, dp = EF. 

Area of triangle = (1/2) × base x height

Base = 5.0 m³ – 2.0 m³ = 3.0 m³

Height = 600 N/m² – 300 N/m² = 300 N/m²

Work done from D to E = (1/2) × 3.0 m³ × 300 N/m² = 450 J.

Work done from E to F:

Length = 5.0 m 3 – 2.0 m 3 = 3.0 m 3  

Width = 300 N/m 2  

Work done from E to F = 3.0 m 3 × 300 N.m 2 = 900 J

Total work done

= 450 j + 900 j

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