## STUDENT CONTEST #3 – WINNERS

**The list of winners and solutions for the second contest have been shared below.**

(This contest ended on **15 January, 2019)**

68 min read

(This contest ended on **15 January, 2019)**

(This contest ended on **15 January, 2019)**

** **

Name | Video | School | Class |

Kartik Gupta | Little Angels Sr. Sec. School, Sonipat | 8 | |

Harsh Fozdar | Children’s Academy, Mumbai | 6 | |

Tejas Bansal | Mount Carmel School, Dwarka, New Delhi | 4 | |

Meemansa Pandey | Mount Carmel School, Anand Niketan, New Delhi | 9 |

Name | School | Class/Grade |

Jayden Antony Thomas | Peepal Prodigy School, Coimbatore | 5 |

Urvi Agrawal | The Heritage School, Kolkata | 5 |

Shivam Pandey | Children’s Academy, Mumbai | 5 |

Debanjan Pan | The Heritage School, Kolkata | 5 |

Bhavya Yadav | Delhi Public School, Ferozepur | 4 |

Dibyanshu Mohanty | B. D. M. International, Kolkata | 5 |

Name | School | Class/Grade |

Divyanshi Gupta | Salwan public school, Delhi | 7 |

Pranav Kurle | Ganges Valley School, Hyderabad | 8 |

Srijan Satpathi | B.D.M International, Kolkata | 7 |

Dhruv Gupta | Salwan public school, Delhi | 6 |

Rucha Kulkarni | Universal High Dahisar, Vadodra | 6 |

Damandeep Singh | Delhi public school, Ferozepur | 7 |

Kunal Krishna | B.D.M International, Kolkata | 6 |

Ankit Bandyopadhyay | Salwan Public School, Delhi | 8 |

Ashwina T Shankar | Salwan Public School, Delhi | 8 |

Shrishti Sumanta | Salwan Public School, Delhi | 8 |

Abhimanyu Singh | Salwan Public School, Delhi | 7 |

Aditya Sinha | Thakur Public School, Mumbai | 6 |

Palaash Jadav | Children’s Academy, Mumbai | 6 |

Name | School | Class/Grade |

Sachin Kumar N | Peepal Prodigy School, Coimbatore | 9 |

Astha Garg | Salwan public school Trans Delhi Signature City, Ghaziabad | 9 |

**Challenge: Explain the following concepts by doing 2 simple experiments with household items:**

**i)Air contracts on cooling **

**ii) Friction opposes motion**

**Solution: **

Experiments should be designed for the following:

- Illustrate that air contracts on cooling:

On cooling the particles of air come closer to each other. Due to this, the volume of air becomes less, the air becomes more dense and heavier. This can be demonstrated by multiple experiments, for instance: keeping an airtight empty bottle in the refrigerator or in cold water for a few minutes. The bottle shrinks in size showing that air contracts on cooling

2. Illustrate that friction opposes motion:

Friction tends to oppose any **relative** motion between objects, acting at the surfaces in contact. To determine the direction think about what would happen if there was no friction. One of the simple ways to demonstrate friction is to slide a box on a tile and rough ground. As the ground is rough, it offers more friction, opposes the motion and the box slides for lesser distance than it would on the tile.

**Question: **

**Teach the concept of “Factors and Multiples”. Explain using two real-life examples.**

*Note: Don’t include common factors, common multiples, LCM or HCF. Only focus on factors and multiples.*

**Solution:**

**Factors** of a number are those numbers which can completely divide the number.

For ex: 20 can be completely divided by 20, 10, 5, 4, 2, 1. Hence, there are 6 factors of 20.

We can observe the following points here.

- Factors of a number can’t be greater than the number as we can see 20 is the greatest factor of 20.
- Factors of a number are finite.

**Multiples **of a number are those numbers which can be completely divided by the number.

For ex: 20 can divide 20, 40, 60, 80, ……… . Here, 20, 40, 60 and 80 are multiples of 20.

We can observe the following points here.

- The smallest multiple of the number is the number itself.
- There are infinite multiples of a number.

Based on the above definitions we can relate factors and multiples also.

For ex: Factors of 20 are 20, 10, 5, 4, 2, 1. So, 20 is completely divided by 20, 10, 5, 4, 2, 1.

Which means 20 is a multiple of 20, 10, 5, 4, 2, 1.

Similarly, 20, 40, 60 and 80 are multiples of 20. So, 20 is a factor of all 20, 40, 60 and 80.

**Examples of factors and multiples.**

Example 1: There are 100 students in a class. Your class teacher wants to make groups such that each group will have an equal number of students but groups should not consist of more than 8 students. How many students can be in a group?

To decide the answer, we need to calculate factors of 100 which is less than 8.

Factors of 100 (less than 8) = 1, 2, 4 and 5

So, we can make groups of 2, 4 or 5 students

Example 2: Chairs are arranged in a theatre in rows as shown in the figure. Each row consists of the same number of chairs. Can there be 230 chairs in the room?

In each row, there are 6 chairs. The total number of chairs will be multiple of

6. Here, 230 is not a multiple of 6 so, there can’t be 230 chairs.

**QUESTION: **

- Define the second law of reflection.
- Take three sticks of equal length. Assume one as an incident ray, second as normal and the third as a reflected ray. Arrange these three sticks such that it follows the first law of reflection but it does NOT follow the second law of reflection.

**SOLUTION: **

The second law of reflection states that the incident ray, the reflected ray, and the normal to the surface of the mirror all lie in the same plane. Furthermore, from the first law, we know that angle of reflection ‘’ is equal to the angle of incidence ‘’. Both angles are measured with respect to the normal to the mirror.

The incident ray, reflected ray and normal all lie in the same plane i.e. the plane of incidence. The second law states the importance of the plane of incidence. In the absence of a particular plane of incidence, there could be multiple reflected rays at same angle of reflection. Thus the reflected ray could not be specified uniquely without using the second law of reflection.

The three stick arrangement was supposed to be built to cater to the understanding of importance of second law. Thus the importance of second law must be clearly explained with the aid of the three stick arrangement as suggested in the question.

__Let’s first see the correct method of adding two decimals__ :

While adding two decimal numbers, the most important thing is to make sure that only digits with the same place value get added. To do it, we can follow the two methods given below:

Method I: Vertically align the two numbers such that the decimal points of both the numbers are in the same line. For example, to add 10.4 + 2.57

Method II: Convert the two decimal numbers into like decimals first and then add the numbers by writing them one below the other. For example, to add 10.4 + 2.57

Number of digits after decimal in 10.4 is one while the number of digits after decimal point is two. So we add a zero in 10.4 to make it 10.40, we have not changed the number as **10.4 = 10.40**

__Problem I:__** 12.43 + 2.34 = (12+2) . (43+34) = 14.77**

Anurag added whole number parts separately and decimal parts separately as shown above and put decimal in between resultant whole number part and decimal part. He is getting the correct answer using this method. it is because they are like decimals. But does that mean that we can follow this method to add all the like decimals?

The answer is No, let’s try using the same method for problem II and III, and see why it is not correct.

__Problem II:__** 1.12 + 2.5 = (1+2).(12+5) = 3.17**

When we use the same method to solve this problem, we can see that in addition to the decimal parts, the digits with different place values are getting added. In 1.12, 2 is at Hundredths place value but it is getting added to 5 which is at Tenths place value. So Anurag’s method is wrong.

__Problem III:__** 2.56 + 1.62 = (2+1).(56+62) = 3.118**

When this problem is solved with the same method, we can see that although both the numbers are like decimals still, we are getting the wrong answer. As we know when we add two digits and their sum crosses 9, a digit with a new place value is generated.

As we can see when we separately add the parts using Anurag’s method. We are ignoring the carry overstep and it is giving us a wrong answer.

The correct method of adding them is as shown.

For example, when we add 5 + 9 ( both has place values Ones), we get 14 which has both Ones and Tens values. Since we can only keep digits with the same place value below each other, we carry over the Tens place value digit to its respective column.

**Question: **

Comment on the following statements in the context of circular motion.

- Is it possible to have a non-uniform motion that is periodic?
- Is it possible to have a uniform motion that is non
**–**periodic?

If you think the above statements are true, then walk around a circular path of any suitable radius to demonstrate the above motions i.e. non-uniform periodic motion and uniform non-periodic motion. *(Please note that uniform motion implies constant speed)*

**Solution**:

Non-uniform motion can be periodic. The example below substantiates the aforementioned statement.

Two bodies R and S cover the same circular path twice in a total time of 70 minutes. Fig. I for both the bodies(R and S) shows time taken to cover the path for the first time along the circle & fig. II shows the time taken to cover the path for the second time. Motion for both the bodies is periodic in this case as each of them completes one cycle in 35 minutes irrespective of their varied speeds. The motion of R is non-uniform while the motion of S is uniform.

This above case can also be analyzed by observing the motion of the pendulum. Its motion is non-uniform but nonetheless, it is periodic.

Uniform motion can be non-periodic:

If a person travels straight with uniform motion and does not retrace his path, then motion, in that case, will be uniform and non-periodic.

**Question: **

You have to construct a quadrilateral ABCD. In which of the following cases can you draw a unique quadrilateral. Draw a rough sketch for each case on a page or whiteboard and explain why are you able/unable to draw a unique quadrilateral in given cases.

Case I: AB = 5 cm, AD = 7 cm, ∠DAB = 70° and ∠ABC = 120°

Case II: AB = 6 cm, BC = 5 cm, CD = 8 cm, AD = 7 cm and ∠ABC = 60°

Case III: AB ‖ CD, AB = 6 cm, BC = 7 cm ∠ABC = 70° and ∠DAB = 80°

**Solution: **

**Case I:**

To construct a unique quadrilateral, we need to **locate all the four vertices **of the quadrilateral.

Given: AB = 5 cm, AD = 7 cm, ∠DAB = 70° and ∠ABC = 120°

**Step 1:** Draw AB = 5 cm. In this case, we are able to locate vertices A and B.

**Step 2:** Draw AD = 7 cm. To fix the orientation of AD with AB, we need ∠DAB, which is equal to 70°.

Now, we are able to locate vertices A, B and D. We need to locate vertex C for unique quadrilateral.

**Step 3: **Given ∠ABC = 120°

We are still unable to locate vertex C, as the length of BC is not known. We have no other information through which we can locate vertex C.

So, in this case we are unable to draw unique quadrilateral.

**Case II:**

To construct a unique quadrilateral, we need to **locate all the four vertices **of the quadrilateral.

Given: AB = 6 cm, BC = 5 cm, CD = 8 cm, AD = 7 cm and ∠ABC = 60°

**Step 1: **Draw AB = 6 cm, ∠ABC = 60° and BC = 5 cm.

So, we are able to locate vertices A, B, and C.

**Step 2:** To locate vertex D, we have CD = 8 cm and AD = 7 cm

So, Assuming C as a center, draw an arc of radius 8 cm and assuming A as the center, draw an arc of radius 7 cm. The intersection point will be vertex D.

So, in this case, we are able to draw a unique quadrilateral.

**Case III:**

To construct a unique quadrilateral, we need to **locate all the four vertices **of the quadrilateral.

**Case III: **AB ‖ CD, AB = 6 cm, BC = 7 cm ∠ABC = 70° and ∠DAB = 80°

**Step 1: **Draw AB = 6 cm, ∠ABC = 70°, BC = 7 cm and ∠DAB = 80°

In this case, we are able to locate vertices A, B and C. We need to locate vertex D.

**Step 2: **Since AB ‖ CD, so ∠BCD + ∠ABC = 180°. Hence, ∠BCD = 110°

Now, we will be able to locate Point D by drawing ∠BCD = 110°

So, in this case, we are able to draw a unique quadrilateral.

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