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Class 10 Physics

# Model Question

## Sample Paper 1

These sample papers have been taken from CBSE (www.cbse.nic.in)

• Questions 1 and 2 are one-mark questions
• Questions 3 to 5 are two-mark questions
• Questions 6 to 15 are three-mark questions
• Questions 16-21 are five-mark questions
• Questions 22 to 27 are two-marks questions, and are based on practical skills

### Section – A

Question 1: Mendel took tall pea plants and short pea plants and produced F1 progeny through cross-fertilisation. What did Mendel observe in the F1 progeny?

Answer: Mendel observed that all the plants in the F1 generation were tall plants.

Question 2: Name two constituents of biogas.

Answer: Methane and Carbon dioxide

Question 3: Carbon has four electrons in its valence shell. Which type of compounds can be formed by carbon atom and why? Give any one example of such compounds.

Answer: Carbon needs to gain or lose 4 electrons in order to make an ionic compound. But doing so requires a huge amount of energy. Hence, carbon makes compounds by sharing its electrons and thus makes covalent compounds. CCl4 (Carbon tetrachloride) is one such example. In this compound, carbon shares its valence electrons one each with one atom of chlorine.

Question 4: Different parts of brain are associated with specific functions. Name the part of human brain which performs the following functions:

1. Sensation of feeling full: Hind brain
2. Vomiting: Pons
3. Picking up a pencil
4. Riding a bicycle

Answer: (a) Hypothalamus, (b) Medulla, (c) Cerebrum and cerebellum, (d) Cerebrum and cerebellum

Question 5: An object is kept 40 cm in front of a concave mirror of focal length 20 cm. Find the position and nature of the image. Draw a ray diagram to show the formation of image in this case.

Answer: Given, u = -40 cm, f = -20 cm, v = ?

Image distance can be calculated as follows:

1/v-1/u=1/f

Or, 1/v=1/f+1/u

Or, 1/v=1/(-20)+1/(-40)

Or, 1/v=(-2-1)/(40)=-1/(40)

Or, v=-40

So, image is formed at 40 cm in front of mirror.

Negative sign shows it is an inverted image.

Image distance is same as object distance, hence image is of same size as object.

As it is formed in front of mirror so it is a real image

Question 6: In the reaction:

MnO2 + 4HCl → MnCl2 + 2H2O + Cl2

(a) Name the compound (i) oxidised, (ii) reduced.

Answer: HCl is getting oxidized and MnO2 is getting reduced

(b) Define oxidation and reduction on its basis.

Answer: Removal of hydrogen is called oxidation, while removal of oxygen is called reduction.

Question 7: 1 g of solid sodium chloride is taken in a clean and dry test tube and 2 mL of conc. sulphuric acid is added to it. If the gas evolved is tested first with dry and then with wet blue litmus paper, in which case will the litmus paper change colour? Give reason for your answer. What inference can be drawn about the nature of the evolved gas? Support your answer with chemical equation for the reaction.

Answer: Following reaction takes place when sodium chloride reacts with sulphuric acid:

2NaCl + H2SO4 → Na2SO4 + 2HCl

HCl dissociates into ions only in the presence of water. So, dry litmus paper does not show color change but wet litmus paper (blue) changes to red. This shows that HCl is acidic.

OR

(a) For the preparation of cakes, baking powder is used. If at home your mother uses baking soda instead of baking powder, how will it affect the taste of the cake and why?

Answer: When baking soda is heated, it gives sodium carbonate; as shown by following equation:

2NaHCO_3→Na_2CO_3+CO_2+H_2O

Formation of sodium carbonate, results in bitter taste in the cake, which is not desirable for a tasty cake.

(b) How is baking soda be converted into baking powder?

Answer: Baking soda is converted into baking powder by addition of a edible oil (generally tartaric acid).

(c) What makes the cake soft and spongy?

Answer: Production of carbon dioxide results in formation of a large number of bubbles in the cake mixture. This results in cake becoming soft and spongy.

Question 8: What happens when hydrogen is added to a vegetable oil in the presence of nickel? Name the reaction and write one difference between the physical property of the vegetable oil and the product obtained in this reaction. Write the role of nickel in this reaction.

Answer: When hydrogen is added to a vegetable oil in presence of nickel, the vegetable oil becomes saturated. This reaction is called hydrogenation and is generally used to produce margarine from vegetable oil. While the vegetable oil is a viscous liquid, margarine is semi-solid. Nickel is the catalyst in this reaction.

Question 9: What is geotropism? Draw a labelled diagram of a potted plant showing positive geotropism and negative geotropism.

Answer: The growth in a plant part in response to the gravity is called geotropic movement, or geotropism. Following figure shows positive and negative geotropism in a plant. Roots are positive geotropism while stem is showing negative geotropism.

Question 10: What are acquired traits? Why are these traits generally not inherited over generations? Explain.

Answer: Traits; which are acquired due to interaction with the environment; are called acquired traits. Acquired traits are not inheritable because such traits do not change the genotype of an individual.

Question 11: We wish to obtain an equal sized inverted image of a candle flame on a screen kept at distance of 4 m from the candle flame.

(a) Name the type of lens that should be used.

Answer: Convex Lens

(b) What should be the focal length of the lens and at what distance from the candle flame the lens be placed.

Answer: The focal length of the lens should be 2 m and candle flame should be at 4 m, i.e. twice the focal length. We know that when an object is placed at 2F1 of a convex lens, image is formed on 2F2. The image is of same size, real and inverted.

(c) Draw a labelled diagram to show the image formation in this case.

Answer:

OR

A 5 cm tall object is placed at a distance of 30 cm from a convex mirror of focal length 15 cm. Find the position, size and nature of the image formed.

Answer: Given, f = 15 cm, u = -30 cm, h1 = 5 cm

Image distance can be calculated as follows:

1/v-1/u=1/f

Or, 1/v=1/f+1/u

Or, 1/v=1/(15)+1/(-30)

Or, 1/v=(2-1)/(30)=1/(30)

Or, v=30 cm

We know that v/u=(h_2)/(h_1)

Or, (h_2)/5=(30)/(30)

Or, h_2=5 cm

Image is of same size, erect and virtual.

Question 12: Consider the following circuit:

What would be the readings of the ammeter and the voltmeter when key is closed? Give reason to justify your answers.

Answer: Let us find equivalent resistance for this combination of resistors

For series combination, equivalent resistance is given as follows:

R=R_1+R_2+R_3

Or, R=5+8+12=25 Ω

Using Ohm’s law we can find the current:

V = IR

Or, I=V/R

Or, I=(6)/(25)=0.24 ampere

Now, reading on voltmeter can be calculated as follows:

V=IR

Or, V=0.24xx12=2.88 V

OR

What is electrical resistivity? Derive its SI unit. In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 100 mA. If the length of the wire is doubled, how will the current in the circuit change? Justify your answer.

Answer: If l is the length and A is the area of cross section of a conductor, then resistance R is given by following:

R∝l/A

Or, R=ρl/A

Here, ρ is the constant of proportionality and is called electrical resistivity of the material of conductor.

Above equation can also be written as:

ρ=(RA)/l

Since SI unit of resistance is Ω that of area is m2 and that of length is m hence,

ρ=(Ωxx\m^2)/m=Ω m

We know that resistance varies directly as length of the conductor. So, doubling the length of the wire will result in resistance becoming double. This will reduce the current to half of earlier value, because current varies inversely as resistance.

Question 13: Calculate the total cost of running the following electrical devices in the month of September, if the rate of 1 unit of electricity is Rs. 6.00.

(i) Electric heater of 1000 W for 5 hours daily.

Answer: There are 30 days in September

So, monthly consumption =30xx5 h\xx\1000 W=150000 Wh

=150 kWh=150 unit

Total cost =6xx150=Rs. 900

(ii) Electric refrigerator of 400 W for 10 hours daily.

Answer: Monthly consumption =30xx10 h\xx\400 W=120000 Wh

=120 kWh

Total cost =6xx120=Rs. 720

Question 14: While teaching the chapter “Our Environment” the teacher stressed upon the harmful effects of burning of fossil fuels, plastic, paper etc. The students noticed the extensive use of plastic and polythene in daily life, which can be avoided and the surroundings can be kept clean. They decided to make their school “plastic and polythene” free and motivated each other for its minimum use.

(a) Why should the use of polythene and plastic be reduced in daily life?

Answer: Polythene and plastic are non-biodegradable and a significant portion of them cannot be recycled. This creates the problem of accumulation of plastic waste in our surroundings which is detrimental to the environment. So, we need to reduce the use of polythene and plastic in our daily life.

(b) In what ways the students would have avoided the use of plastic and polythene in their school?

Answer: The students could avoid the use of plastic and polythene in school in following ways:

• Using jute bags to bring any extra item to school.
• Asking the canteen staff to use paper plates instead of plastic plates.
• Asking parents to buy metallic water bottles, and lunch boxes instead of plastic ones.

(c) How the students would have motivated each other for the success of their decision?

Answer: Students can apply various ways and means to motivate each other for the success of their decision. This can be done through active dialogue, through suggestions and feedbacks. This can also be done by organizing some street plays to raise awareness level.

Question 15: What is “Sustainable Management of Natural Resources”? Why is it necessary? Which one out of reuse and recycle, would you practise in your daily life and why?

Answer: The practice of managing natural resources so that current generations needs should be met without jeopardizing the needs of the future generation is called Sustainable Management of Natural Resources. It is necessary in order to ensure that natural resources shall be available for many generations to come. I would prefer to reuse rather than recycle. Recycling required usage of some resources (fuel and some raw materials), but reuse does not involve usage of any other resource. So, ‘reuse’ is always better in terms of conservation of resources.

Question 16:

(a) List in tabular form any three chemical properties on the basis of which metals and non-metals are differentiated.

Answer:

MetalsNon-metals
Metals react with oxygen to make basic oxides.Non-metals react with oxygen to make acidic oxides.
Some metals displace hydrogen from dilute acidNon-metals do not react with dilute acids.
Metals make ionic compoundsNon-metals form covalent compounds.

(b) State two ways to prevent the rusting of iron.

Answer: Painting and galvanization

Question 17: Explain giving justification for the trends in the following properties of elements, on moving from left to right in a period, in the Modern periodic Table.

(a) Variation of valency.

Answer: Valency first increases up to 4 on moving from left to right in a period and then decreases up to zero. This happens because number of valence electrons increases from 1 to 8. Valency is 1 for elements with valence electrons 1 or 7, is 2 for those with valence electrons 2 or 6, and so on.

(b) Change of atomic radius.

Answer: Atomic size and atomic radii decrease from left to right in a period. This happens because of decrease in electropositivity.

(c) Metallic to non-metallic character.

Answer: Metallic character decreases from left to right in a period. This happens because of decrease in electropositivity.

(d) Electronegative character.

Answer: With reduction in atomic size, a tendency to gain electrons increases across a period. Hence, electronegative character increases from left to right in a period.

(e) Nature of oxides.

Answer: Oxides become less basic and more acidic when we move from left to right in a period. This happens because of reduction in metallic character.

OR

(a) What was the basis of Mendeleev’s classification of elements?

Answer: Atomic mass

(b) List two achievements of Mendeleev’s periodic tables.

Answer: Following are two achievements of Mendeleev’s periodic table:

• Prediction of elements to be discovered in future.
• Correct position of noble gases which were not known during Mendeleev’s time.

(c) List any two observations which posed a challenge to Mandelẽev’s periodic law.

Answer: Following observations posed a challenge to Mendeleev’s periodic law:

• Position of isotopes could not be clarified in Mendeleev’s Periodic Table
• Mendeleev placed many elements in wrong order of their increasing atomic masses in order to place elements having similar properties in similar group.

Question 18:

(a) Write the reaction that occurs when glucose breaks down anaerobically in yeast.

Answer: C_6H_(12)O_6→2C_2H_5OH+2C\O_2

(b) Write the mechanism by which fishes breathe in water.

Answer: Fishes breathe through gills. Gills absorb oxygen (dissolved in water) from water and then oxygen goes to the blood.

(c) Name the balloon likes structures present in lungs. List its two functions.

Answer: Balloon like structures in lungs are called alveoli. Two functions of alveoli are as follows:

• Removal of carbon dioxide from blood
• Addition of oxygen to blood

(d) Name the respiratory pigment and write its role in human beings.

Answer: Haemoglobin is the respiratory pigment and it is responsible for carrying oxygen.

OR

(a) Name the process and explain the type of nutrition found in green plants. List the raw materials required for this process. Give chemical equation for the mentioned process.

Answer: Plants follow autotrophic mode of nutrition and the process by which a green plant makes food is called photosynthesis. The raw materials required for photosynthesis are:

• Carbon dioxide,
• Water
• Sunlight

Chemical Equation for Photosynthesis:

6CO_2+6H_2O→C_6H_(12)O_6+6O_2

(b) Write three events that occur during this process.

Answer: Three events during photosynthesis are as follows:

1. Sunlight activates chlorophyll
2. Splitting of water molecule
3. Reduction of carbon dioxide to make carbohydrate

Question 19:

(a) What is variation? How is variation created in a population? How does the creation of variation in a species promote survival?

Answer: Differences in traits through successive generations are called variations. Variation is created due to some inaccuracies during DNA replication. Chances of variations are higher during sexual reproduction rather than during asexual reproduction. Variations can be either useful or harmful. Useful variations help an organism survive the onslaught of changed environment. Thus, most often useful variations are passed on to the next generation. Thus, variations help in survival of species.

(b) Explain how, offspring and parents of organisms reproducing sexually have the same number of chromosomes.

Answer: Gametes are produced through meiosis. We know that number of chromosomes becomes half after meiosis. Thus, gametes have haploid number of chromosomes. When male and female gametes fuse during fertilization, the number of chromosomes in zygote becomes diploid. This helps in maintaining the number of chromosomes in offspring of sexually producing organisms.

Question 20:

(a) What is presbyopia? State its cause. How is it corrected?

Answer: The power of accommodation of the eye usually decreases with ageing. For most people, the near point gradually recedes away. They find it difficult to see nearby objects comfortably and distinctly without corrective eye-glasses. This defect is called Presbyopia. It arises due to the gradual weakening of the ciliary muscles and diminishing flexibility of the eye lens. Sometimes, a person may suffer from both myopia and hypermetropia. Such people often require bifocal lenses. A common type of bi-focal lenses consists of both concave and convex lenses. The upper portion consists of a concave lens. It facilitates distant vision. The lower part is a convex lens. It facilitates near vision.

(b) Why does the Sun appear reddish early in the morning? Explain with the help of a labelled diagram.

Answer: Red colour scatters the least and hence travels the farthest. During early morning, the sunlight has to travel a longer distance to reach us. During this, the blue colour is scattered almost completely and the sun and the sky appear reddish.

Question 21:

(a) What are magnetic field lines? How is the direction of magnetic field at a point in a magnetic field determined using field lines?

Answer: The imaginary lines of magnetic field around a magnet are called field line or field line of magnet. Outside the magnet, the direction of magnetic field line is taken from north pole to South Pole. Inside the magnet, the direction of magnetic field line is taken from south pole to north pole.

(b) Two circular coils ‘X’ and ‘Y’ are placed close to each other. If the current in the coil ‘X’ is changed, will some current be induced in the coil ‘Y’? Give reason.

Answer: When a conductor is set to move inside a magnetic field or a magnetic field is set to be changing around a conductor, electric current is induced in the conductor. When the current in the coil ‘X’ is changed it changes the magnetic field around ‘X’. Due to this, some current is induced in coil ‘Y’ by way of magnetic induction.

(c) State ‘Fleming’s right hand rule”.

Answer: Fleming’s Right Hand Rule: If the right hand is stretched in a way that the index finger, middle finger and thumb are in mutually perpendicular directions, then the thumb shows the direction of movement of the conductor, index finger shows the direction of magnetic field and the middle finger shows the direction of induced current in the conductor. The directions of movement of conductor, magnetic field and induced current can be compared to three mutually perpendicular axes, i.e. x, y and z axes.

### Section – B

Question 22: In the laboratory of a school, the sample of hard water is not available “to study the comparative cleaning capacity of a sample of soap in soft and hard water”. Which salt from the laboratory can be added to tap water to make it hard? In the experiment how is cleaning capacity of soap compared?

Answer: Salts of calcium and magnesium can be added to tap water to make it hard water. Cleaning capacity of soap is compared by comparing the time it takes the foam to disappear in case of different samples of water.

Question 23: In a school laboratory the students are studying the properties of ethanoic acid through certain experiments. How can they test its acidic nature? Give two tests.

Answer: Acidic nature of ethanoic acid can be tested by following methods:

• By using blue litmus paper: Blue litmus changes to red in presence of acid.
• By using sodium hydroxide to carry out neutralization reaction. If the beaker (containing the reaction mixture) becomes warm, it indicates reaction has taken place.

Question 24: Draw a labelled diagram of the experimental set up for the study of liberation of carbon dioxide gas during respiration.

Answer: Do It Yourself: You can make the diagram of a test tube a straw; showing air being blown by mouth through the straw. Alternately, you can make the diagram of conical flask with germinating seeds kept in it. The flask should be connected to a test tube containing lime water.

Question 25: Draw diagrams showing reproduction in yeast in proper sequence.

Answer:

Question 26: While tracing the path of a ray of light passing through a rectangular glass slab a student tabulated his observations. If in his experiment for two readings he takes two values of ∠i as 40 º and 50º, what should be the correct values of ∠e and ∠r in each case?

Answer: We know that emergent ray is parallel to the incident ray in case of light entering a particular medium and exiting it.

So, ∠i=∠e

OR

Draw a labelled ray diagram for the formation of image by a convex lens of focal length 15 cm when the object is placed at a distance of 25 cm from the lens. Determine the size of the image formed, if size of the object is 4 cm.

Answer:

Given, u = -25 cm, f = 15 cm

Image distance can be calculated as follows:

1/v-1/u=1/f

Or, 1/v+1/(25)=1/(15)

Or, 1/v=1/(15)-1/(25)

Or, 1/v=(5-3)/(75)=2/(75)

Or, v=(75)/2=37.5 cm

Now, size of image can be calculated as follows:

v/u=(h_2)/(h_1)

Or, (37.5)/(-25)=(h_2)/4

Or, h_2=(37.5)/(-25)xx4=-6 cm

Question 27: You have following material:

An ammeter (0-1A), a voltmeter (0-3V), a resistor of 20 Ω, a key, a rheostat, a battery of 3 V and seven connecting wires.

Using this material draw a labelled circuit diagram to study the dependence of potential difference (V) across a resistor on the current (I) passing through it.

Answer: Do it yourself

CHAPTER LIST

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# Math Help - trignometry

1. ## trignometry

Hi! can you solve this sum for me---
PQRS is a quadrilateral.The sides RS and QR are the same length.The sides QP and RS are parallel.Calculate:
1)angle SQR
2)angle PSQ
3)length PQ
4)length PS
5)Area of PQRS

2. Hello, dolly kohli!

Obviously, some information is missing . . .

can you solve this sum for me --- sum?

Sides RS and QR are the same length.
Sides QP and RS are parallel.

Calculate:

$1)\;\angle SQR\qquad 2)\;\angle PSQ\qquad 3)\;\text{ length }PQ \qquad 4)\;\text{ length }PS \qquad 5)\;\text{ Area of }PQRS$

Code:
P                             Q
o  *  *  *  *  *  *  *  *  *  o
θ  * *
* θ *
*              *     *
*         *       *
*    * θ       *
o  *  *  *  o
S           R

Since $RS = QR,\;\Delta QRS$ is isosceles.
. . Hence: . $\angle SQR \:=\:\angle QSR \:=\:\theta$

Since $QP \parallel RS,\;\angle PQS \:=\:\angle QSR \:=\:\theta$

Since $P$ can be anywhere to the upper-left on that line,
. . we cannot determine $\angle PSQ$

Nor can we find the lengths of $PQ$ and $PS$, or the area of $PQRS$.


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Calculating optimal level of negative externality

I am trying to solve the following question(s):

Let $$h \geq 0$$ represent a negative externality of a firm's production on one (representative) consumer. The consumer has a quasi-linear utility function and attaches a utility of $$\phi(h) = -2h^2$$ to the externality. The firm's profit function is $$\pi(h)=120-2(h-10)^2$$.

1. Compute the pareto optimal level of $$h$$, $$h^o$$.

2. Compute the firm's optimal level of $$h$$, $$h^f$$.

3. Suppose the consumer rather than the firm can determine the specific externality levelh with which the firm operates. Compute the optimal level of $$h$$, $$h^c$$, from the consumer's perspective.

My approach so far:

1. To find $$h^o$$ I maximize the aggregate surplus/welfare of h. $$\max_{h \geq0} \pi(h) + \phi(h)$$ $$W(h)=120-2(h-10)^2 - 2h^2 = 0$$ $$\frac{dW}{dh}= -4(h-10)-4h = -8h + 40 = 0$$ $$8h=40 \rightarrow h^o=5$$

2. To find $$h^f$$ $$FOC\ of\ \pi(h): \frac{d\pi}{dh} =-4(h-10) = -4h+40 = 0 \rightarrow h^f=10$$

3. To find $$h^c$$ $$FOC\ of\ \phi(h): \frac{d\phi(h)}{dh} = -4h = 0 \rightarrow h^c=0$$

Is this correct?

Almost correct. Setting $$W(h)=0$$ is wrong (but inconsequential for the solutions). Checking the SOC for completeness should be included, but is somewhat obvious here. Correctness of 3. holds only under the assumption that the consumer owns no shares of the firm, which seems to hold in this exercise.


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# Current Affairs 4th Class

#### Problem-Solving

Problem - Solving       OBJECTIVES   ·         Students will be develop the ability to understand what the goal of the problem is and what rules could be applied that represent the key to solving the problem. ·         They will develop abstract thinking and creative approach.       INTRODUCTION   The process of working through details of a problem to reach a solution. Problem solving may include mathematical or systematic operations and can be a gauge of an individual's critical thinking skills.     Four Stages of Problem Solving Stage 1: Understand and explore the problem; Stage 2: Find a strategy; Stage 3: Use the strategy to solve the problem; Stage 4: Look back and reflect on the solution.             Example 1: Jay is 85 cm tall. His father is twice as tall as Jay. How tall is Jay's mother if she is 15 cm shorter than his father? (a) 1 m 55 cm              (b) 1 m 66 cm    (c) 1 m 72 cm              (d) 1 m 90 cm   Ans.    (a) Explanation: Jay's height                  = 85 cm His father's height        $=85\times 2=170$cm His mother's height      = 170 - 15 = 155 cm                                     = 1 m 55 cm.       Example 2: There are 18 pupils in a group. There are 10 boys and the rest are girls 7 pupils are wearing glasses. If 3 girls are wearing glasses, how many boys are not wearing glasses?                                                (a) 3                             (b) 6             (c) 8                             (d) 9   Ans.    (b) Explanation: more...

#### Estimation

Estimation       OBJECTIVES   ·      Students will be able to generate a range of possible outcomes. ·      They will be able to judge the size, amount and cost of something.       INTRODUCTION   Estimation is a rough calculation of the value, number, quantity, or extent something.       Example 1:  2 litres is most likely the amount of liquid in a ______. (a) Bathtub (b) Pond (c) Large bottle of soda (d) Swimming pool   Ans.    (c) Explanation: 2 litres is most likely the amount in a large bottle of soda.       Example 2: What is the closest estimate of how much longer the back of truck is than the front? (a) 60 feet                    (b) 70 feet        (c) 50 feet                    (d) 40 feet   Ans.    (c) Explanation: Nearest Estimation is 70 - 20 = 50 feet.         Example 3: How many 250 ml cartons of milk would it take to fill the 11 carton? (a) 2                             (b) 3              (c) 4                             (d) 5   Ans.   (c) Explanation:       Example 4: The amount of water in a bath tub is about 50 _______. (a) mililitres                               (b) litres         (c) centilitres                            (d) kilograms   Ans. more...

#### Logical Diagram

Logical Diagram       OBJECTIVES   ·         Students will be able to find out the relation between some times of a group by diagrams. ·         They will be able to compare and contrast groups of things.       INTRODUCTION   A Venn Diagram is a visual brainstorming tool used to compare and contrast two (sometimes three) different things. Comparing is looking at traits that things have in common, while contrasting is looking at how they differ from each other.   A Venn Diagram is made up of two large circles that intersect with each other to form a space in the middle. Each circle represents something that you want to compare and contrast. Where the two circles intersect, you would write traits that the two things have in common. In either side of the intersecting space, you would write the differences among the two things.       TYPE-I: Different Types of Questions Based on Venn Diagrams     CASE 1: When one group of items is completely included in the second group of items and the second, again completely belongs to the third group, they are represented as shown.     Examples 1 Tree, Forest, Leaf   Explanation:              Venn diagram would be as follows: Clearly, leaf is apart of tree and tree is a part of forest.       CASE 2: If the items evidently belong to three different groups, i.e., they are not correlated with each other in any way. They are represented as shown.     Example 2  Whale, Crocodile, Bird Explanation: They all belong to different categories.         CASE 3: If the three items are partly related to each other, they are represented as shown. more...

#### Geometrical Shapes

Geometrical Shapes       OBJECTIVES   ·      To identify different objects with represent some kind of geometrical shapes which we are coming across our day to day life. (Both 2D and 3D) ·      To recognize difference parts of various geometrical shapes such as vertex and sides of a particular geometrical pattern. ·      Use of different geometrical shapes for creating Tangrams. ·      Knowledge of line of symmetry. (Horizontal and vertical)           GEOMETRIC SHAPES   Geometric Shape is defined as a set of points or vertices and sides connecting to the point to form a closed entry. There are various kinds of geometrical shapes. Basing upon the number of vertices and sides they are identified.             2d Geometrical Shapes
·     Triangle has three sides. ·     It has three vertices.
·     Squares have 4 equal sides and 4 right angles. ·     They have 4 lines of symmetry. ·     All squares belong to the rectangle family. ·     All squares belong to the rhombus family. ·     more...

#### Embedded Figure

Embedded Figure       OBJECTIVES   ·      It will measure cognitive functioning and style of students. ·      Students will be able to find shapes within the large more complex image.       INTRODUCTION   A figure is said to be embedded in figure (X), if figure (X) contains a part of that figure.       Types of Questions     Type I: Identify the Small Part hidden in given figure:   In such type of problem, figure (X), is given, followed by four parts, such that one of them is hidden in figure (X), students have to identify that part.     Example 1: Identify which shape is hidden in the Figure (X). (a)                    (b) (c)                        (d)   Ans.    (b)          Type II: Identify the Figure in Which Given Part is Hidden   In such problems, a figure X is given which is followed by four alternative of complex figures in such a way that figure (X) is hidden or embedded in one and only one of them. Student have to identify that particular figure in which the figure (X) is hidden.     Example 2: (a)                 (b) (c)                  (d) Ans.    (a)

#### Mirror/Water Image

Mirror/Water Image       OBJETIVES           ·         Students will learn how different objects are seen when they get reflected both in mirror and in Water.          INTODUCTION   The reflection of object into the mirror is called its mirror image. It is obtained by inverting an object laterally. If we combine the original figure and mirror image together they form symmetry.       MIRROR IMAGES     Mirror Images of Capital letters     ·         The capital letters which have the same mirror images are: A, H, I, M, 0, T, U, V, W, X, Y         Mirror Images of Small letters              ·         The small letters which have the same mirror images are:                         i, l, o, v, w, x       Mirror Images of Numbers   ·      0 and 8 numbers have the same mirror images.     Mirror Image of Clock Time For mirror image of clock time, the given time shall be subtracted from 12 00 or 11.60.     Example 1:   What will be the mirror image of clock time 3:40? Explanation: By using simple trick                         WATER IMAGE     INTRODUCTION more...

#### Logical Reasoning

Logical Reasoning       OBJECTIVES   ·      To identify relationships, similarities and difference between shapes and patterns. ·      To recognize visual sequences and relationships between objects and remembering these.       INTRODUCTION   Visual reasoning is the process of manipulating one's mental image of an object to reach a certain conclusion.         Types of Questions   The chapter includes: 1.    Grouping of figures. 2.    Paper folding and cutting. 3.    Merging and splitting of images. 4.    Finding difference between two images. 5.    Clock wise and anticlock wise rotation of objects. 6.    Tessellation puzzle. 7.    Completion of image. 8.    Corresponding image. 9.    Different sections of image. 10.  Symmetry 11.  Rule detection         Paper Folding and Cutting   In this type of questions, there are three or four figures given in one line. Each figure followed a pattern. Each figure consists of a dotted line along which it is to be folded and the arrow shows the side which it is to be folded.   After folding, the paper is cut or punched and then students have to visualize the correct figure that will be produced when the paper is unfolded.   Direction (Example 1): In the following example, a piece of paper is folded and cut and then unfolded. One of more...

#### Alphabet Test

Type I: How to arrange the numbers given to the corresponding letter so that they form a meaningful word? Arrange the letters in the different combinations given in options to find a meaningful word.   Type II: Finding the word from the options that can/cannot be formed from the given letters. Students have to observe carefully all the letters given in the question and find the combination of letters from the given options that contains or does not contains letters.   Type III: How to form a meaningful word and find the category of word? Arrange the given jumbled letters in a meaningful word and identify the category in which formed word lies.   Type IV: How to identify the combination of letters that has same number of skipped letters between the adjacent letters? Write the alphabetical series in order, then observe carefully all the options.     EXAMPLE   1.         In the question, some letters are given which are numbered 1, 2, 3, 4, 5 and 6 followed by four options containing combinations of these numbers. Find the combination of numbers so that letters arranged accordingly to form a meaningful word. R F W E L O 1 2  3  4 5  6  (a) 2, 5, 6, 3, 1, 4                                 (b) 2, 5, 6. 3, 4, 1                     (c) 1, 5, 6, 4, 3, 1                                 (d) 2, 5, 6, 1, 3, 4     Explanation (b):     First go through each option: (a) 256314 = FLOWRE;                                 (b) 256341= FLOWER (c) 256431 = FLOEWR;                                 (d) 256134= FLORWE Meaningful word: FLOWER.'. Combination: 2, 5, 6, 3, 4, 1     2.       Which word cannot be made from the letters of the given word? TEMPERAMENT (a) METER                  (b) PETER                   (c) TENT                     (d) TESTER   Explanations (d): S of the word TESTER is not in the given word.

#### Patterns

Type I: How to find the missing term or next term in (number or letter) series to continue the given series?
• Identify the (descending or ascending) order of series. Observe the pattern using operations: addition, subtraction, multiplication, and division, skip counting and reverse counting.
• Identify the order of alphabetical series either from A to Z or Z to A.
• Numbering of alphabets either starts from A or starts from Z.
• Skipping letters.
Type II: How to find the missing number in the pattern?
• To find the missing term in the given pattern, identify the rule followed in rest of the given terms in pattern using mathematical operations (+,- ,x, ~), skip counting and reverse counting.
Type III: How to find the missing part in the figure pattern?
• Complete the figure pattern by drawing its incomplete part in the pattern.
EXAMPLE   1.    In the number pattern given below, what are the values of p and Q respectively?   (a) 27,                          (b) 343, 27                   (c) 27,                          (d) 49, 27   Explanation (a):     Pattern followed in above series is: $2\times 2\times 2$, $(3\times 3\times 3)$, $4\times 4\times 4,$$5\times 5\times 5,$$6\times 6\times 6,$ $(7\times 7\times 7)$     2.   Which is the missing domino? (a)                     (b)                      (c) more...

#### Coding-Decoding

How to decode the questions given in coded language? In these type of questions, some particular words are assigned certain substituted names. Then a question is asked that is to be answered in the substituted code language.     EXAMPLE   1.     If' Wrist watch' is called 'Nail'; 'Nail' is called 'Phone' and 'Phone' is called 'Newspaper', then what do we wear on wrist? (a) Wrist watch                                                            (b) Nail                                                (c) Phone                                                         (d) Newspaper   Explanation (b): Wrist watch is worn on wrist and Wrist watch is called Nail.     2.     If B means C, C means D and D means E, which alphabet is a vowel? (a) B                            (b) C                            (c) D                            (d) E   Explanation (c): E is a vowel and D means E.

#### Trending Current Affairs

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text:
# Error Amplifier Gain

I was studying the "SG3525" IC and its functions, and I encountered this issue:

In the "Error Amplifier" section of the Datasheet, there is a parameter that named "DC trans-conductance" which it is "1.5mS" typically. At the bottom of the page it is mentioned that this parameter is related to the DC open loop voltage gain( Av = gm * RL ).

But on next page there is a diagram related to open loop gain vs frequency. at this diagram, the value of the open loop gain at frequency closed to dc is not equal to the relation in the end of previous page ( Av = gm * RL ).

Is this diagram relevant to that DC-gain-relationship? ( Av = gm * RL ) If yes, then why are they not equal?

The graph's displayed gain at DC does match the given equation. But the graph expresses gain in dB.

Consider the case of $$\R_L = 30k\Omega\$$.

$$A_v = gm * R_L = (1.5 mS) * (30 k\Omega) = 45 V/V$$ $$A_v(dB) = 20 log(A_v(V/V)) = 20 log (45) = 33 dB$$

The line for $$\R_L\ = 30 k\Omega\$$ conveniently intersects the vertical axis around 33dB.

• It is true. I made a mistake. In the footnote it is written that the relation is equal to the DC open-loop voltage gain that it is expressed in dB at the table. thank you so much kevin – Mohammad Hussain Asadzadeh Feb 14 at 17:29


text:
# The Moon's orbital plane

1. Aug 30, 2009

### Leo Klem

According to Newton's law, the attraction exerted by the Sun on the Moon is greater than twice the attraction exerted by the Earth on its satellite, whatever the Moon's position during its motion.
Nevertheless, the Moon's orbital plane around the Earth forms a 5.13 degree angle with the ecliptic's plane. The texts of astronomy I could so far access do either omit explaining this fact or clearly admit that current perturbation analysis does not provide any complete and clear explanation on the topic. It seems to be one of the anomalies detected within the solar system.
Somebody might remark that it's a "neglible" aspect, whereas the fact seems to imply that the Moon orbits the Earth irrespective of the solar attraction. In this connection, there is also to consider that the centre of mass of the Earth-Moon system does also orbit the Sun out of the ecliptic, and not according to any plane curve as instead required by the dynamics of central motions.
My question: is there any updated study relevant to the issue?

2. Aug 30, 2009

### D H

Staff Emeritus
Both statements are true. So what? It appears that you think the first statement means the Moon should not be orbiting the Earth.
The Sun/Earth/Moon system comprises three bodies. There is no nice, simple closed form solution to the general three body problem, even for three point masses. There are several other bodies in the solar system besides the Sun, Earth, and Moon. One has to resort to numerical techniques. These numerical techniques could predict the state of the solar system at any point in the future -- if we knew the exact current state (mass, position, and velocity) of all of the bodies in the solar system and if those bodies were point masses and if the solar system was isolated gravitationally from the rest of the universe and if our current best model of gravitation represents exactly how gravity works. Taking these ifs one point at a time,
• We do not know the exact state of the solar system. Our measurements of the masses, positions, and velocities of the solar system bodies are quite accurate, but they are not perfect. The solar system is a chaotic system, which means the future state is highly sensitive to errors in the current state.

• The solar system bodies are not point masses. That the Earth is not a point mass is particular important when it comes to the Moon's orbit about the Earth. The Earth's rotational angular momentum is being transferred to the Moon's orbit via the tides. The rate at which this angular momentum transfer occurs depends strongly on the configuration of the Earth. Right now, the Americas and Eurasia+Africa form two north-south barriers to the tides. This makes the angular momentum transfer considerably higher right now than the average rate over the last billion years or so.

• The solar system is a part of the Milky Way. Nearby stars can perturb the orbits of the planets. Because the solar system is chaotic, even the smallest perturbation will, over time, grow to become huge. To accurately predict the state of the solar system for billions of years means we would have to know and be able to predict the motions of the stars with which the solar system bodies will interact over that time frame.

• Finally, does general relativity truly explain exactly how things work? Probably not. The fervent hope of physicists who are working on things like string theory or quantum gravity is that it is not.

3. Aug 30, 2009

### Leo Klem

It doesn't seem an explanation to me. By the way, due to its relative size, the Earth-Moon system might be viewed as a unique body with respect to the Sun's mass and distance. The point, however, is that the Earth-Moon system's center of mass doesn't describe a plane orbit, whereas it is actually assumed that the Earth's centre runs on the ecliptic's plane. In addition, there is to note that the Moon's orbit around the Earth lies on a plane too.
This “anomaly” could – for example - be considered in association with the planet-satellite system of Saturn, where all the planet’s rings and satellites lie on the planet’s equatorial plane, i.e., with 26.731 degree inclination with respect to the planet’s orbital plane. This fact shows that even the combined strength of the gravitational influence of Sun and Jupiter (the nearest planet to Saturn, whose mass is 2.5 times the mass of the other planets altogether) doesn’t alter the amazing regularity of co-planar orbits of Saturn’s satellites.

Thus, if the huge complication of the overall astronomic/cosmic system, as recalled by you, can provide a satisfactory justification for the “macro-anomalies” I've pointed out, do you thing - for instance - of "the Pioneer's micro-anomaly" as of a case that is worth so much theoretical concern?

4. Aug 30, 2009

### D H

Staff Emeritus
An explanation of what? What do you perceive to be the problem?

Note well: My crackpot meter is starting to go offscale high.

5. Aug 30, 2009

### Staff: Mentor

Why does a pendulum on earth swing back and forth when it should be orbiting the sun?

6. Sep 3, 2009

### Krisia

I think there is an error. According to Newton’s law, the Earth’s gravity strength undergone by the Moon is approximately 4.5 times that exerted by the Sun. The Earth’s gravity should prevails over the Sun’s (Hill sphere) up to a distance of about 1.5 million km. (The Earth-Moon mean distance is 384,400 km, and the radius of the Moon’s Hill sphere should be about 60,000 km).
Nevertheless, the non-coplanar orbits of natural-regular satellites with respect to the planets’ orbital planes is an intriguing issue. It seems that natural-regular satellites (at variance with “stone-satellites”) prefer keeping their orbits co-planar to the respective planet’s equatorial plane (Uranus being the most spectacular example), except the Moon whose orbital plane has 5.22 degree inclination to the ecliptic and 23.53 degree inclination to the Earth’s equatorial plane.
Natural-regular satellites, which show spherical/spheroid shapes, seem strangely suggesting (against Newton’s law) that their own gravity fields differ from those determined (?) by “stone-satellites”.

7. Sep 3, 2009

### D H

Staff Emeritus
Leo was asking about the gravitational acceleration undergone by the Moon, not the Earth. His statement regarding the strength of the gravitational force of the Sun and Earth on the Moon was correct.

Lunar gravitational acceleration toward the Sun: $$\frac{\mu_{\text{sun}}}{(1\,\text{AU})^2} \approx 5.9\,\text{mm}/\text{s}^2$$
Lunar gravitational acceleration toward the Earth: $$\frac{\mu_{\text{earth}}}{(384,400\,\text{km})^2} \approx 2.7\,\text{mm}/\text{s}^2$$

The implication of Leo's comments are incorrect. While he never made his objection clear, he certainly thinks that we don't know why the Moon orbits the Earth.

Now you are on the right track regarding why the Moon does orbit the Earth, even though the gravitational force of the Sun on the Moon is greater than that of the Earth.

The Hill sphere is one measure of the boundary beyond which an orbit will no longer be stable. An alternative is Laplace's sphere of gravitational influence. Both the Hill sphere and sphere of influence look at a restricted three body problem from the perspective of where the perturbative gravitational force from the primary body equals the gravitational force from the secondary body. They differ in that the Hill sphere looks at things from the perspective of a rotating frame in which the third body is stationary while the sphere of influence looks at things from the perspective of a non-rotating frame.

Orbits will most likely become unstable well inside the Hill sphere / sphere of influence. The reason for this is that these surfaces mark the boundaries where the perturbative force from the primary equals the force from the secondary. Calling one force that is equal to another a perturbation is a bit of a misnomer. Perturbation forces should be small. An object at the edge of either sphere almost certainly will escape the secondary body. A more conservative estimate of where orbits become unstable is a third to a half of the Hill sphere / sphere of influence. Note well: the Moon is well inside these more conservative boundaries.

No mystery here either. The planets are not spherical bodies. They are instead oblate spheroids due to their rotation about their axes. On top of that, their rotation axes precess due to gravitational torque exerted by the Sun (and the Moon in the case of the Earth). The dynamical oblateness (J2) couples with this precession to keep moons that orbit close to a planet in that planet's equatorial plane.

8. Sep 3, 2009

### Leo Klem

Thank you Krisia, thank you DH. I owe both of you - and other readers - my apologies, since I’m afraid that Krisia is right. I made a blunder in copying figures from my pocket calculator.
Sun’s and Earth’s gravity accelerations exerted on the Moon are 5.94607*10-4m/s2, and 2.70017*10-3m/s2, respectively. The ratio of the latter to the former is approximately 4.54, as per Krisia’s statement. DH has perhaps incurred a copying mistake similar to mine. Sorry.

9. Sep 3, 2009

### D H

Staff Emeritus
You were right the first time, Leo. The gravitational force on the Moon by the Sun is indeed more than twice that exerted by the Earth.

By the Sun: http://www.google.com/search?q=G*(solar+mass)/(1AU)^2

Now let's look at the things from the perspective of an Earth-centered versus Sun-centered frame. From the perspective of an Earth-centered frame, the third body acceleration of the Moon toward the Sun is about 0.03 mm/s^2.

A rule of thumb is that perturbative effects are an order of magnitude smaller than primary effects. Since the third body acceleration of the Moon toward the Sun is about two orders of magnitude smaller than the acceleration of the Moon toward the Earth, this third body acceleration can definitely be categorized as a perturbative effect.

On the other hand, from the perspective of a Sun-centered frame, the third body acceleration of the Moon toward the Earth is about 2.7 mm/s^2, which is about half of the acceleration of the Moon toward the Sun. The third body acceleration of the Moon toward the Earth from a Sun-centerered perspective does not qualify as a perturbative effect. The effect is quite significant.

Google calculator results again:
Third body effect, Earth-centered: http://www.google.com/search?q=G*(solar+mass)+*(1/(1AU-384e3km)^2-1/(1AU)^2)
Third body effect, Sun-centered: http://www.google.com/search?q=G*(earth+mass)+*(1/(384e3km)^2-1/(1AU)^2)

Last edited: Sep 3, 2009
10. Sep 3, 2009

### Count Iblis

Also note that what matters is the fictitious tidal force exerted by Sun on the Earth-Moon system, not the real gravitational force. The Earth-Moon system is a freely falling system as it orbits the Sun. You have to work in this accelerating reference frame to see what the influence of the Sun is.

11. Sep 3, 2009

### D H

Staff Emeritus
That is exactly what I calculated in post #9, Count. See the google calculator links. I also calculated the third body effect in the accelerating Sun-centered frame. To recapitulate, the third body effect (aka tidal gravity) from the perspective of an Earth-centered frame of the Sun on the Moon is small compared to the primary gravitational force exerted by the Earth; this third body effect definitively qualifies as perturbative. The third body effect from the perspective of a Sun-centered frame of the Earth on the Moon is of the same order of magnitude as the primary gravitational force exerted by the Sun; this does not qualify as perturbative.

In short, one can look at the Moon as being in orbit around the Earth with the Sun being but a perturbative effect on that Earth-centered orbit or one can look at the Moon as being in orbit around the Sun but with the Earth as having something that is much more than a perturbative influence on that Sun-centered orbit.

12. Sep 3, 2009

### Leo Klem

I should definitively avoid calculations by means of pocket calculator, especially when I deem the calculation easy. The remark from Krisia made me doubt, and my doubt led me to a further mistake. Yes DH, I've verified my initial statement was correct, and I can also share in principle the conclusion quoted above.
According to others' perturbation analysis, however, it seems that there is not sufficient stuff to explain the orbit of the center of mass of the Earth-Moon system around the Sun. If the Earth's orbit lies on the ecliptic plane, then the center of the Earth-Moon mass cannot lie on a plane because of the Moon's orbit inclination to the ecliptic. Which (from the Newtonian point of view of central motion mechanics) makes things more complicated than desired, so that also the possible explanations become too elaborate.
In my opinion, the "three-body or multi-body problem" in Newtonian gravitation should be viewed as the true limit of the theory, rather than a question of analytical complexity. It seems the case to remind ourselves of the way in which Newton's gravitation law was derived, to accept the idea that explanations by use of that law have insuperable limits, as it was made evident by the problem of the perihelion precession.
Let me close here by thanking all of you.

13. Sep 3, 2009

### D H

Staff Emeritus
The answer to your concern is simple: The Earth's orbit does not lie on the ecliptic plane. It oscillates around the ecliptic plane, primarily due to the Moon.

Regarding the rest of your post: Just because a differential equation does not have a nice, clean solution in the elementary functions does not mean that the differential equation is not soluble. A solution can be found to any degree of accuracy desired.

Even if it did have a solution in the elementary functions as does the two body problem, that does not mean the exact answer is going to pop out like magic. Suppose you know the orbital elements for some body at some point in time. To find the cartesian state at some time other than the epoch, you need to
• Compute the mean motion, which if you are doing using normal computer/calculator arithmetic is not exact,
• Compute the mean anomaly, which once again is not exact,
• Compute the true anomaly, which requires numerical estimation techniques,
• Compute the sine and cosine of the true anomaly, which once again requires numerical estimation techniques,
• Compute the cartesian coordinates, which once again is not exact.

There are ways to get extra precision beyond that on your calculator, but the result will not be exact. It can, however, with a bit of computation horsepower be made as precise as one desires. That of course assumes you know the orbital elements exactly.

Which you don't.


text:
Problem regarding Information Technology Management.

Asked By 30 points N/A Posted on -

Why are the qualification requirements for the IT Management Series, 2210 covered under two separate individual qualification standards?

SHARE
Answered By 10 points N/A #115552

Problem regarding Information Technology Management.

Previously Alternative A and B requirements were in one document, titled Individual Occupational Requirements for GS-2210: Information Technology Management Series, which had to be used in conjunction with the Group Coverage Qualification Standard for Administrative and Management Positions. In an effort to streamline and improve usability, all Alternative A requirements were combined into an individual qualification standard, Information Technology Management Series, 2210 (Alternative A). All qualification requirements for Alternative B were combined into a separate individual qualification standard, Information Technology Management Series, 2210 (Alternative B).


text:
# Can't execute jar from directory with umlaut

Okay so this is quite an exotic question and I am unsure whether it really fits on this page. But as it involves LaTeX I'll post it here.

I have written a Java program that I am calling from inside LaTeX via \immedeiate\write18{java -jar <myJar>}. This works great except for this edge case:

1. I am in a directory that contains an umlaut in it's name (e.g. ä)
2. I am using lualatex to compile the document
3. (I am using Linux) My System is KDE Neon 5.16 with a 4.18 kernel. I am using TexLive 2018 and my lualatex version is 1.07.0

If all of these conditions are met the call fails with the message Error: LinkageError occurred while loading main class <name> java.lang.ExceptionInInitializerError: null.

I double-checked that I can call the jar from that directory via terminal and the call works when using e.g. pdflatex to compile the document (as soon as the umlaut is removed it also works with lualatex).

I also verified that I can write a lua script that calls the jar via os.execute(...) which is what lualatex uses via it's shellesc package to execute \write18 commands. However as soon as I do the same command from inside LaTeX (via \directlua) I get the error message stated above.
I also checked with print(_VERSION) that lualatex uses the same lua version I tested my external script in.

Does anybody know what the problem here might be? Are there some further things lualatex does to its lua-environment that could be the problem?
Any thought on this is welcome.

EDIT: A friend of mine does not have this issue on a windows machine (though I am not a hundred percent sure he is actually using lualatex as well)

EDIT2: Here's an MWE:

\documentclass{article}

\begin{document}
Some content so TeX won't complain
\directlua{os.execute("java -jar <PathToArbitraryJarFile>")}
\end{document}

Put above code in a folder called Testär (or something else with an umlaut) and compile the document with lualatex

EDIT 3:
I tested it on my new machine and at least now I can only reproduce the error when using latexmk with the -pdflatex=lualatex option.

This problem only and exclusively comes up in the case of a directory with an umlaut in its name (never checked other special characters tbh) ...

• Could you please show a MWE to reproduce the issue? – TeXnician Dec 14 '18 at 17:04
• @TeXnician done – Raven Dec 14 '18 at 17:42
• Which linux did you use in which version? Which tex distribution do you use in which version? Or is your issue already solved? How? – Mensch Jun 23 at 23:36
• Apparently this occurs only if using latexmk though... – Raven Jun 29 at 17:51


text:
2019
Том 71
№ 11

# Volume 44, № 2, 1992

Anniversaries (Ukrainian)

### On the 80th Aniversary of Birthday of V. V. Gnedenko, Member of the Academy of Sciences pf the Ukraine

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 147-148

Article (Russian)

### Uniqueness of solutions of mixed problems and a Cauchy problem for parabolic equations of high order with unbounded coefficients

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 149–155

New uniqueness classes of generalized solutions (generalized Tacklind classes) of initial-boundary-value problems are found for linear and quasilinear divergent parabolic equations of high order with coefficients increasing at infinity.

Article (Russian)

### Stefan problem with a kinetic and the classical conditions at the free boundary

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 155–166

The Stefan problem is considered with the kinetic condition u+=u=ɛk(y, τ)-ɛv at the phase interface, where k(y, τ) is the half-sum of the principal curvatures of the free boundary and v is the speed of its shifting in the direction of a normal. The solvability of a modified Stefan problem in spaces of smooth functions and the convergence of its solutions as ɛ → 0 to a solution of the classical Stefan problem are proved.

Article (Russian)

### Behavior of solutions of the Dirichlet problem for a second-order quasilinear elliptic equation of general form close to a corner point

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 167–173

The Dirichlet problem for the uniformly elliptic equation $$a_{ij} (x,u,u_x )u_{x_i x_j } + a(x,u,u_x ) = 0$$ is considered in a bounded plane region. It is assumed that there is a corner point on the boundary of the region (the origin), and that the coefficients of the equation satisfy minimal smoothness conditions and appropriate conditions of growth on the gradient (not greater than quadratic). For a smooth solution, it is shown that, in a neighborhood of the corner point,$$u(x) = O(|x|^{\pi /\omega } ),\nabla u(x) = O(|x|^{\pi /\omega - 1} ),$$ where $ω$ is the angle in which the two arcs of the boundary of the region intersect at the origin.

Article (Russian)

### Solutions with singularities of a certain equation of mathematical physics

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 174-177

A theorem for the representation of solutions with singularities of logarithmic type of a certain equation of mathematial physics in terms of quasiconformal mappings is obtained.

Article (Russian)

### Families of functions with equiabsolutely continuous integrals

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 178–184

The approximation of functions from Hardy classes by bounded analytic functions is investigated. A theorem is proved, characterizing the sets of functions with equiabsolutely continuous integrals as limit points of the family of bounded subsets of the space H.

Article (Russian)

### Properties of the solutions of multilinear elliptical systems of nondivergence type

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 184–191

Results on the existence of finite and infinite singular points are obtained for multilinear elliptical systems that satisfy an analogue of the Cordes condition.

Article (Russian)

### Convergence of solutions of variational inequalities with two-sided obstructions in perforated regions

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 191–197

Conditions and the type of convergence of the solutions of elliptical variational inequalities with two-sided obstructions in perforated regions are established.

Article (Ukrainian)

### Averaging evolutionary equations perturbed by random processes with jumps

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 197–207

Weak convergence of measures generated by solutions of an evolutionary equation dependent on a small parameter to the unique solution of the martingale problem corresponding to the stochastic evolutionary equation is proved. The coefficients of the initial equation depend on random Markov processes with jumps.

Article (Russian)

### Approximation of analytic functions by polynomials “close” to polynomials of best approximation

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 208–214

The rate of approximation of analytic functions at interior points of compact sets with connected complement by polynomials “close” to polynomials of best approximation is investigated.

Article (Russian)

### Certain classes of extension of a lacunary Hermitian operator

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 215-232

Certain classes of extensions of a lacunary Hermitian operator are described in terms of abstract boundary conditions. The connection between the asymptotic behavior of eigenvalues of an extension near the boundary of lacuna and the asymptotic of the negative spectrum of the corresponding boundary operator is found.

Article (Russian)

### Remarks on the $L_p — L_q$ estimates of solutions of the Klein-Gordon equation

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 233–245

We clarify some difficulties in the proofs of estimates, which were established earlier, in the paper by Marshall, Straus and Wainger devoted to the Lp —Lq estimates of solutions to the Klein-Gordon equation.

Article (Russian)

### Asymptotic behavior of integral functionals of diffusion processes with periodic coefficients

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 245–252

A central limit theorem is proved for functionals of integral form and with its help is established the local asymptotic normality of the logarithm of the likelihood ratio for diffusion processes with periodic coefficients.

Article (Russian)

### Searching for partially ordered structures

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 253–260

General methods are developed to search for minimal and dead-end payoff operators for a class of sources endowed with a partial order structure. It is shown that the results can be used in the design of experiments with automata and in the minimization of Boolean functions.

Article (Russian)

### Multiplicative inequalities in domains with noncompact boundary

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 260–268

Exact embedding theorems of the multiplicative type are established for functions of Sobolev spaces defined in a domain Ω ⊂R n,n⩾2, whose boundary is not compact. The main condition on the domain is of the isoperimetric type.

Brief Communications (Russian)

### The Stefan thermodiffusion problem in the presence of convection

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 269–274

The Stefan thermodiffusion problem is considered, taking into account convective motions in the liquid phase. The solvability of this problem is proved in spaces of smooth functions.

Brief Communications (Russian)

### On strong Carleman means of multiple trigonometric Fourier series

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 275–279

We prove the strong Carleman summability of the Fourier series of continuous functions on the $m$-dimensional torus, with partial sums constructed over polyhedra of a certain class.

Brief Communications (Russian)

### On the behavior of solutions of quasilinear elliptic equations of second order in unbounded domains

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 279–283

Analogues of the well known in the theory of analytic functions Phragmén-Lindelöff theorem are formulated for the solutions of a wide class of quasilinear equations of elliptic type. Examples are given which illustrate the sharpness of the obtained results for solutions of equations of the form div(|∇u|α−2∇u)=f(x, u), where the function f(x, u) is locally bounded in IRn+1,f(x, 0)=0,uf(x, u)⩾a¦u¦1+q,a>0,α>1,α-1>q⩾0, n⩾2.

Brief Communications (Russian)

### Convergence of diffusion processes

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 284–289

For stochastic diffusion equations with coefficients depending on a parameter, necessary and sufficient conditions of the weak convergence of solutions to the solution of a stochastic diffusion equation are obtained.

Brief Communications (Russian)

### A criterion for differentiability in the sense of Belinskii and its consequences

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 289–294

There are obtained necessary and sufficient conditions for differentiability in the sense of Belinskii of quasiconformal mappings at a point.

Chronicles (Russian)

### Conferences on nonlinear problems of mathematical physics and problems with free boundaries

Ukr. Mat. Zh. - 1992. - 44, № 2. - pp. 295-297


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# Mean Absolute Error in Random Forest Regression

I am new to the whole ML scene and am trying to resolve the Allstate Kaggle challenge to get a better feeling for the Random Forest Regression technique.

The challenge is evaluated based on the MAE for each row.

I've run the sklearn RandomForrestRegressor on my validation set, using the criterion=mae attribute. To my understanding this will run the Forest algorithm calculating the mae instead of the mse for each node.

After that I've used this: metrics.mean_absolute_error(Y_valid, m.predict(X_valid)) in order to calculate the MAE for each row of data.

What I would like to know is if the logic I'm following is sound. Am I making a fundemental mistake or missing something here? Should I have used the default MSE based Regressor and then calculate the MAE of each row using the mean_absolute_error function?

This is probably too late but...

The short answer is yes - your logic is sound. If you are going to be evaluated on MAE you want the Random Forest algorithm to use the same metric when it is building its trees (i.e. at each step it will look for the split that leads to the highest reduction in MAE).

If you read the documentation, you will find out that default criteria is mse and mae is an optional criteria.

To verify things, you can dig into documentation and sometimes in codes too. To my best knowledge, sklearn's documentation is very well written, it is easily readable.


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# If Electric field is constant in a region, does it imply potential is also constant?

I am a bit confused about these concepts- when is electric field or electric potential constant so i would appreciate if anyone could brief me about these things as well. Here i am referring to conservative electrostatic fields.

• if the potential is constant the electric field is zero... Apr 9, 2016 at 15:25
• Potential is constant in the direction perpendicular to the electric field. Aug 13, 2017 at 15:44

Electric field lines are always at right angles to equipotential lines or surfaces.

The electric field is minus the potential gradient.

So in the diagram showing a uniform electric field a positive charge would experience a downward force in the direction of decreasing electric potential.

In this case the magnitude of the electric field is $\frac {20}{5} = 4$ N/C.

If the potential is constant then the electric field is zero.
If the rate of change of potential with distance is constant then the electric field strength is constant.

The electrical field is related to a force concept: it describes the force per unit charge. The potential is related to a potential energy concept: it's the added electrical potential energy per unit charge.

So, just as the force is the negative gradient (or in 1-dimension, the negative slope) of the potential energy function, the electric field is the negative gradient/slope of the potential: $$\vec{E}=-\nabla \phi,$$ where $\phi$ is the potential function due to nearby charges.

Related to your specific question, if the electric field is constant, then the slope of the potential is a constant which means that the potential is changing linearly. If the potential is constant, then the slope of the potential is zero, which means the electric field is zero. An extra charge added to an otherwise constant potential region will experience no electrical force.

When the electric potential is constant through out a given region of space the electric field is zero .Because the electric field is the force perunit positive charge. And electric potential is the potential difference. So when the electric potential is constant then there is no force acts and no change in wordone accurs so that's why the electric field is zero

It is an interesting question:

Firstly, the Electric field "$$E$$" is the slope of the potential, i.e., $$E=-{\frac{dV}{dx}}$$. Therefore "constant electric field" means the potential is either increasing or decreasing at a constant rate (along the space).

Secondly, $$E$$ is a physically measurable quantity but $$V$$ is not. You can never know the absolute value of potential $$V$$, although you know the value of $$E$$ which is constant in your case. It is because,

$$V=-\int Edx$$ $$=-Ex+C$$

Note that we have an integrating constant "$$C$$", i.e., you can always add any constant value with your solution for the potential. It further means $$V$$ must be measured with respect to some reference and the measured value always depends on that reference.

An example: Let's consider a parallel plate capacitor in which Potential of one plate is $$V_1$$ and another plate is $$V_2$$ and they are located at position $$x_1$$ and $$x_2$$ respectively. Also let the distance between two plates, $$dx=x_2-x_1$$=2cm, and electric field (which is ideally constant in such capacitor) $$E=5V/cm$$.

What is the value of potential that satisfies the case? Lets check,

if $$V_1=0V$$ and $$V_2=10V$$ then $$E={\frac{dV}{dx}}={\frac{V_2-V_1}{dx}}={\frac{10-0}{2}}=5V/cm$$

Now lets add some constant value with the potentials. i.e.,

if $$V_1=5V$$ and $$V_2=15V$$ then $$E={\frac{V_2-V_1}{dx}}={\frac{15-5}{2}}=5V/cm$$

if $$V_1=10V$$ and $$V_2=20V$$ then $$E={\frac{V_2-V_1}{dx}}={\frac{20-10}{2}}=5V/cm$$

See!!!! All of those set of ($$V_1$$,$$V_2$$) is giving the same "Constant" electric field. Therefore you can know the potential of one plate with respect to the potential of another, but can never know the absolute value of both.

This is true not only for capacitor but also for every other cases. At this point the potential of "Earth" is usually taken to be the reference in order to measure the potential at any point. It is because earth is a very large object and it's absolute potential does not get affected by any of our action.


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 Mathematics , 2009, Abstract: An association scheme is called skew-symmetric if it has no symmetric adjacency relations other than the diagonal one. In this paper, we study 4-class skew-symmetric association schemes. In J. Ma [On the nonexistence of skew-symmetric amorphous association schemes, submitted for publication], we discovered that their character tables fall into three types. We now determine their intersection matrices. We then determine the character tables and intersection numbers for 4-class skew-symmetric pseudocyclic association schemes, the only known examples of which are cyclotomic schemes. As a result, we answer a question raised by S. Y. Song [Commutative association schemes whose symmetrizations have two classes, J. Algebraic Combin. 5(1) 47-55, 1996]. We characterize and classify 4-class imprimitive skew-symmetric association schemes. We also prove that no 2-class Johnson scheme can admit a 4-class skew-symmetric fission scheme. Based on three types of character tables above, a short list of feasible parameters is generated.
 Diogo Oliveira e Silva Mathematics , 2012, Abstract: We establish the nonexistence of extremizers for a local Fourier restriction inequality on a certain class of planar convex curves whose curvature satisfies a natural assumption. We accomplish this by studying the local behavior of the triple convolution of the arclength measure on the curve with itself, and show in particular that every extremizing sequence concentrates at a point on the curve.
 Mathematics , 2008, Abstract: Let X be a pseudocyclic association scheme in which all the nontrivial relations are strongly regular graphs with the same eigenvalues. We prove that the principal part of the first eigenmatrix of X is a linear combination of an incidence matrix of a symmetric design and the all-ones matrix. Amorphous pseudocyclic association schemes are examples of such association schemes whose associated symmetric design is trivial. We present several non-amorphous examples, which are either cyclotomic association schemes, or their fusion schemes. Special properties of symmetric designs guarantee the existence of further fusions, and the two known non-amorphous association schemes of class 4 discovered by van Dam and by the authors, are recovered in this way. We also give another pseudocyclic non-amorphous association scheme of class 7 on GF(2^{21}), and a new pseudocyclic amorphous association scheme of class 5 on GF(2^{12}).
 Mathematics , 2010, DOI: 10.1016/j.ejc.2010.09.006 Abstract: Suzuki (1998) showed that an imprimitive Q-polynomial association scheme with first multiplicity at least three is either Q-bipartite, Q-antipodal, or with four or six classes. The exceptional case with four classes has recently been ruled out by Cerzo and Suzuki (2009). In this paper, we show the nonexistence of the last case with six classes. Hence Suzuki's theorem now exactly mirrors its well-known counterpart for imprimitive distance-regular graphs.
 Sho Suda Mathematics , 2009, Abstract: It is shown that linked systems of symmetric designs with $a_1^*=0$ and mutually unbiased bases (MUB) are triply regular association schemes. In this paper, we characterize triple regularity of linked systems of symmetric designs by its Krein number. And we prove that maximal MUB carries a quadruply regular association scheme and characterize the quadruple regularity of MUB by its parameter.
 Contributions to Discrete Mathematics , 2008, Abstract: E. R. van Dam gave an example of primitive non-amorphous association schemes in which every nontrivial relation is a strongly regular graph, as a fusion scheme of the cyclotomic scheme of class $45$ on $￥GF(2^{12})$. The aim of this paper is to present a new example of primitive non-amorphous association schemes in which every nontrivial relation is a strongly regular graph, as a fusion scheme of the cyclotomic scheme of class $75$ on $￥GF(2^{20})$. We also propose an infinite family of parameters of association schemes containing both of these two examples.
 Physics , 2013, Abstract: We present a fully conservative, skew-symmetric finite difference scheme on transformed grids. The skew-symmetry preserves the kinetic energy by first principles, simultaneously avoiding a central instability mechanism and numerical damping. In contrast to other skew-symmetric schemes no special averaging procedures are needed. Instead, the scheme builds purely on point-wise operations and derivatives. Any explicit and central derivative can be used, permitting high order and great freedom to optimize the scheme otherwise. This also allows the simple adaption of existing finite difference schemes to improve their stability and damping properties.
 Ajay C. Ramadoss Mathematics , 2005, Abstract: Here, we utilize facts about the big Chern classes discovered by M. Kapranov and independently by M. V. Nori to prove certain results about the nonexistence of certain morphisms from Grassmannian to Grassmannian in characteristic 0. In particular, we show that the filtration F_. on CH^l(X) where F_rCH^l(X) is the linear span of ch_l(V) for all vector bundles V of rank <=r is nontrivial as a theory by showing that for a fixed l>=2, for infinitely many r, for a Grassmannian of r dimensional quotient spaces in an n dimensional vector space (n sufficiently large) over a field of charactersitic 0, ch_l(Q) is in F_rCH^l(G(r,n)) but not in F^(r-1)CH^l(G(r,n)) where Q is the universal quotient bundle of the Grassmannian G(r,n). Separately but using similar methods, we show that for r>=2, if Q denotes the universal quotient bundle of the Grassmannian of r dimensional quotients G(r,n), n>=2r+1, then the class [\psi^pQ] can never be equal to the class of a genuine vector bundle in K(G). In parallel, we obtain a formula for the bid Chern classes in terms of the ch_l. This leads to the identification of proper subfunctors of the Hodge functors H^q(X, \Omega^p) in characteristic 0.
 Proyecciones (Antofagasta) , 2011, DOI: 10.4067/S0716-09172011000100005 Abstract: in this paper mainly important properties of skew lattices and symmetric lattices is obtained. a necessary and sufficient condition for skew lattice to be symmetric is obtained. maximal element of a skew lattice is also obtained.
 K. V. R Srinivas Proyecciones (Antofagasta) , 2011, Abstract: In this paper mainly important properties of skew lattices and symmetric Lattices is obtained. A necessary and sufficient condition for skew lattice to be symmetric is obtained. Maximal element of a skew lattice is also obtained.
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# nLab Thibault Damour

Damour made the theoretical prediction (Buonanno-Damour 98, see Damour 10) of the gravitational wave-signal emitted by inspiralling relativistic binaries which has later been observed by LIGO.

## Selected writings

On U-duality:

On prediction of gravitational wave profiles emitted by black hole relativistic binaries (as later detected by LIGO:

category: people

Last revised on August 19, 2019 at 04:48:27. See the history of this page for a list of all contributions to it.