RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H.

Other Exercises

Question 1.
Solution:
Arranging the given data in ascending order :
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
We see that 6 occurs in maximum times.
Mode = 6 Ans.

Question 2.
Solution:
Arranging in ascending order, we get:
15, 20, 22, 23, 25, 25, 25, 27, 40
We see that 25 occurs in maximum times.
Mode = 25 Ans.

Question 3.
Solution:
Arranging in ascending order we get:
1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9
Here, we see that 9 occurs in maximum times.
Mode = 9 Ans.

Question 4.
Solution:
Arranging in ascending order, we get:
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
Here, we see that 50 occurs in maximum times.
Modal score = 50 scores Ans.

Question 5.
Solution:
Arranging in ascending order, we get:
10, 10, 11, 11, 12, 12, 13, 14,15, 17
Here, number of terms is 10, which is even
∴ Median = \(\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]\)
= \(\frac { 1 }{ 2 } \) (5th term + 6th term)
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q5.1

Question 6.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q6.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q6.2

Question 7.
Solution:
Writing its cumulative frequency table
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q7.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q7.2

Question 8.
Solution:
Writing its cumulative frequency table
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q8.1
Here, number of items is 40 which is even.
∴ Median = \(\frac { 1 }{ 2 } \left[ \frac { 40 }{ 2 } th\quad term+\left( \frac { 40 }{ 2 } +1 \right) th\quad term \right]\)
= \(\frac { 1 }{ 2 } \) (20th term + 21th term)
= \(\frac { 1 }{ 2 } \) (30 + 30) = \(\frac { 1 }{ 2 } \) x 60 = 30
Mean= \(\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } \) = \(\frac { 1161 }{ 40 } \) = 29.025
∴Mode = 3 median – 2 mean = 3 x 30 – 2 x 29.025 = 90 – 58.05 = 31.95

Question 9.
Solution:
Preparing its cumulative frequency table we get:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q9.1
Here number of terms is 50, which is even
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q9.2

Question 10.
Solution:
Preparing its cumulative frequency table :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q10.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q10.2

Question 11.
Solution:
Preparing its cumulative frequency table we have,
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q11.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q11.2

Question 12.
Solution:
Preparing its cumulative frequency table we have,
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q12.1
Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H are helpful to complete your math homework.

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Practical Based Questions for Class 9 Science Chemistry

Practical Based Questions for Class 9 Science Chemistry

Question 1.
Mention the temperature of the following in degree Celsius and Kelvin Scales

  1. Ice and ice cold water
  2. Boiling water and steam.

Answer:

  1. 6°C and 273K
  2. 100°C and 373K.

Question 2.
Rima took fine chalk powder, egg albumin, starch powder and alum powder in four test tubes. A, B, C and D respectively. After adding water to all the four test tubes, identify the test tubes as true solution, suspension and colloid. (CBSE 2013)
Answer:
Test tube A : Suspension; Test Tube B : Colloidal solution; Test Tube C : Colloidal solution; Test Tube D : True solution.

Question 3.
A student added water to sand and starch in different test tubes. How will you differentiate between the two on the basis of transparency ? (CBSE 2013)
Answer:
The solution of sand in water is a suspension and is opaque. On the other hand, the solution of starch in water is colloidal and is transleucent. This means that the light can pass partially through the second solution. (CBSE 2013)

Question 4.
List two precautions you must take while finding the melting point of ice.. (CBSE 2013)
Answer:

  1. The bulb of the thermometer must not be dipping in ice. It must touch ice.
  2. The constant stirring must be done during the process of melting.

Question 5.
Dipti was asked to prepare four separate mixtures in four beakers A, B, C and D by mixing sugar, fine sand, thin paste of starch and chalk powder respectively in water and then categorise each as stable or unstable. What will be correct categorization ?
Answer:
Practical Based Questions for Class 9 Science Chemistry image - 1

Question 6.
Identify two clear and transparent solutions from the following mixtures :
(a) Milk and water
(b) Sugar and water
(c) Starch powder and water
(d) Glucose and water.
(CBSE 2013)
Answer:
(b) Sugar and water
(d) Glucose and water.

Question 7.
In an experiment to determine the boiling point of water, state the reason for the following precautions :

  1. The bulb of the thermometer should not touch the sides of the beaker.
  2. While boiling water, pumice stone should be added. (CBSE 2013)

Answer:

  1. The sides of the glass beaker are at higher temperature than the contents of the beaker. Therefore, the reading given by the thermometer touching the sides of the beaker does not give correct result.
  2. Pieces of pumice stone are added to boiling water in order to check any bumping.

Question 8.
Four students A, B, C and D were given funnels, filter paper, test tubes, test tube stands, common salt, chalk powder, starch and glucose powder. They prepared the true solution, suspension and colloidal solutions. Test tubes were arranged as shown in the figure. Observe the filtrate obtained in the test tubes and residue on filter paper. Conclude about filtrate residue and type of solution. (CBSE 2013)
Answer:
Practical Based Questions for Class 9 Science Chemistry image - 2

Question 9.
If in the determination of melting point of ice, the ice is contaminated with some non-volatile impurity like common salt, how is the melting point of ice affected ? (CBSE 2013)
Answer:
Melting point of ice gets lowered. Impurities always lower the melting point temperature of solid.

Question 10.
A mixture of sand, powdered glass and common salt is dissolved in water and then filtered. Name the substance left on the filter paper. Name the substance in the filtrate. (CBSE 2013, 2016)
Answer:
Sand and powdered glass are left as residue on the filter paper. Common salt (sodium chloride) solution in water constitutes the filtrate.

Question 11.
In an experiment to determine the boiling point of water, mention two important precautions. (CBSE 2013)
Answer:

  1. Constant stirring must be done so that heating may be uniform.
  2. Bulb of the thermometer must not dip in water.

Question 12.
Four students A, B, C and D are asked to prepare colloidal solutions. The following diagrams show the preparation done by them. Name the student who will be able to prepare colloidal solution. Write two properties of colloidal solutions. (CBSE 2014)
Practical Based Questions for Class 9 Science Chemistry image - 3
Answer:
Only student A’ has prepared the colloidal solution. Egg as such does not mix with water. Only white of an egg forms colloidal solution on stirring in water. Both sugar and common salt form true solution in water. The two properties of colloidal solution are :

  1. Colloidal solutions are of heterogeneous nature consisting of dispersed phase and dispersion medium.
  2. Colloidal solutions show Tyndall effect.

Question 13.
Rima took fine chalk powder, egg albumin, starch powder and alum powder in four test tubes A, B, C and D respectively. After adding water to all the four test tubes, identify the test tubes as true solution, suspension and colloid. (CBSE 2014)
Answer:
Test tube (A) contains a suspension
Test tube (B) contains a colloidal solution
Test tube (C) contains a colloidal solution
Test tube (D) contains a true solution.

Question 14.
What is a suspension ? Give two characteristics of suspension ? (CBSE 2015)
Answer:
A suspension may be defined as a heterogeneous mixture in which the solid particles are spread throughout the liquid without dissolving in it. They settle as precipitate if the suspension is left undisturbed for sometime.
Two characteristics: 

  1. A suspension is of heterogeneous nature. There, are two phases. The solid particles represent one phase while the liquid in which these are suspended or distributed forms the other phase.
  2. The particle size in a suspension is more than 100 nm .

Question 15.
List two properties of a true solution. How would you prepare a true solution. List two different solutes which will form a true solution. (CBSE 2015)
Answer:

  1. A true solution is always homogeneous in nature.
  2. A true solution is transparent in nature.

A true solution can be prepared by dissolving water soluble solute such as sugar/sodium chloride in water. Both sugar and sodium chloride form true solution when dissolved separately in water.

Question 16.
While doing an experiment to determine the melting point of ice, state the role of glass stirrer and mention the correct position of bulb of the thermometer. (CBSE 2015)
Answer:
Stirring the contents of the beaker with a glass stirrer keeps the heating uniform. The position of the thermometer should be such the bulb is just touching the ice cubes.

Question 17.
Rima took fine chalk powder, egg albuminm starch powder and alum powder in four test tubes A, B, C and D respectively. After adding water to all the four test tubes, identify the test tubes as true solution, suspension and colloid. (CBSE 2015)
Answer:
True solution is formed in test tube D. Test tube A contains suspension while colloidal solutions are formed in test tubes B and C.

Question 18.
While determining the melting point of ice, it was observed that even when ice cubes were being moderately heated using the gas burner, the temperature did not rise for sometime till the whole ice melted. Give the possible reason.  (CBSE 2015)
Answer:
Once the melting of ice, starts upon heating with the help of a gas burner, the temperature becomes constant till the entire ice melts to form water. In fact the heat energy now absorbed is used up to overcome the intermoleular foces or is used as latent heat of fusion. Therefore, the temperature does not rise.

Question 19.
In an experiment to verify the law of Conservation of Mass in a chemical reaction, four students A, B, C and D noted down the following observations for the difference in the mass of apparatus before and after the chemical reaction;
A – 4 g ; B- 8 g ; C – zero g ; D – 10 g
Which student has made the correct observation and why ?
Answer:
Student C has made the correct observation since according to the law, there is no change in mass of reactants and products taking part in a chemical reaction.

Question 20.
When 15g of copper sulphate react with 15g of sodium hydroxide, 20g of sodium sulphate along with copper hydroxide is formed. What is the mass of copper hydroxide formed ?
Answer:
The chemical reaction taking place is :
Practical Based Questions for Class 9 Science Chemistry image - 4
On the basis law of conservation of mass :
Mass of copper hydroxide (x) = (15 g + 15 g) – (20 g) = 10 g

Question 21.
To verify the law of conservation of mass in a chemical reaction, four students A, B, C and D perfomed the following chemical reactions in the school laboratory.
(A) Added zinc granules to dil hydrochloric acid
(B) Heated lead nitrate (solid) in a test tube
(C) Dipped Mg ribbon in copper sulphate solution
(D) Added barium chloride solution to sodium sulphate solution.
Which student according to you is likely to obtain the best results and why ?
Answer:
Student (D) is likely to obtain the best results because the reaction is completed immediately and a white precipitate is formed.

Question 22.
16.8 g of sodium hydrogen carbonate are added to 12.0 g of acetic acid. The residue left weighed 20.0 g. What is the mass of CO2 escaped in the reaction ?
Answer:
The chemical reaction taking place is :
Practical Based Questions for Class 9 Science Chemistry image - 5
Mass of CO2 released = (16.8 g +12.0 g ) – (20.0 g) = 8.8 g.

Question 23.
While studying the properties of cathode rays in a discharge tube, a student placed a mica wheel mounted on an axle in the path of the rays. What would happen to the. wheel ? What conclusions can be drawn from it ? (CBSE 2016)
Answer:
The mica wheel would start rotating around the axle. This shows that the cathode rays are made of material particles (electrons).

NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom

NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom

Multiple Choice Questions

Question 1.
Which of the following correctly represents the electonic distribution in the Mg atom ?
(a) 3, 8, 1         (b) 2, 8, 2
(c) 1, 8, 3         (d) 8, 2, 2.
Correct Answer:
(b).

More Resources

Question 2.
Rutherfords alpha (a) particles scattering experiment resulted in to discovery of :
(a) Electron
(b) Proton
(c) Nucleus in the atom
(d) Atomic mass.
Correct Answer:
(c).

Question 3.
The number of electrons in an element X is 15 and the number of neutrons is 16. Which of the following is the correct representation of the element ?
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 1
Correct Answer:
(a) Mass no. = no. of electrons (protons) + no. of neutrons = 15 + 16 = 31.

Question 4.
Daltons atomic theory successfully explained :
(i) Law of conservation of mass
(ii) Law of constant composition
(iii)Law of radioactivity
(iv) Law of multiple proportion
(a) (i), (ii) and (iii)         (b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)       (d) (i), (ii) and (iv).
Correct Answer:
(d) Except for the law of radioactivity, Dalton’s atomic theory explained all other laws which have been listed.

Question 5.
Which of the following statements about Rutherford’s model of atom are correct ?
(i) Considered the nucleus as positively charged
(ii) Established that the α-particles are four times as heavy as a hydrogen atom
(iii) Can be compared to solar system
(iv) Was in agreement with Thomson’s model
(a) (i) and (iii)         (b) (ii) and (iii)
(c) (i) and (iv)         (d) only (i).
Correct Answer:
(a) The statements (i) and (iii) are both correct.

Question 6.
Which of the following are true for an element ?
(i) Atomic number = number of protons + number of electrons
(ii) Mass number = number of protons + number of neutrons
(iii) Atomic Mass = number of protons + number of neutrons
(iv) Atomic number = number of protons = number of electrons
(a) (i) and (ii)           (b) (i) and (iii)
(c) (ii) and (iii)         (d) (ii) and (iv).
Correct Answer:
(d) The statements (it) and (iv) are both correct.

Question 7.
In the Thomson’s model of atom, which of the following statements are correct ?
(i) The mass of the atom is assumed to be uniformly distributed over the atom
(ii) The positive charge is assumed to be uniformly distributed over the atom
(iii) The electrons are uniformly distributed in the positively charged sphere
(iv) The electrons attract each other to stabilise the atom
(a) (i), (ii) and (iii)             (b) (i) and (iii)
(c) (i) and (iv)                   (d) (i), (iii) and (iv).
Correct Answer:
(a) Except for the statements (iv), all other statements are correct.

Question 8.
Rutherford’s α-particle scattering experiment showed that
(i) electrons have negative charge
(ii) the mass and positive charge of the atom is concentrated in the nucleus
(iii) neutron exists in the nucleus
(iv) most of the space in atom is empty Which of the above statements are correct ?
(a) (i) and (iii)        (b) (ii) and (iv)
(c) (i) and (iv)        (d) (iii) and (iv).
Correct Answer:
(b) The statements (ii) and (iv) are correct.

Question 9.
The ion of an element has 3 positive charges. Mass , number of the atom is 27 and the number of neutrons is 14. What is the number of electrons in the ion ?
(a) 13      (b) 10
(c) 14      (d) 16.
Correct Answer:
(b) The ion has 13 protons (27-14) and 10 electrons. It is Al3+ ion.

Question 10.
Identify the Mg2+ ion from the given figure where n and p represent the number of neutrons and protons respectively
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 2
Correct Answer:
(d) Since the ion has 10 electrons, it is Mg2+ ion.

Question 11.
In a sample of methyl ethanoate (CH3COOCH3) the two oxygen atoms have the same number of electrons but different number of neutrons. Which of the following is the correct reason for it ?
(a) One of the oxygen atoms has gained electrons
(b) One of the oxygen atoms has gained two neutrons
(c) The two oxygen atoms are isotopes
(d) The two oxygen atoms are isobars.
Correct Answer:
(c) The two oxygen atoms are the isotopes. These are the different atoms of the same element which have same number of electrons (or atomic number) but different number of neutrons.

Question 12.
Elements with valency 1 are :
(a) always metals
(b) always metalloids
(c) either metals or non-metals
(d) always non-metals.
Correct Answer:
(c) Elements with valency 1 may be metals (+1 valency) or non-metals (-1 valency)

Question 13.
The first model of an atom was given by :
(a) N. Bohr
(b) E. Goldstein
(c) Rutherford
(d) J.J. Thomson.
Correct Answer:
(d).

Question 14.
An atom with 3 protons and 4 neutrons will have a valency of :
(a) 3         (b) 7
(c) 1         (d) 4.
Correct Answer:
(c) The atom has also 3 electrons with electronic configuration 2, 1. It has valency of 1.

Question 15.
The electronic distribution in an aluminium atom is :
(a) 2, 8, 3         (b) 2, 8, 2
(c) 8, 2, 3         (d) 2, 3, 8.
Correct Answer:
(d).

Question 16.
Which of the following do not represent Bohr’s model of an atom correctly ?
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 3
(a) (i) and (ii)           (b) (ii) and (iii)
(c) (ii) and (iv)          (d) (i) and (iv).
Correct Answer:
(c) The first shell (K) cannot have more than two electrons in it.

Question 17.
Which of the following statements is always correct ?
(a) An atom has equal number of electrons and protons.
(b) An atom has equal number of electrons and neutrons.
(c) An atom has equal number of protons and neutrons.
(d) An atom has equal number of electrons, protons and neutrons.
Correct Answer:
(a) The number of electrons or protons in an atom are always equal.

Question 18.
Atomic models have been improved over the years. Arrange the following atomic models in the order of their chronological order :
(i) Rutherford’s atomic model
(ii) Thomson’s atomic model
(iii) Bohr’s atomic model
(a) (i), (ii) and (iii)
(b) (ii), (iii) and (i)
(c) (ii), (i) and (iii)
(d) (iii), (ii) and (i).
Correct Answer:
(c). It is the correct order.

Short Answer Questions

Question 19.
An element has one proton, one electron and no neutron. Name the element. How will you represent it ?
Answer:
The element is called protium or hydrogen (H). It can be represented as 11H.

Question 20.
Write any two observations to show that atoms are divisible.
Answer:
The discovery of electrons and protons led to the opinion that the atoms are divisible. Even neutrons were also discovered at a later stage.

Question 21.
Will 35Cl and 37Cl have different valencies ? Justify your answer.
Answer:
The atomic number (Z) of the element (Cl) is 17. Its electronic distribution is 2, 8, 7. It has valency (8 – 7) = 1 which is the same for both the species, which act as isotopes.

Question 22.
Why did Rutherford select a gold foil in his α-scattering experiment ?
Answer:
Gold is a highly malleable. It could be converted into very fine foils or leaves. Moreover, the nucleus of the element is very heavy. It could not be displaced by the impact of fast moving α-particles.

Question 23.
Find out the valency of the atoms represented by the Fig. (a) and (b).
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 4Answer:
(a) The electronic configuration of the atom is 2, 8, 8. It has completely filled K, L, M shells. Its valency is zero. The atom belongs to the element Argon (Ar).
(b) The electronic configuration of the atom is 2, 7. It has seven electrons in the valence shell. Its valency is (8 – 7) equal to one. The atom belongs to the element fluorine (F).

Question 24.
One electron is present in the outermost shell of the atom of an element X. What would be the nature and value of charge on the ion formed if this electron is removed from the outermost shell ?
Answer:
Since the atom of the element X has lost one electron from the outermost shell, it has +1 unit charge.

Question 25.
Write down the electron distribution of chlorine atom. How many electrons are there in the L shell ? (Atomic number of chlorine is 17).
Answer:
The electronic distribution in chlorine (Z = 17) is K(2), L(8) and M(7). The L-shell has eight electrons.

Question 26.
In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed ?
Answer:
With six (6) electrons in the outermost shell, the atom of the element needs two (2) electrons more to have eight (8) electrons in the outermost shell. The ion (x2-) has -2 units charge.

Question 27.
What information do you get from the following electronic distribution about the atomic number, mass number and valency of atoms X, Y and Z ? Give your answer in a tabular form.
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 5
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 6

  • In atom X, valency is equal to number of valence electrons (2, 3) = 3
  • Atom Y has electronic configuration 2, 6. Valency is eight – No. of valence electrons (8-6)
  • In atom Z, valency is equal to number of valence electrons (2, 8, 5), or eight – no. of valence electrons
    (8 – 5) = 3.

Question 28.
In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement ? Justify your answer.
Answer:
The statement is wrong. In an atom, number of protons and electrons are always equal. These can be different only in an ion (positive or negative).

Question 29.
Calculate the number of neutrons present in the nucleus of an element 3115X which is represented as .
Answer:
Mass no. of the element (A) = 31
Atomic no. of the element (Z) =15
No. of neutrons (n) = A – Z = 31 – 15 = 16.

Question 30.
Match the names of the Scientists given in column A with their contributions towards the understanding of the atomic structure
as given in column B.
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 7
Answer:
(a) — (iii) ;
(b) — (iv);
(c) — (i) ;
(d) — (ii)
(e) — (vi) ;
(f) — (vii) ;
(g) — (v).

Question 31.
The atomic numbers of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements ?
Answer:
The elements having same mass number but different atomic numbers are known as isobars.

Question 32.
Complete the following table on the basis of information available in the symbols given below :
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 8
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 9

Question 33.
Helium atom has 2 electrons in its valence shell but its valency is not 2. Explain.
Answer:
Helium (He) atom with atomic number (Z) equal to two has only one shell (K-shell). It can have a maximum of two electrons which it has. Therefore, the valence shell is complete and valency of the atom is zero.

Question 34.
Fill in the blanks in the following statements :
(a) Rutherfords α-particle scattering experiment led to the discovery of the .
(b) Isotopes have same but different
(c) Neon and chlorine have atomic numbers 10 and 17 respectively. Their valencies will be
and respectively.
(d) The electronic configuration of silicon is and that of sulphur is
Answer:
(a) nucleus
(b) atomic numbers, mass numbers
(c) zero and one
(d) 2, 8, 4 and 2, 8, 6.

Question 35.
An element X has a mass number 4 and atomic number 2. Write the valency of this element.
Answer:
The element X with mass number 4 and atomic number 2 is helium (He). It has 2 electrons (maximum possible) in its only shell which is K-shell. Therefore, the valency of the element is zero.

Long Answer Questions

Question 36.
Why do Helium, Neon and Argon have zero valency ?
Answer:
All the three elements have completely filled valence shells, according to Bohr Bury scheme. Therefore, their valencies are zero.

Question 37.
The ratio of the radii of hydrogen atom and its nucleus is 105. Assuming the atom and the nucleus to be spherical,
(i) what will be the ratio of their sizes ?
(ii) If atom is represented by planet earth ‘Re’ = 6.4 x 106 m, estimate the radius of the nucleus.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 10

Question 38.
Enlist the conclusions drawn by Rutherford from his α-ray scattering experiment.
Answer:
He made the following conclusions from the experiment.

  1. As most of the alpha particles passed through undeflected, this means that they did not come across any obstruction in their path. Thus, most of the space in an atom is expected to be empty.
  2. As a few alpha particles suffered minor deflections and a very few major deflections, this means that these must have met with some obstructions in their path.
  3. This obstruction must be :
    1. Very small : Only a few particles were obstructed by it.
    2. Massive : Each alpha particle has 4 u mass and is quite heavy. It could easily pass through a light obstruction by pushing it aside.

Question 39.
In what way is the Rutherfords atomic model different from Thomsons atomic model ?
Answer:
According to Thomsons model, an atom may be regarded as a positively charged sphere containing protons in which the negatively charged protons are supposed to be studied or embedded. He gave no clue about the nucleus and extranuclear portion. This was given for the first time by Rutherfords model atom with the help of α-ray scattering experiment.

Question 40.
What were the drawbacks of Rutherfords model of an atom ?
Answer:
There are infact, two main drawbacks of Rutherford Model atom

  1. Rutherford model of atom could not explain the stability of the atom.
  2. Rutherford model of atom could not explain as to how the electrons are distributed in the extra nuclear portion in an atom.

Question 41.
What are the postulates of Bohr’s model of an atom ?
Answer:
The main postulates of the theory are listed :

  1. In the extra nuclear portion of an atom, the electrons revolve in well defined circular paths known as orbits.
  2. These circular orbits are also known as energy levels or energy shells.
  3. These have been designated as K, L, M, N, O, … (or as 1, 2, 3, 4, 5, …) based on the energy present.
  4. The order of the energy of these energy shells is :
    K<L<M<N<0 <…. or 1< 2< 3 < 4<5 <….
  5. While revolving in an orbit, the electron is not in a position to either lose or gain energy. In other words, its energy remains stationary. Therefore, these energy states for the electrons are also known as stationary states.

Question 42.
Show diagramatically the electronic distribution in a sodium atom and a sodium ion and also give their atomic number.
Answer:
The atomic number of sodium (Na) is 11. The electronic distribution in the atom is K(2) L(8) and L(1). Sodium
ion (Na+) is formed by removal of one electron from the atom. Its electronic distribution is K(2) and L(8). This means that Na+ ion has the elecronic configuration of Ne atom which is inert gas atom. The electronic distribution of the atom and ion are shown as follows :
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 11

Question 43.
In the Gold foil experiment of Geiger and Marsden, that paved the way for Rutherfords model of an atom, 1.00% of the α-particles were found to deflect at angles > 50°. If one mole of α-particles were bombarded on the gold foil, compute the number of α-particles that would deflect at angles less than 50°.
Answer:
Percentage (%) of particles deflected at an angle more than 50° = 1%
Percentage (%) of α-particles deflected at an angle less than 50° = 100 – 1 = 99%
NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 12

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RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G.

Other Exercises

Question 1.
Solution:
Arranging in ascending order, we get:
2,2,3,5,7,9,9,10,11
Here, number of terms is 9 which is odd.
∴ Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 9+1 }{ 2 } \) th term = 5th term = 7 Ans.
(ii) Arranging in ascending order, we get: 6, 8, 9, 15, 16, 18, 21, 22, 25
Here, number of terms is 9 which is odd.
∴ Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 9+1 }{ 2 } \) th term = 5th term = 16 Ans.
(iii) Arranging in ascending order, we get: 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here, number of terms is 11 which is odd.
∴ Median = \(\frac { 11+1 }{ 2 } \) th term = \(\frac { 12 }{ 2 } \) th term = 6th term = 16 Ans.
(iv) Arranging in ascending order, we get:
0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here, number of terms is 13, which is odd.
Median = \(\frac { 13+1 }{ 2 } \) th term = \(\frac { 14 }{ 2 } \) th term = 7th term = 4 Ans.

Question 2.
Solution:
Arranging in ascending order, we get 9, 10, 17, 19, 21, 22, 32, 35
Here, number of terms is 8 which is even
∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } th\quad term+\left( \frac { 8 }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [4th term + 5th term] = \(\frac { 1 }{ 2 } \) (19 + 21) = \(\frac { 1 }{ 2 } \) x 40 = 20
(ii) Arranging in ascending order, we get:
29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here number of terms is 10 which is even
∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) (60 + 63) = \(\frac { 1 }{ 2 } \) x 123 = 61.5 Ans.
(iii) Arranging in ascending order we get
3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
Here number of terms is 12 which is even.
∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 12 }{ 2 } th\quad term+\left( \frac { 12 }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) (6th term + 7th term) = \(\frac { 1 }{ 2 } \) (15 + 17)= \(\frac { 1 }{ 2 } \) x 32
= 16 Ans.

Question 3.
Solution:
Arranging the given data in ascending order, we get :
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
∴ Median = \(\frac { 15+1 }{ 2 } \) th term = \(\frac { 16 }{ 2 } \) th term = 8th term = 23
∴ Median score = 23 Ans.

Question 4.
Solution:
Arranging in ascending order, we get:
143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1
Here, number of terms is 9 which is odd.
Median = \(\frac { 9+1 }{ 2 } \) th term = \(\frac { 10 }{ 2 } \) th term = 5th term = 147.3 cm
Hence, median height = 147.3 cm Ans.

Question 5.
Solution:
Arranging in ascending order, we get:
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Here number of terms is 8 which is even
∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } th\quad term+\left( \frac { 8 }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \)[4th term + 5th term]
= \(\frac { 1 }{ 2 } \) (13.4 + 14.3) = \(\frac { 1 }{ 2 } \) (27.7) = 13.85
∴ Median weight = 13.85 kg. Ans.

Question 6.
Solution:
Arranging in ascending order, we get:
32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Here, number of terms is 10 which is even.
∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [5th term + 6th term ] = \(\frac { 1 }{ 2 } \) (40 + 44) = \(\frac { 1 }{ 2 } \) x 84 = 42 .
∴ Median age = 42 years.

Question 7.
Solution:
The given ten observations are 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41
These are even
∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [5th term + 6th term ] = \(\frac { 1 }{ 2 } \)(x + 1 + x + 3) = \(\frac { 1 }{ 2 } \)(2x + 4)
= x + 2
But median is given = 24
∴ x + 2 = 24 => x = 24 – 2 = 22
Hence x = 22.

Question 8.
Solution:
Preparing the cumulative frequency table, we have:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q8.1
Here, number of terms (n) = 41, which is odd,
Median = \(\frac { 41+1 }{ 2 } \) th term = \(\frac { 42 }{ 2 } \) th term = 21st term = 50 (∵ 20th to 28th term = 50)
Hence median weight = 50 kg Ans.

Question 9.
Solution:
Arranging first in ascending order, we get:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q9.1
Now preparing its cumulative frequency table
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q9.2
Here, number of terms is 37 which is odd.
Median = \(\frac { 37+1 }{ 2 } \) th term = \(\frac { 38 }{ 2 } \) th term = 19 th term = 22 (∵18th to 21st = 22)
Hence median – 22 Ans.

Question 10.
Solution:
first arranging in ascending order we get
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q10.1
Now preparing its cumulative frequency table,we find:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q10.2
Here, number of terms is 43, which if odd.
Median = \(\frac { 43+1 }{ 2 } \) th term = \(\frac { 44 }{ 2 } \) th term = 22nd term = 25 25 (∵ 11th to 26th = 25)

Question 11.
Solution:
Arranging in ascending order,we get
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q11.1
Now preparing its cumulative frequency table, we find :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q11.2
Here, number of terms = 50 which is even
∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 50 }{ 2 } th\quad term+\left( \frac { 50 }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) (154 + 155) = \(\frac { 1 }{ 2 } \) (309) = 154.5 (∵ 22nd to 25th = 154, 26th to 34th= 155)

Question 12.
Solution:
Arranging in ascending order, we get:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q12.1
Now, preparing its cumulative frequency table.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q12.2
Here, number of terms is 60 which is even
∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 60 }{ 2 } th\quad term+\left( \frac { 60 }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) (30th term + 31st term)
= \(\frac { 1 }{ 2 } \) (20 + 23) = \(\frac { 1 }{ 2 } \) x 43 = 21.5 (∵ 18th to 30th term = 20, 31st term to 34th = 23)
Hence median = 21.5 Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D

RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10D.

Other Exercises

Question 1.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 1
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 2
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 3

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 4

Question 3.
Solution:
x² + px – q² = 0
Comparing it with ax² + bx + c = 0
a = 1, b = p, c = -q²
Discriminant (D) = b² – 4ac
= (p)² – 4 x 1 x (-q²)
= p² + 4 q²
p and q are of two powers
p² + 4q² is always greater than 0
The roots are real for all real values of p and q.

Question 4.
Solution:
The quadratic equation is 3x² + 2kx + 27 = 0
Comparing it with ax² + bx + c = 0
a = 3, b = 2k, c = 27
Discriminant (D) = b² – 4ac
= (2k)² – 4 x 3 x 27
= (2k)² – 324
Roots are real and equal
(2k)² – 324 = 0
⇒ (2k)² – (18)² = 0
⇒ (k)² – (9)² = 0
⇒ (k + 9) (k – 9) = 0
Either k + 9 = 0, then k = -9
or k – 9 = 0, then k = 9
Hence, k = 9, -9

Question 5.
Solution:
The quadratic equation is
kx (x – 2√5) x + 10 = 0
kx² – 2√5 kx + 10 = 0
Comparing it with ax² + bx + c = 0
a = k, b = -2√5 k, c = 10
D = b² – 4ac = (-2 k)² – 4 x k x 10 = 20k² – 40k
Roots are real and equal.
D = 0
20k² – 40k = 0
⇒ k² – 2k = 0
⇒ k (k – 2) = 0
Either, k = 0 or k – 2 = 0, then k = 2
k = 0, k = 2

Question 6.
Solution:
The quadratic equation is 4x² + px + 3 = 0
Comparing it with ax² + bx + c = 0
a = 4, b = p, c = 3
D = b² – 4ac = p² – 4 x 4 x 3 = p²- 48
Roots are real and equal.
D = 0
⇒ p² – 48 = 0
⇒ p² = 48 = (±4√3)²
⇒ P = ± 4√3
P = 4√3, p = -4√3

Question 7.
Solution:
The quadratic equation is 9x² – 3kx + k = 0
Comparing it with ax? + bx + c = 0
a = 9, b = -3k, c = k
D = b² – 4ac = (-3k)² – 4 x 9 x k = 9k² – 36k
Roots are real and equal.
D = 0
9k² – 36k = 0
⇒ 9k (k – 4) = 0
Either, k = 0 or k – 4 = 0, then k = 4
The value of k is non-zero.
k = 4

Question 8.
Solution:
(i) The equation is (3k + 1) x² + 2(k + 1) x + 1 = 0
Comparing it with ax² + bx + c = 0
a = (3k + 1), b = 2(k + 1), c = 1
D = b² – 4 ac
= [2(k + 1)]² – 4(3k + 1) x 1
= 4k² + 4 + 8k – 12k – 4
= 4k² – 4k
= 4k (k – 1)
Roots are real and equal.
Either, k = 0 or k – 1 = 0, then k = 1
k = 0, k = 1
(ii) x² + k(2x + k – 1) + 2 = 0
⇒ x² + 2kx + (k² – k + 2) = 0
Here, a = 1, b = 2k, c = (k² – k + 2)
Discriminant (D) = b² – 4ac = (2k)² – 4 x 1 x (k² – k + 2)
= 4k² – 4k² + 4k – 8
= 4k – 8
Roots are real and equal.
D = 0
⇒ 4k – 8 = 0
⇒ k = 2
Hence, k = 2

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 5

Question 10.
Solution:
The given quadratic equation is
(p + 1) x² – 6(p + 1) x + 3(p + 9) = 0, p ≠ -1
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 6

Question 11.
Solution:
-5 is a root of 2x² + px – 15 = 0
x = -5 will satisfy it
Now, substituting the value of x = -5
⇒ 2(-5)² + p(-5) – 15 = 0
⇒ 50 – 5p – 15 = 0
⇒ 35 – 5p = 0
⇒ 5p = 35
⇒ P = 7
In quadratic equation p(x² + x) + k = 0
⇒ 7 (x² + x) + k = 0 (p = 7)
⇒ 7x² + 7x + k = 0
Comparing it with ax² + bx + c = 0
a = 7, b = 7, c = k
D = b² – 4ac = (7)² – 4 x 7 x k
= 49 – 28k
Roots are real and equal.
49 – 28k = 0
⇒ 28k = 49
k = \(\frac { 49 }{ 28 }\) = \(\frac { 7 }{ 4 }\)

Question 12.
Solution:
3 is a root of equation x² – x + k = 0
It will satisfy it
Now, substituting the value of x = 3 in it
(3)² – (3) + k = 0
⇒ 9 – 3 + k = 0
⇒ 6 + k = 0
⇒ k = -6
Now in the equation, x² + k (2x + k + 2) + p = 0
x² + (-6)(2x – 6 + 2) + p = 0
⇒ x² – 12x + 36 – 12 + p = 0
⇒ x² – 12x + (24 + p) = 0
Comparing it with ax² + bx + c = 0
a = 1, b = -12, c = 24 + p
D = b² – 4ac
= (-12)² – 4 x 1 x (24 + p)
= 144 – 96 – 4p = 48 – 4p
Roots are real and equal.
D = 0
48 – 4p = 0
⇒ 4p = 48
⇒ p = 12
Hence, p = 12

Question 13.
Solution:
-4 is a root of the equation x² + 2x + 4p = 0
Then it will satisfy the equation
Now, substituting the value of x = -4
(-4)² + 2(-4) + 4p = 0
16 – 8 + 4p = 0
⇒ 8 + 4p = 0
⇒ 4p = -8
⇒ p = -2
In the quadratic equation x² + px (1 + 3k) + 7(3 + 2k) = 0
⇒ x² – 2x (1 + 3k) + 7(3 + 2k) = 0
Comparing it with ax² + bx + c = 0
a = 1, b = -2 (1 + 3k), c = 7 (3 + 2k)
D = b² – 4ac
= [-2(1 + 3k)]² – 4 x 1 x 7(3 + 2k)
= 4(1 + 9k² + 6k) – 28(3 + 2k)
= 4 + 36k² + 24k – 84 – 56k
= 36k² – 32k – 80
Roots are equal.
D = 0
⇒ 36k² – 32k – 80 = 0
⇒ 9k² – 8k – 20 = 0
⇒ 9k² – 18k + 10k – 20 = 0
⇒ 9k (k – 2) + 10(k – 2) = 0
⇒ (k – 2) (9k + 10) = 0
Either, k – 2 = 0, then k = 2
or 9k + 10 = 0, then 9k = -10 ⇒ k = \(\frac { -10 }{ 9 }\)
k = 2, k = \(\frac { -10 }{ 9 }\)

Question 14.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 7

Question 15.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 8

Question 16.
Solution:
The quadratic equation is 2x² + px + 8 = 0
Comparing it with ax² + bx + c = 0
a = 2, b = p, c = 8
D = b2 – 4ac = p² – 4 x 2 x 8 = p² – 64
Roots are real.
D ≥ 0
p² – 64 ≥ 0
⇒ p² ≥ 64 ≥ (±8)²
p ≥ 8 or p ≤ -8

Question 17.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 9
Roots are equal
D = 0
⇒ 4(α – 12) (α – 14) = 0
⇒ α – 14 = 0 {(α – 12) ≠ 0}
⇒ α = 14
Hence, α = 14

Question 18.
Solution:
9x² + 8kx + 16 = 0
Comparing it with ax2 + bx + c = 0
a = 9, b = 8k, c = 16
D = b² – 4ac
= (8k)² – 4 x 9 x 16 = 64k² – 576
Roots are real and equal.
D = 0
64k² – 576 = 0
64k² = 576
⇒ k² = 9 = (±3)²
k = 3, k = -3

Question 19.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 10
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 11
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 12

Question 20.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 13
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 14

Question 21.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 15
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 16
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 17

Question 22.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 18

Hope given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.