TS SSC Physical Science Important Questions
TS SSC Physical Science Important Questions and Answers prepared by Net Explanations. TS Xth Exam all Chapter Important Questions with Answers also known as Study Material.
Chapter 1 Reflection of light at curved surfaces
Multiple choice questions
1.) The Geometric center of the the mirror is called as
a)Principal axis. b) center of curvature.
c) Pole d) Radius of curvature.
Ans : c) pole
2.) If we place an object between the concave mirror and focal point,then the position of the image is
a) At infinity. b) At C.
c) Behind the mirror. d) Between F and C
Ans : Behind the mirror
3.) If we place an object beyond the centre of curvature of a concave mirror, then the formed image will be
a) Virtual. b)Real
c) Effect. d) Non of these.
Ans : b) Real
4.) Which mirror is used for making of solar cooker.
a) Concave. b) Convex.
c) d) Non of these.
Ans : a) Concave
5.) When parallel rays are incident of convex mirror, after the reflection the rays will become
a) Converge. b) Diverge.
c) Passes through the lens. d) It goes back the way it comes.
Ans : b) Diverge.
6.) If we place an object on the centre of curvature of a concave mirror then the position of the image is
a)At infinity b) At C c) Beyond C d) At focus
Ans : b) At C
7.) If we place an object beyond the centre of curvature of the concave mirror then the size of the image is
a) Enlarge b) Same size c) Diminished d) Point
Ans : c) Diminished
Two mark questions
1.) Mention any two rules of sign convention.
Ans :
- Always measure the distance from pole of the mirror.
- When we measure the object height (ho) and image height (hi) then the height will be positive if height is measured upwards from the axis and height will be negative if height is measured downwards from the axis.
2.) What is focal point and focal length of the concave mirror.
Ans : The rays coming from the source of light is parallel to principal axis of the concave mirror, after reflection all rays are converged at a point this point is called focal point(F).
The distance between the geometric center of the mirror and the focal point of the mirror is called focal length (f).
Five mark questions
1.) Derive the equation of magnification.
Consider a concave mirror, let OO’ be the the object and ho be the height of the object. A ray incident from O’ to pole of the to pole of the mirror P, and it makes an angle θ ,here angle of incidence and angle of reflection both are same which is θ, II’ be the image where hi be height of the image, C be the centre of curvature and F be the focal point.
Let us consider two an triangle in the shown figure Δ OO’P and Δ PII’
We can see that both are similar triangles
Then. PI / PO = II’ / OO’
v / u = hi / ho …………..(1)
But According to sign conversation
Object distance u will be PO = -u
Image distance v will be PI = -v
Object height h will be OO’ = ho
Image height h will be II’ = -hi
Substitute the above values in the equation (1)
-v / -u = -hi / ho
-v / -u = hi / ho
Therefore magnification, m = -v / u
m = hi / ho
We can find the magnification in the two ways
m = – image distance / object distance
m = -v / u
m = height of the image / height of the object
m = hi / ho
2.) An object is placed at distance 22 cm from the concave mirror and height of the image will be 5 cm and the magnifiacation is 1.6 then find the image distance and height of the object.
Ans : Given that,
Object distance u = -22 cm
Height of the image hi = -5 cm
Magnification m = 1.6
w.k.t magnification m = – image distance / object distance = -v /u
m × u = – v
1.6 × -22 = – v
-35.5 = – v
Image distance, v = 35.2 cm
w.k.t magnification m = height of the image / height of the object = hi / ho
ho = hi / m
ho = -5 / 1.6
ho = -3.125 cm
Therefore the height of the object is 3.125 cm.
3.) An object is placed at a distance 17 cm from the concave mirror and image will be formed at a distance 25.5 cm then find the focal length of the mirror.
Ans : Given that,
Object distance, u = -17 cm
Image distance , v = -25.5 cm
According to mirror formula
1 / f = (1 / u) + (1 / v)
1 / f = (1 / -17) + (1 / -25.5)
1 / f = (-25.5 – 17)/ 433.5
= -42.5 / 433.5
1/ f = 0.09808
f = -10.2 cm
Therefore required focal length of the mirror is 10.2 cm.
Chapter 2 Chemical Equations
Multiple choice questions.
1.) Fill in the blank
2_____(s) → sunlight→ 2Ag(s)+ Cl2(g)
a)Ag2Cl b) AgCl2
c)AgCl d) Ag2Cl2
Ans : c) AgCl
2.) What type of reaction is this
C(s) + O2 (g) → CO2 (g) +Q
a) Endothermic reaction. b) Exothermic reaction
c)both (a) and (b). d) Non of these.
Ans : b) Exothermic reaction.
3.) What type of reaction is this
N2 (g) + O2 (g) → 2NO(g) – Q
a) Endothermic reaction. b) Exothermic reaction
c) both (a) and (b). c) Non of these.
Ans : a) Endothermic reaction.
4.) What is the meaning of downward arrow mark, if this symbol present in chemical equation after the chemical reaction.
a) Gas evolved. b) Reverseble reaction.
c) Precipitate formed. d) both (a) and (b).
Ans : c) Precipitate formed.
5.) What is the meaning of upward arrow mark, if the symbol present in chemical equation after the chemical reaction.
a)Gas evolved. b) Reversible reaction.
c) Precipitate formed. d) both (a) and (c )
Ans : a) Gas evolved.
6) .In an chemical equation, a chemical substance has the aq next to it, what is the meaning of aq.
a) Liquid. b) Gaseous
c) Solid state. d) Aqueous.
Ans : d) Aqueous.
7.) In chemical equation where we write the reactants
a) left side. b) right side
c) middle. d) both (a) and (b)
Ans : a) left side.
8.) Fill in the blank.
C3H8 + ______ → 3CO2 + 4H2O
a) 5O2 b)4O2
c) 3O2 d) 2O2
Ans : a) 5O2
9) Fill in the blank.
Calcium oxide + water → _________
a) Calcium dioxide. b) Calcium hydroxide
c) Calcium oxide. d) Calcium salicylate.
Ans : b) Calcium hydroxide.
10.) In chemical equation in which side we write the products of the reaction.
a) Middle. b) Left side
c) Right side. d) both (a) and (b)
Ans : c) Right side.
Two mark questions
1.) Mention any two relationship which we get from balanced chemical equation.
Ans :
- Mass – mass relationship.
- Volume – volume relationship.
- Mass – volume relationship.
2.) Balanced the equations.
- Al + Fe2O3 → Al2O + Fe
- Na + H2O → 2NaOH + H2
Ans:
- 2Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe(s)
- 2Na (s) + 2H2O (l) → 2NaOH (aq) + H2(g)
3.) Write any two characteristics to make the chemical equation more informative.
Ans:
1.) Endothermic or exothermic heat changes in the chemical equation.
2.) If any gas evolved in the chemical equation.
Five mark questions
1.) How we interpret a balanced chemical equation.
Ans:
1.) From the symbols and formulae of the balanced chemical equation,we get the information of reactants and products of the chemical equation.
2.) The balanced chemical equation gives the ratio of reactants and products of the molecules.
3.) Balanced chemical equation gives the molar ratio of reactants and products if the chemical substance masses are expressed in terms of grams.
4.) In balanced chemical equation we can find out the number of molecules and atoms of different substances using the Avagadro’s number and molar mass from the equation.
5.) From the balanced chemical equation we get information about relative mass of reactants and products.
1. Mass – Mass relationship.
2. Mass – Volume relationship.
3. Volume – Volume relationship.
4. Mass – Volume – number of molecules relationship etc,.
2.) If propane (C3H8) react with a oxygen, then explain how we write the balanced chemical equation.
Ans : We know that propane is an colourless and odourless gass which is used as cooking fuel and heating.
First we write an skeleton equation which is,
Propane + Oxygen → Carbon dioxide + Water
When propane react with oxygen it gives carbon dioxide with water as product.
An equation which contains molecular formulae of substances even chemical equation is unbalanced then it is known as skeleton equation.
Step 1 :
C3H8 + O2 → CO2 + H2O ………….(1)
Step 2 :
Elements |
Number of atoms | |
LHS |
RHS |
|
C | 3 ( in C3H8 ) | 1 ( in CO2 ) |
H | 8 ( in C3H8 ) | 2 ( in H2O ) |
O | 2 ( in O2 ) | 3 ( in CO2 , H2O ) |
Step 3 : Observe the equation (1) , 3 carbon atoms are present in LHS of the equation but only 1 carbon atom is present in RHS, to balance the carbon atoms in both sides add coefficient 3 in CO2 of RHS side , then
C3H8 + O2 → 3CO2 + H2O ……………(2)
Step 4 : Next we observe the equation (2), 8 hydrogen atoms are present in LHS of the equation but only 2 hydrogen atoms are present in RHS, to balance the hydrogen atoms in both sides add coefficient 4 in H2O of RHS side, then
C3H8 + O2 → 3CO2 + 4H2O ………….(3)
Step 5 : Next we observe the equation (3), 2 oxygen atoms are present in LHS of the equation but 10 oxygen atoms are present in RHS , to balance the oxygen atoms in both sides add coefficient 5 in O2 of LHS side, then
C3H8 + 5O2 → 3CO2 + 4H2O
Which is a balanced equation when propane react with oxygen.
Chapter 3 Acids, Bases and Salts
Multiple choice questions
1.) Fill in the blank.
Metal oxide + Acid → Salt + ______
a) Acid. b) Water
c) Metal dioxide. d) Non of these
Ans : b) Water
2.) Fill in the blank.
Acid + Metal → salt + _____
a) Metal oxide. b) metal dioxide
c) water. d) hydrogen
Ans : d) hydrogen
3.) Fill in the blank
2HCl + Zn → ZnCl2 + _______
a)H2 b) 2Cl
c)Zn d) 2Zn
Ans : a) H2
4.) CaCO3 + _______ + CO2 → Ca(HCO3)2
a)H2O b)HCO
c)HCa d) 2H2O
Ans : a)H2O
5.) The common name of the sodium carbonate is
a) Baking soda b) Bleaching powder
c) Washing soda. d) Plaster of Paris.
Ans : c) Washing soda.
6.) The common name of the sodium hydrogen carbonate.
a) Baking soda. b) Washing soda.
c) Bleaching powder. d) Plaster of Paris.
Ans : a) Baking soda
7.) What is common name of calcium sulphate hemihydrate.
a) Baking soda. b) Washing soda.
c) Bleaching powder. d) Plaster of Paris.
Ans : d) Plaster of Paris.
8.) Fill in the blank.
2_______ + Zn → Na2ZnO2 + H2
a.) NaZn. b) NaOH
c) NaO. d)Na2OH
Ans : b) NaOH
9.) Fill in the blank.
Plaster of Paris + 1½_____ → Gypsum
a) Calcium. b) Water
c) Carbon. d) Hydrogen
Ans : b) Water
10.) Fill in the blank.
2NaHCO3 → Heat → _______ + H2O + CO2
a)Na2OH3 b)Na2CO3
c)Na3CO2 d) Na3OH2
Ans : b) Na2CO3
Two mark questions
1.) Write any two uses of Bleaching powder.
Ans :
- In the chemical industries bleaching powder is used as oxidising agent.
- In the preparation of chloroform bleaching powder is used as reagents.
2.) Write any two uses of baking soda.
Ans :
- The baking soda is used to clean the tooth for teath whitening.
- The baking soda is used in cooking the foods like cake, Idli, dosa.
3.) Write any two uses of washing soda.
Ans :
- In industrial level the manufacturer of soap, glass and paper the washing soda is used.
- The washing soda is also used in daily life or domestic purposes.
Five mark questions.
1.) Write five points about plaster of Paris.
Ans : The chemical name of the plaster of Paris is calcium sulphate hemihydrate, which is normally used to make sculpture and it is used to support the fractured bones in hospital level. In generally when gypsum ( CaSO4•2H2O) heated in 373K it loses all moisture or water after the reaction we get plaster of Paris but when plaster of Paris is connect with water or moisture it will become so hard, that’s way we keep plaster of Paris in a water- free or moisture free area.
CaSO4•½H2O + 1½H2O → CaSO4•2H2O
2.) Write five points about baking soda.
Ans : The chemical name of the baking soda is sodium hydrogen carbonate (NaHCO3), which is normally we used to cooking or baking food purposes.
When sodium chloride react with water, carbon dioxide and ammonia after the reaction it gives ammonium chloride and sodium hydrogen carbonate (Baking soda)
NaCl + H2O + CO2 + NH3 → NH4Cl + NaHCO3
The pH value of baking soda is 9 which means it is base, and when it is react with acids like vinegar it shows it’s basic nature
Chapter 4 Refraction of Light at Curved Surfaces
Multiple choice questions
1.) Fill in the blank.
Lens formula 1/f = _______
a)1/v – 1/u b) 1/u -1/v
c)1/v + 1/u d) v-u / uv
Ans : a) 1/v – 1/u
2.) Fill in the blank.
Lens maker’s formula 1/f = ______(1/R1-1/R2)
a)(n+1) b) (n-1)
c) (1-n) d) (1+n)
Ans : b) (n-1)
3.) If we place an object beyond the C2 of the Convex lens then the position of the image will be.
a) Between F2 and C1. b) Between F1 and C2
c) Between F2 and C2. d) Between F1 and C1
Ans : d) Between F1 and C1
4.) If we place an object at C2 of the convex lens then the position of the image will be.
a) On F1 b)On C1
c) On F2 d) Between F1 and C1
Ans : b) On C1
5.) Complete the formula
(n2 / v ) – ( n1 / v ) = _______R
a)(n1+n2) b)(n1-n2)
c)(n2+n1) d)(n2-n1)
Ans : d)(n2-n1)
6.) Complete the formula
(n2 – n1)/ R = __________
a)u-v / n2-n1. b)( un1 – vn2 )/ uv
c) (un2 – vn1)/uv. d) (vn2 – un1)/uv
Ans : c) ( un2 – vn1 ) / uv
7.) What we called the centre of the sphere which contains the part of the curved surface.
a) Pole. b) Centre of curvature c) Radius of curvature. d) Principal axis
Ans : b) Centre of curvature.
8.) The two curved surface in which what we called the line joining the points C1 and C2
a) Pole.b) Centre of curvature c) Radius of curvature. d) Principal axis
Ans : d) Principal axis
9.) What we called the distance between curved surface and center of curvature.
a) Optical center. b) Center of curvature.
c) Radius of curvature. d) Principal axis
Ans : c) Radius of curvature.
10.) What we called the distance between focal point and optical center.
a) Focal length. b) Focal point
c) Pole. d) Principal axis
Ans : a) Focal length
Two mark questions
1.) Draw a ray diagram and mention the size of the image, when object placed at the centre of curvature.
Ans:
When we place an object at centre of curvature C2 on the principal axis, then we get real and inverted image at C1 which has same size of the object.
2.) Draw a ray diagram and mention the size of the image, when object placed between the centre of curvature and focal point.
Ans :
3.) Draw a ray diagram and mention the size of the image, when object located at the focal point.
Ans :
When object located at the focal point on the principal axis then we get an image at infinite distance so we cannot imagine the size and nature of the image.
Five mark questions
1.) Derive the equation of the magnification for convex lens.
Ans :
Let OO’ be the object and ho be the height of the object F1 and F2 be the focal points and u be the object distance from object O to pole P and v be the image distance from pole P to image I.
In the diagram the two triangles
∆OO’P and ∆PII’ are similar triangles
Therefore OO’/OP = II’ / IP
IP / OP = II’ / OO’. …………(1)
According to sign convention
Object distance u will be OP = u
Image distance v will be PI = -v
Height of the object ho will be OO’ = ho
Height of the image hi will be II’ = -hi
Substituting the above values in equation (1)
-v / u = -hi / ho
v / u = hi / ho
Therefore, m = v / u = hi / ho
We can write magnification in two ways
Magnification, m = v / u = image distance / object distance
Magnification, m = hi /ho = height of the image / height of the object
2.) Write a position of the image and characteristics of the image for different positions of object placed in convex lens.
Ans :
- If we place an object at infinite then the image will be formed at focal point F1 and it will be point image.
- If we place an object at beyond C2 then the image will be formed between F1 and C1 , the image will be inverted, Diminished and real in size.
- If we place an object at C2 then the will be formed on C1, the image will be inverted, real and same size.
- If we place an object between F2 and C2 then the image will be formed beyond C1, the image will be inverted, real and same size.
- If we place an object at F2 then the image will be formed at infinity so the size of the image is cannot defined.
- If we place an object between F2 and P then the image will be formed beyond the F2, the image will be erect, virtual and magnified in size.
Chapter 5 Human Eye and Colourful world
Multiple choice questions
1.) Complete the formula
Power of lens, P =_____
a.) 1/C. b) 1/f
c) f/C. d) C/f
Ans : b) 1/f
2.) What is the name of the angle which is present in between the incident ray and emergent ray.
a.) Angle of incidetion. b) Angle of deviation.
c) Angle of emertion. d) Angle of reflection.
Ans : b) Angle of deviation.
3.) Complete the formula
n = sin [ (A+D)/2] / _____
a) sin (2/A) b) sin (A/2)
c) sin (2A/2). d) sin (D/2)
Ans : b) sin (A/2)
4.) The splitting of white light into different colours of light is called.
a) emersion. b) diffusion
c) dispersion d) Non of these
Ans : c) dispersion
5.) The relation between speed of wave, wavelength and frequency is
a) υ =v λ. b) v = υ λ
c) λ =v υ. d) Non of these
Ans : b) v = υ λ
6.) The process of adjusting focal length is called.
a) accommodation. b) dispersion.
c) myopia. d) near sightedness
Ans : a) accommodation.
7.) A person can see distant object clearly but cannot see object at near distance, then the name of the defect of the vision is
a) myopia. b) presbyopia
c) Hypermetropia. d) Non of these
Ans : c) Hypermetropia.
8.) A person cannot see object at long distance but can see nearby object clearly, then the name of the defect of vision is
a) myopia. b) presbyopia
c) Hypermetropia d) Non of these
Ans : a) myopia
Two mark questions
1.) A doctor adviced a person to use a some power of lens, which has focal length 38 cm , then find the power of the lens.
Ans : Given that,
Focal length, f = 38 cm = 0.38 m
w.k.t Power of lens, P = 1/f
= 1/0.38
P = 2.63D
Therefore the power of lens is 2.63 D
2.) A doctor adviced a person to use 1.5 D power of lens then find the focal length of the lens.
Ans : Given that,
Power of lens, P = 1.5 D
w.k.t power of lens, P = 1/f
f = 1/1.5
= 0.6666 m
f = 66.6 cm
Therefore the focal length of the lens is 66.6 cm
3.) Mention the three common defect of vision.
Ans :
- Myopia
- Hypermetropia
- Presbyopia
Five mark questions
1.) An object placed at a distance 1m from the eye then find the focal length of the eye.
Ans : Given that,
Object distance u = -1 m = -100 cm
w.k.t distance between eye lens and ratina is 2.5 cm
Then image distance v = 2.5 cm
Using the formula, 1/f = 1/v – 1/u
1/f = (1/ 2.5) – (1/-100)
= (1/ 2.5)+(1/100)
= (100+2.5)/250
= 102.5/250
1/f = 0.41
f = 2.43 cm
Therefore the focal length of the eye is 2.43 cm when object is placed at a distance 1m.
Chapter 6 Structure of atom
Multiple choice questions
1.) What is the name of the number which described by a set of three numbers n,l,ml in a atom of each electron.
a) quantum number. b) configuration number.
c) atomic number. d) orbital number
Ans : a) quantum number
2.) No two electrons of the same atom can have all four quantum number the same. Who’s principle is thiy.
a) Aufbau principle. b) Paulis exclusion principle.
c) Hand’s rule. d)Non of these.
Ans : b) Paulis exclusion principle
3.) In an atom what we call the distribution of electrons in shells, subshells and orbits.
a) principle energy. b) electronic configuration.
c) quantum number. d)Non of these.
Ans : b) electronic configuration.
4.) Name the atom which has electronic configuration 1s²2s²2p²
a) Barium. b) Boron
c) Carbon. d) Oxygen
Ans : c) Carbon
5.) Which principle state that the lowest energy orbitals are filled first.
a)Paulis exclusion principle. b) Aufbau principle.
c) Hand’s rule. d)Non of these
Ans : b) Aufbau principle.
6.) What is called that the probability of finding the electron is maximum in the region of space arround the nucleus.
a) Shell. b)Subshell
c) Orbital. d)Non of these
Ans : c) Orbital.
7.) Who’s rule is this, that is electron pairing is only possible when orbital of same energy is available or empty.
a)Pauli’s rule. b) Aufbau’s rule
c) Hand’s rule. d) Non of these.
Ans : c) Hand’s rule.
8.) Fill in the blank.
Maximum number of electrons in any shell is ______, where ‘n’ is the principal quantum number.
a.) 2n. b) 2n²
c) 2n d) n²n
Ans : b) 2n2
Two mark questions
1.) Write the two rules which helps the predict electronic configuration.
Ans:
- Assigne the electrons to orbits in increasing order of the value (n+1)
- In Subshell of same value (n+1) electrons are assigned to the subshell first in lower ‘n’.
2.) Write about Hand’s rule.
Ans: Hand’s rule state that the electron paring is only possible when orbital of same energy is available or empty orbital.
For example configuration of Nitrogen (N) atom, who’s atomic number is 7 is 1s²2s²2p³ . The first two electron go into 1s orbit and next two electron go into 2s orbit but next 3 electron go into separate 2p orbit with 3 elements having same spin.
Five mark questions.
1.) Explain about Aufbau principle.
Ans : Aufbau means building up. In this principle lowest energy orbitals of atoms are filled first and maximum electrons present in any shell is in the form of 2n² where n is the principal quantum number.
Here maximum electrons present in subshell (s,p,d or f) is equal to 2(2l+1) where l = 0,1,2,3……….
Therefore the subshell has maximum 2,6,10 and 14 electrons.
The group state electronic configuration build up in the lowest energy orbitals of atoms are fille first and electrons are add equal to the atom number, this principle is called Aufbau principle the orbitals are filled in increasing order of energy.
The two main rule is helps to predict electronic configuration.
- Assign the electrons to orbits in increasing order of values (n+1)
- In subshell of same value (n+1) electrons are assigned the subshell first in lower’n’
Ascending order of energies of various atomic orbitals is given below
1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<………
Consider an atom Boron (B) its atomic number is 5 then we can write its electronic configuration like below.
B(Z = 5). 1s²2s²2p1
Chapter 7 Classification of Elements – The Periodic Table
Multiple choice questions
1.) The horizontal rows in the periodic table is called.
a) Periods. b) Groups
c) Nobel gases. d) Halogen family
Ans : a) Periods
2.) The vertical columns in the periodic table is called.
a) Periods. b) Groups
c) Halogen family. d) Noble gases
Ans : b) Groups
3.) In a neutral gaseous atom what is the energy which is required to remove an electron from the outermost orbit of the atom.
a) Atomic energy. b) Nuclear energy.
c) Ionisation energy. d) Orbital energy.
Ans : c) Ionisation energy.
4.) Complete the formula
_______ = ( Ionisation energy + Electron affinity ) / 2
a) Electropositivity. b) Electronagetivity
c) Electroneutrolity. d) Electrogravity.
Ans : b) Electronagetivity
5.) Atomic radii of the metals are called
a) nuclear. b) electronic radii
c) atomic radii. d) metallic radii
Ans : d) metallic radii
6.) Second group elements in the periodic table is called
a.) Noble gases. b) Alkali earth metals.
c) Non metals. d) Non of these.
Ans : b) Alkali earth metals.
7.) How many groups are present in the modern periodic table
a) 18. b)7
c) 17. d)6
Ans : a)18
8.) How many periods are present in modern periodic table.
a)18. b)6
c)7. d)17
Ans : c)7
9.) Fill in the blank.
________ elements are called lanthanide.
6s b) 4f
c) 5d d) 6p
Ans : b) 4f
Two mark question
1.) Mention two limitations of mendeleeff’s periodic table and give one example.
Ans :
Anomalous pair of elements.
Ex: Tellurium (atomic weight 126.9) precedes iodine ( atomic weight 127.6)
Dissimilar elements placed together.
Ex : Alkali metals like Li,Na,K, etc,.
2.) State mendeleeff’s periodic law and modern periodic law.
Ans :
Mendeleeff’s periodic law : The physical and chemical properties of the elements are the periodic function of their atomic masses or atomic weights.
Modern periodic law : The physical and chemical properties of the elements are the periodic function of their electronic configuration or atomic number.
Five mark questions
1.) Write about Ionisation energy of the element.
Ans : In an neutral gaseous atom the energy required to remove the electron which is present in outermost shell of the atom this energy is known as Ionisation energy. Here in the neutral gaseous atom the energy required to remove the first electron which is present in outermost shell of the atom is called first ionisation energy and in the neutral gaseous atom the energy required to remove electron from uni – positive ion of the element is called second ionisation energy and so on.
M(g) + IE1 → M+(g) + e– ( IE1 = first ionisation energy)
M+(g) + IE2 → M+2(g) + e– ( IE2 = second ionisation energy )
Ionisation energy of the elements is depend on.
- Shielding effect
- Neuclear energy
- Penetration power of the orbitals
- Atomic radius
- Stable configuration
2.) Write about metals and Non metals in the modern periodic table.
Ans : The elements which are have three and less than three electrons in their outermost shell are called metals and the elements which are have five and more electrons in there outermost shell are called non metals.In the periodic table the group 3rd to 12th is called d – Block and in this d – Block the elements are called transition metal as we move from left to right in the periodic table the metallic chemical of d – block is decreases gradually . Lanthanoid and antinoids are also belongs to 3rd group but they are not there, so we called these are inner transition elements. Some elements are have both the properties of metals and non metals these metals are called metaloids. They possess properties like metals but are brittle like non metals these are semiconductors for example Boron and silicon.
Chapter 8 Chemical Bonding
Multiple choice questions.
1.) The force of attraction between any two atoms, then it is known as.
a) Physical bond. b) Chemical bond
c) Biological bond. d) Static bond
Ans : b) Chemical bond
2.) Who proposed the electrostatic bond.
a) Lewis. b) Mendeleeff
c) d) Kossel
Ans : d) Kossel
3.) In periodic table 18th group elements are called.
a) Metals. b) Non metals
c) Metalloids. d) Noble gases.
Ans : d) Noble gases.
4.) An element which losses its electron to complete its outermost shell then it is known as.
a) Cation. b) electroion
c) anion. d)Non of these.
Ans : a) Cation
5.) An element which gain electron from another element to complete its outermost shell then it is known as.
a) Cation. b) electroion
c) anion. d) Non of these.
Ans : c) anion
6.) Fill in the blank.
13Al3+ → 2,__
a.) 16. b) 8
c) 10 d) 18
Ans : b) 8
7.) Fill in the blank
16S2- → 2,__,__
a)16,2. b) 2,16
c) 8,8. d) 18,18
Ans : c)8,8
8.) The element Krypton (Kr) has K,L,M,N shell then how many electrons present in N shell of Krypton.
a.) 2. b) 18
c) 8. d) 10
Ans : c) 8
9.) The valence electron present in the element Argon is.
a)2. b)8
c)18. d)10
Ans : b) 8
Two mark questions
1.) Mention the factors which are depend on gain electron to form anion or lossing electron to form cation.
Ans :
- Ionisation potential.
- Atomic size.
- Electro negative
- Electron affinity.
2.) Draw the dot structure of following.
- NaCl
- CO2
Ans:
3.) Write about formation of methane.
Ans : The formation of methane which is CH4 the carbon atom contributes 4 electrons to form a bond in which each electron form a bond with each hydrogen atom, whereas to form CH4 hydrogen contributes single electrons in which each electron form bond with carbon atoms electron, Thus finally four C-H covalent bond from to form CH4
Five mark questions.
1.) Write about the facts which are based on ionic bond.
Ans : Kossel proposed the ionic bond which is also known as electrostatic bond and below facts are based on ionic bond.
- When transfer one electrons of an atom to another atom which are dissimilar, then ionic bond will be form.
- In the periodic table highly reactive metals which are also called alkali metals placed in left side and the highly reactive non metals which are also called as halogens placed in right side.
- Nobel gases are chemically inactive and stable because except helium all noble gases have 8 electrons in there outermost shell or valence shell.
- Elements which have 1,2 or 3 electrons in there outermost shell gradually they lose electron and attain eight electrons in outermost shell like noble gas and form stable positive ions called cations.
Ex : 11Na → 2,8,1 ; Na+ → 2,8
- Elements which have 5,6 or 7 electrons in there outermost shell gradually they gain electrons and attain eight electrons in there outermost shell like noble gas and form stable nagetive ions called anion.
Ex : 17Cl → 2,8,7 ; Cl– → 2,8,8
Chapter 9 – Electric Current
Multiple choice questions
1.) An human body get current shock of 0.010 A then the effect coused on human body is.
a) Can be felt.
b) lightlt painful
c) Causes involuntary muscle contractions
d) Causes loss of muscle control.
Ans : c) Causes involuntary muscle contractions.
2.) Four resistors 10,15,20,25 are connected in series then the equivalent resistance is,
a) 50. b)60
c) 80. d) 70
Ans : d) 70
3.)Two resistors 1Ω and 3Ω are connected in parallel then the equivalent resistance is
a)0.75. b) 0.65
c)0.55. d) 0.85
Ans : a) 0.75
4.) Fill in the blank
Current , I = V / ___
5.) Volt / Ampere is the same as
a) Coulomb. b) Joule’s
c) Ohm. d) Kelvin
Ans : c) Ohm
6.)The material in which electrons are in motion in the form of resistance then the material is known as.
a.) Ammeter. b) Resistor
c) Galvanometer. d) Voltmeter
Ans : b) Resistor
7.) The obstruction to the motion of the electrons in the conductor is called.
a) Ohm. b) Resistance
c) Ampere. d)Volt
Ans : b) Resistance
8.) Fill in the blank.
Power = Potential difference × ______
a) current. b) resistance
c) work. d) energy
Ans : a) current
9.) One Kilo Watt Hour (1KWH) is equal to
a)3.6×10-6 J b) 3.6×106 J
c) 3.6 × 106 d) 3.6 × 106 s
Ans : b)3.6 × 106 J
Two mark questions
1.) Write any two rule of sign convention in Kirchhoff’s loop law.
Ans :
- When we move from positive terminal side of the battery to nagetive terminal side then the emf of the battery is taken as nagetive.
- But when we move from nagetive terminal side of the battery to positive terminal side then the emf of the battery is taken as positive.
2.) State and draw the Kirchhoff’s junction law.
Ans : Kirchhoff’s junction law state that sum of the current are enter the junction is equal to sum of the current are leaving the junction.
Five mark questions
1.) Derive the formula for equivalent resistance when resistores are connected in parallel
Ans :
Let I be the current and I1, I2 , I3 be the branches of current I and A be the Ampere and R1, R2,R3 are the resistance are connected in parallel to each other as shown in figure.
According to Ohm’s law,
When current through R the current I = V / Req ………..(1)
Where V be the potential difference across the circuit.
Then current through R1 is , I1 = V / R1 ……….(2)
Then current through R2 is , I2 = V / R2 ……….(3)
Then current through R3 is , I3 = V / R3 ………..(4)
We know that current I = I1+I2+I3 ……….(5)
Substituting equation (1),(2),(3),(4) in equation (5)
V / Req = (V/R1)+(V/R2)+(V/R4)
1 / Req = (1/R1)+(1/R2)+(1/R3)
The equivalent resistance of the resistor R and R are connected in parallel then. 1/Req = (1/R1)+(1/R2)
Req = (R1R2)/ (R1+R2)
2.) Derive the equation for equivalent resistance where resistors are connected in series.
Ans :
Let I be the current flow through the circuit and V be the potential difference across the circuit in which V1,V2,V3 are the branches of V. Let R1,R2,R3 are the resistores which are connected in series.
We know that Ohm’s law
The potential difference across the circuit is V = I Req ………..(1)
The potential difference across the resistor R is V1 = IR1 ………….(2)
The potential difference across the resistor R is V2 = IR2 ………….(3)
The potential difference across the resistor R is V3 = IR3 …………..(4)
w.k.t potential difference V = V1+V2+V3 ………..(5)
Substituting equation (1),(2),(3),(4) in equation (5)
Then IReq = IR1 + IR2 + IR3
IReq = I(R1+R2+R3)
Req = R1+R2+R3
The equivalent resistance of resistores R1 and R2 are connected in series then Req = R1 + R2
Chapter 10 Electromagnetism
Multiple choice questions.
1.) Fill in the blank.
Induced EMF = Change in flux /______
a.) Number of turns in the coil
b.) Time
c.) Magnetic field
d.) Current in Ampere
Ans : b) Time
2.) Magnetic field can obtaied from.
a) Voltmeter.
b) ammeter
c) Current carrying wire.
d) Rheostat.
Ans : c) Current carrying wire.
3.) Fill in the blank.
Whenever there is a continuous change of magnetic flux liked with a closed coil, a current is generated in the ________
- a) Magnet. b)Coil
c)Electromagnet d) Non of these
Ans : b) Coil
4.) Complete the formula
Induced EMF, ε = N ( ∆ Φ / ____)
a) ∆t b) ∆v
c) ∆l d) ∆B
Ans : a) ∆t
5.) Complete the formula.
Force F = qvB_____
a)sinθ. b)cosθ
c)secθ. d)tanθ
Ans : a)sinθ
6.) Complete the formula.
Force F = I____sinθ
a)(L+B). b)(L-B)
c)LB. d)(L/B)
Ans : c)LB
7.) Faraday notice that change in magnetic flux through the coil is responsible for generation of ________ in the coil
a) Magnetic field. b) Current c) Magnetic flux. d) Number of turns
Ans : b) Current.
8.) Fill in the blank.
In a closed loop, generated induced EMF is equal to rate of change of ____ passing through it.
a) Current. b) Magnetic field. c) Magnetic flux. d) Number of turns in a coil.
Ans : c) Magnetic flux.
9.) Fill in the blank.
________= (Induced EMF) / Change in flux
a) time. b) magnetic field
c) current. d) magnetic flux.
Ans : a) time
10.) Fill in the blank.
Magnetic flux density = magnetic flux / _______
a) time. b)area
c) magnetic field. d) circuit.
Ans : b)area
Two mark questions
1.) Write about right hand thumb rule.
Ans : According to right hand thumb rule when we grab an electronic wire which contain current then our thumb will be represent the direction of current where as curled fingers represents the direction of magnetic field. When current flow through in downwards then the direction of field lines are clockwise direction, whereas if current flow through in upwards then the direction of field lines are anti-clockwise direction.
2.) If a conductor has length 2 m which is moving with a speed of 20 m/s in the direction of perpendicular to the direction of magnetic of induction 0.16 T then find the induced EMF between the ends of the conductor.
Ans : Given that,
Length of the conductor, l =2m
Conductor which is moving with a speed, v = 20 m/s
Magnetic field of induction, B = 0.16 T
Induced EMF between the ends of the conductor,ε =?
W.k.t ε = Blv
ε = 0.16×2×20
ε = 6.4 V
Therefore the induced EMF between the ends of the conductor is 6.4 V.
Five mark questions.
1.)
Given that a rectangular coil and find the following.
a.) Find the net force of the coil ABCD.
b.) To make the coil rotate continuously , what we do ?
Ans :
a.) In a given rectangular coil the current flow through in the direction of A,B,C to D ,we know that current flow equally in opposite direction then the force will be zero. We can observe that the force acting on A to B and C to D is equal and the direction of current is opposite they carry equal current but opposite direction, so the force of A to B and C to D is zero and similarly the force acting on B to C and D to A is equal but the direction of current is opposite so the force B to C and D to A also zero for the same reason,so the net force of the rectangular coil ABCD is zero.
b.) In a given rectangular coil ABCD the coil rotate and after half cycle of rotation it will become reversed and rotate opposite direction. If the current in the coil change its direction in every half rotation then the coil rotate continuously.
Chapter 11 Principles of Metallurgy
Multiple choice questions
1.) If we get matals by extracting the ore then the process is known as
a)Metal doping. b) Metal extracting.
c) Metallurgy. d) Ore breaking.
Ans : c) Metallurgy
2.) From which ore we can obtain Magnesium metal
a) Magnetite. b) Magnesite
c) Magnemite. d) Magnelite
Ans : b) Magnesite.
3.) An metal ore is converted into oxide by heating it strongly in excess of air, then the process is known as.
a) Extracting. b) Dressing
c) Refining. d) Roasting
Ans : d) Roasting
4.) Complete the reaction.
Fe2O3 + 3CO → 2Fe + _______
a) 2CO3 b)3CO2
c) 3CO3 d)2CO2
Ans : b) 3CO2
5.) Find the temperature in which below chemical reaction will be done.
TiCl4 + 4Na → Ti + 4NaCl
a) 750°C. b)850°C
c) 950°C. d)800°C
Ans : b)850°C
6.) If we arrange the metal in the order of decreasing of their reactivity then it is known as.
a) Reactive series. b) Decreasing series.
c) Activity series. d) Arranging series
Ans : c) activity series.
7.) What substance is used to remove gangue or impurities from metal ores.
a) Field. b)Sand
c) Flux. d) soil
Ans : c) Flux
8.) In which process the carbonate is converted to its oxide.
a) Calcination. b)Flux
c) Roasting. d) Furnace
Ans : a) Calcination.
9.) In which process the metal ore is heated in the absence of air.
a) Roasting. b) Flux
c) Calcination. d) Furnace
Ans : a) Calcination
10.) In the earth crust we found metallic compound with some impurities called.
a) Proteins. b) Vitamins
c) Carbohydrates. d) Minerals
Ans : d) Minerals.
Two mark questions.
1.) Name the three stages of extraction of metal.
Ans :
- Concentration or Dressing the metal.
- Extraction of crude metal.
- Refining or purification of the metal.
2.) Write about Roasting of sulphides ore.
Ans : Generally reverberatory furnace is used for roasting. An process of heating the metal ores in presence of oxygen below the metals melting point,this pyrochemical process is known as roasting. The process of getting Sulphur dioxide from Zinc sulphide ore also are in solid state, below reaction shows that when Zinc sulphide ore is heated with oxygen it gives Zinc oxide with sulphur dioxide as product.
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
Five mark questions.
1.) Write a note on metals which are present in nature.
Ans : We get metals from earth crust, most of the metals are found in nature with combined form, these are more reactive metals, these metals obtained with some impurities called minerals where as least reactive metals are not obtaied with impurities like gold (Au) and silver (Ag), and metals are extracted from minerals without any economic loss then it is called ores. Some metals also found in sea like sodium chloride (salt) and magnesium chloride. For example Alluminium which is naturally found in the form of Bauxite in the earth crust, we extract aluminium from bauxite are without any economic loss then it is profitable.
2.) Complete the following.
1.) 2C + O2 → ________
2.) Fe2O3 + 3CO → _______
3.) _____ → CaO + CO2
4.) CaO + SiO2 → _______
5.) ________→ MgO + CO2
Ans :
1.) 2C + O2 → 2CO
2.) Fe2O3 + 3CO → 2Fe + 3CO2
3.) CaCO3 → CaO + CO2
4.) CaO + SiO2 → CaSiO3
5.) MgCO3 → MgO + CO2
Chapter 12
Carbon and its Compounds
Multiple choice questions.
1.) The conversion of starches and sugers to C2H5OH then the process is known as.
a) Saponification. b) Esterification.
c) Fermentation. d) Non of these.
Ans : c) Fermentation.
2.) Facts are esters of higher fatty acids and trihydroxy alcohol also it is known as.
a) Glycerol. b) Micelle
c) Yeast. d) Colloid
Ans : a) Glycerol.
3.) Which alcohol compound is called absolute 100% alcohol.
a) Ethanol. b) Methanol
c) Propanol. d) Butanol
Ans : a) Ethanol.
4.) The colour of the ethanol.
a) Pinkish brown. b) Greenish brown.
c) Reddish brown. d) Colourless.
Ans : d) Colourless.
5.) What type of smell does ethanol have ?
a)Woody odour. b) Sweet odour
c)Fruity odour. d) Minty odour.
Ans : b) Sweet odour.
6.) Boiling point of the ethanol.
a)56.8°C. b)100°C
c) 78.3° C. d) 92.5 °C
Ans : c) 78.3°C
7.) The general molecular formula for Alkenes.
a)CnH2n b)CnH2n-2
c)CnH2n+1 d)CnH2n+2
Ans : a)CnH2n
8.) The general molecular formula for Alkynes.
a)CnH2n+1 b)CnH2n-1
c) CnH2n+2 d)CnH2n-2
Ans : d) CnH2n-2
9.) The general molecular formula for Alkanes.
a)CnH2n+2 b) CnH2n-1
c) CnH2n+1 d) CnH2n-2
Ans : a)CnH2n+2
10.) An hydrocarbon group contains triple bond between carbon atoms then it is,
a) Alkyne. b) Alkanes
c) Alkenes. d) Aldehydes.
Ans : a) Alkynes
Two mark questions
1.) Write the three rules which are use to numbering carbon atoms.
Ans :
- Counting the number of corbon from left to right or from right to left in this reason we can find out the position of subtituents by sum of the numbers and functional groups should be minimum as possible.
- The carbon function group should be given as lowest number even if it is not obey rule number (1).
- If an carbon chain contain any –CHO or –COOH groups then give it as number 1 even if it is not obey rule (1) and (2).
2.) Complete the following reactions.
- 2CH3COOH + 2Na → ____+ H2
- 2CH3COOH + Na2CO3 → _____ + H2O + CO2
Ans :
- 2CH3COOH + 2Na → 2CH3COONa + H2
- 2CH3COOH + 2Na2CO3 → CH3COONa + H2O + CO2
Five mark questions.
1.) Give example for each reaction in the form of chemical equation.
1.) Combustion.
2.) Oxidation reaction.
3.) Addition reaction.
4.) Substitution reaction.
Ans :
1.) Combustion reaction.
2C2H6 + 7O2 → 4CO2 + 6H2O + Energy
2.) Oxidation reaction.
CH3CH2OH (Ethanol) → (alkaline KMnO4 + Heat) / (Acidified K2Cr2O7 + Heat) → CH3CHO (Ethanal) → CH3COOH (Ethanoic acid)
3.) Addition reaction.
CH3C≡C–CH3 (But-2-yne) → (Ni Catalyst) / H2 → CH3CH=CH–CH3 (But-2-ene) → ( Ni Catalyst) / H2 → CH3–CH2–CH2–CH3 (Butane)
4.) Substitution reaction.
CH4 + Cl2 → sunlight → CH3Cl + HCl
Methyl chloride. Hydrogen chloride
CH3Cl + Cl2 → sunlight → CH2Cl2 + HCl
Methalene chloride
CH2Cl2 + Cl2 → sunlight → CHCl3 + HCl
Chloroform
CHCl3 + Cl2 → sunlight → CCl4 + HCl
Carbon Tetrachloride
2.) Give two structural example for each.
1.) Alkane.
2.) Alkene.
3.) Alkyne.
Ans :
1.) Alkene.
1.) Methane → CH4 → H–CH4–H
2.) Ethane → C2H6 → H–(CH2)2–H
2.) Alkene.
1.) Propene → C2H6 → CH3–CH=CH2
2.) Butene → C4H8 → CH3–CH2–CH=CH2
3.) Alkyne.
1.Ethyne → C2H2 → HC≡CH
2.) Pentyne → C5H8 → CH3–CH2–CH2–C≡CH