Maharashtra Board Class 10 Science 1 Sample Paper Set 6 with Answers

Maharashtra Board SSC Class 10 Science 1 Sample Paper Set 6 with Answers Solutions Pdf Download.

Maharashtra Board Class 10 Science 1 Model Paper Set 6 with Answers

Time: 2 Hours
Total Marks: 40

Note:

1. All questions are compulsory.
2. Use of a calculator is not allowed.
3. The numbers to the right of the questions indicate full marks.
4. In case of MCQs (Q. No. 1(A)) only the first attempt will be evaluated and will be given credit.
5. For each MCQ, the correct alternative (A), (B), (C) or (D) with subquestion number is to be written as an answer.
   For Eg.: (i) (A), (ii) (B), (iii) (C)
6. Scientifically correct, labelled diagrams should be drawn wherever necessary.

Question 1.
(A) Choose the correct alternative. [5]
i. Law of gravitation gives the gravitational force between ___________. (C)
(A) the earth and a point mass only
(B) the earth and Sun only
(C) any two bodies having some mass
(D) two charged bodies only
Answer:
(C) any two bodies having some mass

ii. Satellite revolving around the earth at a height of 400 km above the earth’s surface has a time period of about ___________. (B)
(A) 24 hours
(B) 90 minutes
(C) 2 hours
(D) > 24 hours
Answer:
(B) 90 minutes

iii. What happens when a piece of zinc metal is added to copper sulphate solution? (B)
(A) Copper sulphide is formed.
(B) Colourless solution of zinc sulphate is formed.
(C) Copper sulphate solution is not affected at all.
(D) Hydrogen sulphide gas is evolved.
Answer:
(B) Colourless solution of zinc sulphate is formed.

iv. If two ice cubes are pressed against each other, they stick together firmly after some time because ___________. (B)
(A) temperature of the ice cubes reduces.
(B) the melting point of ice reduces to below 0 °C due to the pressure applied.
(C) temperature around the ice cubes reduces.
(D) the melting point of ice becomes 0 °C due to pressure.
Answer:
(B) the melting point of ice reduces to below 0 °C due to the pressure applied.

v. ___________ does not occur during the formation of rainbow. (D)
(A) Refraction
(B) Dispersion
(C) Reflection
(D) Induction
Answer:
(D) Induction

(B) Answer the following. [5]
i. State true or false.
Higher the temperature of air, higher is the amount of water vapour needed to saturate it.
ii. Find the odd one out.
INSAT, GSAT, EDUSAT, IRNSS
iii. Match the following:

Element	Group	Period
a. Lithium	1. 1	i. 1
b. Aluminium	2. 3	ii. 2
	3. 12	iii. 3
	4. 13	iv. 4

iv. Complete the given analogy and state the relation.

v. Name the defect of eye shown in the diagram.

Answer:
(B) i. True
ii. IRNSS
iii.

Element	Group	Period
a. Lithium	1. 1	ii. 2
b. Aluminium	4. 13	iii. 3

iv.

When light ray passes from rarer medium to denser medium, it bends towards the normal whereas when light ray passes from denser medium to rarer medium, it bends away from the normal.
v. Nearsightedness or myopia

Question 2.
(A) Give scientific reasons. (Attempt any 2) [4]
i. Value of g is zero at the centre of the earth.
ii. Rainbow is the combined effect of the refraction, dispersion, and total internal reflection.
iii. Carbon is a tetravalent atom.
(B) Answer the following. (Attempt any 3) [6]
i. Complete the following table for the image formation by a convex lens.

ii. Write short note on: Position of isotopes in the Mendeleev’s and the modem periodic table.
iii. An electric bulb is connected to 250 volts. If the current of 0.5 A passes through it, what is the power of the bulb?
iv. Give four examples of homopolymers.
v. Distinguish between: mass and weight of an object.
Answer:
i. a. The acceleration due to gravity (g) on earth’s surface is given as, g = \(\frac{GM}{R^2}\)
The value of g depends on the mass M of the earth and the radius R of the earth.
b. As we go inside the earth, our distance from the centre of the earth decreases and no longer remains equal to the radius of the earth (R).
c. Along-with the distance, the part of the earth which contributes towards the gravitational force felt also decreases, decreasing the value of (M).
d. Due to combined result of change in R and M, value of g becomes zero at the centre of the earth.

ii. a. The rainbow appears in the sky after a rainfall.
b. Water droplets present in the atmosphere act as small prism.
c. When sunlight enters these water droplets it gets refracted and dispersed.
d. This dispersed light gets totally reflected inside the droplet and again is refracted while coming out of the droplet.
e. As a combined effect of all these phenomena, the seven coloured rainbow is observed.

iii. a. Carbon is a non-metal having atomic number 6.
b. The electronic configuration of carbon is (2, 4).
c. There are four electrons present in the outermost shell of a carbon atom. So, its valency is 4.
Hence, carbon is a tetravalent atom.

(B) i.

ii. a. Isotopes were discovered long time after Mendeleev developed the periodic table.
b. All isotopes of the same element have different atomic masses but same atomic number. They also have the same chemical properties.
c. In Mendeleev’s periodic table, elements are arranged in increasing order of their atomic masses such that chemically similar elements are placed together in a group. So, it was difficult to place them in Mendeleev’s periodic table.
d. In modern periodic table, elements are arranged in increasing order of their atomic numbers. Hence, all the isotopes of an element occupy the same position in the modern periodic table.

iii. Solution:
Given: Potential difference (V) = 250 V,
Current (I) = 0.5 A
To find: Power (P)
Formula: P = VI
Calculation: From formula,
P = 250 × 0.5
∴ P = 125 W
Ans: Power of electric bulb is 125 W.

iv. Examples of homopolymers:
Polyethylene, polystyrene, teflon, polypropylene
(Note: In ’Give examples’, students are expected to write minimum 4 examples.)

v.

Mass	Weight
a. Mass is the quantity of matter contained in an object.	Weight is the force with which the earth attracts an object.
b. Mass remains same everywhere.	Weight of a body keeps on changing from place to place.
c. Mass is measured in kilogram (kg).	Weight is measured in newton (N).
d. Mass is a scalar quantity.	Weight is a vector quantity.
e. Mass of an object can never be zero.	Weight of an object becomes zero at the centre of the earth.

[Any four points]

Question 3.
Answer the following. (Attempt any 5) [15]
i. Select the appropriate options and complete the following paragraph.
(refractive index, velocity, same, direction, ratio, different, frequency, product)
The extent of change in the ___________ of light ray is different for different media and depends ___________ upon the of the medium. The value of refractive index depends on the ___________ of the light in the medium. The refractive index of second medium with respect to the first is given by the ___________ of the magnitude of velocity of light in first medium (v₁) to that in second medium (v₂). Velocity of light in a medium depends on the ___________ of light and thus different colours travel with different velocity in a medium. Therefore, refractive index of a medium is ___________ for different colours.
ii. The reaction of sodium chloride solution with silver nitrate solution is shown in the following figure:

a. Name the products of the reaction.
b. Write chemical equation of the reaction involved.
c. Does the reaction follow law of conservation of mass? Justify your answer. Complete the following table.
iii. Complete the following table.

iv. Explain Hall’s process used for concentration of bauxite ore.
v. Doctor has prescribed a lens having power + 1.5 D. What will be the focal length of the lens? What is the type of the lens and what must be the defect of vision?
vi. A copper block of 3.1 kg is heated in a furnace from 20 °C to 520 °C and then placed on thick ice block. Calculate specific heat of copper if mass of ice melted is 1.8 kg. (Given: Latent heat of fusion of ice = 333 kJ/kg)
vii. Following diagram represents one of the methods to concentrate ores. Label ‘A’, ‘B’, ‘C’ and ‘D’ in the following diagram. Explain the diagram.

viii. Complete the following table:

Name of the compound	Structural formula of the compound	Name of the functional group
Propanone
Methanol
Methanal

Answer:
i. The extent of change in the direction of light ray is different for different media and depends upon the refractive index of the medium. The value of refractive index depends on the velocity of the light in the medium. The refractive index of second medium with respect to the first is given by the ratio of the magnitude of velocity of light in first medium (v₁) to that in second medium (v₂). Velocity of light in a medium depends on the frequency of light and thus different colours travel with different velocity in a medium. Therefore, refractive index of a medium is different for different colours.

ii. a. The products are silver chloride (AgCl) and sodium nitrate (NaNO₃).
b. The reaction is AgNO_3(aq) + NaCl_(aq) → AgCl ↓ + NaNO_3(aq)
c. Yes, the reaction follows law of conservation of mass. There is no change in the weight of the conical flask before and after the reaction.

iii.

iv. a. In this method, the powdered bauxite ore is soaked in aqueous Na2CO3 which dissolves the main ingredient of ore, alumina to form soluble sodium aluminate. The undissolved gangue that is left behind is filtered off.

b. The filtrate is warmed and neutralized by passing carbon dioxide through it resulting in the precipitation of aluminium hydroxide.

c. The precipitate of aluminium hydroxide obtained in the process is filtered, washed, dried and calcined at 1000 °C to get pure alumina.

v. Solution:
Given: Power (P) = + 1.5 D
To find: Focal length (f)
Formu/a: P = 1/f
Calculation: From formula,
f = \(\frac{1}{1.5}\)
= +0.67m
SMART TIP
Note the unit of focal length is metre if power is measured in dioptres.

The prescribed lens is convex lens as the focal length is positive. Convex lens has positive focal length and it is used for the correction of farsightedness or hypermetropia. Thus, the defect in vision is hypermetropia (farsightedness).
Ans: The prescribed lens is convex lens of focal length +0.67 m and the defect in vision is hypermetropia (farsightedness).

vi. Solution:
Given :
Mass of copper block (m_c) = 3.1 kg.
Change in temperature (∆T) = 520 – 20
= 500 °C.
Mass of ice melted (m_ice) = 1.8 kg.
Latent heat of fusion of ice (L_melt) = 333 kJ/kg
To find: Specific heat of copper (c_c)
Formulae:
i. Heat gained by copper block,
Q₁ = m_c c_c ∆T
ii. Heat required to melt the ice,
Q₂ = m_ice(L_melt)_ice
SMART TIP
There is no change in temperature during melting of ice. Hence, ∆T term is not considered while calculating heat required to melt the ice.
Calculation : From formula (i),
Q₁ = 3.1 × c_c × 500
= 1550 c_c
From formula (ii),
Q₂ = 1.8 × 333 kJ
= 1.8 × 333 × 10³ J

According to principle of heat exchange, Q₁ = Q₂
1550 c_c = 1.8 × 333 × 10³
∴ c_c = \(\frac{1.8\times333\times10^3}{1550}\) = 386.7 J/kg°C.
Ans: Specific heat of copper is 386.7 J/kg °C.

vii. Labelled diagram:

Explanation:
a. The magnetic separation process is based on the differences in magnetic properties of the ore particles.
b. This machine consists of a nonmagnetic conveyor belt moving over two rollers, one of which is magnetic in nature while other is nonmagnetic. Two collector vessels are placed below magnetic roller.
c. Powdered ore is released over the belt near nonmagnetic roller. The nonmagnetic particles of ore are carried further with belt and fall in the collector vessel which is placed away from magnetic roller. At the same time, the magnetic particles of the ore are attracted by the magnetic roller and fall in the collector vessel near the magnetic roller.

viii.

Question 4.
Answer the following. (Attempt any 1) [5]
i. Correct the diagram given below of AC electric generator. Label the corrected diagram and explain its construction and working.

ii. In the following table, six elements A, B, C, D, E and F (here letters are not the usual symbols of the elements) of the modem periodic table with their atomic numbers are given.

a. Which of these is an inert gas?
b. Which of these is a halogen?
c. Which of these are metals?
d. If B combines with F, what would be the formula of the compound formed?
e. Write the electronic configuration of C and E.
Answer:
i. Corrected labelled diagram of electric generator (AC):

Construction: The main components of A.C. generator are:
a. Rectangular coil: A large number of turns of insulated copper wire wound on iron core in rectangular shape forms a coil ABCD as shown in figure.
b. Strong magnets: The coil is placed in between two pole pieces (N and S) of a strong magnet. This provides a strong magnetic field.
c. Rings: The two ends of the coil are connected to two rings R1 and R2. These rings rotate along with the coil.
d. Brushes: Two carbon brushes B1 and B2 are used to press the rings.
e. Axle: The two rings have resistive coating in their inner surfaces and are tightly fitted on the axle and the function of the axle is to rotate with the coil.

Working:
a. When the axle is rotated with the help of a machine from outside, the coil ABCD starts rotating.
b. On rotating the axle, the branch AB move upwards and branch CD move downwards hence coil ABCD rotates clockwise.
c. According to Fleming’s right hand rule, electric current flows in the direction A→B→C→D. Therefore, current flows from B₂ to B₁ in the external circuit through galvanometer.
d. After half rotation, the branch AB and CD exchange their position and the induced current flows as D→C→B→A.
e. Since, branch BA is in contact with brush B1 and branch DC is in contact with B2, current flow from B₁ to B₂ in the external circuit i.e., in the direction opposite to the previous half rotation.
f. This repeats after every half rotation and in this way, alternate current is produced using AC generator.
(Note: In ’Correct the diagram and explain’, students are expected to draw a new
corrected labelled diagram and write a detailed explanation.)

ii. a. G is an inert gas (Neon) because its electronic configuration is (2, 8).
b. F is a halogen (chlorine) because its atomic number is 17 and electronic configuration is (2, 8, 7).
c. A, B and C are metals.
d. B with atomic number 11 will have electronic configuration (2, 8, 1). It will lose one electron. F with atomic number 17 will have electronic configuration (2, 8, 7). It will gain one electron.
∴ Compound formed will have formula BF.
e. Electronic configuration of C = (2, 8, 2) and that of E = (2, 6).
SMART TIP
Write down electronic configurations of all. Based on electronic configuration, one can predict if the element is inert gas (complete octet), halogen (7 valence electrons) or metals (1, 2 or 3 valence electrons). It helps to predict formula when metal and non-metal combine.

SSC Maharashtra Board Science 1 Question Paper with Solutions