Maharashtra Board Class 10 Science 1 Sample Paper Set 5 with Answers

Maharashtra Board SSC Class 10 Science 1 Sample Paper Set 5 with Answers Solutions Pdf Download.

Maharashtra Board Class 10 Science 1 Model Paper Set 5 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.
Choose the correct alternative. [5]
i. Stars twinkle because of the _________. (D)
(A) explosions occurring in stars from time to time.
(B) absorption of light in the earth’s atmosphere.
(C) motion of stars.
(D) changing refractive index of the atmospheric gases.
Answer:
(D) changing refractive index of the atmospheric gases.

ii. According to Newton’s first law of motion, higher the mass _________ is the inertia. (B)
(A) lower
(B) higher
(C) zero
(D) double
Answer:
(B) higher

iii. Which of the following metals does not react with cold or hot water but reacts with steam? (D)
(A) Potassium
(B) Calcium
(C) Magnesium
(D) Iron
Answer:
(D) Iron

iv. A student obtains a clear image of a candle on a screen with the help of a convex lens. He now wants to focus on a distant tree and obtain a clear image. In which direction should he move the lens in order to get a clear image? (C)
(A) Behind the screen
(B) Away from the screen
(C) Towards the screen
(D) At a very long distance from the screen.
Answer:
(C) Towards the screen

v. Water expands on reducing its temperature below _________°C. (B)
(A) 0
(B) 4
(C) 12
(D) 5
Answer:
(B) 4

(B) Answer the following. [5]
i. Find the odd one out.
Red colour, Yellow colour, Violet colour, White colour
ii. Match the physical quantities given in column I to units used to express them given in column II.

Column I	Column II
a. Specific latent heat	1. cal/g °C
b. Heat	2. cal/g
	3. J

iii. State true or false. If false, write the correct sentence.
The gravitational force acting on two bodies will be affected when they are dipped in water.
iv. Complete the analogy.
The first artificial satellite : Sputnik :: The first Indian satellite : _________.
Name the family of nonmetals having valency one.
Answer:
i. White colour
ii.

Column I	Column II
a. Specific latent heat	2. cal/g
b. Heat	3.J

iii. False
The gravitational force acting on two bodies will not be affected when they are dipped in water.
iv. The first artificial satellite : Sputnik :: The first Indian satellite : Aryabhatta.
v. Halogens

Question 2.
(A) Give scientific reasons. (Attempt any 2) [4]
i. Why is it beneficial to use satellite launch vehicles made of more than one stage?
ii. Carbon forms a large number of compounds.
iii. Why can one sense colours only in bright light?
(B) Answer the following. (Attempt any 3) [6]
i. Write a note on space expeditions.
ii. Light travels with a velocity 2.8 × 10⁸ m/s in a medium. On entering second medium its velocity becomes 1.4 × 10⁸ m/s. What is the refractive index of the second medium with respect to the first medium?
iii. Distinguish between metals and nonmetals (on the basis of their chemical properties).
iv. State the characteristics of a homologous series.
v. Two tungsten bulbs of power 50 W and 60 W work on 220 V potential difference. If they are connected in parallel, how much current will flow in the main conductor?
Answer:
(A)
i. a. Fuel in the launch vehicle forms the major part of the total weight of the launch vehicle.
b. Thus, the vehicle has to carry large amount of weight in the form of fuel with it.
c. For the launch vehicles with more than one stage, the fuel in the first stage is used to give the initial thrust to the vehicle. After this fuel is exhausted, the empty fuel tank along with the engine is detached from the main body of the vehicle. This reduces weight of the vehicle step by step.
Hence, it is beneficial to use satellite launch vehicles made of more than one stage.

ii. a. Carbon is a tetravalent atom with the property of catenation. Thus, it forms compounds in which the carbon atoms are arranged in the form of straight chains, branched chains or rings.
b. It can form single and multiple covalent bonds with other carbon atoms.
c. Carbon is tetravalent. Thus, it can form four covalent bonds with carbon or other elements like oxygen, nitrogen, sulphur, halogens and phosphorus.
d. One more characteristics of carbon, which is responsible for large number of carbon compounds is isomerism.
Hence, carbon forms a large number of compounds.

iii. a. The retina in our eyes is made of different cells viz., rod-shaped and cone-shaped.
b. The rod shaped cells respond to intensity of light and send the information to brain. While cone shaped cells send information of different colours of light to the brain.
c. Rod like cells are sensitive to faint light, whereas, the conical cells do not respond to faint light.
Hence, one can sense colours only in bright light.

(B)
i. a. Study missions planned for establishing artificial satellites in the earth’s orbit, using it for research or for the benefit of life, or sending spacecraft to the various components of the solar system or outside are called as space expeditions.
b. Need and importance of these space expeditions is as follows:
1. Due to space expeditions accessing internet is very easy. As a result, world has come closer. It has become possible to contact a person in any part of the world or collect any information within seconds.
2. Use of satellites can give us advance alerts about various natural calamities, so that, we can take proper precautions and reduce the damage.
3. During war, the aerial surveillance can be used to get information about the actions of enemies.
4. Due to space expeditions, it is also possible to explore the fossil reserves and minerals in the earth.

ii. Solution:
Given: Velocity of light in first medium (v₁) = 2.8 × 10⁸ m/s,
Velocity of light in second medium (v₂) = 1.4 × 10⁸ m/s
To find: Refractive index of second medium with respect to first medium (¹n₂)

Ans: Refractive index of second medium with respect to first medium is 2.

iii.

Metals	Nonmetals
a. Generally, metals form basic or amphoteric oxides.	Generally, nonmetals form acidic or neutral oxides.
b. Metals such as Na, K, Ca, Al, Fe, etc. displace hydrogen from water or steam.	Nonmetals generally do not react with water.
c. Metals which are highly reactive and displace hydrogen from dilute acids.	Nonmetals generally do not displace hydrogen from dilute acids.
d. Metals have strong tendency to lose electrons and hence, they behave as reductants.	Nonmetals have strong tendency to accept electrons and hence, they behave as oxidants.

iv. Characteristics of homologous series:
a. The general molecular formula of all the compounds in the series is the same.
b. All the members have the same functional group.
c. Consecutive members of the series differ from one another by -CH₂– unit (methylene group), and their molecular mass differs by 14 u.
d. Physical properties like melting point, boiling point, density, etc., generally show a gradual change with increase in molecular mass in the series.
e. Chemical properties of the members of the series show similarity because of the presence of the same functional group in them.
[Any four points]

v. Solution:
Given : Potential difference (V) = 220 V, Power of bulb 1 (P₁) = 50 W,
Power of bulb 2 (P₂) = 60 W,
To find : Total current (I)
Formula:
a. P = VI
b. I = I₁ + I₂
Calculation : Since both the bulbs are connected in parallel, the potential difference will be same.

Ans: The total current flowing in the main conductor is 0.5 A.

Question 3.
Answer the following. (Attempt any 5) [15]
i. If 80 g steam of temperature 97 °C is released on an ice slab of temperature 0 °C, how much ice will melt? How much energy will be transferred to the ice when the steam will be transformed to water?
(Given: Latent heat of melting the ice = L_meit = 80 cal/g, latent heat of vaporisation of water = L_vap = 540 cal/g)
ii. Explain the term with example: Balanced equation
iii. Complete the diagram and explain:

iv. Label the following diagram and explain the process.

v. Complete the following table:

Reaction	Reductant	Oxidant
CuO + H2 → Cu + H2O
Mg + O2 → 2MgO
Fe + S → FeS

vi. Explain with neat ray diagram, the formation of image in a concave lens.
vii. Select the appropriate options and complete the following paragraph.
(ethyne, ethene, ethane, propane, saturated, unsaturated, oxygen, hydrogen)
Carbon compounds contain many elements. The compounds which contain carbon and _________ as the only two elements are called hydrocarbons. Hydrocarbons are the simplest and the fundamental organic compounds. The smallest hydrocarbon is methane formed by combination of one carbon atom and four hydrogen atoms. _________ having molecular formula C₂H₆, is an example of _________ hydrocarbon. _________ (C₂H₄) and _________ (C₂H₂) are _________ hydrocarbons.
viii. The figure shows the elliptical orbit of a planet about the Sun S. An ellipse is the curve obtained when a cone is cut by an inclined plane. It has two focal points. The sum of the distances to the two focal points from every point on the curve is constant. F₁ and F₂ are the two focal points of the ellipse. The shaded area CF₁D is twice the shaded area AF₁B. t₁ is the time taken by the planet to move from C to D and t₂ is the time to move from A to B.

a. Which laws do we understand from the above diagram and description?
b. State the law regarding areas swept by the line joining the planet and the Sun.
c. Out of the following points P, Q, R, B; at which point will the velocity of the plant be maximum?
Answer:
i. Solution:
Given:
Mass of steam (m_s) = 80 g, Change in temperature (∆T) = 97 – 0 = 97 °C
We know that: Latent heat of melting of ice = L_melt = 80 cal/g
Latent heat of vaporisation of water = L_vap = 540 cal/g
Specific heat of water C_w = 1 cal /g °C
To find : a. Energy transferred (Q)
b. Mass of ice that melts (m_i)
Formulae : a. Heat released during conversion of steam into water at 97 °C (Q₁) = m_s × L_vap
b. Heat released during decrease of temperature of water form 97 °C to 0 °C (Q₂) = m_s × C_w × ∆T
c. Heat gained by ice (Q) = m_i × L_melt
Calculation: From formula (a),
Q₁ = 80 × 540 cal
From formula (b),
Q₂ = 80 × 1 × (97 – 0) = 80 × 97 cal
According to principle of heat exchange, Total heat gained by ice
Q = Q₁ + Q₂
= 80 × 540 + 80 × 97
= 80 × (540 + 97)
= 80 × 637
= 50960 cal
This energy would cause m_i mass of ice to melt,
From formula (c),
m_i × L_melt = 50960
m_i = \(\frac{50960}{80}\) = \(\frac{80\times637}{80}\) = 637 g
Ans: Energy transferred to ice is 50960 cal and it will melt 637 g of ice.

ii. a. A chemical equation in which the number of atoms of the elements in the reactants is same as the number of atoms of those elements in the products is called a balanced chemical equation.
b. It is written by using the symbols and the formulae of the reactants and products. The reactants are written on the left hand side of the arrow and products are written on the right hand side of arrow.
c. A balanced chemical equation is in accordance the law of conservation of mass.
d. Unbalanced equation is balanced by applying proper factors to appropriate reactants or products.
E.g. NaOH + H₂SO₄ → Na₂SO₄ + H₂O
The above reaction is an unbalanced equation.
The above equation can be balanced by applying a factor of ‘2’ to NaOH and H₂O.
Thus, the equation is
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

iii. Diagram:

Explanation:
a. Light ray is incident from air at point X and enters water.
b. Water being optically denser than air, light ray bends towards normal (N₁) such that ∠i > ∠r₁.
c. Refracted ray travels along straight line path in water and reaches point Y on the plane mirror at the bottom.
d. Mirror reflects the light according to laws of reflection. Hence, ∠r₁ = ∠r₂.
e. Reflected ray travels along straight line path in water and reaches point Z at surface of water.
f. As it enters air, it bends away from normal (N₂) such that ∠r₂ < ∠e.

iv. Labelled diagram:

Explanation:
a. The froth floatation process is based on two opposite properties of the particles, hydrophilic and hydrophobic.
b. The metal sulphide particles are hydrophobic. Due to this property, they get wetted mainly with oil. On the other hand, gangue particles are hydrophilic and get wetted by water.
c. In this method, the finely divided metal ore is added in a big tank containing large amount of water. To this, certain vegetable oil (pine oil, eucalyptus oil, etc.) is added for the formation of froth.
d. In the tank, compressed air is circulated through the water. Bubbles are formed due to this blown air. A rotating agitator at the centre of the floatation tank agitates the mixture and draws air into it to form bubbles. As a result, foam is formed which carries metal sulphide particles. The foam rises to the surface of water and floats. Therefore, this method is called froth floatation process.
e. This method is used to separate metal sulphide ores such as zinc blende (ZnS) and copper pyrite (CuFeS2).

v.

Reaction	Reductant	Oxidant
CuO + H2 → Cu + H2O	H2	CuO
Mg + O2 → 2MgO	Mg	O2
Fe + S → FeS	Fe	S

SMART TIP
Reductant gains O or any other electronegative atom. Oxidant losses O or gains any electropositive atoms.

vi. A concave lens always forms a virtual, erect and diminished image of a given object.

a. Consider an object AB placed between F₁ and 2F₁ in front of a concave lens.
b. The incident ray AP is parallel to the principal axis. After refraction it travels along path PQ. When PQ is extended backwards, it passes through focus F₁.
c. Now, the incident ray AO passes undeviated through optical centre O. This ray meets the backward extension of ray PQ at A’ i.e., A’ is the image of A.
d. As point B lies on principal axis, its image B’ will lie on principal axis beneath the point A’.
In this way, A’B’, the virtual, erect and diminished image of object AB is formed by a concave lens.

vii. Carbon compounds contain many elements. The compounds which contain carbon and hydrogen as the only two elements are called hydrocarbons. Hydrocarbons are the simplest and the fundamental organic compounds. The smallest hydrocarbon is methane formed by combination of one carbon atom and four hydrogen atoms. Ethane having molecular formula C₂H₆, is an example of saturated hydrocarbon. Ethene (C₂H₄) and ethyne (C₂H₂) are unsaturated hydrocarbons.

viii. a. From the given diagram, we understand Kepler’s laws of planetary motion.
b. Kepler’s second law: The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
c. At point P, the velocity of the planet will be maximum.

Question 4.
Answer the following. (Attempt any 1) [5]
i. The modem periodic table consists of seven horizontal rows called the periods and eighteen vertical coluihns called the groups. In addition to these seven rows, lanthanide and actinide series are shown separately at the bottom of the periodic table. In the following modem periodic table, some elements are indicated by alphabets: A, B, C, D, E, F, G, H and I. Based on the positions of these elements and the periodic trends, answer the below given questions.
a. Identify metalloids.
b. Which elements are transition elements?
c. Among elements ‘D’ and ‘C’ which element is more electronegative? Why?
d. If element ‘A’ and element ‘I’ react, what will be the molecular formula of the product formed?
e. Among elements ‘A’, ‘D’ and ‘I’, which element has the highest nonmetallic character?
ii. Tara, after completing her exams, went to her grandmother’s house along with her younger brother Rahul. One day, Rahul came home after playing and switched on the fan as well as air conditioner. At the same time, his grandmother was preparing a milkshake for him using a mixer. Suddenly, there was a big sound and the electricity of the house got shut down. Tara then called an electrician. He said the main reason of supply to shut down was overloading. Both Tara and Rahul asked few questions to the electrician as given below. What answers would the electrician had given to these questions?
a. What is overloading?
b. How can the effects of overloading be avoided?
c. Is overloading and short circuiting same? When does short circuiting take place?
d. What happens to the resistance of the circuit during a short circuit?
e. What happens to the flow of electric current during a short circuit?
Answer:
a. Elements ‘H’ and ‘C are metalloids.
b. Elements ‘G’ and ‘B’ belong to d-block.
c. Element ‘D’ is more electronegative than element ‘C’ as the electronegativity decreases while moving down a group.
d. The molecular formula of the product formed is AI₂.
e. Element ‘I’ has the highest nonmetallic character among ‘A’, ‘D’ and ‘I’.

ii. a. The flow of a large amount of current in the circuit beyond the permissible value (of current) is called overloading.
b. The effects of overloading can be avoided by not connecting many appliances (especially of high power rating) at a time in the circuit.
c. No, overloading and short circuiting are not same. Short circuiting takes place when a live wire and a neutral wire come in direct contact with each other.
d. During a short circuit, the resistance of the circuit becomes very small.
e. A huge amount of current flows through the circuit during a short circuit.

SSC Maharashtra Board Science 1 Question Paper with Solutions