SSC Maharashtra Board Science 1 Question Paper 2019 with Solutions

Maharashtra Board SSC Class 10 Science 1 Question Paper 2019 with Solutions Answers Pdf Download.

SSC Science 1 Question Paper 2019 with Solutions Pdf Download Maharashtra Board

Time : 2 Hours
Max. Marks : 40

Note:

1. All questions are compulsory.
2. Draw scientifically, technically correct labelled diagram wherever necessary.
3. Start writing each main question on new page.
4. Figure to the right indicate full marks.
5. For each MCQs (i.e. Q. No. 1-B) evaluation would be done for first attempt only.
6. For each MCQ correct answer must be written along with its alphabet.
   Eg :
   1. (a) ……..,
   2. (b) ………,
   3. (d) …….. .

1. (A) Answer the following questions :

Question 1.
Write proper answer in the box :

Answer:

Question 2.
In Dobereiner’s triads Li, Na, K, the atomic masses of lithium and potassium are 6.9 and 39.1 respectively, then what will be the atomic mass of sodium ?
Answer:
Average atomic mass of sodium = Atomic mass of \(\frac{\mathrm{Li} \text { and } \mathrm{K}}{2}\)
= \(\frac{6.9+39.1}{2}\) = \(\frac{46}{2}\) = 23

Question 3.
State whether the given statement is true or false :
A concave lens is a converging lens.
Answer:
False. Concave lens is a diversing lens

Question 4.
By considering first correlation complete the second correlation :
Hubble telescope : 569 km high from earth surface Revolving orbit of Hubble telescope :
Answer:
Revolving orbit of hubble telescope = 2π (Re + h)
= 2π (6400 + 569) km
≈ 43,805 km
Revolving orbit of hubble telescope : 43,805 km

Question 5.
Find the odd man out :
Tinning, Anodization, Alloying, Froth floatation.
Answer:
Froath Floatation.

(B) Choose the correct alternative : [5]

Question 1.
The reaction of iron nail with copper sulphate solution is ……… reaction.
(a) combination
(b) decomposition
(c) displacement
(d) double displacement
Answer:
(c) Displacement reaction.

Question 2.
Observe the following diagram and choose the correct alternative :

(a) The intensity of magnetic field in A is larger than in B.
(b) The intensity of magnetic field in B is less than in A.
(c) The intensity of magnetic field in A and B is same.
(d) The intensity of magnetic field in A is less than in B.
Answer:
(d) The intensity of magnetic field in A is less than in B.

Question 3.
A ray of light makes an angle of 50° with the surface S₁ of the glass slab. Its angle of incidence will be ……….
(a) 50°
(b) 40°
(c) 140°
(d) 0°
Answer:
(b) 40°.

Question 4.
Water expands on reducing its temperature below. ………°C.
(a) 0
(b) 4
(c) 8
(d) 12
Answer:
(b) 4°C.

Question 5.
The carbon compound is used in daily life is ……..
(a) Edible oil
(b) Salt
(c) Carbon dioxide
(d) Baking soda
Answer:
(a) Edible oil.

2. Attempt any five of the following questions : [10]

Question 1.
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:
Given, P₁ = 50 W, P₂ = 60 W and V = 220 V
To find I = ? ‘
Formulae : P = VI
We know : Total Power (P) = P₁ + P₂
= 50 + 60 = 110 W
We know, P = VI
⇒ I = \(\frac{P}{V}\) = \(\frac{110}{220}\) = \(\frac{1}{2}\)
The current of 0.5 A will flow in the main conductor.

Question 2.
Give scientific reason :
In the electric equipment producing heat e.g., iron, electric heater, boiler, toaster etc., and alloy such as nichrome is used, not pure metals.
Answer:
The appliances like iron, electric heater works on the principle of heating effect of electric current. At high temperature, pure metal can get oxidised and also pure metals have low resistivity.

An alloy such as Nichrome has high resistivity and can be heated to a high temperature without undergoing oxidation. Thus, nichrome is preferred over other pure metals and are used for making coil for devices working on heating effect of electric current.

Question 3.
A metal ball of mass 5 kg falls from a height of 490 m. How much time it will take to reach the ground ? (g = 9.8 m/s²)
Answer:
Given, Mass (M) = 5 kg.
Distance travelled by the ball (s) = 490 m
Initial velocity of the ball (u) = 0
Gravitational acceleration (g) = 9.8 m/s²
Newton’s second equation of motion gives
s = ut + \(\frac{1}{2}\)gt²
∴ 490 = 0 × t + \(\frac{1}{2}\) × 9.8 × t² [∵ Initial velocity (u) = 0]
⇒ 490 = 4.9 t²
⇒ t² = \(\frac{490}{4.9}\)
⇒ t² = 100
∴ t = 10 sec.
The ball takes 10 seconds to reach the ground.

Question 4.
Write names of first four homologous series of alcohols :

Answer:

Question 5.
Observe the following figure and complete the table :

Points	Answers
(i) Position of the object (ii) Position of the image (iii) Size of the image (iv) Nature of the image

Answer:

Points	Answers
(i) Position of the object	Between Fi and O
(ii) Position of the image	On the same side of the lens as that of object and between 2F1 and F1.
(iii) Size of the image	Very large
(iv) Nature of the image	Virtual and erect

Question 6.
Out of sodium and sulphur which is a metal ? Explain its reaction with the oxygen.
Answer:
Out of sodium and sulphur; sodium is a metal. So, sodium reacts with oxygen to form sodium oxide.

Question 7.
A tapping vessel opens in a tank like container that is tapering on the lower side. The tank has an outlet for water on the upper side and a water inlet on the lower side. Finely ground ore is released in the tank. A forceful jet of water is introduced in the tank from lower side and gangue particles and pure ore are separated by this method.
(i) The above description is of which gravitation separation method ?
(ii) Draw labelled diagram of this method.
Answer:
(i) The description given in question is of hydraulic separation method.
(ii) Digram of hydraulic separation method :

3. Attempt any five of the following questions : [15]

Question 1.
What would be the value of ‘g’ on the surface of the earth if its mass was twice and its radius half of what it is now ?
Answer:
Given, Mass of the earth (M) = 2M
Radius of the earth (R’) = \(\frac{R}{2}\)
Gravitational acceleration = g’
Formula : g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
Thus, g’ = \(\frac{G \times M^{\prime}}{\left(R^{\prime}\right)^2}\) = \(\frac{\mathrm{G} \times 2 \mathrm{M}}{\left(\frac{\mathrm{R}}{2}\right)^2}\)
⇒ g’ = \(\frac{G \times 2 M \times 4}{R^2}\)
∴ g’ = 8g = 8 × 9.8
∴ g = 78.4 m/s²
The value of ‘g’ on the earth surface if its mass was twice and its radius was half of original value is, 78.4 m/s² i.e., 8 times the original value of g.

Question 2.
Write merits of Mendeleev’s periodic table.
Answer:
Merits of Mendeleev’s periodic table are as follows :

(i) Atomic masses of some elements were revised to give them proper place in the periodic table in accordance with their properties. For example, the previously determined atomic masses of beryllium 14.0 g was changed to the correct value 9.4 and beryllium was placed before boron.

(ii) Mendeleev kept vacant places in the periodic table for elements not discovered till then. Three of these unknown elements were given the names eka-boron, eka- aluminium and eka-silicon from the known neighbours and their atomic masses were indicated as 44, 68 and 72 respectively. Their properties were also predicted.

Later on, these elements were discovered and named as scandium (Sc), gallium (Ga) and germanium (Ge) respectively. The properties of these elements matched well with those predicted by Mendeleev.

(iii) There was no place reserved for noble gases in Mendeleev’s original periodic table. However when noble gases such as helium, neon and argon were discovered towards the end of nineteenth century, Mendeleev created the ‘zero’ group without disturbing the original periodic table in which the noble gases were fitted very well.

Queston 3.
Study the following chemical reaction and answer the questions given below :

(i) Identify and write the type of chemical reaction.
(ii) Write the definition of the above type of chemical reaction.
(iii) Write the names of reactants and products of the above reaction.
Answer:
(i) Double displacement reaction.
(ii) The reaction in which the ions in the reactants are exchanged to form a precipitate are called double displacement reactions.
(iii)

Question 4.
Explain the following temperature Vs. time graph :

Answer:

1. The graph shown is a temperature-time graph.
2. AB shows that at constant temperature of 0°C ice changes to water.
3. The temperature at which ice changes to water is called melting point of ice i.e. 0°C.
4. Now, water at 0°C starts heating and it’s temperature gradually increases to 100°C.
5. BC shows increase in temperature from 0°C to 100°C without changes in state.
6. CD shows that, at constant temperature of 100°C water changes to vapour.
7. The temperature at which water changes in vapour is called boiling point of water i.e., 100°C.

Question 5.
Surabhi from std. X uses spectacle. The power of the lenses in her spectacle is 0.5 D. Answer the following questions from the given information :
(i) Identify the type of lenses used in her spectacle.
(ii) Identify the defect of vision Surabhi is suffering from.
(iii) Find the focal length of the lenses used in her spectacle.
Answer:
Given, power of lenses in spectacle used by Surabhi is 0.5 D.

(i) Convex lenses are used in Surabhi’s spectacles.
(ii) Far sightedness or hypermetropia.
(iii) We know, Power (P) = \(\frac{100}{f(\mathrm{~cm})}\) = \(\frac{100}{0.5}\) = 200 cm

Question 6.
Complete the following table :

Answer:

Common Name	Structural Formula	IUPAC Name
1. Ethylene	cH2 = cH2	Ethene
2. Acetic acid	cH3-cooH	Ethanoic acid
3. Methyl alcohol	CH3-OH	Methanol

Question 7.
What is meant by space debris ? Why there is need to manage the debris ?
Answer:
Space debris : In addition to the artificial satellite, some other objects are also revolving around the earth like non-functional satellites, parts of the launcher detached during launching in earth’s orbit are called space debris. This debris can be harmful to the artificial satellites. It can collide with these artificial satellites or space crafts and damage them. Thus, space debris should be managed.

4. Answer any one of the following questions : [5]

Question 1.
Taking into consideration the period of the elements given below, answer the following questions :
(i) Arrange the above elements in a decreasing order of their atomic radii.
(ii) State the period to which the above elements belong.
(iii) Why this arrangement of elements is similar to the above period of modem periodic table ?
(iv) Which of the above elements have the biggest and the smallest atom ?
(v) What is the periodic trend observed in the variation of atomic radius while going from left to right within a period ?
Answer:
(i)

(ii) Elements O, B, C, N, Be, Li belong to 2nd period.
(iii) Yes, this arrangement of elements is similar to the above period of modern periodic table. As we move from left to right within a period the atomic number increases one by one, meaning the positive charge on the nucleus increases by one unit at a time, but the electrons are added to the same orbit thereby increasing the pull towards the nucleus which decreases the size of the atom.
(iv) Li has the biggest atom and O has the smallest atom.
(v) Atomic radius decreases from left to right within a period due to increase in the effective nuclear charge.

Question 2.
The observation made by Swarali while doing the experiment are given below. Based on these write answers to the questions :
Swarali found that the light ray travelling from the denser medium to rarer medium goes away from the normal. If the angle of incidence (i) is raised by Swarali, the angle of refraction (r) went on increasing. However, after certain value of the angle of incidence the light ray is seen to return back into the denser medium.
(i) What is the specific value of ∠i called ?
(ii) What is the process of reflection of incident ray into denser medium called ?
(iii) Draw the diagrams of three observations made by Swarali.
Answer:
(i) Critical Angle
(ii) Total internal reflection
(iii)

Observation 1 : When i₁ < i_C, then refraction takes place and refracted ray moves away from normal i.e., r₁ > i₁.
Observation 2 : When i₁ = i_C then refracted ray becomes perpendicular to normal i.e., r₂ = 90°.
Observation 3 : When i₃ > i_C then total internal reflection takes place i₃ = r₃, where, i_C is the critical angle.

SSC Maharashtra Board Science 1 Question Paper with Solutions