Lakhmir Singh & Manjit Kaur 2020 Solutions for Class 10 Science Chapter 7 - Model Test Paper 1

Answer:

In general use, appliances are connected in parallel to each other. In that case, the resistance is indirectly proportional to the power of the appliance. Hence, the lamp with smaller power i.e. 40 W will have higher resistance. 

Answer:

The largest cell present in human body is the female egg/ovum.

Answer:

An electromagnet is a rod of magnetic material (say soft iron) placed inside a solenoid coil. On passing current in the solenoid coil, the iron rod begins to behave as a magnet. The strength of magnetization can be increased by increasing the amount of current flow or by increasing the number of turns in the coil. 


 

Answer:

The water of crystallization is the number of water molecules that combine chemically in definite molecular proportion with the concerned salt in the crystalline state. This water is responsible for the geometric shape and colour of the crystal. For eg: ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$

Copper sulphate crystals are blue in colour. However, when copper sulphate is heated in a china dish for some time, the blue colour disappears and it becomes white-grey in colour. This white grey substance is the anhydrous copper sulphate. The reaction is written as follows:

${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}\stackrel{∆}{\to }{\mathrm{CuSO}}_4+5{\mathrm{H}}_2\mathrm{O}$
 

Answer:

(a) The tissue which transports soluble products of photosynthesis in a plant is phloem.
(b)The tissue which transports water and minerals in a plant is xylem.

Answer:

The chemical formula for plaster of paris is calcium sulphate hemihydrate $({\mathrm{CaSO}}_4.\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\mathrm{H}}_2\mathrm{O})$.
Plaster of Paris is prepared by heating of gypsum at 373 K.

$2{\mathrm{CaSO}}_4.2{\mathrm{H}}_2\mathrm{O}\stackrel{373 \mathrm{K}}{\to }\underset{\mathrm{Plaster} \mathrm{of} \mathrm{paris}}{2{\mathrm{CaSO}}_4.\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\mathrm{H}}_2\mathrm{O}}+3{\mathrm{H}}_2\mathrm{O}$

When we mix plaster of Paris with the appropriate amount of water, it will give gypsum chemically known as calcium sulphate dihydrate.

$2{\mathrm{CaSO}}_4.\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\mathrm{H}}_2\mathrm{O}+3{\mathrm{H}}_2\mathrm{O}\to \underset{\mathrm{Gypsum}}{2{\mathrm{CaSO}}_4.2{\mathrm{H}}_2\mathrm{O}}$


 

Answer:

(a) Carbon forms a large number of compounds because of the following reasons:
(i) Catenation property, i.e. self linking property of carbon to form long chains.
(ii) Small size, helps in forming double and triple bonds.
(iii) The tendency to form rings.
(iv) High valency, allows forming bonds with other atoms.

(b) Blackening of vessels is due to deposition of waste material in the holes of the burner. These depositions prevent complete combustion of fuel and create black soot. Thus, the holes of the burner need to be cleaned and unclogged to give the undisturbed flow of gas.

(c) Synthetic detergents are made up of chemical compounds like fatty acid salts (aliphatic) or aromatics like linear alkyl benzene. These are non-biodegradable compounds which cannot be degraded and result in polluting our environment.

 

Answer:

Rusting of iron: When iron is exposed to air and moisture for a very long time, it gets covered with a brown flaky substance. This brown flaky substance is called rust and the process is called rusting.

$4\mathrm{Fe}+3{\mathrm{O}}_2+{\mathrm{xH}}_2\mathrm{O}\to 2{\mathrm{Fe}}_2{\mathrm{O}}_3.{\mathrm{xH}}_2\mathrm{O}$

Activity -

• Take three boiling tubes and label them A, B, and C. Place three clean iron nails (free from rust) in each of the tube.
• Add water to tube A in such a manner that some parts of the nails are exposed to air.
• Add water to test tube B, in such a way that the nails are completely submerged in water. Then, pour some amount of oil in the tube.
• Add anhydrous calcium chloride to test tube C and then seal it. Anhydrous calcium chloride absorbs all the moisture present inside the tube so that the nails are only exposed to oxygen and no moisture.
• Seal all the tubes with the help of rubber corks. Leave the tubes undisturbed for a few weeks and observe the changes. 

Conditions for rusting of iron 

Itwill be observed after a few weeks that the nails in tube A have rusted, while those in test tubes B and C have not. 

Explanation:

In tube A, both moisture and oxygen were available and the nails started rusting after two weeks. 

In tube B, iron nails were only in contact with water, and oxygen was not allowed to dissolve in water because of the layer of oil. As nails did not rust even after two weeks when only water was available, it shows that rusting cannot take place if only water is available.

In tube C, only oxygen is present as anhydrous calcium chloride absorbed all the moisture present inside the boiling tube. As the nails did not rust even after two weeks when only oxygen was available, it shows that rusting cannot take place if only oxygen is available.

This shows that the presence of both oxygen and water is essential for rusting to take place.

Answer:

(a) Ferrous sulphate is green in colour. On heating, the green crystals are converted into a dirty-yellow anhydrous solid as it loses water of crystallization.

(b) When ferrous sulphate is strongly heated anhydrous material loses sulphur dioxide and sulphur trioxide leaving a reddish-brown iron oxide.
$2{\mathrm{FeSO}}_4\to {\mathrm{Fe}}_2{\mathrm{O}}_3+{\mathrm{SO}}_2+{\mathrm{SO}}_3$
 

Answer:

(a) The resistance of a conductor depends on the following factors:

(i) Shape of the conductor
(ii) Size of the conductor
(iii) Material of the conductor
(iv) Temperature of the conductor

(b) Initial resistance = 20 Ω

When the length of the resistance is increased, it's volume remains constant. 

According to relation,

$$\begin{gathered} R\propto \frac{l}{A}\propto \frac{l^2}{V} \\ \because V = \mathrm{Constant} \\ \therefore R\mathit{ }\mathit{\propto }\mathit{ }{l}^{\mathit{2}} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{R_{\mathrm{i}}}{R_{\mathrm{f}}}=\frac{l_{\mathrm{i}}^2}{l_{\mathrm{f}}^2} \\ \Rightarrow \frac{20 \mathrm{\Omega }}{R_{\mathrm{f}}}=\frac{l_{\mathrm{i}}^2}{(2l_{\mathrm{i}}{)}^2}=\frac{1}{4} \\ \Rightarrow R_{\mathrm{f}}=80 \mathrm{\Omega } \end{gathered}$$

Answer:

(a) When a charged particle enters at right angles into a uniform magnetic field, it feels a force, $F=q(v \times B)$, also known as Lorentz force. Here, q is the charge of the particle, v is it's velocity vector and B is the magnetic field strength vector in that region.

Clearly, the direction of the force is perpendicular to both, the velocity vector and the magnetic field strength vector. In the given case, the cross product of velocity and magnetic field points vertically out of the page. The direction of charge is also given the same. Hence, the charged particle is positive in nature.

(b) The object should be placed between the focus and radius of curvature of the concave mirror. Then it will form an inverted and enlarged image. In that case, a magnification of '-3' can be obtained.

(c) Refractive index of a medium is the measure of the speed of light in that, medium with respect to the vacuum. It is defined as the ratio of the speed of the light in a vacuum to the speed of the light in the given medium. So, "The refractive index of carbon disulphide is 1.63" means that the speed of light in carbon disulphide is $\frac{1}{1.63}$ times the speed of the light in vacuum.

Answer:

(a) Nearsightedness (myopia) is a common vision defect, in which a person can see only nearby objects clearly, but distant objects appear to be blurry. This defect arises due to the decresed focal length of the eye lens or due to elongation of the eyeball.

(b) An eye having Myopia:

 

(c) Correction of myopia using a concave lens.

 
 

Answer:

(a) Biotic and abiotic components are the two main components of our environment.
The living parts of the ecosystem are referred to as biotic components. For example: plants, animals humans etc.
The non-living parts of the ecosystem are referred to as abiotic components. For example: land, air etc.

(b) Xylem is the water-conducting tissue of plants. If the xylem of a plant is removed, then the conduction of water will not take place. As a result of which the plant will die.

(c) Neuron and muscular tissues are involved in control and coordination in multicellular organisms.

Answer:

Collecting rainwater for future use by storing it in storage reservoirs is called rainwater harvesting. In rainwater harvesting, the rainwater that falls on the roof and other parts of a building is channeled through pipes to an underground reservoir. These collecting pipes run along the sides of the roof and join together at a point to form the main pipe. This pipe flows into the tank present in the ground. The collected water may be allowed to seep inside the ground to recharge ground water or can be stored in tanks for future use.

Rainwater harvesting is one of the methods to deal with water shortage and meet the demands of water for the growing population. It also ensures water conservation and all round availability of water for various purposes.

Answer:

Natural resources are the resources present in nature, which can be used for our various needs. The examples of such resources are forests, wildlife, water, coal, petroleum, etc.

Two factors that work against the equitable distribution of these resources are:

(i) human greed

(ii) corruption and the lobby of the rich and the powerful

Answer:

(a) The potential difference across the ends of a conductor is maintained by connecting a battery or a dry cell at the ends of the conductor.

(b) When 1 Joule of work is done to move a 1 Coulomb charge from one point to another, then the potential difference between the two points is said to be 1 Volt (V).

(c) According to Ohm's Law, the current flowing in a circuit is directly proportional to the applied potential difference. 

Mathematical expression for Ohm's Law: $V \propto I$ or V = IR (Where 'R' is the constant of proportionality and is known as the resistance of the circuit) 

(d) Unit of resistance is Ohm, denoted by the symbol $\mathrm{\Omega }$.

(e) According to Ohm's law, V = IR

$\Rightarrow R=\frac{V}{I}=\frac{110 \mathrm{V}}{5 \mathrm{A}}=22 \mathrm{\Omega }$

Hence, the resistance of the heater is $22 \mathrm{\Omega }$.

Now, if the potential difference is changed to 220 V, then the current in the circuit = $I=\frac{V}{R}=\frac{220 \mathrm{V}}{22 \mathrm{\Omega }}=10 \mathrm{A}$
 

Answer:

(a) Electric lamps should be connected in parallel so that every lamp gets its separate current. In that condition, switching on and off in a room will not affect the other lamps in the building. Advantage of this type of connection is, if any appliances fail in the circuit or stop working, then it won't affect the other appliances, as in parallel connected every appliance get their separate current from the main circuit. 

(b) Heating effect of current is utilized in the working of an electric fuse. A fuse is made such that it can allow only a particular amount of current to flow in the circuit. So, when a current more than the rating of the fuse starts flowing in the circuit, the wire of the fuse melts and it breaks the circuit. Therefore, no current flows further in the circuit, and no other appliances get any damage.

(c) (i)The usual colour of the earth wire is either green or yellow.
(ii) While the usual colour for the live wire is brown.

(d) An electric short-circuit occurs when the live wire comes in contact with the neutral in any circuit. A short-circuit can result in a flow of sudden high amount of current in the circuit, which may damage low rating appliances.

(e) The main purpose of earthing is to provide an immediate path to excess flow of current due to a fault in the circuit. This is done to prevent us from getting a shock. For this, the earthing wire is connected directly to the earth, so that whenever a high current flows in the circuit, it can be discharged directly to the ground. 

Answer:

(a) Electronic configuration of magnesium is 2, 8, 2 and of chlorine is 2, 8, 7.

(b) Electron dot structure of magnesium and chlorine.


(c) Electronic configuration of magnesium is 2, 8, 2, so valence shell contains two electrons. To attain noble gas configuration magnesium will lose two electrons.
Electronic configuration of chlorine is 2, 8, 7, so valence shell contains seven electrons. To attain noble gas configuration chlorine will gain one electron.
So, the valency of magnesium is two while valency of chlorine is one. 

                                                      

(d) Ionic bond is formed in Magnesium chloride.

(e) Ionic compounds have a high melting point because of strong inter electronic attraction forces between them.

Answer:

(i) The letter "b" represents halogen as "b" belongs to 17 group where valence electrons are 7.

(ii) The letter "e" represents noble gas as "e" belongs to 18 group where valence electrons are 8 and their octet is full.

(iii) Electronic configuration of a is 2, 8, 4
Electronic configuration of "b" is 2, 8, 7.
Elements "a" and "b" will form a bond by sharing of the electron, so a covalent bond will be form between them.

(iv) Electronic configuration of a is 2, 8, 2.
Electronic configuration of "b" is 2, 8, 7.
Elements "c" and "b" will form a bond by transfer of electrons and thus will form an ionic bond.

(v) The letter "d" will form a divalent anion. 
Electronic configuration of d will be 2,8,6. It has 6 valence electrons in its outermost shell, so it will gain 2 electrons and become a divalent anion. 

 

Answer:

(a) Biodegradable wastes - Paper, vegetable wastes
Non-biodegradable wastes - Plastic, glass objects

(b) Biotic components - Plants and animals
Abiotic components - Land and air

(c) Decomposers include microorganisms such as bacteria and fungi that obtain nutrients by breaking down the remains of dead plants and animals.

Role of decomposers -

• They help in the breakdown of organic matter or biomass of dead plants and animals into simple inorganic raw materials such as CO₂, H₂O and nutrients.
• They help in the natural replenishment of soil.
• They help in keeping the environment clean.

Answer:

The human female reproductive system consists of ovary, oviduct, uterus and vagina.
​

Functioning of human female reproductive system is as follows -

Answer:

An organic compound A of molecular formula C₂H₆O is a constituent of wine and beer, so it must be an alcohol containing molecule, i.e. ethanol (CH₃CH₂OH).

When ethanol is treated with alkaline potassium permanganate, it oxidizes ethanol and forms compound "B" which is ethanoic acid.

${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+{\mathrm{KMnO}}_4\stackrel{\mathrm{alkaline}}{\to }{\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{H}}_2\mathrm{O}$

Due to the formation of ethanoic acid, blue litmus turns red. 

Answer:

(a) Compound "A" must contain a carbonate as it gives brisk effervescence with hydrochloric acid. So, compound "A" must be sodium bicarbonate (${\mathrm{Na}}_2{\mathrm{CO}}_3$).

(b) When dilute hydrochloric acid is added to a sodium compound A, then brisk effervescence of gas is produced which means carbon dioxide gas is released during the reaction. Carbon dioxide on passing through lime water, turns lime water milky.

${\mathrm{Na}}_2{\mathrm{CO}}_3+2\mathrm{HCl}\to 2\mathrm{NaCl}+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

(c) When carbon dioxide is treated with lime water (calcium hydroxide), it forms insoluble compound "C" which is calcium carbonate.

$\mathrm{Ca}(\mathrm{OH}{)}_2+{\mathrm{CO}}_2\to {\mathrm{CaCO}}_3+{\mathrm{H}}_2\mathrm{O}$

(d) When an excess of carbon dioxide is passed through an aqueous solution of calcium carbonate, it forms compound "D" which is calcium bicarbonate.

${\mathrm{CaCO}}_3+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2\to \mathrm{Ca}({\mathrm{HCO}}_3{)}_2$

Answer:

A diagram of the path of a ray passing through a triangular glass prism:

Answer:

(a) He should connect all the three resistances in a parallel connection. Then the net resistance of the parallel combination will be 1 ohm.

$$\begin{gathered} \frac{1}{R_{\mathrm{net}}}=\frac{1}{R_1}+\frac{{\displaystyle 1}}{{\displaystyle R_2}}+\frac{{\displaystyle 1}}{{\displaystyle R_3}}=\frac{1}{3}+\frac{{\displaystyle 1}}{{\displaystyle 3}}+\frac{{\displaystyle 1}}{{\displaystyle 3}} \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle R_{\mathrm{net}}}}=\frac{3}{3}=1 \mathrm{\Omega } \\ \Rightarrow R_{\mathrm{net}}\mathit{=}1\mathit{ }\mathrm{\Omega } \end{gathered}$$

(b) All three given resistors connected in parallel:

 

Answer:

(a) upward bending of shoot (or stem) - phototropism
(b) downward bending of root - geotropism

Answer:

In this case there are two possibilities -

Possibility I - The genetic makeup of the tall parent can be TTWW.
Possibility II - The genetic makeup of the tall parent can be TtWW. 

According to possibility I, if the tall parent has the genotype TTWW, then all the offsprings that are obtained will be tall bearing white flowers.

According to possibility II, if the tall parent has gentotype TtWW, then half the offsprings will be short bearing white flowers.

oSo, we can say that, the genotype of the tall parent would be TtWW.