Solutions of Most Important Science Questions For Class 10 CBSE Board Exam-Part 2

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Q1.Explain the function of fuse in a domestic electric circuit ? An
electric oven having power rating 2000 W, 220V is used in an electric
circuit, having a fuse of 5A rating. What is likely to happen when the
oven is switched on ? Explain.

Answer. The use of fuse is to protect the electrical appliances by the breaking of the circuit when the circuit is overloaded or excess current enters to the fuse.

In the given question we have

Power of oven(P) = V× I

$I=\frac{P}{V}=\frac{2000}{220}$

I = 9.09 A

The fuse of 5A rating means it can allow the maximum current flow of 5A,any excess to this current (in this case 9.09A) melt it because H =I²RT,where H is heat,R is resistance and I is current so the circuit will be broken down and electrical appliances would be kept safe,but there may be chances of short circuit both of the wire of earth and fase stucked to each other because of excess heat , thus it results the current flows in both the wire, this extra current results larger amount of heat so a fire could be taken place.

Q2. For the circuit diagram given below: Calculate :

(i) Value of current flowing through each resistor.

(ii) The total current in the circuit.

(iii) The total effective resistance of the circuit.

Solutions.

(i) Since all the resistors are arranged in Parallel combination, therefore, the voltage across all resistors will be 6V

So applying ohm’s law V= IR

Let the current flowing through 5Ω, 10Ω and 30Ω are I₁, I₂and I₃respectively

$I_{1}=\frac{V}{R}=\frac{6}{5}=1.2A$

$I_{2}=\frac{V}{R}=\frac{6}{10}=0.6A$

$I_{3}=\frac{V}{R}=\frac{6}{30}=0.2A$

(ii) Total electric current in the circuit ,I = I₁+ I₂+ I₃=1.2 + 0.6 + 0.2 = 2 A

(iii) Let the total resistance is = R

1/R = 1/5 + 1/10 + 1/30

1/R = (6 + 3 +1)/30 = 10/30 = 1/3

R = 3 Ω

So total effective resistance of the circuit =3Ω

Q3. A resistance wire is stretched to double its length so that its
area of cross-section is halved. How will its resistance change ? Will
any change be observed in its resistivity? Justify your answer.

Answer.As we know the relationship among resistance(R), length of the conductor(l), area of cross-section of the conductor(A) and resistivity of the substance the conductor is made of(ρ) is following.

$R= \rho \frac{l}{A}....\left ( i \right )$

Let the resistance of the conductor becomes R’ after its length changes to 2l and area of the cross-section to A/2, so

$R'= \rho \frac{2l}{A/2}$

$R'= \rho \frac{4l}{A}....\left ( ii \right )$

Dividing eq.(i) by (ii)

$\frac{R}{R'} =\rho \frac{l}{A}\times \frac{A}{4\rho l}$

R’ = 4R

Therefore the resistance of the conductor will become 4 times of the resistance as it was before.

The resistivity(ρ) of the substance is constant because substance remains the same in this change.

Q4. Give reasons for the following :

(i) Ionic compounds have high melting point and boiling point.

(ii) Ionic compounds conduct electricity in the molten state.

(iii) Ionic compounds are solids at room temperature and are somewhat hard.

Answer. Ionic compounds(like NaCl) are formed when one of the atom of the substance loss electrons forming cation(+ion) and atom of another substance gains the electrons forming anion(–ion), these anion and cation stucked to each other due to strong force of attraction between them ,obviously it will be needed high energy to overcome this strong electrostatic force between them,so their meling and boiling point is high.

(ii) The flow of electricity occurs due to the movement of ions but in solid-state, there is no movement of ions in ionic compound. When ionic compounds are in the molten stage their ions become free to move,it is thats why in the molten stage ionic compound conducts electricity.

(iii) An ionic compound has a strong bond between their constituent cation and anion,it is required high temperature to separate them, the room temperature is very less (i.e almost 25°C) turning them into liquid. In this temperature, they are harder because of this strong force of attraction.

Q5. A compound which is prepared from gypsum has the property
of hardening when mixed with right quantity of water.

(i) Identify the compound and write its chemical formula.

(ii) Write the chemical equation for its preparation.

(iii) List any two uses of the above compound.

Answer.(i) The compound that prepared from the gypsum is Plaster of Peris(PoP),its chemical formula is CaSO4.1/2.H2O.

(ii) When water is added to PoP in a proper quantity it turns into a hard substance called gypsum. PoP is produced by heating up gypsum.

$\underset{Gypsum}{CaSO_{4}.2H_{2}O}\xrightarrow[Heat]{371K}\underset{PoP}{CaSO_{4}.\frac{1}{2}H_{2}O}$

(iii) (a) The property of Plaster of Paris turning to gypsum(hard substance) when water is added on it is used in manufacturing toys, statue, decorative items.

(b) Plaster of Paris is used widely in the hospitals in setting up fractured or broken bones in their place.

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

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NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

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NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals