Skip to main content

Solved Electricity numerical for class 10

Question: Two bulbs have ratings 100 W, 220 V and 60 W, 220 V respectively. Which one has a greater resistance?
Answer: P=VI=  V2/R For the same V, R is inversely proportional to P. Therefore, the bulb 60 W, 220 V has a greater resistance.

Question: A torch bulb has a resistance of 1 Ω when cold. It draws a current of 0.2 A from a source of 2 V and glows. Calculate
(i) the resistance of the bulb when glowing and
(ii) explain the reason for the difference in resistance.

Answer: (i) When the bulb glows:
V = I R ---- Ohm's law R = V/I = 2/.2 =10 Î©
(ii) Resistance of the filament of the bulb increases with increase in temperature. Hence when it glows its resistances is greater than when it is cold.

Question: Calculate the resistance of 1 km long copper wire of radius 1 mm. (Resistivity of copper = 1.72 x 1 0-8

Answer: L = 1 km = 1000 m
R = 1 mm = 1 x 1 0-3
A = p r2






Question: When a potential difference of 2 V is applied across the ends of a wire of 5 m length, a current of 1A is found to flow through it. Calculate: (i) the resistance per unit length of the wire (ii) the resistance of 2 m length of this wire (iii) the resistance across the ends of the wire if it is doubled on itself.

Answer: (i) V = I R ----- Ohm's law




(ii) Resistance of 2 m length of the wire = 0.4 x 2

(iii) When the wire is doubled on itself:
(1) the area of cross-section is doubled. If A is the original C.S. area, now it is 2 A.
(2) The length becomes half i.e. 



Comments

  1. it was very usesfull
    I just need someone to help me with how i should approach this question. I have NO idea how i should solve it please help!
    Two identical particles, each having charge +q, are fixed in space and separated by a distance d. A third point charge -Q is free to move and lies initially at rest on the perpendicular bisector of the two charges a distance x from the midpoint between the two fixed charges.
    the question is b) How fast will the charge -Q be moving when it is at the midpoint between the two fixed charges, if initially it is released at a distance a<<d from the midpoint?
    pls try to send reply soon to -ishwaryasun@gmail.com

    ReplyDelete

Post a Comment

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get d...

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination(Solved)

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination Law of conservation of mass Law of constant proportion Empirical formula 1. If 10 grams of CaCO 3 on heating gave 4.4g of CO 2 and 5.6g of CaO, show that these observations are in agreement with the law of conservation of mass.(Based on Law of conservation of mass) Solution:  Mass of the reactants = 10g ;  Mass of the products = 4.4 + 6.6g = 10g Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. 2. 1.375 g of cupric oxide was reduced by heating and the weight of copper that remained was 1.098g.  In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition . The weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state? Solution: in first experiment: Copper oxide = 1....

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page ...

Structure of Matter class07

The language of chemistry class 7 Basic science soled questions. Q. What is chemistry? Ans: Chemistry is a branch of science in which we study the composition, structure, properties and change of matter. Q. What is matter? Ans: Anything that is around us having some mass and occupies space is called matter. Q. What are the different classifications of matter? Ans: Matters are classified into element, compound and mixture. Q. What is element? Ans: Elements are the purest form of substance (matter) made up of same kinds of particles. The smallest unit of element is an atom. eg. Hydrogen, oxygen, carbon. Q. What is compound?            See full post   The language of chemistry class 7 Basic science Class 07 Basic sciences Chapter the Language of Chemistry A. Answer these questions 1. What does the formula of a substance tell us? Answer: Formula of a compound or an element represents the name and number of atoms present ...