Skip to main content

Ch.1- Matter in our surroundings 9th Solved Question

Picture
Q.1 Define matter.

Ans: Anything that occupies space and has mass and is felt by senses is called matter.

Q.2 what happens if you put copper sulphate crystals in water?

Ans: Copper sulphate crystals mixed between the spaces of molecules of water and disappear.

Q.3 A substance has a definite volume but no definite shape? State whether this substance is a solid, a liquid or a gas.



Ans: Liguid

Q.4 Arrange the following substances in increasing order of force of attraction between the particles. 
(a) Milk (b) Salt (c) Oxygen.

Ans: Oxygen < Milk< Salt

Q.5 A substance has neither a fixed shape nor a fixed volume . State whether it is a solid , a liquid or a gas.

Ans: Gas

Q.6 The melting point of a substance is below the room temperature. Predict its physical state.

Ans: Ice

Q.7 What is vapour ?

Ans: A substance that found in gaseous state only at room temperature is called Vapour

Q.8 Name the temperature at which the solid and liquid states of substance can exist together.

Ans; 0 degree C

Q.9 What is the effect of pressure on boiling point?

Ans: boiling point  increases with  increase in pressure.

Q.10 Name any two substances which sublime.

Ans: Camphor and Naphthalene, iodine and Ammonium chloride iodine

Q.11 Define Condensation.

Ans: The change of solid state from liquid state on cooling is also known as Condensation.

Q.12 For any substance, why does the temperature remain constant during the change of state?

Ans: On increasing the temperature of solids, the kinetic energy of the particles increases which is used to overcome the forces of attraction between the particles therefore the temperature remain constant during the change of state.

Q.13 Which is the slow process , Evaporation or Boiling ?

Ans: Evaporation

Q.14 State the effect of surface area on rate of evaporation.

Ans: If the surface area is increased, the rate of evaporation increases.

Q.15 Why are we able to sip hot tea faster from saucer rather than from a cup?

Ans: This is because the rate of evaporation is faster from saucer.

Q.16 What is the SI unit of temperature?

Ans: Kelvin

Q.17 Kelvin scale of temperature is regarded as better scale than Celsius. Why?

Ans: As it has wide range of measurement and Kelvin scale of temperature has always positive sign , hence regarded as better scale than Celsius.

Q.18.  Convert 10oC into Kelvin scale.

Ans: 270 + 10 = 280 K

Q.19 Pressure on the surface of a gas is increased. What will happen to the inter particle forces?

Ans: come closer and compressed

Q.20 A gas can exert pressure on the walls of the container. Assign reason.

Comments

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

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 dissolved in it

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 cm. Therefor

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

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

Electricity numerical for class 10 CBSE Trend Setter 50 Problems

1. The current passing through a room heater has been halved. What will happen to the heat produced by it? 2. An electric iron of resistance 20 ohm draws a current of 5 amperes. Calculate the heat produced in 30 seconds. 3. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater. 4. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination in half a minute. 5. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute. 6. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor? 7. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Ca