Skip to main content

CBSE_NCERT_ 6th Chapter _Living things and surrounding: Objective formative questions(MCQ's)

Fill in the blanks : Chapter:Habitat
(i)In the sea plants and animal surrounded by -------------------------(saline water)

(ii) The place where living beings live is called their -------------(Habitat)

(iii) In Cactus plant, ---------------are reduced to spines (Leaves)

(iv) Global warming increases the concentration of ------in the atmosphere(carbon dioxide)

(v) ------------are examples for Decomposers (Bacteria and fungus)

(vi) Living component of a habitat is called as -------------- (Biotic)

(vii) Anything which produce and response in an organism is called as -----------(Stimulus)

(viii) A potted plant is kept in a card board box with a small slit of sun light exhibits ----- Movement 
Phototrophic)

(ix) If we blow through lime water it turns milky. This proves that exhaled air contains ------ (carbon 
dioxide)

(x)-------- are plants that grow in dry desert area (Xerophytes)

(xi)--------- are plants that grow in moist area (Mesophytes)

(xii) )--------- are plants that grow in soil rich in salt (helophytes)

(xiii) ---- help in pollination and seed dispersal (wind)

(xiv)--------- are plants that best grow in full sunlight (heliophytes)

(xv) --------- are plants that best grow in low sunlight (sciophytes)

(xvi) Roots are ---------------phototropic (negatively)

(xvii) Earthworms breathe through their ---------------( moist skin)

(xviii) ------are those animals which can live on land and also in water (Amphibians)

(xix) ------ is an example of a non moveable animal .( coral)

(xx) -------are those animals which lay eggs example- crow, lizard (Oviparous)

(xxi) -------are those animals which do not lay eggs & give birth to young ones for example- lion, dog. 
viviparous)

(xxii) ---------- habitat comes under aquatic habitat (Marine)

(xxiii) _________are certain changes in the surroundings that cause an organism to respond to them. Stimuli)

(xxiv) -------------- help aquatic animal to float in water { Aerenchyma (air spaces in their bodies)}

(xxv) Some fungi grow in soils of burnt areas called -----------------. (Pyrophilous)

More post related to this chapter

6th Living things and Habitat and adaptation-1


6th Living things and Habitat and adaptation-2


6th Living things and Habitat and adaptation -3

Read and Download

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