TISSUE

1.     How are simple tissues different from complex tissues in plants?

Simple tissue	Complex tissue
These tissues consist of only one type of cells.	These tissues are made up of more than one type of cells.
The cells are more or less similar in structure and perform similar functions.	Different types of cells perform different functions. For example, in the xylem tissue, tracheids help in water transport, whereas parenchyma stores food.
Three types of simple tissues in plants are parenchyma, collenchyma, and sclerenchyma.	Two types of complex permanent tissues in plants are xylem and phloem.

2.     Differentiate between parenchyma, collenchyma and sclerenchyma, on the basis of their cell wall.

Parenchyma	Collenchyma	Sclerenchyma
Cell walls are relatively thin, and the cells in parenchyma tissues are loosely packed.	The cell wall is irregularly thickened at the corners, and there is very little space between the cells.	The cell walls are uniformly thickened, and there are no intercellular spaces.
The cell wall in this tissue is made up of cellulose.	Pectin and hemicellulose are the major constituents of the cell wall.	An additional layer of the cell wall composed mainly of lignin is found.

3.     What are the functions of the stomata?

Functions of the stomata:

(i) They allow the exchange of gases (CO₂ and O₂) with the atmosphere.

(ii) Evaporation of water from the leaf surface occurs through the stomata. Thus, the stomata help in the process of transpiration.

PARAMECIUM

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PARAMECIUM

Difference between Collenchyma & Sclerenchyma

Difference between Collenchyma & Sclerenchyma

Collenchyma	Sclerenchyma
1. Living cells contain cytoplasm.	Dead cells are empty.
2. Cell walls are cellulosic.	Cell walls are lignified.
3. Lumen of the cell is wide	Lumen of the cell is narrow
4. The thickening of the cell wall is not uniform.	Thickening of cell wall is uniform.
5. It provides mechanical support & elasticity to the plant body.	It is chiefly a mechanical tissue.

SCLERENCHYMA

SCLERENCHYMA

Primarily meant for providing mechanical support and protection to different parts of the plant body.

Hence, sclerenchyma occurs in almost all the parts of the plant body, including root, fruit and seed.

It provide hardness and stiffness to the plant

The sclerenchyma cells are usually elongated, with tapering ends. These are matured dead cells. Its wall are thickened by deposition of lignin hence are refered as lignified walls. The walls are so think that there is hardly any internal space. Such cells are commonly described as sclerenchyma fibres. In some cases, the cells are shorter and irregular in shape. Such cells are commonly described as sclereids or stone cells.

EPIDERMIS

      Q: What is the role of epidermis in plants?

Ans: Epidermis present on the outer surface of the entire plant body. Its function is to protect. Hence the cells of the epidermal tissue form a continuous layer without any intercellular space. Most cells are relatively flat. Usually there outer and side wall are thicker than the inner wall (Why? – Its main function is protection hence outer wall is thicker). On aerial parts of plants it secrets a waxy water-resistant layer to protect against the loss of water, mechanical injury and invasion by parasitic fungi.It performs the following important functions:

(i) It protects the plant body  

(ii) It protects the plant against mechanical injury

(iii) It allows exchange of gases through the stomata

IX TEST: FORCE AND IMPROVEMENT IN FOOD RESOURCES

Answer of some questions of test conducted on 29.06.2015

Q. 1       Write C.G.S unit of force?   

Dyne                                                                                                    

Q. 2       State Newton’s first Law of motion.        

It states that an object at rest will stay at rest and an object in motion will stay in motion with the same speed and direction unless acted upon by unbalanced force.                                                         

Q. 3       Why a cricket player moves his hand backward while catching the ball?

By doing so the player increases the time period in which the momentum of the ball is going to decrease. According to Newton's second law of motion, the rate of change of momentum decreases. This results in lesser force applied to hands of player i.e. magnitude of the force decreases

F = (mv-mu) / t.         

Here force is indirectly proportional to the time. So lesser impact on the hands of player.

If action is always equal to the reaction, explain how a horse can pull a cart?

Q. 4       A stone of 1 kg is thrown with a velocity of 20 m/s  across the frozen surface of a lake and comes to    rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?                   

Solution:

Given

Mass of stone, ms                   = 1 kg

Initial velocity, u                      = 20m/s

Final velocity, v                       = 0 m/s

Distance travelled, s               = 50 m

We observe that all values are in SI            

To         calculate: 

Force,                                       f = ma

We are to calculate acceleration, a

Using 3rd law of motion

v² –u² = 2as

(0)² – (20)² = 2 x a x 50

A = -4 m/s²

Now, F = ma

     = 1 x – 4

     = - 4 N…………………..Ans

                                                     

Q. 5       If action is always equal to the reaction, explain how a horse can pull a cart?             

According to Newton's third law of motion, action force is equal to reaction but acts on two different bodies and in opposite directions. When The horse pushes the ground on backward direction is the action whereas the ground pushes the horse in forward direction is the reaction The force is able to overcome friction force of the cart and it moves.

Q. 6       A boy of mass 40 kg jumps with a horizontal velocity 5m/s onto a stationery cart with frictionless wheels. The mass of the cart is 3 kg. What is his velocity as the cart starts moving? Assume that there is no external unbalanced force working in horizontal direction

Solution:

Let v be the velocity of the boy on the cart as the cart starts moving.

The total momenta of the boy and cart before the interaction

40 × 5 + 3 × 0      =        200 kg m s^–1.

Total momenta after the interaction

(40 + 3)  × v       =        43 v kg m s^–1.

According to the law of conservation of momentum, the total momentum is conserved during the interaction. That is,

43 v = 200

⇒ v = 200/43 = + 4.65 m s^–1.

Q. 7       A bullet of mass 4 g when fired with a velocity of 50 m/s, can enter a wall up to a depth of 10 cm. How much will be the average resistances offered by the wall?

Solution :

(a)First of all write the given quantities

(b) Check if their units are in the same system, if not convert the units.

Here

Mass of bullet, m_b                        = 4 g    (this in in gram ; change it to SI system i.e. into kg.)                                                                    = 4/1000 =.004 kg

Initial velocity of bullet, u_b       = 50 ms-1

Depth of penetration, s             = 10 cm (this in in cm ; change it to SI  i.e. into meter.)                                                                         = 10/100 = 0.1m

It is given that ball can enter the wall up to 10cm. Hence it stops after penetrating into the wall after 10 cm.

So, final velocity of bullet, v_b      = 0

How much will be the average resistances offered by the wall ?

This resistance the force.

We know that force,                  F = ma

Here                                              m =0.004 kg, but a (acceleration is not known)

We should calculate the acc.(a) using 3^rd law of motion

v² –u² = 2as ;

(0)² – (50)²  = 2 x a x 0.1

a = -12500 m/s²

Now , F = ma

     = 0.004 x – 12500

     = - 50 N…………………..Ans

Q. 8       Deduce Newton’s 1st law mathematically by using Newton’s 2nd law?              

                                                      SECTION-II(BIOLOGY)

Q. 9       Name two different species of Indian cattle. What type of food should be given to cattle? Explain   

Species are

(a)  Cow :        Bos indicus

(b)  Buffalo :  Bos bubalis

Food Requirement: Cattle need balanced diet containing all nutrients in proportionate amount. Food it fed for following purposes.

For maintenance: This type of food keep the animal healthy

For milching: Food required during lactation for higher and good quality yield.

To fulfil above requirement animals are given foods like:

(a)  Roughage: It contains low nutrients contents but is very source of fibres  and carbohydrate. It is dry, coarse and fibrous food. These are dried straw of rice, wheat and leaves of cereal crops. Dry and green silage etc.

(b)  Concentrates: It is rich nutrient diet containing proteins, vitamins, fats and minerals like oilseeds and oil cakes. Grains of gram, maize, bajra and jawar is also fed.

Apart from these, certain feed additives containing micronutrients keeps the animal healthy and promote milk output of dairy animals.

                     

Q. 10  Name Italian bee variety. Why it is considered ideal for honey production?   

Apis mellifera                     

Apis mellifera is considered ideal for honey production because:

(a)  More honey collecting capacity  

(b)  Breeds well

(c)   Remains in hives for a longer duration

(d)  Can protect itself from other

(e)  Gentle in nature

Q. 11  Why are improved poultry breeds developed? Describe the desirable traits for which new varieties are developed?                                                                                                  

           Solve yourself

Q. 12  Describe composite fish culture  system ? What is the major problem in fish farming? How is it overcome? 

In this usually a combination of 5 to 6 species of fish are selected for culturing in the same pond. These are selected in in such a way that these have different food habits and prefer different levels of water body to capture their food. Hence they don’t compete for food. As a result food available in all parts of the pond is utilized. Different combinations of fish are:

Indigenous:     Catla, Rohu and Mrigal

Exotic:              Silver carp, Grass carp and Common carp

These are paired as

Catla and Silver carp :            being surface feeder.

Rohus:                                      Feed in middle zone

Grass carps:                              feed on weeds.

Mrigals and Common carp:    bottom feeders

Together these species can use all the food in the pond without competing each other.

Major Problem:

Many of these fish breed only during monsoon and their seed (egg) mixes. So there is lack of availability of good quality seed.

How to overcome:

During breeding fish are kept in separate pond and they are induced to breed using hormonal stimulation. It ensures the availability of pure seed as there is no mixing now.

Q. 13  What does the quality of honey depend upon?   

            Solve yourself                                                                    

IX: TEST 5 CELL,TISSUE,POLYNOMIAL, HERON’S

1.       [1] Which tissue makes up the husk of coconut?

2.       [1] Name the meristematic tissue, which increases the thickness of plants.

3.       [1] Name types of simple tissues.

4.       [1] Who presented the cell theory for the first time? What was it?

5.       [4] How many types of elements together makeup the xylem &  phloem tissue? Name them.

6.       [3] Differentiate between parenchyma, collenchyma and sclerenchyma, on the basis of their cell wall.

7.       [3] What is the utility of tissues in multi-cellular organisms?

8.       [3] What are the types of plastids? Write their names and fuctions.

9.       [3] Write any four differences between the plasma membrane and cell wall.

10.    [3] If  (x³ + ax² + bx + 6 )     has x – 2 as a factor BUT it leaves a remainder 3 when  divided by  ( x – 3 ), find the values of  a and b                       

11.    [3] Find the product:   (x – 2y + 3) (x²+ 4y² + 2xy – 3x + 6y + 9)                      

12.    [3] If x = 2y +  6 , find the value of (x³ – 8y³ -36xy -216)  

13.    [3]A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of Rs 5 per cm2.

14.    [3] A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

15.    [3] Sanya has a piece of land which is in the shape of a rhombus . She wants her one daughter and one son to work on the land and produce different crops. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get for their crops?

16.    [2] A park, in the shape of a quadrilateral ABCD, has  C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?