Important Questions

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 8 Cell: The Unit of Life Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 8 Cell: The Unit of Life

Very short answer type questions

Question 1.
Give one difference between the characteristic feature of Gram positive and Gram negative bacteria.
Answer:
Gram positive bacteria retain crystal violet dye and stain blue in colour, while Gram negative bacteria loose dye colour on washing.

Question 2.
What is the name given to the infoldings of plasmalemma in fungal cell below the wall?
Answer:
Mesosomes.

Question 3.
What does ‘S’ refers in a 70S and 80S ribosome? [NCERT Exemplar]
Answer:
The ‘S’ refers to Svedberg units of sedimentation coefficient. The sedimentation coefficient is a measure of the speed of the sedimentation for a particular cell organelle in ultracentrifuge.

Question 4.
What are inclusion bodies?
Answer:
Reserve material in prokaryotic cells are stored in the cytoplasm in the form of inclusion bodies.

Question 5.
In which organelle the proteins required for functioning of nucleus are formed?
Answer:
Proteins required are formed in cytoplasm.

Question 6.
Which is considered to be the main arena of cellular activities in plant and animal cells?
Answer:
Cytoplasm is considered to be the main arena of cellular activities.

Question 7.
Mention a single membrane bound organelle, which is rich in hydrolytic enzymes. [NCERT Exemplar]
Answer:
Lysosome.

Question 8.
What is the significance of vacuole in a plant cell? [NCERT Exemplar]
Answer:
The vacuole in a plant cell helps to maintain osmotic pressure for turgidity and osmosis. It also stores useful as well as waste substances.

Question 9.
Why are the mitochondria and plastids called semi-autonomous particles?
Answer:
These are called so, because they are not dependent upon nuclear DNA and cytoplasmic ribosomes for the synthesis of proteins, while other organelles are dependent.

Question 10.
What is referred to as satellite chromosome? [NCERT Exemplar]
Answer:
Sometimes, few chromosomes have non-staining secondary constrictions at a constant location. This gives appearance of a small fragment called satellite. The chromosome having satellite are known as satellite chromosomes.

Question 11.
What do you meant by 9+2 pattern of organisation?
Answer:
It means that 9 microtubule doublets surround two single microtubules.

Question 12.
Define the microbodies.
Answer:
Membrane bound minute vesicles that contain various enzymes are called microbodies. These are present in both plant and animal cells.

Short answer type questions

Question 1.
Differentiate between prokaryotic and eukaryotic cells.

Prokaryotic Cell	Eukaryotic Cell
•» Nuclear membrane absent	Nuclear membrane present
•» Cell organelles absent (except ribosome)	Cell organelles present
•» Endomembrane system absent	Endomembrane system present
•» Example: bacteria	Example: RBC, neuron

Question 2.
Describe passive transport, osmosis and active transport in plasma membrane.
Answer:
Passive Transport: The plasma membrane is selectively permeable to some molecules present on either side of it. Many molecules can move briefly across the membrane without any requirement of energy and this is called the passive transport.

Osmosis: Neutral solutes may move across the membrane by the process of simple diffusion along the concentration gradient, i.e., from higher concentration to the lower. Water may also move across this membrane from higher to lower concentration. Movement of water by diffusion is called osmosis.

Active Transport: As the polar molecules cannot pass through the nonpolar lipid bilayer, they require a carrier protein of the membrane to facilitate their transport across the membrane. A few ions or molecules are transported across the membrane against their concentration gradient, i.e., from lower to the higher concentration. Such a transport is an energy dependent process, in which ATP is utilised and is called active transport, e.g., Na⁺/K⁺ Pump.

Question 3.
What is endomembrane system in a cell?
Answer:
Certain cell organelles function in coordinated manner. Their v coordination makes the endomembrane system of cell. Following structures comprise the endomembrane system:

• Endoplasmic Reticulum
• Golgi Complex
• Lysosome, and
• Vacuoles.

Question 4.
Cells in the epithelial tissue are held together with very little intercellular material (matrix). Specialised junctions provide both structural and functional links between individual cells. Three types of cell junctions have been recognised.
(a) Name the three types of cell junctions and write their functions.
(b) What value is shown by such an arrangement?
Answer:
(a) The three types of cell junctions are:

1. Tight junctions,
2. Gap junctions and
3. Adhering junctions

Functions of cell junctions are ‘

• Tight junctions prevent the leaking of substances across a tissue.
• Gap junctions facilitate the cells to communicate with each other by connecting their cytoplasm.
• Adhering junctions perform cementing of the adjacent cells to keep them together.

(b) No individual can function alone in a society. We need some form of interaction and help from others; so lend a helping hand to the needy.

Question 5.
Give a brief description of ribosomes.
Answer:
Ribosomes are the granular structures first observed under the electron microscope as dense particles by George Palade (1953). They are composed of ribonucleic acid (RNA) and proteins and are not surrounded by any membrane. The eukaryotic ribosomes are 80S while the prokaryotic ribosomes are 70S. Here ‘S’ stands for the sedimentation coefficient; it indirectly is a measure of density and size. Both 70S and 80S ribosomes are composed of two subunits.

Question 6.
Write a short note on-
(i) Cytoskeleton,
(ii) Cilia & Flagella
Answer:
(i) Cytoskeleton: An elaborate network of filamentous proteinaceous structures present in the cytoplasm is collectively referred to as the cytoskeleton. The cytoskeleton in a cell are involved in many functions such as mechanical support, motility, maintenance of the shape of the cell.

(ii) Cilia and Flagella: Cilia (sing.: cilium) and flagella (sing.: flagellum) are hair-like outgrowths of the cell membrane. Cilia are small structures which work like oars, causing the movement of either the cell or the surrounding fluid. Flagella are comparatively longer and responsible for cell movement. The prokaryotic bacteria also possess flagella but these are structurally different from that of the eukaryotic flagella.

Long answer type questions

Question 1.
Write the functions of the following: [NCERT Exemplar]
(i) Centromere
(ii) Cell wall
(iii) Smooth ER
(iv) Golgi apparatus
(v) Centrioles
Answer:
(i) Functions of Centromere
(a) It is a narrow non-stainable area which join chromatids together to form a chromosome.
(b) The centromere thus, keep the two chromatids of a chromosome in an intact stage.
(c) This is an essential step for chromosomes of a cell during cell division whether it may be mitosis or meiosis.

(ii) Functions of Cell Wall
(a) It helps in providing a definite shape to the cell and also protects protoplasm against any mechanical injury, i.e., damage and infection.
(b) It also helps in cell-to-cell interaction.
(c) It provides barrier to undesirable macromolecules and attack of pathogens.

(iii) Functions of Smooth Endoplasmic Reticulum
(a) It provides mechanical support to colloidal complex of cytoplasmic matrix.
(b) It holds various cell organelles in position.
(c) It conducts information from outside to inside of cell.

(iv) Functions of Golgi Apparatus
(a) It performs the function of packaging material.
(b) It acts as an important site for the formation of glycoproteins and glycolipids.
(c) It helps in the production of complex carbohydrates other than glycogen and starch.
(d) It helps in the formation of cell wall.

(v) Functions of Centrioles
(a) Formation of new centrioles from pre-existing one’s during cell division.
(b) They form basal bodies, which in turn form cilia and flagella.
(c) They form spindle fibres that give rise to spindle apparatus during cell division .in animal cells.

Question 2.
Is there a species or region specific type of plastids? How does one distinguish one from the other? [NCERT Exemplar]
Answer:
Plastids are both region or species specific. These are as follows :
(i) Proplastids: These are colourless, rounded but amoeboid plastid precursors, found in meristematic and newly formed cells of plants. It has a double membrane envelope that surrounds a colourless matrix, containing DNA, ribosomes and reserve food. A few vesicles and lamellae also occur in the matrix.

(ii) Leucoplasts: These are colourless plastids that occur in non-green plant cells commonly near the nucleus. They are as follows:
(a) Amyloplasts: These leucoplasts store starch, e.g., tuber of potato, grain of rice and wheat.
(b) Elaioplasts: These store fats, e.g., Rose.
(c) Aleuroplasts: They are protein storing plastids, e.g., Castor endosperm.

(iii) Chromoplasts: These are non-photosynthetic coloured plastids which synthesise and store carotenoid pigments. They appear, orange, red or yellow in colour. These mostly occur in ripe fruits (tomato and chillies) carrot roots, etc.

(iv) Chloroplasts: These are photosynthetic plastids, which are green in colour and found in the leaves of all green plants. They have lamellae organised in the form of grana.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 7 Structural Organisation in Animals Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 7 Structural Organisation in Animals

Very short answer type questions

Question 1.
A muscular fibre tapers at both the ends and does not show striations. Name the muscle fibre. [NCERT Exemplar]
Answer:
Smooth or non-striated muscle fibre.

Question 2.
Name few called the specialised connective tissues?
Answer:
Bones, cartilage and blood are the special types of connective tissues.

Question 3.
How does a gap junction facilitate the intercellular communication? [NCERT Exemplar]
Answer:
A gap junction facilitates the cells to communicate with each other by connecting the cvtoplasm of the adjacent cells.

Question 4.
Which tissue forms the ligaments? What is their function?
Answer:
Ligaments are formed of yellow elastic connective tissue. The ligaments join the bones together.

Question 5.
On which segment of the body, male and female genital apertures are present in earthworm?
Answer:
Male – 18th segment; Female – 14th segment.

Question 6.
Why the body segmentation in earthworm is called metameric segmentation ?
Answer:
In metameric segmentation, the external segmentation corresponds to the internal segmentation of the body.

Question 7.
In earthworm, from which segment intestine starts and where it ends?
Answer:
In earthworm, intestine starts from the 15th segment onwards and continues till the last segment.

Question 8.
Where the sclerites are present in cockroach? [NCERT Exemplar]
Answer:
In the exoskeleton all over the body.

Question 9.
Which month part of cockroach is comparable to our tongue? [NCERT Exemplar]
Ans.
Hypopharynx.

Question 10.
Name the upper lip and lower lip of the cockroach.
Answer:
Upper lip – Labrum
Lower lip – Labium

Question 11.
Why the blood of cockroach is not responsible for transporting respiratory gases?
Answer:
Because the respiratory pigment is absent in their blood.

Question 12.
List the parts of blood vascular system of cockroach.
Answer:
Haemocoel, heart and blood.

Short answer type questions

Question 1.
Answer:

Smooth Muscle	Skeletal Muscle	Cardiac Muscle
1. Spindle shaped cells	Striated unbranched cells	Striated branched cells
2. Found in muscles of internal organs	Found	Found in the heart
3. Control involuntary actions	Control voluntary actions	Control heart’s pumping which is involuntary

Question 2.
What do you understand by special junctions between cells? Which type of special junction are found in epithelial tissues?
Answer:
All cells are held together with intercellular material. These materials form junctions between cells.
There are three types of cell junctions found in epithelial tissue.
(a) Tight Junctions: These junctions help in stopping leakage of substances across a tissue.
(b) Adhering Junctions: These junctions keep neighbouring cells together.
(c) Gap Junctions: These junctions connect cytoplasm of adhering cells, and facilitate exchange of materials.

Question 3.
Explain fertilization and development in Pheretima.
Answer:
Fertilization and Development in Pheretima

• A mutual exchange of sperm occurs between two worms during mating. One worm has to find another worm and they mate juxtaposing opposite gonadal openings exchanging packets of sperms called spermatophores.
• Mature sperm and egg cells and nutritive fluid are deposited in cocoons produced by the gland cells of clitellum. Fertilisation and development occur within the cocoons which are deposited in soil.
• The cocoon holds the worm embryos. After about 3 weeks each cocoon produces two to twenty baby wopns with an average of four. Earthworms development is direct, i.e., there is no larva formed.

Question 4.
Describe mouth parts of the cockroach. ,
Answer:
Mouth Parts of Cockroach: Anterior end of the head bears appendages forming biting and chewing type of mouth parts. The mouth ‘ parts consist of following structures:
(a) Labrum (upper lip),
(b) A pair of mandibles,
(c) A pair of maxillae and
(d) Labium (lower lip).
Apart from these a median flexible lobe, acting as tongue (hypopharynx), lies within the cavity enclosed by the mouth parts.

Question 5.
Explain in brief the nervous system of cockroach.
Answer:
Nervous System of Cockroach

• The nervous system of cockroach consists of a series of fused, segmentally arranged ganglia joined by paired longitudinal connectives on the ventral side.
• Three ganglia lie in the thorax, and six in the abdomen. The nervous system of cockroach is spread throughout the body.
  The head holds a bit of a nervous system while the rest is situated along the ventral (belly-side) part of its body.

Question 6.
Explain digestion in frogs.
Answer:
Digestion in Frog

• In the stomach digestion of food takes place by the action of HCl and gastric juices secreted from the walls of the stomach.
• Partially digested food called chyme is passed from stomach to the first part of the intestine, the duodenum. The duodenum receives bile from gall bladder and pancreatic juices from the pancreas through a common bile duct.
• Bile emulsifies fat and pancreatic juices digest carbohydrates and proteins. Final digestion takes place in the intestine.
• Digested food is absorbed by the numerous finger-like folds in the inner wall of intestine called villi and microvilli.

Question 7.
Explain in brief the central nervous system of frog.
Answer:
Central Nervous System of Frog ,

• Brain is enclosed in a bony structure called brain box or skull (cranium).
• The brain is divided into fore-brain, mid-brain and hind-brain.
• Forebrain includes olfactory lobes, paired cerebral hemispheres and unpaired diencephalon.
• The midbrain is characterised by a pair of optic lobes.
• Hind-brain consists of cerebellum and medulla oblongata.
• The medulla oblongata passes out through the foramen magnum and continues into spinal cord, which is enclosed in the vertebral column.
• There are ten pairs of cranial nerves arising from the brain.

Question 8.
How do different senses work in frog? Explain in brief. ,
Answer:
Sense Organs in Frog : Frog has different types of sense organs which are as follows:
(a) Sensory papillae or organs of touch,
(b) Taste buds.
(c) Nasal epithelium for the sense of smell,
(d) Eyes for vision and
(e) Tympanum ‘with internal ears for hearing.

Out of these, eyes and internal ears are well-organised structures and the rest are cellular aggregations around nerve endings. Eyes in a frog are a pair of spherical structures situated in the orbit in skull. These are simple eyes. External ear is absent in frogs and only tympanum can be seen externally. The ear is an organ of hearing as well as balancing.

Long answer type questions

Question 1.
Which features distinguish blood from lymph?
Answer:
Differences between blood and lymph are given below :

Blood	Lymph
1. It contains plasma, erythrocytes, leucocytes and platelets.	It contains plasma and leucocytes.
2. The presence of haemoglobin imparts red colour to it.	It is colourless as haemoglobin is absent.
3. Its plasma contains more protein, calcium and phosphrous as compared to lymph.	Its plasma has fewer protein and less calcium and phosphorus than blood.
4. Contains moderate amount of CO2 and other metabolic waste.	Contains excessive amount of CO2 and other metabolic waste.

Question 2.
In which segment does the following structures lies in the earthworm’s body?
(i) Spermathecae
(ii) Pharynx
(iii) Gizzard
(iv) Intestine
(v) Septal nephridia
(vi) Ovary
(vii) Testes
(viii) Typhlosole
(ix) Lateral heart
(x) Pharyngeal nephridia
Answer:
(i) Spermathecae – 6th, 7th, 8th, 9th
(ii) Pharynx – 4th
(iii) Gizzard – 8th
(iv) Intestine – 15th to last
(v) Septal nephridia – 15th to last
(vi) Ovary – 13th
(vii) Testes – 10th, 11th
(viii) Typhlosole – 26th-95th
(ix) Lateral heart – 7th, 9th
(x) Pharyngeal nephridia – 4th-6th

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 16 Digestion and Absorption Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 16 Digestion and Absorption

Very short answer type questions

Question 1.
What do we call the type of teeth attachment in which each tooth, is embedded in a socket of jaws bones? [NCERT Exemplar]
Answer:
Thecodont dentition.

Question 2.
Out of the three types of salivary glands, which is situated in the upper palate on either side of cheek.
Answer:
Parotid gland (type of salivary gland) is situated in the upper palate on the either side of the cheek.

Question 3.
If a person is suffering from reflux oesophagitis, which part of his alimentary canal is not functioning properly?
Answer:
In reflux oesophagitis, the oesophagus does not function properly.

Question 4.
HCl is secreted in stomach. Give the name of the cells that secrete it.
Answer:
Oxyntic cells (parietal cells) are the stomach epithelial cells, that secretes gastric acid, i.e., HCl.

Question 5.
Crypts of Lieberkuhn are found in which part of the alimentary canal.
Answer:
These are found in small intestine portion of alimentary canal.

Question 6.
Give the name of the enzymes involved in the breakdown of nucleotides into sugars and bases? [NCERT Exemplar]
Answer:
Nucleosidases.

Question 7.
Trypsinogen is an inactive enzyme of pancreatic juice. An enzyme, enterokinase activates it. Which tissue/cells secrete this enzyme? How is it activated? [NCERT Exemplar]
Answer:
The cells of duodenum secrete enzyme enterokinase. It is activated by food in the duodenum.

Question 8.
Mention the function of saliva other than digestion.
Answer:
Apart from taking part in digestion, saliva also helps to lubricate the food for swallowing.

Question 9.
State the source of trypsin and the food constituent which this enzyme hydrolyses.
Answer:
Pancreatic juice is the source of trypsin which hydrolyses proteins into peptides.

Question 10.
If the bile duct is completely blocked. How would it affect the digestion of food?
Answer:
If the bile duct is blocked completely, the bile will fail to reach the small intestine and the digestion of fats gets affected.

Question 11.
By which type of mechanism, amino acids are absorbed in our body?
Answer:
Small amounts of amino acids are absorbed by active transport and some are absorbed by the facilitated transport.

Question 12.
As fatty acids and glycerol are not absorbed directly. Name the form in which fatty acids are converted to get absorbed?
Answer:
Fatty acids and glycerols are not absorbed directly into the bloodstream. Thus, they are absorbed in the form of small droplets called micelles.

Question 13.
Which type of absorption takes place in large intestine?
Answer:
Absorption of water, some minerals and drugs takes place in the large intestine.

Question 14.
What happens in the condition when bile from the liver crystalises?.
Answer:
When bile from the liver get crystallised, person form stones in the body.

Short answer type questions

Question 1.
What is the role of tongue indigestion?
Answer:
Tongue helps in mixing the food properly with salivary enzymes. Moreover, tongue has tastebuds which give the sense of different tastes. Eating is a complex process which needs involvement of olfactory and visual senses as well. Alongwith tongue all these senses help in picking the right food.

Question 2.
What is the function of large intestine?
Answer:
Functions of large intestine are as follows:

• Absorption of some water, minerals and certain drugs.
• Secretion of mucus. Mucus helps in adhering the waste particles together and lubricates it for easy passage.

Question 3.
What is digestive waste and how is it removed from the body?
Answer:
After digestion and absorption of food is over the residue left makes the digestive waste. The digestive wastes, solidified into coherent faeces in the rectum initiate a neural reflex causing an urge or desire for its removal. The egestion of faeces to the outside through the anal opening (defaecation) is a voluntary process and is carried out by a mass peristaltic movement.

Question 4.
How does the nervous system control the activities of gastrointestinal tract?
Answer:
The sight, smell and/or presence of food in the oral cavity can stimulate the secretion of saliva. Gastric and intestinal secretions are also stimulated by similar neural signals. Muscular activities of alimentary canal is coordinated by both local and CNS neural mechanisms. Hormonal control of secretion of digestive enzymes is carried out by local hormones.

Question 5.
Write a short note on-Disorders of digestive system.
Answer:
Disorders of Digestive System
(i) Inflammation of intestinal tract
This is the most common disorder of the digestive system.
It is caused by infections by bacteria or viruses and also by parasites like roundworm, hookworm, pinworm, etc.

(ii) Jaundice
It is the infection and inflammation of the liver.
Bile pigments are present in the blood and cause yellowing Of eyes, skin, etc.

(iii) Vomiting
It is the egestion of contents of the stomach through the mouth.

(iv) Diarrhoea
The abnormal frequency of bowel movement and increased liquidity of the faecal discharge is known as diarrhoea.
It reduces the absorption of food.

(v) Constipation
In constipation, the faeces are retained within the rectum as the bowel movements occur irregularly.

(vi) Indigestion
In indigestion, the food is not properly digested leading to a feeling of fullness.
The causes of indigestion are inadequate enzyme secretion, anxiety, food poisoning, overeating, spicy food, etc.

Long answer type questions

Question 1.
As a result of intestinal disease, parts of the alimentary canal are sometimes surgically removed. Discuss the effect of this removal on lifestyle and digestive function:
(i) the stomach
(ii) the colon.
Answer:
(i) Removal of the stomach, which is the main organ for digestion of protein, leads to a change in the patient’s diet. It is the stomach where proteins are first broken down into polypeptides by pepsin in gastric juice before they can be acted on by pancreatic enzymes and intestinal enzymes. The patient must lower down the amount of protein in his diet. Animal meat especially red meat, which is rich in proteins must be avoided.

(ii) Removal of the colon, which is responsible for absorption of water from undigested food, result in loss of water (dehydration) inpatient. The patient needs to drink plenty of fluids to replace water loss. In most cases, where the whole colon is removed, a surgical procedure.is done to attach the small intestine to the rectum to allow for recta elimination of liquid stools. A small pouch is created in the lower abdomen to collect the stool. The patient has to learn to regulate his bowel movements.

Question 2.
A person had roti and dal for his lunch. Trace the changes that will take place during its complete passage through the alimentary canal. [NCERT Exemplar]
Answer:
Changes that will take place in food (roti and dal) through the passage of alimentary canal are given below:

• The food substances are first masticated by the teeth in the mouth, where carbohydrate part of the food is digested by the action of salivary amylase enzyme secreted by the salivary glands.
• This partially digested food reaches the stomach, where it receives acidic HCl and mainly the protein part of the food is digested by the action of proteolytic enzymes.
• The lipid part of the food is digested by the bile secreted by the gall bladder.
• In the small intestine, particularly in the duodenum, this semi-digested food is finally digested by the digestive enzymes present in the intestinal and pancreatic juices.
• After digestion, the broken down products of food, i.e., amino acids, glycerol, starch, etc., are observed mainly in the small intestine.
• The undigested remains of food will finally pass through the anus.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 6 Anatomy of Flowering Plants Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 6 Anatomy of Flowering Plants

Very short answer type questions

Question 1.
The product of photosynthesis is transported from the leaves to various parts of the plant and stored in some cells before being utilised. What are the cells/tissues that store them?
Answer:
Parenchyma.

Question 2.
What is the function of phloem parenchyma? [NCERT Exemplar]
Answer:
It takes part in lateral conduction of food and supply of water from xylem.

Question 3.
What is the epidermal cell modification in plants which prevent [NCERT Exemplar]
Answer:
Cuticularised trichoblasts or epidermal hair which produce a stationary layer of air over the surface that reduces isolation and rate of transpiration.

Question 4.
The cells of this tissue are living and show angular wall thickenings. They also provide mechanical support. The tissue is: [NCERT Exemplar]
(a) xylem
(b) sclerenchyma
(c) collenchyma
(d) epidermis
Answer:
(c) The collenchyma is a simple permanent tissue, which provide mechanical support.

Question 5.
In which vascular bundles, phloem lies on the outer side and xylem towards the inner side or central axis?
Answer:
Collateral vascular bundles.

Question 6.
What is present on the surface of the leaves which helps the plant prevent loss of water but is absent in roots? [NCERT Exemplar]
Answer:
Cuticle layer and wax.

Question 7.
What constitutes the cambial ring? [NCERT Exemplar]
Answer:
Fusion of interfascicular and intrafascicular cambium strips.

Question 8.
The cross-section of a plant material showed the following features when viewed under the microscope (a) Vascular bundles were radially arranged (b) Four xylem strands with exarch condition of protoxylem. To which organ should it be assigned? [ NCERT Exemplar]
Answer:
Dicot root.

Question 9.
What do hardwood and softwood stand for? [NCERT Exemplar]
Answer:
Hardwood contains xylem vessels, i.e., dicot wood. Softwood contains tracheids only, i. e., gymnospermous wood.

Question 10.
Give one basic functional difference between phellogen and phelloderm.
Answer:
Phellogen is meristematic and divides to produce new cells. Phelloderm stores food materials.

Short answer type questions

Question 1.
Differentiate among parenchyma, collenchyma and sclerenchyma.

Parenchyma	Collenchyma	Sclerenchyma
1. Living cells.	1. Living cells.	1. Dead cells.
2. Forms	2. Found below epidermis.	2. Usually found in epidermis.
3. Chloroplast present.	3. Chloroplast present sometimes.	3. Chloroplast absent.
4. Performs many vital functions.	4. Provides mechanical support to growing parts.	4. Provides mechanical support to organs.

Question 2.
Mention key differences between xylem and phloem.
Answer:
Differences between xylem and phloem

Xylem	Phloem
1. Composed of tracheids, vessels and xylem parenchyma.	1. Composed of sieve tubes, companion cells and phloem parenchyma.
2. Facilitate conduction of water and minerals from roots.	2. Facilitate conduction of food from leaves.

Question 3.
What is the difference between simple tissue and complex tissue in plants?
Answer:
Simple tissues are composed of similar cells. Complex tissues are composed of dissimilar cells. Simple tissues provide bulk and mechanical support to plants. Complex tissues are meant for transportation of substances and they also provide mechanical support.

Question 4.
What is the difference between monocot and dicot leaves?
Answer:
Stomata are found on both surfaces in monocot leaves, while they are found on ventral surface only, in dicot leaves. In monocot leaves venation is parallel, which is evident by similar size of vascular bundles. In dicot leaves venation is reticulate, so vascular bundles are of various sizes.

Question 5.
What are the differences between meristematic tissues and permanent tissues?
Answer:
Differences between meristematic tissues and permanent tissues

Meristematic Tissues	Permanent Tissues
1. Cells keep on dividing.	1. Cells stop dividing.
2. Growth is the basic function.	2. Protection is the basic function.
3. Found in tips of roots and stem.	3. Found in girth and periphery.

Long answer type questions

Question 1.
Distinguish between following:
(i) Exarch and endarch condition of protoxylem
(ii) Stele and vascular bundles
(iii) Protoxylem and metaxylem
(iv) Interfascicular cambium and intrafascicular cambium
(v) Open and closed vascular bundles
(vi) Stem hair and root hair [NCERT Exemplar]
Answer:
(i) In exarch condition protoxylem towards the periphery. In endarch
condition protoxylem towards the centre.
(ii) Stele is the arrangement of vascular tissues and vascular bundle is a group of xylem and phloem.
(iii) Protoxylem is an early formed xylem and metaxylem is late formed xylem.
(iv) Interfascicular cambium is formed by permanent tissues. Intrafascicular cambium is present in between the primary xylem and primary phloem of a vascular bundle.
(v) Open vascular bundles have intrafascicular cambium and show secondary growth. Closed vascular bundles do not have intrafascicular cambium.
(vi) Stem hair are multicellular, whereas root hair are unicellular.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 17 Breathing and Exchange of Gases Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 17 Breathing and Exchange of Gases

Very short answer type questions

Question 1.
Why do cells need a constant supply of oxygen?
Answer:
Cells continuously need oxygen for the metabolic reactions that releases energy from molecules. This energy is used by cells for various functions of body.

Question 2.
Give the name of the organ that produces sound.
Answer:
Larynx.

Question 3.
A fluid filled double membranous layer surrounds the lungs. Name it. [NCERT Exemplar]
Answer:
This layer is called pleural membrane and pleural fluid is found in between the two.

Question 4.
Which component of the respiratory system help in generation of pressure gradient for breathing? [NCERT Exemplar]
Answer:
The diaphragm and a specialised set of muscles external and internal intercostals between the ribs.

Question 5.
Define Residual Volume (RV). [NCERT Exemplar] Or State the volume of air remaining in the lungs after a normal breathing. [NCERT Exemplar]
Answer:
Residual Volume (RV): It is the volume of air remaining in the lungs even after a forcible expiration. It is about 1100-1200 ml.

Question 6.
What is the RQ for glucose and proteins?
Answer:
Glucose RQ= 1
Proteins RQ = 0.85

Question 7.
Why is it advisable to do nasal breathing rather than mouth breathing?
Answer:
Nasal breathing is advisable because it is healthier as the air inhaled gets filtered in the nose so cleaner air goes to the lungs.

Question 8.
A major percentage of O₂ (97%) is transported by RBCs in the blood. How does the remaining 3% of O₂ transported? [NCERT Exemplar]
Answer:
In dissolved state through plasma.

Question 9.
What is the amount of O₂ supplied to tissues through every 100 mL of oxygenated blood under normal physiological conditions? [NCERT Exemplar]
Answer:
Around. 5 mL.

Question 10.
Cigarette smoking causes emphysema. Give reason. [NCERT Exemplar]
Answer:
Cigarette smoking cause damages to the alveolar walls due to alveolar sacs remaining filled with air leading to decreased respiratory surface for exchange of gases.

Question 11.
What causes snoring?
Answer:
Partial blocking of upper respiratory tract by tongue leading to turbulence of airflow causes a rough rattling inspiratory noise called snoring.

Question 12.
Give the name of some common respiratory disorders.
Answer:
Asthma, emphysema, occupational lung disorders (e.g., Silicosis and asbestosis).

Short answer type questions

Question 1.
Explain the structure of thoracic chamber.
Answer:
Anatomically, thoracic chamber is an air-tight chamber. The thoracic chamber is formed dorsally by the vertebral column, ventrally by the sternum, laterally by the ribs, and on the lower side by the dome-shaped diaphragm. The anatomical setup of lungs in thorax is such that any change in the volume of the thoracic cavity will be reflected in the lung (pulmonary) cavity. Such an arrangement is essential for breathing, as we cannot directly alter the pulmonary volume.

Question 2.
Explain various steps of respiration.
Answer:
Steps of Respiration

• Breathing or pulmonary ventilation by which atmospheric air is drawn in and CO₂-rich alveolar air is released out.
• Diffusion of gases (O₂ and CO₂) across alveolar membrane.
• Transport of gases by the blood.
• Diffusion of O₂ and CO₂ between blood and tissues.
• Utilization of O₂ by the cells for catabolic reactions and resultant release of CO₂.

Question 3.
Give a brief account of exchange of gases during respiration.
Answer:
Alveoli are the primary sites of exchange of gases. Exchange of gases also occur between blood and tissues. O₂ and CO₂ are exchanged in these sites by simple diffusion mainly based on pressure/concentration gradient.
Solubility of the gases as well as the thickness of the membranes involved in diffusion are also some of the important factors that can affect the rate of diffusion. Pressure contributed by an individual gas in a mixture of gases is called partial pressure and is represented as pO₂ for oxygen and pCO₂ for carbon dioxide.

Question 4.
What do you understand by occupational respiratory disorders?
Answer:
In some industries, workers may inhale harmful dust. For example, workers in stone-crushing plant may inhale silica dust. When there is long-term exposure to such situations, lungs can develop inflammation, which leads to fibrosis and ultimately to serious damage to lungs.

Long answer type questions

Question 1.
Explain the mechanism of expiration under normal conditions.
Answer:
Expiration: It is a passive process by which CO₂ is expelled out from the lungs. It takes place when the intra-pulmonary pressure is higher than the atmospheric pressure.
The movement of following muscles are involved :
(a) Diaphragm: The muscle fibres of the diaphragm relax making it convex, decreasing volume of the thoracic cavity.
(b) Internal intercostal muscles: These muscles contract thus, pulling the ribs downward and inward, decreasing the thoracic volume.

The overall volume of the thoracic cavity thus decreases thereby reducing the pulmonary volume.
Ribs and sternum returned to original position (lowered)
As a result, the intrapulmonary pressure increases slightly above the atmospheric pressure. This inturn causes the expulsion of the air from the lungs. The process of expiration is simpler than inspiration.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 18 Body Fluids and Circulation Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 18 Body Fluids and Circulation

Very short answer type questions

Question 1.
Mention the total amount of normal leucocyte count in human.
Answer:
6000-8000 per cubic mm is the normal leucocyte count in human.

Question 2.
A person has a blood group AB positive. What does it mean?
Answer:
AB positive means that a person has both A and B antigens and also has Rh factor in his blood.

Question 3.
Comment. Blood is called river of life.
Answer:
It is called so, because blood plasma helps in transportation of materials like nutrients, gases, wastes, hormones, etc., within the body, which is very essential for the survival of life.

Question 4.
Why is blood group identification not needed for serum identification?
Answer:
Because serum does not have the coagulation/clotting factor.

Question 5.
Due to developmental abnormality, the wall of the left ventricle of an infant’s heart is as thin as that of right ventricle. What would be its specific effect in circulation of blood It may not be able to develop sufficient pressure to pump blood in all distant parts.

Question 6.
Why are veins provided with valves along their length?
Answer:
Valves are presept to prevent the backward flow of blood.

Question 7.
Where does the cardiac impulse originates?
Answer:
The cardiac impulse originates in cardiac muscle fibres and is not brought to the heart by any nerve fibres. The origin of cardiac impulse is said to be myogenic.

Question 8.
Given below are the abnormal conditions related to blood circulation. Name the disorders.
Acute chest pain due to failure of O₂ supply to heart muscles. Increased systolic pressure. [NCERT Exemplar]
Answer:
(a) Angina,
(b) High blood pressure.

Question 9.
Heart failure is called congestive heart failure. Why?
Answer:
The congestion of lungs is a symptom of heart failure. Thus, it is also called congestive heart failure.

Question 10.
Indicate the blood vessel that transports hormones from the hypothalamus to the anterior pituitary.
Answer:
Hypophyseal portal vein.

Question 11.
From where does the hepatic portal system brings the blood?
Answer:
Hepatic portal system brings blood from the alimentary canal, pancreas and spleen to the liver.

Question 12.
Write a short note on-coagulation of blood.
Answer:
Coagulation of Blood:
When an injury occurs, there is bleeding from the wound for some time, but soon the blood stops flowing out.
This is because blood exhibits a mechanism called blood coagulation or clotting, to prevent excess loss of blood from the body.

A clot or coagulum is formed which consists of a network of fibres called fibrin in which the dead and damaged corpuscles are trapped. The blood clot seals the injured blood vessel and thereby bleeding stops.

Short answer type questions

Question 1.
What is Rh incompatibility of mother and foetus? What are the complications and necessary precautions involved in this case?
Answer:
Rh Incompatibility of Foetus and Mother: A special case of Rh incompatibility (mismatching) has been observed between the Rh -ve blood of a pregnant mother with Rh +ve blood of the foetus. Rh antigens of the foetus do not get exposed to the Rh -ve blood of the mother in the first pregnancy as the two bloods are well separated by the placenta. However/during the delivery of the first child, there is a possibility of exposure of the maternal blood to small amounts of the Rh +ve blood from the foetus.

In such cases, the mother starts preparing antibodies against Rh in her blood. In case of her subsequent pregnancies, the Rh antibodies from the mother (Rh -ve) can leak into the blood of the foetus (Rh +ve) and destroy the foetal RBCs. This could be fatal to the foetus or could cause severe anaemia and jaundice to the baby. This condition is called erythroblastosis foetalis. This can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the first child.

Question 2.
How is cardiac activity regulated by the nervous system?
Answer:
Normal activities of the heart are regulated intrinsically, i.e., auto regulated by specialised muscles (nodal tissue), hence the heart is called myogenic. A special neural centre in the medulla oblongata can moderate the cardiac function through autonomic nervous system (ANS).

Neural signals through the sympathetic nerves (part of ANS) can increase the rate of heartbeat, the strength of ventricular contraction and thereby the cardiac output. On the other hand, parasympathetic neural signals (another component of ANS) decrease the rate of heartbeat, Speed of conduction of action potential and thereby the cardiac output. Adrenal medullary hormones can also increase the cardiac output.

Question 3.
Describe systemic circulation.
Answer:
The oxygenated blood entering the aorta is carried by a network of arteries, arterioles and capillaries to the tissues from where the deoxygenated blood is collected by a system of venules, veins and vena cava and emptied into the right atrium. This is the systemic circulation. The systemic circulation provides nutrients, O₂ and other essential substances to the tissues and takes CO₂ and other harmful substances away for elimination.

Question 4.
What is stroke volume? What is its relation with cardiac output?
Answer:
During one cardiac cycle or one heartbeat, the volume of blood pumped by the heart is called stroke volume. This is normally 70 mL. In one minute the heart beats about 72 times and the amount of blood pumped per minute is called cardiac output. This is usually 4900 mL ~ 5 litres.

Question 5.
Write a short note on-disorders of circulatory system.
Answer:
Disorders of Circulatory System
(i) Hypertension
It is commonly known as high blood pressure, to indicate a blood pressure that is higher than the normal, i.e., 120/80 mm Hg.
A sustained high blood pressure of 140/90 mm Hg or higher, is called hypertension.
It leads to heart diseases and affects the functioning of vital organs like kidneys and brain.

(ii) Coronary artery disease (CAD)
It is a disorder which affects the blood vessels (coronary arteries) that supply blood to the heart muscles.
Atherosclerosis is a form of CAD, which is caused by the deposition of cholesterol on the wall lining of the lumen of blood vessels.

It makes the lumen narrow and reduces the blood flow to the heart.
When the cholesterol deposits on the wall of blood vessels become calcified and hardened, the condition is called arteriosclerosis; such blood vessels lose their elasticity and become stiff, apart from having narrow lumen.

(iii) Angina pectoris
It is commonly called angina and occurs due to any condition that affects the blood flow to the heart muscle.
Due to this, enough of oxygen is not supplied to the heart muscle and a symptom of acute chest pain appears.
It can occur in men and women of any age.

(iv) Heart failure
It is the condition or state of the heart when it cannot pump sufficient blood to meet the needs of the body.
More often the cause for this condition is congestion of lungs; hence it is called congestive heart failure.
Heart failure is different from cardiac arrest, where the heart stops beating and heart attack, where the heart muscle is damaged suddenly due to insufficient blood supply.

Long answer type questions

Question 1.
Explain different types of blood groups and donor compatibility making a table. [NCERT Exemplar]
Answer:
Two groupings, i.e., the ABO and Rh are widely used all over the world. ABO grouping is based on the presence or absence of two surface antigens (chemicals that can induce immune response) on the RBCs, i.e., A and B. Similarly, the plasma of different individuals contain two natural antibodies (proteins produced in response to antigens).
Blood Groups and Donor Compatibility:

Blood Group	Antigen on RBCs	Antibody in Plasma	Donor’s Group
A	A	Anti-B	A, 0
B	B	Anti-A	B, 0
AB	A, B	Nil	AB, A, B, 0
0	Nil	Anti-A, B	0

From the above-mentioned table it is evident that group ‘O’ blood can be donated to persons with any other blood group and hence ‘O’ group individuals are called ‘universal donors’. Persons with ‘AB’ group can accept blood from persons with AB as well as the other groups of blood. Therefore, such persons are called ‘universal recipients’.

Question 2.
Describe briefly the internal structure of human heart with neat and labelled diagram.
Answer:
Draw a diagram to show the internal structure of human heart. Internal Structure: Internally, the chambers of heart, i.e., two auricles (atria) and ventricles are separated by different septa and valves.
(a) Auricles (Atria): These are the upper two thin-walled and smaller chambers. These serve to receive the blood, therefore are called receiving chambers (right atrium and left atrium). Both the right and the left atria are separated by a thin, muscular wall known as interatrial septum. Right Atrium: This right chamber deals with only impure (deoxygenated) blood. It receives impure blood from various parts of the body, through two major veins, i.e., superior and inferior vena cava. It also receives blood from the walls of the heart itself (through a coronary sinus).

(b) Left Atrium: This chamber is meant to deal with only pure (oxygenated) blood. It receives blood (pure) from lungs through two pulmonary veins (i.e., one from the each lung).
Ventricles: These are lower two chambers of the heart, that pumps the blood away from the heart. Thus, function as pumping chambers. Both the right and the left ventricles are separated by the interventricular septum. The atrium and the ventricle of the same side are also separated by another septum, a thick fibrous tissue called atri oventricular septum (i.e., AV septum).

(a) Right Ventricle: It receives impure blood from right atrium and pumps to pulmonary artery, which further takes this blood to lungs for purification.

(b) Left Ventricle: It receives pure (oxygenated) blood from left atrium and pumps its pure blood to aorta (largest artery in the pathway), which in turn takes this blood to whole body and organs.

Cardiac Valves: Apart from septum, heart is also separated by the various valves. These valves act as a door-like structure in the heart that serves to maintain the unidirectional flow of blood.

Different valves present in the heart are given below :

• Tricuspid Valve: It is formed by three muscular flaps or cusps to guard the opening between the right atrium and the right ventricle.
• Bicuspid Valve: (Mitral valve) It is the type of valve that guards the opening between the left atrium and the left ventricle.
• Semilunar Valve: The opening of the right and the left ventricles into the pulmonary artery and the aorta respectively are provided with the semilunar valves.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 5 Morphology of Flowering Plants Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 5 Morphology of Flowering Plants

Very short answer type questions

Question 1.
In Swampy areas like Sunderbans in West Bengal, plants bear special kind of roots called [NCERT Exemplar]
Answer:
Pneumatophores. These are respiratory roots present in mangrove plants which grow in saline areas.

Question 2.
In aquatic plants like Pistia and Eichhomia, leaves and roots are found near [NCERT Exemplar]
Answer:
Water surface. This helps in balancing the plants over water surface.

Question 3.
Why some tap roots become swollen and fleshy?
Answer:
These roots store food in them.

Question 4.
Why insects attract towards pitcher plants?
Answer:
In pitcher plants, the leaf apex gives rise to a coloured lid for attracting the insects.

Question 5.
When the corolla is described as gamopetalous?
Answer:
The corolla is described as gamopetalous when it has fused petals.

Question 6.
Describe the fruit of Allium cepa (onion).
Answer:
A loculicidal capsule with endospermic seeds.

Question 7.
In epigynous flower, ovary is situated below the [NCERT Exemplar]
Answer:
Thalamus of the flower.

Question 8.
A maize grain is not a seed. Explain.
Answer:
A maize grain is a single seeded fruit called caryopsis or grain, in which the pericarp (fruit wall) is inseparably fused with testa.

Question 9.
How superior and inferior ovaries are indicated by symbol?
Answer:
Superior ovary, e.g., G. Inferior ovary, e.g., \(\overline{\mathrm{G}}\)

Question 10.
What does these symbol indicate ⊕ and ⊕ ?
Answer:
⊕ – Actinomorphic, + -Zygomorphic.

Question 11.
Add the missing floral organs of the given formula of Fabaceae, [NCERT Exemplar]
Answer:

Question 12.
Write the floral formula of Liliaceae.
Answer:

Short answer type questions

Question 1.
Describe the various regions of the root.
Answer:
Regions of the Root: The root is covered at the apex by a thimble-like structure called the root cap. It protects the tender apex of the root as it makes its way through the soil. A few millimeters above the root cap is the region of meristematic activity. The cells of this region are very small, thin-walled and with dense protoplasm. They divide repeatedly.

The cells proximal to this region undergo rapid elongation and enlargement and are responsible for the growth of the root in length. This region is called the region of elongation. The cells of the elongation zone gradually differentiate and mature. Hence, this zone, proximal to region of elongation, is called the region of maturation. From this region some of the epidermal cells form very fine and delicate, thread-like structures called root hairs. These root hairs absorb water and minerals from the soil.

Question 2.
Write the structure and functions of the stem.
Answer:
The Stem: The stem is the ascending part of the axis bearing branches, leaves, flowers and fruits. It develops from the plumule of the embryo of a germinating seed. The stem bears nodes and internodes. The region of the stem where leaves are born are called nodes while internodes are the portions between two nodes. The stem bears buds, which may be terminal or axillary. Stem is generally green when young and later often become woody and dark brown.

The main function of the stem is spreading out branches bearing leaves, flowers and fruits. It conducts water, minerals and photosynthates. Some stems perform the function of storage of food, support, protection and of vegetative propagation.

Question 3.
Explain the structure of leaf.
Answer:
A typical leaf consists of three main parts leaf base, petiole and lamina. The leaf is attached to the stem by the leaf base and may bear two lateral small leaf like structures called stipules. In monocotyledons, the leaf base
expands into a sheath covering the stem partially or wholly. In some leguminous plants the leafbase may become swollen, which is called the pulvinus. The petiole help hold the blade to light. Long thin flexible petioles allow leaf blades to flutter in wind, thereby cooling the leaf and bringing fresh air to leaf surface. The lamina or the leaf blade is the green expanded part of the leaf with veins and veinlets.

There is, usually, a middle prominent vein, which is known as the midrib. Veins provide rigidity to the leaf blade and act as channels of transport for water, minerals and food materials. The shape, margin, apex, surface and extent of incision of lamina varies in different leaves.

Question 4.
Describe the venation of leaf in brief.
Answer:
The arrangement of veins and the veinlets in the lamina of leaf is termed as venation. When the veinlets form a network, the venation is termed as reticulate. When the veins run parallel to each other within a lamina, the venation is termed as parallel. Leaves of dicotyledonous plants generally possess reticulate venation, while parallel venation is the characteristic of most monocotyledons.

Question 5.
What is the difference between simple leaf and compound leaf?
Answer:
In simple leaf lamina is usually entire and when the lamina is showing incision, the incision do not touch the midrib.
In compound leaf, the incision on lamina reach up to the midrib, which results in number of leaflets. Presence or absence of axillary bud also shows the difference between leaf and leaflets.

Question 6.
What is the difference between valvate and twisted aestivation?
Answer:
In valvate aestivation sepals or petals don’t overlap, while in twisted aestivation sepals or petals slightly overlap.

Question 7.
What is the difference between a mango fruit and a coconut fruit in terms of edible part?
Answer:
Edible part in mango: Mesocarp
Edible part in coconut: Seed.

Long answer type questions

Question 1.
What is the difference and similarity between prop root and stilt root?
Answer:
Difference: Prop roots come out of branches of the main stem and they come from greater heights. Stilt roots come out from the main stem and they come out from just above the ground.
Similarity: Both prop roots and stilt roots give additional mechanical support to the plant.

Question 2.
Write the description of gynoecium in various plants.
Answer:
The description of gynoecium varies in following ways :

• Carpels: Monocarpellary/bicarpellary/tricar pellary/tetracarpellary/ multi carpellary.
• Cohestion: Apocarpous/syncarpous.
• Ovary: Superior/semi-inferior/inferior.
• Placentation: Marginal/axile/parietal/basal/ffee- central/superficial
• Style: Terminal/lateral/gynobasic/stylopodium.
• Stigma: Number, shape-simple, lobed, capitate, branched

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 14 Oscillations Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 14 Oscillations

Very Short Answer Type Questions

Question 1.
What are the basic properties required by a system to oscillate?
Answer:
Inertia and elasticity are the properties which are required by a system to oscillate.

Question 2.
All oscillatory motions are periodic and vice-versa. Is it true?
Answer:
No, there are other types of periodic motions also. Circular motion and rotatory motion are periodic but non-oscillatory.

Question 3.
Give three important characteristics of a SHM.
Answer:
Three important characteristics of an SHM are amplitude, time period (or frequency) and phase.

Question 4.
What is the force equation of a SHM?
Answer:
According to force equation of SHM, F = -kx,
where k is a constant known as force constant.

Question 5.
Under what condition is the motion of a simple pendulum be simple harmonic? (NCERT Exemplar)
Answer:
When the displacement amplitude of the pendulum is extremely small as compared to its length.

Question 6.
A simple pendulum is transferred from earth to the surface of Moon. How will its time period be affected?
Answer:
As value of g on Moon is less than that on earth, in accordance with the relation T = \(2 \pi \sqrt{l / g}\) , the time period of oscillations of a simple pendulum on Moon will be greater.

Short Answer Type Questions

Question 1.
A girl is swinging in the sitting position. How will the period of the swing be changed if she stands up?
Solution:
This can be explained using the concept of a simple pendulum. We know that the time period of a simple pendulum is given by
T = \(2 \pi \sqrt{\frac{l}{g}} \text { i.e., } T \propto \sqrt{l}\)
When the girl stands up, the distance between the point of suspension and the center of mass of the swinging body decreases i.e., I decreases, so T will also decrease.

Question 2.
A particle is subjected to two simple harmonic motions
x₁ = A₁ sinωt
And
x₂ = A₂ sin \(\left(\omega t+\frac{\pi}{\mathbf{3}}\right)\)
Find (i) the displacement at t = 0
(ii) the maximum speed of the particle and
(iii) the maximum acceleration of the particle
Solution:
(i) At t = 0, x₁ = A₁ sin ωt = 0
And
x₂ = A₂ sin \(\left(\omega t+\frac{\pi}{3}\right)=\frac{A_{2} \sqrt{3}}{2}\)
Thus the resultant displacement at t = 0 is

Question 3.
The maximum acceleration of a simple harmonic oscillator Is a₀ and the maximum velocity is v₀. What is the displacement amplitude?
Solution:
Let A be the displacement amplitude and o be the angular frequency of the simple harmonic oscillator.
Then, a₀ = ω²A ……………………………. (i)
and v₀ = ωA …………………………………………………. (ii)
Squaring eq. (ü) and dividing from eq. (j), we get
\(\frac{v_{0}^{2}}{a_{0}}=\frac{\omega^{2} A^{2}}{\omega^{2} A}\) = A or A = \(\frac{v_{0}^{2}}{a_{0}}\)

Question 4.
A particle performs SHM on a rectilinear path. Starting from rest, it travels x₁ distance in first second, and in the next second, it travels x₂ distance. Find out the amplitude of this SHM.
Solution :
Because the particle starts from rest, so its starting point will be extreme position.
Thus, the displacement of the particle from the mean position after one second
A-x₁ = A cos ωt = A cos ω ……………………………… (i) [puttingt =1 s]
where A is the amplitude of the SHM and for next second
A – (x₁ + x₂) = Acosωt
= Acos2ω = A[2cos²ω-1]
[putting t = 2s]
[ ∵ cos 2 ω =
2 [cos² ω -1] ……………………………………………. (ii)
From eqs. (i) and (ii), we have

Question 5.
Apartide is executing SHM. If ν₁ and ν₂ are the speeds of the particle at distance x₁ and x² from the equilibrium position, show that the frequency of oscillations is
f = \(\frac{1}{2 \pi}\left(\frac{v_{1}^{2}-v_{2}^{2}}{x_{2}^{2}-x_{1}^{2}}\right)^{1 / 2} \)
Solution:
The displacement of a particle executing SHM is given by
x = Acosωt
\(\frac{d x}{d t}\) = – ωAsin ωt
∴ velocity,ν = \(\frac{d x}{d t}\)
or ν²=A²ω²sin2ωt

Subtracting eq. (ii) from eq. (i), we get

Question 6.
Define the restoring force and it characteristics in case of an oscillating body.
Answer:
A force which takes the body back towards the mean position in oscillation is called restoring force. Characteristic of Restoring force: The restoring force is always directed towards the mean position and its magnitude of any instant is directly proportional to the displacement of the particle from its mean position of that instance.

Long Answer Type Questions

Question 1.
A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s^-1 and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.
(i) Wifi there be any change in weight of the body, during the oscillation?
(ii) If answer to part (i) is yes, what will be the maximum and minimum reading In the machine and at which position? (NCERT Exemplar)
Solution:
This is a case of variable acceleration. In accelerated motion, weight of body depends on the magnitude and direction of acceleration for upward or downward motion.
(i) Hence, the weight of body changes.
(ii) Considering the situation in two extreme positions, as their acceleration is maximum in magnitude.

Wehave mg-N=ma
Note at the highest point, the platform is accelerating downward.
⇒ N=mg – ma

But a = ω²A (in magnitude)
∴ N = mg – mω²A
where, A = amplitude of motion
Given, m = 50 kg, frequency v = 2 s^-1
∴ ω = 2πv = 4πrad/s
A = 5cm = 5 x 10^-2 m
∴ N = 50 x 9.8 – 50 x (4π²) X 5 x 10^-2
= 50 [9.8-16π² x 5 x 10^-2]
= 50 [9.8 – 7.89] = 50 x 1.91 = 95.5N

When the platform is at the lowest position of its oscillation,

It is accelerating towards mean position that is vertically upwards. Writing the equation of motion
N – mg = ma = mω²A
or N = mg + mat2A = m [g + ω²A]
Putting the data

Now, the machine reads the normal reaction.
It is clear that maximum weight = 884 N (at lowest point)
minimum weight = 95.5 N (at top point)

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 1 Physical World Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 1 Physical World

Very short answer type questions

Question 1.
Why do we call Physics an exact Science?
Answer:
Most of measurement in Physics are made with high precise and accuracy, so it is called an exact Science.

Question 2.
Give two approaches to study physics.
Answer:
Two approaches to study physics are unification and reduction.

Question 3.
Name the scientific principle behind the technology of steam engine.
Answer:
Laws of thermodynamics is the scientific principle behind the technology of steam engine.

Question 4.
Give one major discovery resulted due to basic laws of electricity and magnetism.
Answer:
Wireless communication technology was a major discovery due to laws of electricity and magnetism.

Question 5.
What is the range of weak nuclear force?
Answer:
The range of a weak nuclear force is of the order of 10^-16 m.

Question 6.
Give an example of achievement in unification.
Answer:
Unified celestial and terrestrial mechanics showed that the same laws of motion and the law of gravitation apply to both the domains.

Question 7.
Give an example for conservation law of energy.
Answer:
A freely falling body under gravity is an example of conservation law of energy.

Short answer type questions

Question 1.
Give the salient features of Einstein’s theory.
Answer:
According to Einstein

• Mass and energy are interconvertible.
• Space and time are interconnected.

Question 2.
Name the phenomena/fields with which microscopic domain of physics deals. Which theory explains these phenomena?
Answer:
The microscopic domain of physics deals with the constitution and structure of matter at atomic and nuclear scale.
The Questionuantum theory is currently accepted, as the proper framework for explaining microscopic phenomena.

Question 3.
Name three important discoveries of physics, which have revolutionised modem chemistry.
Answer:
Three important discoveries of physics, which have revolutionised modem chemistry are :

1. study of radioactivity,
2. quantum theory
3. study of isotopes and determination of their masses by mass spectrographs.

Question 4.
Name four fundamental forces in nature.
Answer:
Four fundamental forces present in nature are:

• Gravitational force
• Electromagnetic force
• Weak nuclear force
• Strong nuclear force.

Question 5.
Name three important discoveries of physics, which have contributed a lot in development of biological sciences.
Answer:
The most important discoveries of physics, which have contributed a lot in development of biological sciences are :

• Ultrasonic waves.
• X-rays and neutron diffraction technique.
• Electron microscope.
• Radio isotopes.

Question 6.
Briefly explain how physics is related to technology?
Answer:
Progress in the field of science and technology is interrelated. Sometimes technology gives rise to new physics and at other times physics generates new technology. The discipline of thermodynamics arose mainly to understand and improve the working of heat engines. Similarly discovery of basic laws of electricity and magnetism led to development of wireless communication technology. Therefore, we can conclude that physics and technology are closely related.

Long answer type questions

Question 1.
How Physics is related to other sciences?
Answer:
Physics is so important branch of science that without the knowledge of Physics, other branches of science cannot make any progress. This can be seen from the following:

(a) Physics in relation to Mathematics: The theories and concepts of Physics lead to the development of various mathematical tools like differential equations, equations of motion etc.

(b) Physics in relation to Chemistry: The concept of interaction between various particles lead to understand the bonding and the chemical structure of a substance. The concept of X-ray diffraction and radioactivity had helped to distinguish between the various solids and to modify the periodic table.

(c) Physics in relation to Biology: The concept of pressure and its measurement has helped us to know the blood pressure of a human being, which in turn is helpful to know the working of heart. The discovery of X-rays has made it possible to diagonose the various diseases in the body and fracture in bones. The optical and electron microscopes are helpful in the studies of various organisms. Skin diseases and cancer can be cured with the help of high energy radiations like X-rays, ultraviolet rays.

(d) Physics in relation to Geology: The internal structure of various rocks can be known with the study of crystal structure. Age of rocks and fossils can be known easily with the help of radioactivity i. e., with the help of carbon dating.

(e) Physics in relation to Astronomy: Optical telescope has made it possible to study the motion of various planets and satellites in our solar system.
Radio telescope has helped to study the structure of our galaxy and to discover pulsars and quasars (heavenly bodies having star like structure). Pulsars are rapidly rotating neutron stars. Doppler’s effect predicted the expAnswer:ion of universe. Kepler’s laws are responsible to understand the nature of orbits of the planets around the sun.

(f) Physics in relation of Meterology: The variation of pressure with temperature leads to forecast the weather.

(g) Physics in relation to Seismology: The movement of earth’s crust and the types of waves produced help us in studying the earthquake and its effect.

Question 2.
Write short note on origin and Fundamental forces in nature.
Answer:
These are the. following four basic forces in nature:
(a) Gravitational forces
(b) Electromagnetic forces
(c) Strong force or nuclear forces
(d) Weak forces.
Some of the important features of these forces are discussed below:

(a) Gravitational forces: These are the forces of attraction between any two bodies in the universe due to their masses separated by a definite distance. These are governed by Newton’s law of gravitation given by

where, m₁, m₂ are the masses of two bodies
r = distance between them
G = Universal gravitational constant
= 6.67 × 10¹¹ Nm²kg²

Characteristics of Gravitational Forces

• They are always attractive. They are never repulsive. They exist between macroscopic as well as microscopic bodies.
• They are the weakest forces in nature.
• They are central forces in nature i. e., they set along the line joining the centres of two bodies.
• They are conservative forces.
• They obey inverse square law i.e.,F ∝ \(\frac{I}{r^{2}}\) they vary inversely as the
  square of the distance between the two bodies.
• They are long range forces i.e., gravitational forces between any two bodies exist even when their distance of separatioji is quite large.
• The field particles of gravitational forces are called gravions. The concept of exchange of field particles between two bodies explains how the two bodies interact from a distance.

(b) Electromagnetic forces: They include the electrostatic and magnetic forces. The electrostatic forces are the forces between two static charges while magnetic forces are the forces between two magnetic poles. The moving charges give rise to the magnetic firce. The combined action of these forces are called electromagnetic forces.
Characteristics of Electromagnetic Forces

• These forces are both attractive as well as repulsive.
• They are central forces in anture.
• They obey inverse sQuestionuare law.
• They are conservative forces in nature.
• These forces are due to the exchange of particles known as photons which carry no charge and have zero rest mass.
• They are 10 times stronger as compared to gravitational forces and 1011 times stronger than the weak forces.

(c) Strong forces: They are the forces of nuclear origin. The particles inside the nucleus are charged particles (protons) and neutral particles (neutrons) which are bonded to each other by a strong interaction called nuclear force or strong force.
Hence they may be defined as the forces binding the nucleons (protons and neutrons) together in a nucleus. They are responsible for the stability of the atomic nucleus. They are of three types :

1. n-n forces are the forces of attraction between two neutrons.
2. p-p forces are the forces of attraction between two protons.
3. n-p forces are the forces of attraction between a proton and a neutron.

Characteristics of Strong Forces

• They are basically attractive in nature and become repulsive when the distance between nucleons is less than 0.7 fermi.
• They obey inverse square law.

(d) Weak forces: They are defined as the interactions which take place between elementary particles during radioactive decay of a radioactive substance. In β – decay, the nucleus changes into a proton, an electron and a particle called anti-neutrino (which is uncharged). The interaction between the electron and the anti-neutrino is known as weak interaction or weak force.

Characteristics of Weak Forces

• They are 10²⁵ times stronger than the gravitational forces.
• They exist between leptons and leptons, leptons and mesons etc.
  (a) and (b) types are the forces that we encouncer in macroscopic world while (c) and (d) types are the forces that we encountered in microscopic world.

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 2 Units and Measurements Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 2 Units and Measurements

Very short answer type questions

Question 1.
Is it possible to have length and velocity both as fundamental quantities? Why?
Answer:
No, since length is fundamental quantity and velocity is the derived quantity.

Question 2.
Which of these is largest: astronomical unit, light year and par sec?
Answer:
Par sec is larger than light year which in turn is larger than an astronomical unit.

Question 3.
Define one Bam. How it is related with metre?
Answer:
One bam is a small unit of area used to measure area of nuclear cross-section.
∴ 1 barn = 10^-28 m²

Question 4.
What is meant by angular diameter of moon?
Answer:
Angular diameter of moon is the angle subtended at a point on the earth, by two diameterically opposite ends of the moon. Its value is about 0.5°.

Question 5.
Name the device used for measuring the mass of atoms and molecules. (NCERT Exemplar)
Answer:
Spectrograph.

Question 6.
Write the dimensional formula of wavelength and frequency of a wave.
Answer:
Wavelength [λ] = [L]
Frequency [v] = [T^-1]

Question 7.
Obtain the dimensional formula for coefficient of viscosity.
Answer:
Coefficient of viscosity (η) = \(\frac{F d x}{A \cdot d v}\)
= \(\frac{\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]}{\left[\mathrm{L}^{2}\right]\left[\mathrm{LT}^{-1}\right]}\) = [M¹L^-1T^-1]

Question 8.
Write three pairs of physical quantities, which have same dimensional formula.
Answer:

• Work and energy
• Energy and torque
• Pressure and stress

Short answer type questions

Question 1.
Does AU and Å represent the same unit of length?
Answer:
No, AU and Å represent two different units of length.
1 AU = 1 astronomical unit = 1.496 x 10¹¹ m
1Å = 1 angstrom = 10^-10 m

Question 2.
What is common between bar and torr?
Solution:
Both bar and torr are the units of pressure.
1 bar =1 atmospheric pressure = 760 mm of Hg column .
= 10⁵ N/m²
1 torr = 1 mm of Hg column
bar 760 torr

Question 3.
Why has second been defined in term of periods of radiations from cesium-133?
Answer:
Second has been defined in terms of periods of radiation, because

• this period is accurately defined.
• this period is not affeced by change of physical conditions like temperature, pressure and volume etc.
• the unit is easily reproducible in any good laboratoty.

Question 4.
Why parallax method cannot be used for measuring distances of stars more than 100 light ýears away?
Answer:
When a star is more than loo light years away, then the parallax angle is so small that it cannot be measured accurately.

Question 5.
What is the technique used for measuring large time intervals?
Answer:
For measuring large time intervals, we use the technique of radioactive dating. Large time intervals are measured by studying the ratio of number of radioactive atoms decayed to the number of surviving atoms in the
specimen.

Question 6.
Using the relation E = hv, obtain the dimensions of Planck’s constant.
Answer:
We know that dimensional formula of energy E of photon is [M¹L²T^-2
and dimensional formula of frequency is y is [T^-1].
The given relation is E = hv
[h] = \(\frac{[E]}{[v]}=\frac{\left[M^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}\) = M¹L²T^-1

Question 7.
The rotational kinetic energy of a body is given by E = \(\frac {1}{2}\)Iω², where ω is the angular velocity of the body. Use the equation to obtain dimensional formula for moment of inertia I. Also write
its SI unit.
Solution:
The given relation is E = \(\frac {1}{2}\)Iω²

Its SI unit is Joule.

Question 8.
Distinguish between dimensional variables and dimensional constants. Give example too.
Answer:
Dimensional variables are those quantities which have dimensions and whose numerical value may change. Speed, velocity, acceleration etc. are dimensional variables.

Dimensional constants are quantities having dimensions but having a constant value, e.g., gravitation constant (G), Planck’s constant (H), Stefan’s constant (σ) etc.

Question 9.
Dow will you convert a physical quantity from one unit system to another by method of dimensions?
Solution:
If a given quantity is measured in two different unit system, then Q = n₁u₁ = n₂u₂.
Let the dimensional formula of the quantity be [M^aL^bT^c], then we have n₁ [M₁^aL₁^bT₁^c ] = n₂ [M₂^aL₂^bT₂^c]
Here M₁, L₁, T₁ are the fundamental units of mass, length and time in
first unit system and M₂, L₂, T₂

This relation helps us to convert a physical quantity from one unit system to another.

Question 10.
The displacement of a progressive wave is represented by y = A sin (ωt – kx), where x is distance, and t is time. Write the dimensional formula of (i) ω and (ii) k. (NCERT Exemplar)
Solution:
Now, by the principle of homogeneity, i. e., dimensions of LHS and RHS should be equal, hence
[LHS] = [RHS]
⇒ [L] = [A] = L
As ωt – kx should be dimensionless,
[ωt] [kx] = 1
⇒ [ω]T = [k]L= 1
⇒ [ω] = T^-1 and [k] = L^-1

Question 11.
Which of the following time measuring devices is most precise?
(a) A wall clock
(b) A stop watch
(c) A digital watch
(d) An atomic clock
Give reason for your answer. (NCERT Exemplar)
Solution:
A wall clock can measure time correctly upto one second. A stop watch can measure time correctly upto a fraction of a second. A digital watch can measure time up to a fraction of second. An atomic clock can measure time most precisely as its precision is 1 s in 10¹³ s.

Long answer type questions

Question 1.
A large fluid star oscillates in shape under the influence of its own gravitational field. Using dimensional analysis, find the expression for period of oscillation (T) in terms of radius of star (R), Mean density of fluid (ρ) and universal gravitational constant (G).
Solution:
Suppose period of oscillation T depends on radius of star R, mean density of fluid p and universal gravitational constant (G) as
T = kR^a ρ^b G^c,where kis a dimensionless constant
Writing dimentions on both sides of the equation, we have
[M⁰L⁰T¹]=[L]^a[ML^-3]^b[M^-1L³T^-2]^c
= M^{b – c}L^{a – 3b + 3c}T^-2c
Comparing powers of M, L and T, we have
b – c = 0;
a – 3b + 3c = 0 and -2c = 1
On simplifying these equations, we get
c = -1/2,b = -1/2, a = 0
Thus, we have T = kρ^-1/2G^-1/2 = \(\frac{k}{\sqrt{\rho G}}\)

Question 2.
Find an expression for viscous force F acting on a tiny steel ball of radius,r,moving in a viscous liquid of viscosity q with a constant speed υ by the niethod of dimensional analysis.
Solution:
It is given that viscous force F depends on (i) radius r of steel ball, (ii) coefficient of viscosity η of viscous liquid (iii), Speed υ of the ball i.e.,F = kr^aη^bυ^c,where kis dimensionless constant
Writing dimensions on both sides of equation, we have
[MLT^-2] = [L]^a[M¹L^-1T^-1]^b[LT^-1]^c
= [M^aL^{a – b + c}T^{-b -c}]
Comparing powers of M, L and T on two sides of equation, we get
a = 1
a – b + c = 1
-b -c =-2
On solving, these above equations, we get ,
a = 1, b = 1 and c = 1
Hence, the relation becomes
F = krηυ