Important Questions

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 Biology Important Questions Chapter 19 Excretory Products and their Elimination Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 19 Excretory Products and their Elimination

Very short answer type questions

Question 1.
Write the mode of excretion performed by Xenopus.
Answer:
Dual excretion (mainly ammonotelism and partly ureotelism).

Question 2.
In which of the organism antennary glands are found as excretory organ?
Answer:
Crustaceans.

Question 3.
What is the excretory product from the kidney of reptiles? [NCERT Exemplar]
Answer:
Uric acid.

Question 4.
Give the name of vessel of peritubular capillaries that runs parallel to the loop of Henle.
Answer:
Vasa recta.

Question 5.
Give the name of the reservoir of urine in the body.
Answer:
Urinary bladder.

Question 6.
Give the name of cells that are responsible for the formation of filtration slits or slit pores.
Answer:
Podocytes (epithelial cells of Bowman’s capsule).

Question 7.
In which part of excretory system of mammals you can first use the term urine?
Answer:
The filtrate formed by the process of ultrafiltration in the Bowman’s capsule is called glomerular filtrate or primary urine.

Question 8.
What is the ratio of the concentrated filtrate to that of the initial filtrate?
Answer:
The concentrated urine (filtrate) is nearly four times concentrated than the initial filtrate formed.

Question 9.
What will be the effect on the amount of urine released when water is abundant in the body tissues in case of vertebrates?
Answer:
Vertebrates excrete large quantities of dilute urine when water is abundant in the body tissues and vice-versa.

Question 10.
What is the pH of urine? [NCERT Exemplar]
Answer:
It ranges from 4.5-8.2, average pH is 6.0.

Question 11.
What are the two substances responsible for causing the gradient for increasing hyperosmolarity of medullary interstitium?
Answer:
NaCl and urea.

Question 12.
Give the name of the main component that play an important role in the counter-current mechanism.
Answer:
Henle’s loop and vasa recta.

Short answer type questions

Question 1.
Describe the structure of human kidney.
Answer:

• Shape and Size of Kidney: Each kidney of an adult human measures 10-12 cm in length, 5-7 cm in width, 2-3 cm in thickness with an average weight of 120-170 g.
• Structure of Kidney: Towards the centre of the inner concave surface of the kidney is a notch called hilum through which ureter, blood vessels and nerves enter.
• Inner Structure: Inner to the hilum is a broad funnel-shaped space called the renal pelvis with projections called calyces. The outer layer of kidney is a tough capsule. Inside the kidney, there are two zones, an outer cortex and
• an inner medulla. The medulla is divided into a few conical masses (medullary pyramids) projecting into the calyces. The cortex extends in between the medullary pyramids as renal columns called Columns of Bertini.

Question 2.
What is the function of proximal convoluted tubules in the kidney?
Answer:
Function of Proximal Convoluted Tubule (PCT): PCT is lined by simple cuboidal brush border epithelium which increases the surface area for reabsorption. Nearly all of the essential nutrients, and 70-80 percent of electrolytes and water are reabsorbed by this segment. PCT also helps to maintain the pH and ionic balance of the body fluids by selective secretion of hydrogen ions, ammonia and potassium ions into the filtrate and by absorption of HCO from it.

Question 3.
What is the function of Henle’s loop?
Answer:
Function of Henle’s Loop: Reabsorption in this segment is minimum. However, this region plays a significant role in the maintenance of high osmolarity of medullary interstitial fluid. The descending limb of loop of Henle is permeable to water but almost impermeable to electrolytes. This concentrates the filtrate as it moves down. The ascending limb is impermeable to water but allows transport of electrolytes actively or passively. Therefore, as the concentrated filtrate pass upward, it gets diluted due to the passage of electrolytes to the medullary fluid.

Question 4.
What is the function of distal convoluted tubule?
Answer:
Function of Distal Convoluted Tubule (DCT): Conditional reabsorption of Na⁺ and water takes place in this segment. DCT is also capable of reabsorption of HCO and selective secretion of hydrogen and potassium ions and NH₃ to maintain the pH and sodium-potassium balance in blood.

Question 5.
What is the function of collecting duct?
Answer:
Collecting Duct

• Large amounts of water could be reabsorbed from this region.
• This segment also allows the transport of small amounts of urea into the medullary interstitium, to maintain the osmolarity.
• It also plays a role in maintaining pH and ionic balance of the body fluids by selective section of K⁺ and H⁺ ions.

Question 6.
How does reabsorption take place in the excretory system in human?
Answer:
Reabsorption: A comparison of the volume of the filtrate formed per day (180 litres per day) with that of the urine released (1.5 litres), suggest that nearly 99 per cent of the filtrate has to be reabsorbed by the renal tubules. This process is called reabsorption. The tubular epithelial cells in different segments of nephron perform this either by active or passive mechanisms. For example, substances like glucose, amino acids, Na+, etc., in the filtrate are reabsorbed actively whereas the nitrogenous wastes are absorbed by passive transport. Reabsorption of water also occurs passively in the initial segments of the nephron. During urine formation, the tubular cells secrete substances like H⁺, K⁺ and ammonia into the filtrate. Tubular secretion is also an important step in urine formation as it helps in the maintenance of ionic and acid base balance of body fluids.

Question 7.
Which gland releases ADH? What is the role of ADH in excretion?
Answer:
Hypothalamus releases ADH. ADH facilitates water reabsorption from latter part of tubules. This prevents diuresis. Excess fluid loss, through urine is called diuresis.

Question 8.
Write a short note on-disorders of the excretory system.
Answer:
Disorders of the Excretory System
The disorders related to kidneys are :

• Uremia
• Renal calculi and
• Glomerulonephritis

Hemodialysis is the process of removal of nitrogenous wastes from the blood of a uremia patient.
Kidney transplantation is the ultimate method of correcting urinary failure, in which a functioning kidney from a suitable donor is transplanted.

Hemodialysis

• Blood from the artery of an uremia patient is taken, cooled to 0°C and mixed with an anticoagulant like heparin.
• It is put into the cellophane tubes of the artificial kidney, where cellophane is permeable to micromolecules, but not to macromolecules like plasma protein.
• Outside the cellophane tube is the dialysing fluid, which has the same composition as that of plasma except the nitrogenous molecules like urea, uric acid, creatine, etc.
• Hence, the nitrogenous molecules from within the cellophane tubes flow into the dialysing fluid, following concentration gradient, (dialysis)
• The blood coming out of the artificial kidney is warmed to body temperature, mixed with antiheparin and restored to a vein of the patient.

Long answer type questions

Question 1.
The glomerular filtrate in the loop of Henle gets concentrated in the descending and then gets diluted in the ascending limbs. Explain. [NCERT Exemplar]
Answer:

• The gradient of increasing hyperosmolarity of medullary interstitium is maintained by a counter current mechanism and the proximity between the Henle’s loop and vasa recta.
• This gradient is mainly caused by NaCl and urea. The transport of these substances facilitated by the special arrangement of Henle’s loop and vasa recta is called the counter current mechanism.
• NaCl is transported by the ascending limb of Henle’s loop, which is exchanged with the descending limb of vasa-recta. NaCl is returned to the medullary interstitium by the ascending part of vasa recta.
• But, contrarily, the water diffuses into the blood of ascending limb of vasa recta and is carried away into the general blood circulation.
• Permeability to urea is found only in the deeper parts of thin ascending limbs of Henle’s loops and collecting ducts. Urea diffuses out of the collective ducts and enters into the thin ascending limb.
• A certain amount of urea recycled in this way is trapped in medullary interstitium by the collecting tubule. This mechanism helps in the maintenance of a concentration gradient in the medullary interstitium.
• Presence of such gradient helps in an easy passage of water from the collecting tubule, resulting in the formation of concentrated urine (filtrate) i.e., nearly four times concentrated than the initial filtrate formed.

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 13 Kinetic Theory Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 13 Kinetic Theory

Very Short Answer Type Questions

Question 1.
Under what conditions, real gases behave as an ideal gas?
Answer:
At low pressure and high temperature, real gases behave as an ideal gas.

Question 2.
When air is pumped into a cycle tyre, the volume and pressure of the air in the tyre, both are increased. What about Boyle’s law in this case? (NCERT Exemplar]
Answer:
When air is pumped, more molecules are pumped in Boyle’s law is stated for situation where number of molecules remain constant.

Question 3.
What is the minimum possible temperature on the basis of Charles’ law?
Answer:
The minimum possible temperature on the basis of Charles’ law is -273.15°C.

Question 4.
If a vehicle runs on the road for a long time, then the air pressure in the tyres increases. Explain.
Answer:
Due to the presence of friction between the road and tyres, the tyres get heated as a result of which temperature of air inside the tyre increases and hence pressure in tyre also increases.

Question 5.
What is the number of degree of freedom of a bee flying in a room?
Answer:
Three, because bee is free to move along x-direction or y-direction or z-direction.

Question 6.
How degree of freedom of a gas molecule is related with the temperature?
Answer:
Degree of freedom will increase when temperature is very high because at high temperature, vibrational motion of the gas will contribute to the kinetic energy. Hence, there is an additional kinetic energy associated with the gas, as a result of increased degree of freedom.

Question 7.
Is molar specific heat of a solid a constant quantity?
Answer:
Yes, the molar specific heat of a solid is a constant quantity and its value is 3 cal/mol-K.

Question 8.
Name experimental evidence in support of random motion of gas molecules.
Answer:
Brownian motion and diffusion of gases provide experimental evidence in support of random motion of gas molecules.

Question 9.
What is mean free path of a gas?
Answer:
The average distance travelled by a molecule between two successive collisions is known as mean free path of the molecule.

Short Answer Type Questions

Question 1.
State ideal gas equation. Draw a graph to check whether a real gas obeys this equation. What is the conclusion drawn?
Answer:
According to the ideal gas equation, we have PV = µRT
Thus, according to this equation \(\frac{P V}{\mu T}\) = R i.e., value of \( \frac{P V}{\mu T}\) must be a constant having a value 8.31 J mol^-1 K^-1. Experimentally value of \(\frac{P V}{\mu T}\) for real gases was calculated by altering the pressure of gas at different temperatures. The graphs obtained have been shown in the figure.

Here, for the purpose of comparison, graph for an ideal gas has also been drawn, which is a straight line parallel to pressure axis. From the graph it is clear that behaviour of real gases differ from an ideal gas. However, at high temperatures and low pressures behaviour is nearly same as that of an ideal gas.

Question 2.
Explain, why
(i) there is no atmosphere on Moon.
(ii) there is fall in temperature with altitude. (NCERT Exemplar)
Answer:
(i) The Moon has small gravitational force and hence the escape velocity is small. As the Moon is in the proximity of the Earth as seen from the Sun, the Moon has the same amount of heat per unit area as that of the Earth. The air molecules have large range of speeds.

Even though the rms speed of the air molecules is smaller than escape velocity on the Moon, a significant number of molecules have speed greater than escape velocity and they escape. Now, rest of the molecules arrange the speed distribution for the equilibrium temperature. Again, a significant number of molecules escape as their speeds exceed escape speed. Hence, over a long time, the Moon has lost most of its atmosphere.

(ii) As the molecules move higher, their potential energy increases and hence kinetic energy decreases and temperature reduces. At greater height, more volume is available and gas expands. Hence, some cooling takes place.

Question 3.
Two perfect gases at absolute temperatures T₁ and T₂ are mixed. There is no loss of energy. Find the temperature of the mixture if the masses of the molecules are m₁ and m₂ and the number of the molecules in the gases are n₁ and n₂ respectively.
Solution:
According to kinetic theory, the average kinetic energy per molecule of a
gas = \(\frac{3}{2} \) K_BT
Before mixing the two gases,the average K.E. of all the molecules of two gases
= \(\frac{3}{2} \)K_Bn₁T₁ + \(\frac{3}{2} \)K_Bn₁T₂
After mixing, the average K.E. of both the gases
= \(\frac{3}{2} \)k_B (n₁ +n₂)T
where, T is the temperature of mixture.
Since there is no loss of energy,
Hence, \(\frac{3}{2} \)k_B (n₁ +n₂)T = \(\frac{3}{2} k_{B} n_{1} T_{1}+\frac{3}{2} k_{B} n_{2} T_{2}\)
or T = \(\frac{n_{1} T_{1}+n_{2} T_{2}}{\left(n_{1}+n_{2}\right)}\).

Question 4.
At room temperature, diatomic gas molecule has five degrees of freedom. At high temperatures, it has seven degrees of freedom. Explain.
Answer:
At low temperatures, diatomic gas has three translational and two rotational degrees of freedom, so total number of degrees of freedom is 5. But at high temperature, gas molecule starts to vibrate which give two additional degrees of freedom. So the total numbers of degrees of freedom is 7.

Question 5.
What is basic law followed by equipartition of energy?
Answer:
The law of equipartiüon of energy for any dynamical system in thermal equilibrium, the total energy is distributed equally amongst all the degrees of freedom. The energy associated with each molecule per degree of freedom is \(\frac{1}{2}\) k_BT, where kB is Boltzmann’s constant and T is temperature of the system.

Question 6.
On what parameters does the λ (mean free path) depends?
Solution:
We know that,
λ = \(\frac{k T}{\sqrt{2} \pi d^{2} P}=\frac{m}{\sqrt{2} \pi d^{2} \rho}=\frac{1}{\sqrt{2} \pi n d^{2}}\)
Therefore, A depends upon:
(i) diameter (d) of the molecule, smaller the ‘d’, larger is the mean free path λ .
(ii) λ ∝ T i. e., higher the temperature larger is the λ.
(iii) λ ∝ \(\frac{1}{P}\) i.e., smaller the pressure larger is the λ.
(iv) λ ∝ \(\frac{1}{\rho}\) i.e., smaller the density (ρ), larger will be the λ.
(v) λ ∝ \(\frac{1}{n}\) i. e., smaller the number of molecules per unit volume of the gas, larger is the λ.

Question 7.
Although velocity of air molecules is very fast but fragrance of a perfume spreads at a much slower rate. Explain?
Answer:
This is because perfume vapour molecules do not travel uninterrupted, they undergo a number of collisions and trace a zig-zag path, due to which their effective displacement per unit time is small, so spreading is at a much slower rate.

Long Answer Type Questions

Question 1.
Consider an ideal gas with following distribution of speeds:

Speed (m/s)	% of molecules
200	10
400	20
600	40
800	20
1000	10

(i) Calculate υ_rms and hence T(m = 3.0 x 10^-26 kg)
(ii) If all the molecules with speed 1000 m/s escape from the system, calculate newvma and hence T.(NCERTExemplar)
Solution:
This problem is designed to give an idea about cooling by evaporation.
(i) υ²_rms = \(\frac{\sum n_{i} v_{i}^{2}}{\sum n_{i}}\)

(ii)

Question 2.
A box of 1.00 m³ is filled with nitrogen at 1.50 atm at 300 K. The box has a hole of an area Is 0.010 mm². How much time is required for the pressure to reduce by 0.10 atm., if the pressure outside is 1 atm.
Solution:
Given, the volume of the box, V 1.00 m³
Area of hole, a = 0.010 mm³ = 0.01 x 10^-6 m²
Temperature outside = Temperature inside
Initial pressure inside the box = 1.50 atm
Final pressure inside the box = 0.10 atm

Assuming,
υ_ix= Speed of nitrogen molecule inside the box along x-direction.
n₁ = Number of molecules per unit volume in a time interval of Δt, all the particles at a distance (υ_ixΔt) will collide the hole and the wall, the particle colliding along the hole will escape out reducing the pressure in the box.

Let the area of the wall is A, Number of particles colliding in time, Δt = \(\frac{1}{3}\) n₁(υ_ixΔt)A \(\frac{1}{2}\) is the factor because all the particles along x-direction are behaving randomly. Hence, half of these are colliding against the walls on either side.
Inside the box, υ²_ix + υ²_iy + υ²_iz = υ²_rms
⇒ υ²_ix = \(\frac{v_{r m s}^{2}}{3}\) [∵ υ_ix = υ_iy= υ_iz]

If particles collide along hole, they move out. Similarly, outer particles colliding along hole will move in.
Ifa = area of hole
Then, net particle flow in time,
Δt = \(\frac{1}{2}\left(n_{1}-n_{2}\right) \frac{k_{B} T}{m} \Delta t a\) [∵υ_rms = \(\sqrt{\frac{3 k_{B} T}{m}} \)]

[Temperature inside and outside the box are equal]
Let n = number of density of nitrogen
n = \(\frac{\mu N_{A}}{V}=\frac{p N_{A}}{R T}\) [∵ \(\frac{\mu}{V}=\frac{p}{R T}\)]
where, N_A = Avogadro’s number
If after time Δt, pressure inside changes from p₁ to p₂
n’₁ = \(\frac{p_{1}^{\prime} N_{A}}{R T}\)
Now, number of molecules gone out = n₁V -n’₁V

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 3 Motion in a Straight Line Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 3 Motion in a Straight Line

Very short answer type questions

Question 1.
What is the condition for an object to be considered as a point object?
Answer:
An object can be considered as a point object if the distance travelled by it is very large than its size.

Question 2.
For which condition, the distance and the magnitude of displacement of an object have the same values?
Answer:
The distance and the magnitude of displacement of an object have the same values, when the body is moving along a straight line path in a fixed direction.

Question 3.
Speed of a particle cannot be negative. Why?
Answer:
Speed is the distance travelled in unit time and distance cannot be negative.

Question 4.
Is it possible that a body could have constant speed but varying velocity?
Answer:
Yes, a body could have constant speed but varying velocity if only the direction of motion changes.

Question 5.
For which condition, the average velocity will be equal to the instantaneous velocity?
Answer:
When a body moves with a uniform velocity, then
υ_av = υ_inst

Question 6.
Give an example of uniformly accelerated linear motion.
Answer:
Motion of a body under gravity.

Question 7.
Give example of motion where x > 0, υ < 0, a > 0 at a particular instant. (NCERT Exemplar)
Solution:
Let the motion is represented by
x(t) = A + Be^-γt ……………. (i)
Let A>B and γ > 0
Now velocity x(t) = \(\frac{d x}{d t}\) = -Bγe^-γt
Acceleration a(t) = \(\frac{d x}{d t}\) = Bγ²e^-γt
Suppose we are considering any instant t, then from Eq. (i) we can say that
x(t)>0,υ(t)< 0 and a>0

Short answer type questions

Question 1.
Explain how an object could have zero average velocity but non-zero average speed?
Solution:
υ = \(=\frac{\text { Net displacement }}{\text { Total time taken }}\)
and average speed,

If an object moves along a straight line starting from origin and then returns back to origin.
Average velocity = 0
and Average speed = \(\frac{2 s}{t}\)

Question 2.
If the displacement of a body is zero, is distance necessarily zero? Answer with one example.
Answer:
No, because the distance covered by an object is the path length of the path covered by the object. The displacement of an object is given by the change in position between the initial position and final position.

Question 3.
Is earth inertial or non-inertial frame of reference?
Answer:
Since, earth revolves around the sun and also spins about its own axis, so it is an accelerated frame of reference. Hence, earth is a non-inertial frame of reference.
However, if we do not take large scale motion such as wind and ocean currents into consideration, we can say that approximation the earth is an inertial frame.

Question 4.
A person travels along a straight road for the first half with a velocity υ ₁ and the second half with velocity υ ₂. What is the mean velocity of the person?
Solution:

Question 5.
The displacement of a particle is given by at² What is dependency of acceleration on time?
Solution:
Let x be the displacement. Then, x = at²
∴ Velocity of the object, υ = \(\frac{d x}{d t}\) = 2 at
Acceleration of the object, a = \(\frac{d v}{d t}\) = 2 a
It means that a is constant.

Question 6.
What are uses of a velocity-time graph?
Solution:
From a velocity-time graph, we can find out
(i) The velocity of a body at any instant.
(ii) The acceleration of the body and
(iii) The net displacement of the body in a given time-interval.

Question 7.
Draw displacement-time graph for a uniformly accelerated motion. What is its shape?
Solution:
Displacement-time graph for a uniformly accelerated motion has been shown in adjoining fig. The graph is parabolic in shape.

Question 8.
The distance travelled by a body is proportional to the square of time. What type of motion this body has?
Solution:
Let x be the distance travelled in time t. Then,
x ∝ t² [given]
x = kt² [here, k = constant of proportionality]
We know that velocity is given
υ = \(\frac{d x}{d t}\) = 2kt
and acceleration is given by
a = \(\frac{d v}{d t}\) = 2 k [constant]
Thus, the body has uniform accelerated motion.

Long answer type questions

Question 1.
It is a common observation that rain clouds can be at about a kilometre altitude above the ground.
(i) If a rain drop falls from such a height freely under gravity, what will be its speed? Also, calculate in km/h (g = 10m/s²).
(ii) A typical rain drop is about 4 mm diameter. Momentum is mass × speed in magnitude. Estimate its momentum when it hits ground.
(iii) Estimate time required to flatten the drop.
(iv) Rate of change of momentum is force. Estimate how much force such a drop would exert on you?
(v) Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm.
(Assume that umbrella is circular and has a diameter of 1 m and cloth is not pierced through it.) (NCERT Exemplar)
Solution:
Here, height (h) = 1 km = 1000 m, g = 10 m/²
(i) Velocity attained by the rain drop in freely falling through a height h.
υ = \(\sqrt{2 g h}=\sqrt{2 \times 10 \times 1000}\)
= 100√2 m/s
= 100√2 \(\frac{60 \times 60}{1000}\) km/h
= 360√2 km/h ≈ 510 km/h

(ii) Diameter of the drop (d) = 2 r = 4 mm
∴ Radius of the drop (r) = 2 mm = 2 × 10^-3 m
Mass of a rain drop (m) = V × ρ
= \(\frac{4}{3}\) πr³ρ = \(\frac{4}{3} \times \frac{22}{7}\) x (2 × 10^-3)³ × 10³
[ v density of water = 10³ kg/m³ ]
≈ 3.4 × 10^-5 kg
Momentum of the rain drop (p) = mυ
= 3.4 × 10^-5 × 100√2
≈ 4.7 × 10^-3 kg-m/s

(iii) Time required to flatten the drop = time taken by the drop to travel the distance equal to the diameter of the drop near the ground
t = \(\frac{d}{v} \times \frac{4 \times 10^{-3}}{100 \sqrt{2}}\) = 0.028 × 10^-3 s
= 2.8 × 10^-5 s

(iv) Force exerted by a rain drop

= \(\frac{p-0}{t}=\frac{4.7 \times 10^{-3}}{2.8 \times 10^{-5}}\) ≈ 168 N

(v) Radius of the umbrella (R) = \(\frac{1}{2}\) m
∴ Area of the umbrella (A) = πR² = \(\frac{22}{7}\) x (\(\frac{1}{2}\))² = \(\frac{22}{28}=\frac{11}{14}\) ≈ 0.8M²
Number of drops striking the umbrella
simultaneously with average separation of 5 cm or 5 × 10^-2 m
= \(\frac{0.8}{\left(5 \times 10^{-2}\right)^{2}}\) = 320
∴ Net force exerted on umbrella = 320 × 168 = 53760 N

Question 2.
If a body moving with uniform acceleration in straight line describes successive equal distance in time interval t₁, t₂ and t₃, then show that
\(\frac{1}{t_{1}}-\frac{1}{t_{2}}+\frac{1}{t_{3}}=\frac{3}{t_{1}+t_{2}+t_{3}}\)
Solution:
As shown in figure, let three successive equal distances be represented by AB, BC and CD

Let each distance berm. Let υ_A,υ_B,υ_C and υ_D be the velocities at points A, B, C and D respectively.
Average velocity between A and B = \(\frac{v_{A}+v_{B}}{2}\)

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)