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Biology Board Question Paper Solution March 2020 Maharashtra Board

Biology Board Question Paper Solution

Maharashtra State Board - March 2020 (HSC Class 12)

Max. Marks: 70 | Time: 3 Hours

Section-A
Q.1. i.
Which of the following is most appropriate for thalassemia?
  • (A) decrease of either beta (β) or alpha (α) globin chain of HbA
  • (B) decrease of alpha (α) cells of pancreas
  • (C) decrease of WBC count
  • (D) decrease of blood platelets
Answer: (A) decrease of either beta (β) or alpha (α) globin chain of HbA
Q.1. ii.
Injury to _______ causes sudden death.
  • (A) cerebrum
  • (B) pons varolii
  • (C) medulla oblongata
  • (D) diencephalon
Answer: (C) medulla oblongata
Q.1. iii.
Name the smooth muscle of urinary bladder.
  • (A) cardiac muscle
  • (B) detrusor muscle
  • (C) dartos muscle
  • (D) gubernaculum
Answer: (B) detrusor muscle
Q.1. iv.
Identify the cell labelled 'A' in the T.S. of testis :
[Diagram: T.S. of testis showing seminiferous tubule. Label A points to large pyramidal cells extending from the basement membrane to the lumen, supporting developing sperm.]
  • (A) Leydig cell
  • (B) Basement membrane
  • (C) Sperm
  • (D) Sertoli cell
Answer: (D) Sertoli cell
Q.1. v.
_______ represents connecting link between amphibians and reptiles.
  • (A) Seymouria
  • (B) Archaeopteryx
  • (C) Ichthyostegia
  • (D) Archaeornis
Answer: (A) Seymouria
Q.1. vi.
How many meiotic and mitotic divisions are required for the formation of male gametophyte from pollen mother cell?
  • (A) 2 meiotic and 1 mitotic
  • (B) 1 meiotic and 1 mitotic
  • (C) 1 meiotic and 2 mitotic
  • (D) 2 meiotic and 2 mitotic
Answer: (C) 1 meiotic and 2 mitotic
Q.1. vii.
_______ is the common pathway for aerobic and anaerobic respiration.
  • (A) Krebs’ cycle
  • (B) ETS
  • (C) Calvin cycle
  • (D) Glycolysis
Answer: (D) Glycolysis
Q.1. viii.
Find the odd man out with respect to chemoautotrophs:
  • (A) Nitrosomonas
  • (B) Chromatium
  • (C) Thiobacillus
  • (D) Ferrobacillus
Answer: (B) Chromatium
(Reason: Chromatium is a photoautotroph, while others are chemoautotrophs.)
Q.1. ix.
Genotype of blood group ‘AB’ in human is _______.
  • (A) \(I^A I^B\)
  • (B) \(I^B i\)
  • (C) \(I^A I^A\)
  • (D) ii
Answer: (A) \(I^A I^B\)
Q.1. x.
Linker-DNA, connecting two successive nucleosomes, consists of _______.
  • (A) 146 base pairs
  • (B) 200 base pairs
  • (C) 160 base pairs
  • (D) 54 base pairs
Answer: (D) 54 base pairs

HSC Biology

Q.2.
Answer the following questions:

i. Where were the bones of jaws and teeth of Ramapithecus found?

Answer: The fossils (jaws and teeth) of Ramapithecus were found in the Shivalik Hills of India and in Kenya (Africa).

ii. In electrocardiogram, QRS complex stands for:

In electrocardiogram, QRS complex stands for:
[Diagram: ECG wave P-QRS-T]
Answer: The QRS complex stands for ventricular depolarization (spread of impulse from AV node to the wall of ventricles).

iii. Laxman has low secretion of ADH resulting in _______ type of diabetes.

Answer: Diabetes insipidus

iv. Name the region of retina where rods and cones are absent.

Answer: Blind spot (Optic disc)

v. Among biotic components, the micro consumers are called _______.

Answer: Decomposers (or Reducers)

vi. Identify ‘A’ in the chart given below:

Product Plant
(1) Nicotine Nicotiana tabacum
(2) Vincristin, Vinblastin ‘A’
Answer: ‘A’ is Catharanthus roseus (or Vinca rosea).

vii. The genotypic ratio 1:2:2:4:1:2:1:2:1 is obtained in F2 generation. What will be the phenotypic ratio?

Answer: 9 : 3 : 3 : 1

viii. Define the term ‘recessive’.

Answer: A recessive allele is an allele that is not expressed in the presence of an alternative dominant allele. It expresses itself only in the homozygous condition (presence of two identical alleles) or in the absence of a dominant allele.
Section-B

Attempt any eight of the following questions:

Q.3.
Sketch and label angiospermic embryo sac.
Solution: Sketch and label angiospermic embryo sac
[Diagram: Sketch and label angiospermic embryo sac]

(Student should draw the 7-celled, 8-nucleate structure of the female gametophyte)

Labels required:

  • Chalazal end: Contains 3 Antipodal cells.
  • Central part: Large central cell with Secondary nucleus (or two Polar nuclei).
  • Micropylar end: Egg apparatus containing 1 Egg cell (Oosphere) and 2 Synergids.
  • Filiform apparatus: Inside synergids.
Q.4.
To avoid photorespiration, which anatomical peculiarities are shown by C4 plants?
Answer:

C4 plants show a specialized anatomy called Kranz anatomy to avoid photorespiration:

  1. Dimorphic Chloroplasts:
    • Mesophyll cells: Contain granal chloroplasts (with grana).
    • Bundle Sheath cells: Contain large, agranal chloroplasts (without grana).
  2. Concentric Arrangement: The bundle sheath cells form a wreath-like (Kranz) layer around the vascular bundles, surrounded by mesophyll cells.
  3. Thick Walls: Bundle sheath cells have thick walls impervious to gaseous exchange, concentrating CO2 internally.
Q.5.
Enlist the steps involved in rDNA technology.
Answer:
  1. Isolation of DNA (Genetic material) from the donor organism.
  2. Cutting of DNA at specific locations using Restriction Endonuclease enzymes.
  3. Amplification of the gene of interest using PCR (Polymerase Chain Reaction).
  4. Insertion of the Recombinant DNA (rDNA) into the host cell/organism using a vector.
  5. Selection and screening of transformed host cells.
  6. Obtaining the foreign gene product (downstream processing).
Q.6.
Define the terms:
i. Bio-patent
ii. Bio-piracy
Answer:

i. Bio-patent: It is a patent granted by the government to the inventor for biological entities (like strains of microorganisms, cell lines, genetically modified strains), DNA sequences, and biotechnological processes and products.

ii. Bio-piracy: It refers to the use of bio-resources by multinational companies and other organizations without proper authorization from the countries and people concerned without compensatory payment.

Q.7.
Give the flow chart of central dogma.
Answer:
DNA \(\xrightarrow{\text{Transcription}}\) mRNA \(\xrightarrow{\text{Translation}}\) Protein

It can also be represented including Replication:

Replication \(\circlearrowleft\) DNA \(\rightarrow\) mRNA \(\rightarrow\) Polypeptide (Protein)
Q.8.
How will you identify that, F1 hybrid is homozygous or heterozygous? Explain it with a suitable example.
Answer:

We can identify the genotype of an F1 hybrid by performing a Test Cross. In a test cross, the F1 individual is crossed with the homozygous recessive parent.

Example: Consider height in pea plants (T = Tall, t = Dwarf).

  • Case 1 (Heterozygous): If F1 is Hybrid Tall (Tt):
    Cross: Tt (F1) × tt (Recessive parent)
    Progeny: 50% Tall (Tt) and 50% Dwarf (tt). Ratio 1:1.
    Conclusion: F1 is Heterozygous.
  • Case 2 (Homozygous): If F1 were Homozygous Tall (TT):
    Cross: TT × tt
    Progeny: 100% Tall (Tt).
    Conclusion: F1 is Homozygous.
Q.9.
Give any two contrasting traits studied by Mendel.
Answer:

(Any two from the seven pairs)

  1. Stem height: Tall vs. Dwarf
  2. Seed colour: Yellow vs. Green
  3. Seed shape: Round vs. Wrinkled
  4. Pod colour: Green vs. Yellow
Q.10.
Match the pairs and rewrite:
Column I Column II
(1) Mechanical means (a) Saheli
(2) Physiological device (b) Jellies
(3) Chemical device (c) Vasectomy
(4) Permanent Method (d) Diaphragm
Answer:
  • (1) Mechanical means — (d) Diaphragm
  • (2) Physiological device — (a) Saheli (Oral Contraceptive Pill)
  • (3) Chemical device — (b) Jellies (Spermicides)
  • (4) Permanent Method — (c) Vasectomy
Q.11.
Redraw, complete and label the diagram given below, which relates to reflex arc:
Redraw, complete and label the diagram given below, which relates to reflex arc:
[Diagram: Cross section of spinal cord with reflex arc pathway]
Solution:

Simplified Exam Diagram

Note: Students are required to draw this in the exam.

Diagram: Cross section of spinal cord with reflex arc pathway
[Diagram: Cross section of spinal cord with reflex arc pathway]

Detailed Diagram (Reference Only)

Note: For understanding purposes only. Not required for the exam.

Diagram: Cross section of spinal cord with reflex arc pathway
[Diagram: Cross section of spinal cord with reflex arc pathway]

The student needs to draw the transverse section of the spinal cord showing the reflex path. Key labels to include:

  1. Receptor: Skin (where pin prick occurs).
  2. Sensory Neuron (Afferent): Enters via Dorsal root.
  3. Dorsal Root Ganglion: Swelling on dorsal root containing cell body of sensory neuron.
  4. Association Neuron (Interneuron): Inside the Grey matter of spinal cord.
  5. Motor Neuron (Efferent): Leaves via Ventral root.
  6. Effector: Muscle (showing contraction).

Arrows should indicate flow: Skin \(\rightarrow\) Sensory Neuron \(\rightarrow\) Spinal Cord \(\rightarrow\) Motor Neuron \(\rightarrow\) Muscle.

Q.12.
Explain Hardy-Weinberg’s principle, with the help of Punnett square.
Answer:

Principle: It states that allele frequencies in a population remain constant from generation to generation in the absence of other evolutionary influences (like mutation, selection, migration).

The sum of allelic frequencies is 1: \(p + q = 1\)

The genotypic frequencies are given by: \((p + q)^2 = p^2 + 2pq + q^2 = 1\)

Punnett Square:

Gametes p (Dominant allele) q (Recessive allele)
p \(p^2\) (AA - Homozygous Dominant) \(pq\) (Aa - Heterozygous)
q \(pq\) (Aa - Heterozygous) \(q^2\) (aa - Homozygous Recessive)
Q.13.
Complete the following chart and rewrite:
S.NO Type Example
1. Vulnerable species Clouded leopard, Musk deer
2. ________________ Great Indian Bustard, Hawaiian monk seal
3. ________________ Three banded armadillo (Brazil), Short eared rabbit (Sumatra)
Answer:
  1. (Given) Vulnerable species
  2. Endangered species
  3. Intermediate species (Note: According to Maharashtra Board Textbook context)
Q.14.
Complete the tree diagram and write examples of (A) and (B):
Redraw, complete and label the diagram given below, which relates to reflex arc:
[Diagram: Complete the tree diagram and write examples of (A) and (B)]
Types of air pollutants
(A) Fine particles | (B) Coarse particles
Answer:

(A) Fine particles:

  • Size: Less than 5 µm (or 2.5 µm depending on specific text edition) in diameter.
  • Ex: (i) Aerosols
  • (ii) Smoke / Soot / Fumes

(B) Coarse particles:

  • Size: Over 5 µm in diameter.
  • Ex: (i) Carbon particles
  • (ii) Dust
Section-C

Attempt any EIGHT of the following questions:

Q.15.
Give the location and one function of the following receptors:
(i) Mechanoreceptors
(ii) Statoacoustic receptors
(iii) Baroreceptors
Answer:
  • (i) Mechanoreceptors:
    Location: Skin.
    Function: Detect mechanical stimuli like touch, pressure, and pain.
  • (ii) Statoacoustic receptors:
    Location: Inner ear (Internal ear).
    Function: Hearing (Phonoreceptors) and Body Balance/Equilibrium (Statoreceptors).
  • (iii) Baroreceptors:
    Location: Walls of carotid sinus and aortic arch.
    Function: Detect changes in blood pressure.
Q.16.
Classify the following composition of blood plasma given below as per column ‘A’ and complete column ‘B’.
Answer:
Column A Column B
(1) Plasma Proteins (i) Serum albumin, (v) Fibrinogen
(2) Nitrogenous waste (iii) Urea, (vi) Uric acid
(3) Inorganic Salts (ii) Bicarbonates, (iv) Sulphates of sodium
Q.17.
Name the causative agent of malaria. State any two symptoms and two preventive measures of malaria.
Answer:

Causative agent: Protozoan parasite of the genus Plasmodium (e.g., Plasmodium vivax, P. falciparum).

Symptoms (Any two):

  • High fever with chills and shivering.
  • Severe headache and nausea.
  • Profuse sweating followed by lowering of temperature.

Preventive measures (Any two):

  • Use of mosquito nets and insect repellents to avoid bites.
  • Elimination of mosquito breeding grounds (stagnant water).
  • Spraying insecticides to kill adult mosquitoes and larvae.
Q.18.
Identify ‘1’ and ‘2’ in the following diagram:
[Diagram of Vaccine Production]
Write in brief about production of vaccine.
Redraw, complete and label the diagram given below, which relates to reflex arc:
Answer:

Identification:

  • 1: Isolation of Antigen (Separation of specific antigen from the pathogen).
  • 2: Formulation / Mixing (Mixing of antigen with diluent/adjuvant).

(Note: Interpretation based on standard vaccine production flowchart found in textbooks where step 1 is antigen isolation and step 2 is formulation).

Brief about production of vaccine:

Vaccines are produced using biotechnology. The pathogen is cultured and inactivated or attenuated. The specific antigen (protein) responsible for immunity is isolated ('1'). It is then mixed with a suitable diluent or adjuvant ('2') to increase stability and immune response. This mixture forms the final vaccine.

Q.19.
Satish is a colorblind boy. His mother has normal vision but his maternal grandfather is colourblind. His father and maternal grandmother have normal vision. Explain the pattern of inheritance with a suitable chart.
Answer:

Analysis: Colorblindness is an X-linked recessive disorder.

  • Satish is colorblind (\(X^cY\)).
  • Maternal Grandfather was colorblind (\(X^cY\)). He passed his \(X^c\) chromosome to his daughter (Satish's mother).
  • Satish's Mother is phenotypically normal but must be a carrier (\(X^CX^c\)) because she received the affected X from her father.
  • Satish's Father is normal (\(X^CY\)).

Inheritance Chart:

Parents: Carrier Mother (\(X^CX^c\)) × Normal Father (\(X^CY\))

Gametes \(X^C\) (Sperm) Y (Sperm)
\(X^C\) (Egg) \(X^CX^C\) (Normal Daughter) \(X^CY\) (Normal Son)
\(X^c\) (Egg) \(X^CX^c\) (Carrier Daughter) \(X^cY\) (Colorblind Son - Satish)

Pattern of Inheritance: This is an example of Criss-cross inheritance. The gene for colorblindness was passed from the maternal grandfather to his daughter (carrier), and then from the daughter to her son (Satish).

Q.20.
What are the requirements of dairy management? Give one example of each Indian and exotic breed of cow.
Answer:

Requirements of dairy management:

  • Selection of good breeds with high yielding potential and disease resistance.
  • Proper housing (well-ventilated, sufficient water).
  • Scientific feeding (fodder quantity and quality).
  • Hygiene and cleanliness during milking and handling.
  • Regular veterinary checkups.

Examples:

  • Indian breed: Sahiwal, Gir, or Red Sindhi.
  • Exotic breed: Jersey, Holstein-Friesian, or Brown Swiss.
Q.21.
Distinguish between DNA and RNA.
Answer:
Feature DNA (Deoxyribonucleic Acid) RNA (Ribonucleic Acid)
Sugar Contains Deoxyribose sugar. Contains Ribose sugar.
Strands Usually double-stranded (Double Helix). Usually single-stranded.
Nitrogen Bases Contains Adenine, Guanine, Cytosine, and Thymine. Contains Adenine, Guanine, Cytosine, and Uracil.
Function Stores genetic information. Helps in protein synthesis.
Q.22.
What is ‘green revolution’? Give any two examples each of the improved varieties of wheat and rice.
Answer:

Green Revolution: It refers to the drastic increase in the production of food grains (especially wheat and rice) in developing countries due to the introduction of high-yielding varieties (HYV), use of fertilizers, pesticides, and better irrigation techniques.

Examples:

  • Wheat: Sonalika, Kalyan Sona.
  • Rice: Jaya, Ratna (or Padma).
Q.23.
Give microbial source of the following products in industrial production:
(i) Vitamin B12
(ii) Chloromycetin
(iii) Pectinase
Answer:
  • (i) Vitamin B12: Pseudomonas denitrificans (or Propionibacterium shermanii)
  • (ii) Chloromycetin (Antibiotic): Streptomyces venezuelae
  • (iii) Pectinase (Enzyme): Aspergillus niger (or Sclerotinia libertiana)
Q.24.
State the significance of respiration.
Answer:
  1. Energy Release: It releases energy in the form of ATP, which is essential for various metabolic activities of the cell.
  2. Intermediates: It provides carbon skeleton intermediates required for the synthesis of other biomolecules (like amino acids, fatty acids).
  3. Substrate Activation: It converts insoluble complex food substances into soluble simpler forms.
  4. CO2 Balance: It releases CO2, which is used in photosynthesis, helping maintain the balance of gases in the atmosphere.
Q.25.
Explain the mechanism of anaerobic respiration.
Answer:

Anaerobic respiration occurs in the absence of oxygen. It involves two main steps:

  1. Glycolysis (EMP Pathway):
    • Glucose (6C) is broken down into two molecules of Pyruvate (3C).
    • Net gain: 2 ATP and 2 NADH2.
    • This occurs in the cytoplasm.
  2. Fermentation:
    • The pyruvate produced is reduced to other products depending on the organism.
    • Alcoholic Fermentation (in Yeast): Pyruvate \(\rightarrow\) Acetaldehyde + CO2 \(\rightarrow\) Ethanol (Ethyl Alcohol). NADH2 is reoxidized to NAD.
    • Lactic Acid Fermentation (in Muscle/Bacteria): Pyruvate \(\rightarrow\) Lactic Acid.

Overall, it produces very less energy (2 ATP) compared to aerobic respiration.

Q.26.
Describe the role of citizens in solid waste management.
Answer:

Citizens play a crucial role in solid waste management by adopting the following practices:

  • 3R Principle: Following Reduce, Reuse, and Recycle to minimize waste generation.
  • Segregation: separating waste into biodegradable (wet) and non-biodegradable (dry) waste at the source.
  • Composting: Using wet waste (kitchen scraps) to make compost for home gardens.
  • Avoiding Plastics: Reducing the use of single-use plastics and carrying cloth bags.
  • Safe Disposal: Not littering in public places and disposing of hazardous waste (batteries, medicines) separately.
Section-D

Attempt any THREE of the following questions:

Q.27.
Sketch the internal structure of human heart. Label all the valves present in it. Mention the function of any one valve in the heart.
Solution: Sketch of the internal structure of human heart

Sketch Requirements: Draw a vertical section of the heart showing 4 chambers (RA, RV, LA, LV), major blood vessels (Aorta, Pulmonary Artery, Vena Cavae), and septum.

Labels for Valves:

  • Tricuspid Valve: Between Right Atrium and Right Ventricle.
  • Bicuspid (Mitral) Valve: Between Left Atrium and Left Ventricle.
  • Pulmonary Semilunar Valve: At the base of Pulmonary Artery.
  • Aortic Semilunar Valve: At the base of Aorta.
  • Eustachian Valve: (At opening of IVC - usually vestigial).
  • Thebesian Valve: (At opening of coronary sinus).

Function (Any one):

  • Tricuspid Valve: Prevents the backflow of blood from the right ventricle into the right atrium during ventricular contraction.
Q.28.
With the help of a suitable diagrammatic representation explain HSK pathway.
Answer: Sketch of the internal structure of human heart

HSK Pathway (Hatch-Slack Pathway / C4 Cycle):

This pathway occurs in C4 plants (e.g., Maize, Sugarcane) involving two types of cells: Mesophyll and Bundle Sheath.

  1. In Mesophyll Cell:
    • CO2 is accepted by PEP (Phosphoenolpyruvate) in the presence of PEP carboxylase.
    • Product: OAA (Oxaloacetic Acid - 4C compound).
    • OAA is converted to Malic Acid (or Aspartic Acid).
  2. Transport: Malic acid is transported to Bundle Sheath cells.
  3. In Bundle Sheath Cell:
    • Malic acid undergoes decarboxylation to release CO2 and Pyruvate.
    • The released CO2 enters the Calvin Cycle (C3 cycle) to form glucose.
  4. Regeneration: Pyruvate is transported back to Mesophyll cells and regenerated into PEP using ATP.
Q.29.
Describe the process of fertilization in human with the help of four sequential diagrams.
Answer: Process of fertilization in human with the help of four sequential diagrams

Process Description:

  1. Approach of Sperm: Millions of sperms reach the ampulla. Capacitation prepares sperm for fertilization.
  2. Entry of Sperm (Acrosome Reaction): The acrosome releases lysins (Hyaluronidase) to penetrate the Corona Radiata and Zona Pellucida. The sperm head fuses with the oocyte membrane.
  3. Cortical Reaction: Upon entry of one sperm, cortical granules in the egg release enzymes that harden the Zona Pellucida, preventing polyspermy (fertilization membrane formed).
  4. Activation of Ovum: The entry stimulates the secondary oocyte to complete Meiosis II, releasing the second polar body and forming the female pronucleus.
  5. Syngamy (Fusion): The male pronucleus and female pronucleus fuse (Amphimixis) to form a diploid Zygote.

Diagrams required:

  1. Sperms attacking the ovum.
  2. Acrosome reaction and penetration.
  3. Extrusion of polar body and cortical reaction.
  4. Fusion of pronuclei.
Q.30.
What is artificial method of vegetative propagation?
(i) Cutting
(ii) Budding.
Answer:

Artificial Vegetative Propagation: It is the process of growing new plants from vegetative parts of parent plants (root, stem, leaf) using man-made methods.

(i) Cutting:

  • A small piece of any vegetative part of a plant with one or more buds is cut and planted in soil.
  • Stem cutting: e.g., Rose, Sugarcane.
  • Leaf cutting: e.g., Sansevieria.
  • Root cutting: e.g., Blackberry.

(ii) Budding:

  • It is a form of grafting where a single bud (scion) from a desired plant is inserted into a slit in the bark of a rooted stock plant.
  • Common method: T-budding or Shield budding.
  • Example: Rose, Orange, Peach.
Q.31.
Describe the system associated with elimination of urine with the help of a neat, labelled diagram.
Answer:

The system associated with urine elimination is the Human Excretory System.

Components:

  1. Kidneys (Pair): Bean-shaped organs that filter blood to produce urine.
  2. Ureters (Pair): Muscular tubes that carry urine from the renal pelvis of the kidneys to the urinary bladder.
  3. Urinary Bladder: A muscular sac that temporarily stores urine. It has a smooth muscle layer called the Detrusor muscle.
  4. Urethra: A tube leading from the bladder to the exterior for the discharge of urine (micturition).

Diagram Labels Required: Kidney, Renal Artery, Renal Vein, Ureter, Urinary Bladder, Urethra.

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Maharashtra HSC Physics Board Question Paper Solution October 2014

Board Question Paper Solution: October 2014 Physics

Maharashtra State Board - HSC Physics

SECTION – I

Q.1. Attempt any THREE:

i. Draw a diagram showing all components of forces acting on a vehicle moving on a curved banked road. Write the necessary equation for maximum safety speed and state the significance of each term involved in it.

Diagram:

Imagine a vehicle on a banked road inclined at angle \(\theta\).

  • Weight \(mg\) acts vertically downwards.
  • Normal reaction \(N\) acts perpendicular to the road surface.
  • Frictional force \(f_s\) acts downwards along the slope (preventing upward skidding at maximum speed).

Equation for Maximum Safety Speed:

\[ v_{max} = \sqrt{rg \left( \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta} \right)} \]

Significance of terms:

  • \(v_{max}\): Maximum safe velocity of the vehicle.
  • \(r\): Radius of the curved road.
  • \(g\): Acceleration due to gravity.
  • \(\mu_s\): Coefficient of static friction between tires and road.
  • \(\theta\): Angle of banking.
ii. Explain Maxwell distribution of molecular speed with necessary graph.

Explanation:

Maxwell derived the law of distribution of molecular speeds for a gas in thermal equilibrium. According to this distribution:

  • The gas molecules move with all possible speeds ranging from zero to infinity.
  • The number of molecules having very low speeds or very high speeds is very small.
  • The number of molecules increases with speed, reaches a maximum for a particular speed called the "most probable speed" (\(v_{mp}\)), and then decreases.
  • The area under the graph represents the total number of molecules.

Graph: A curve plotted with "Speed of molecules" on the X-axis and "Number of molecules per unit speed interval" (\(dN/dv\)) on the Y-axis. The curve is asymmetrical, starting at the origin, rising to a peak, and tapering off towards the right.

iii. Find the total energy and binding energy of an artificial satellite of mass 800 kg orbiting at a height of 1800 km above the surface of the earth. (G = 6.67 × 10-11 S.I. units, Radius of earth : R = 6400 km, Mass of earth : M = 6 × 1024 kg)

Given:

  • Mass of satellite, \(m = 800 \text{ kg}\)
  • Height, \(h = 1800 \text{ km} = 1.8 \times 10^6 \text{ m}\)
  • Radius of Earth, \(R = 6400 \text{ km} = 6.4 \times 10^6 \text{ m}\)
  • Mass of Earth, \(M = 6 \times 10^{24} \text{ kg}\)
  • \(G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2\)

Orbital Radius: \(r = R + h = 6.4 \times 10^6 + 1.8 \times 10^6 = 8.2 \times 10^6 \text{ m}\)

Total Energy (T.E.):

\[ T.E. = -\frac{GMm}{2r} \] \[ T.E. = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 800}{2 \times 8.2 \times 10^6} \] \[ T.E. = -\frac{32016 \times 10^{13}}{16.4 \times 10^6} \] \[ T.E. \approx -1.95 \times 10^{10} \text{ J} \]

Binding Energy (B.E.):

\[ B.E. = -T.E. = +\frac{GMm}{2r} \] \[ B.E. = +1.95 \times 10^{10} \text{ J} \]

Answer: Total Energy = \(-1.95 \times 10^{10} \text{ J}\), Binding Energy = \(1.95 \times 10^{10} \text{ J}\).

iv. Wavelengths of two notes in the air are \(\left(\frac{70}{153}\right)\) m and \(\left(\frac{70}{157}\right)\) m. Each of these notes produces 8 beats per second with a tuning fork of fixed frequency. Find the velocity of sound in the air and frequency of the tuning fork.

Given:

  • \(\lambda_1 = \frac{70}{153} \text{ m}\)
  • \(\lambda_2 = \frac{70}{157} \text{ m}\)
  • Beat frequency with tuning fork = 8 Hz for both notes.

Formula: \(v = n\lambda \Rightarrow n = \frac{v}{\lambda}\)

Frequencies of the two notes:

\[ n_1 = \frac{v}{\lambda_1} = \frac{v}{70/153} = \frac{153v}{70} \] \[ n_2 = \frac{v}{\lambda_2} = \frac{v}{70/157} = \frac{157v}{70} \]

Since \(\lambda_1 > \lambda_2\), \(n_1 < n_2\). Let \(N\) be the frequency of the tuning fork.
Since both produce 8 beats/sec, the frequencies are \(N \pm 8\).
Since \(n_2 > n_1\), we must have \(n_1 = N - 8\) and \(n_2 = N + 8\).
Thus, \(n_2 - n_1 = 16\).

\[ \frac{157v}{70} - \frac{153v}{70} = 16 \] \[ \frac{v}{70}(157 - 153) = 16 \] \[ \frac{4v}{70} = 16 \] \[ v = \frac{16 \times 70}{4} = 4 \times 70 = 280 \text{ m/s} \]

Now, find the frequencies:

\[ n_1 = \frac{153 \times 280}{70} = 153 \times 4 = 612 \text{ Hz} \]

Since \(n_1 = N - 8\):

\[ N = 612 + 8 = 620 \text{ Hz} \]

Answer: Velocity of sound = 280 m/s, Frequency of tuning fork = 620 Hz.

HSC Physics Board Papers with Solution

Q.2. Attempt any SIX:

i. Draw a diagram showing different stages (cases) of projection for artificial satellite.

This requires drawing the Earth with a launch tower. Different trajectories are shown based on horizontal velocity \(v_h\):

  1. \(v_h < v_c\) (Critical velocity): Satellite falls back to Earth (Parabolic/Elliptical path intersecting Earth).
  2. \(v_h = v_c\): Circular orbit.
  3. \(v_c < v_h < v_e\) (Escape velocity): Elliptical orbit.
  4. \(v_h = v_e\): Parabolic path (escapes).
  5. \(v_h > v_e\): Hyperbolic path (escapes).
ii. State the law of conservation of angular momentum and explain with a suitable example.

Statement: If the resultant external torque acting on a rotating body is zero, its angular momentum remains constant.

\(L = I\omega = \text{constant}\). If \(I\) increases, \(\omega\) decreases, and vice versa.

Example: A ballet dancer or ice skater.
When the dancer folds her arms close to her body, her moment of inertia (\(I\)) decreases. To conserve angular momentum (\(L\)), her angular velocity (\(\omega\)) increases, allowing her to spin faster.

iii. Define the angle of contact and state its any ‘two’ characteristics.

Definition: The angle of contact is defined as the angle between the tangent drawn to the free surface of the liquid and the surface of the solid at the point of contact, measured within the liquid.

Characteristics:

  1. It is constant for a given solid-liquid pair.
  2. It depends on the nature of the liquid and the solid in contact.
  3. It depends on the impurities present in the liquid.
  4. It depends on the temperature of the liquid.
iv. With a neat and labelled diagram, explain Ferry’s perfectly black body.

Description:

  • Fery's black body consists of a double-walled hollow sphere.
  • The space between the walls is evacuated to minimize heat loss by conduction and convection.
  • The inner surface is coated with lampblack (absorptivity \(\approx 98\%\)).
  • There is a small aperture acting as the inlet for radiation.
  • A conical projection is placed directly opposite the aperture to prevent direct reflection of incident light back out of the hole.
  • Any radiation entering the hole undergoes multiple internal reflections and is almost completely absorbed. Thus, the aperture acts as a perfectly black body.
v. A stone of mass 5 kg, tied to one end of a rope of length 0.8 m, is whirled in a vertical circle. Find the minimum velocity at the highest point and at the midway point. (g = 9.8 m/s²)

Given: \(m = 5 \text{ kg}\), \(r = 0.8 \text{ m}\), \(g = 9.8 \text{ m/s}^2\).

1. Minimum velocity at highest point (\(v_H\)):

\[ v_H = \sqrt{rg} = \sqrt{0.8 \times 9.8} = \sqrt{7.84} = 2.8 \text{ m/s} \]

2. Minimum velocity at midway (horizontal) point (\(v_M\)):

\[ v_M = \sqrt{3rg} = \sqrt{3 \times 0.8 \times 9.8} = \sqrt{23.52} \approx 4.85 \text{ m/s} \]
vi. The maximum velocity of a particle performing linear S.H.M. is 0.16 m/s. If its maximum acceleration is 0.64 m/s², calculate its period.

Given: \(v_{max} = A\omega = 0.16 \text{ m/s}\), \(a_{max} = A\omega^2 = 0.64 \text{ m/s}^2\).

Calculation:

\[ \frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega \] \[ \omega = \frac{0.64}{0.16} = 4 \text{ rad/s} \]

Period \(T\):

\[ T = \frac{2\pi}{\omega} = \frac{2 \times 3.142}{4} = \frac{6.142}{4} = 1.571 \text{ s} \]

Answer: Period = 1.57 s.

vii. Water rises to a height 3.2 cm in a glass capillary tube. Find the height to which the same water will rise in another glass capillary having half area of cross section.

Given: \(h_1 = 3.2 \text{ cm}\). Area \(A_2 = \frac{A_1}{2}\).

Relation between Area and Radius: \(A = \pi r^2\).
Since \(A_2 = \frac{A_1}{2} \Rightarrow \pi r_2^2 = \frac{\pi r_1^2}{2} \Rightarrow r_2 = \frac{r_1}{\sqrt{2}}\).

Capillary rise formula: \(h = \frac{2T\cos\theta}{r\rho g} \Rightarrow h \propto \frac{1}{r}\).

\[ \frac{h_2}{h_1} = \frac{r_1}{r_2} \] \[ \frac{h_2}{3.2} = \frac{r_1}{r_1 / \sqrt{2}} = \sqrt{2} \] \[ h_2 = 3.2 \times 1.414 = 4.5248 \text{ cm} \]

Answer: Height = 4.52 cm.

viii. A 36 cm long sonometer wire vibrates with frequency of 280 Hz in fundamental mode, when it is under tension of 24.5 N. Calculate linear density of the material of wire.

Given:

  • \(L = 36 \text{ cm} = 0.36 \text{ m}\)
  • \(n = 280 \text{ Hz}\)
  • \(T = 24.5 \text{ N}\)

Formula: \(n = \frac{1}{2L}\sqrt{\frac{T}{m}}\), where \(m\) is linear density.

\[ n^2 = \frac{1}{4L^2} \frac{T}{m} \Rightarrow m = \frac{T}{4L^2 n^2} \] \[ m = \frac{24.5}{4 \times (0.36)^2 \times (280)^2} \] \[ m = \frac{24.5}{4 \times 0.1296 \times 78400} \] \[ m = \frac{24.5}{40642.56} \] \[ m \approx 6.028 \times 10^{-4} \text{ kg/m} \]

Answer: Linear density \(m \approx 6.03 \times 10^{-4} \text{ kg/m}\).

Q.3. Select and write the most appropriate answer from the given alternatives for each sub-questions:

i. A thin wire of length L and uniform linear mass density ρ is bent into a circular coil. Moment of inertia of the coil about tangential axis in its plane is _______.
  • (A) \( \frac{3\rho L^2}{8\pi^2} \)
  • (B) \( \frac{8\pi^2}{3\rho L^3} \)
  • (C) \( \frac{3\rho L^3}{8\pi^2} \)
  • (D) \( \frac{8\pi}{3\rho L^2} \)
Calculation:
Mass \(M = \rho L\). Length \(L = 2\pi R \Rightarrow R = \frac{L}{2\pi}\).
M.I. about diameter \(I_d = \frac{1}{2}MR^2\).
M.I. about tangent in plane \(I_t = I_d + MR^2 = \frac{3}{2}MR^2\).
\(I_t = \frac{3}{2} (\rho L) \left(\frac{L}{2\pi}\right)^2 = \frac{3\rho L^3}{8\pi^2}\).
ii. The average displacement over a period of S.H.M. is _______. (A = amplitude of S.H.M.)
  • (A) 0
  • (B) A
  • (C) 2A
  • (D) 4A
Reason: Displacement is a vector. The particle returns to the starting point after one period, so net displacement is 0.
iii. In which of the following substances, surface tension increases with increase in temperature?
  • (A) Copper
  • (B) Molten copper
  • (C) Iron
  • (D) Molten iron
Reason: Generally surface tension decreases with temperature. However, for molten copper and molten cadmium, surface tension increases with temperature over a certain range.
iv. The ratio of diameters of two wires of the same material and length is n : 1. If the same load is applied to both the wires then increase in the length of the thin wire is (n > 1) _______.
  • (A) \(n^{1/4}\) times
  • (B) \(n^{1/2}\) times
  • (C) n times
  • (D) \(n^2\) times
Calculation: \(Y = \frac{FL}{A\Delta l} \Rightarrow \Delta l \propto \frac{1}{A} \propto \frac{1}{d^2}\).
\(d_{thick}/d_{thin} = n/1\).
\(\frac{\Delta l_{thin}}{\Delta l_{thick}} = \left(\frac{d_{thick}}{d_{thin}}\right)^2 = n^2\).
v. The co-efficient of reflection of an opaque body is 0.16. Its co-efficient of emission is _______.
  • (A) 0.94
  • (B) 0.84
  • (C) 0.74
  • (D) 0.64
Calculation: Opaque \(\Rightarrow t = 0\).
\(a + r = 1 \Rightarrow a = 1 - r = 1 - 0.16 = 0.84\).
By Kirchhoff's law, \(a = e\), so \(e = 0.84\).
vi. Let velocity of a sound wave be ‘v’ and ‘ω’ be angular velocity. The propagation constant of the wave is _______.
  • (A) \(\sqrt{\frac{\omega}{v}}\)
  • (B) \(\sqrt{\frac{v}{\omega}}\)
  • (C) \(\frac{\omega}{v}\)
  • (D) \(\frac{v}{\omega}\)
Reason: Propagation constant \(k = \frac{2\pi}{\lambda}\). Since \(v = \frac{\omega}{k}\), we get \(k = \frac{\omega}{v}\).
vii. The value of end correction for an open organ pipe of radius ‘r’ is _______.
  • (A) 0.3 r
  • (B) 0.6 r
  • (C) 0.9 r
  • (D) 1.2 r
Reason: The end correction \(e\) at one open end is approx \(0.6r\). While total correction for an open pipe is \(2e = 1.2r\), the phrase "value of end correction" typically refers to the constant \(e\) itself.

Q.4. Attempt any ONE:

Option 1: Distinguish between forced vibrations and resonance. Draw neat, labelled diagrams for the modes of vibration of a stretched string in second harmonic and third harmonic.

Problem: The area of the upper face of a rectangular block is 0.5 m × 0.5 m and the lower face is fixed. The height of the block is 1 cm. A shearing force applied at the top face produces a displacement of 0.015 mm. Find the strain and shearing force. (Modulus of rigidity : η = 4.5 × 1010 N/m²)

Distinguish Forced Vibrations vs Resonance:

  • Forced Vibration: Body vibrates with frequency of external periodic force, not its natural frequency. Amplitude is generally small.
  • Resonance: Special case of forced vibration where external frequency matches natural frequency. Amplitude becomes maximum.

Diagrams:

  • 2nd Harmonic: String vibrates in 2 loops. (Nodes at ends and center, Antinodes in between).
  • 3rd Harmonic: String vibrates in 3 loops.

Problem Solution:

Given: \(A = 0.5 \times 0.5 = 0.25 \text{ m}^2\), \(h = 1 \text{ cm} = 0.01 \text{ m}\), \(x = 0.015 \text{ mm} = 1.5 \times 10^{-5} \text{ m}\), \(\eta = 4.5 \times 10^{10} \text{ N/m}^2\).

1. Shearing Strain (\(\theta\)):

\[ \theta = \frac{x}{h} = \frac{1.5 \times 10^{-5}}{0.01} = 1.5 \times 10^{-3} \]

2. Shearing Force (\(F\)):

\[ \eta = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F/A}{\theta} \] \[ F = \eta A \theta = (4.5 \times 10^{10}) \times (0.25) \times (1.5 \times 10^{-3}) \] \[ F = 4.5 \times 0.25 \times 1.5 \times 10^7 \] \[ F = 1.6875 \times 10^7 \text{ N} \]
OR - Option 2: Define phase of S.H.M. Show variation of displacement, velocity and acceleration with phase...
Problem: A body starts rotating from rest. Due to a couple of 20 Nm it completes 60 revolutions in one minute. Find the moment of inertia of the body.

Problem Solution:

Given: \(\omega_0 = 0\), Torque \(\tau = 20 \text{ Nm}\).

Displacement \(\theta = 60 \text{ rev} = 60 \times 2\pi = 120\pi \text{ rad}\).

Time \(t = 1 \text{ min} = 60 \text{ s}\).

Using kinematic equation: \(\theta = \omega_0 t + \frac{1}{2}\alpha t^2\)

\[ 120\pi = 0 + \frac{1}{2}\alpha (60)^2 \] \[ 120\pi = 1800\alpha \] \[ \alpha = \frac{120\pi}{1800} = \frac{\pi}{15} \text{ rad/s}^2 \]

Using \(\tau = I\alpha\):

\[ I = \frac{\tau}{\alpha} = \frac{20}{\pi/15} = \frac{300}{\pi} \approx 95.49 \text{ kg m}^2 \]

SECTION – II

Q.5. Attempt any THREE:

i. In a biprism experiment... distance between 4th bright band on one side and 4th dark band on the other side...

Given: \(\lambda = 4800 \text{ Å} = 4.8 \times 10^{-7} \text{ m}\). \(d = 3 \text{ mm} = 3 \times 10^{-3} \text{ m}\).

Distances: Slit to Biprism = 15 cm. Biprism to Eyepiece = 85 cm.
Total distance \(D = 15 + 85 = 100 \text{ cm} = 1 \text{ m}\).

Fringe width \(X = \frac{\lambda D}{d}\).

Position of 4th bright band from center: \(y_{B4} = 4X\).

Position of 4th dark band (on other side): \(y_{D4} = (4 - 0.5)X = 3.5X\).

Total distance = \(y_{B4} + y_{D4} = 4X + 3.5X = 7.5X\).

\[ \text{Distance} = 7.5 \times \frac{4.8 \times 10^{-7} \times 1}{3 \times 10^{-3}} \] \[ \text{Distance} = 2.5 \times 4.8 \times 10^{-4} \] \[ \text{Distance} = 12 \times 10^{-4} \text{ m} = 1.2 \text{ mm} \]
ii. Six capacitors of capacities 5, 5, 5, 5, 10 and X µF are connected as shown... Find X if balanced, and resultant between A and C.

Analysis: The circuit is a bridge. Outer arms are AB, BC, AD, DC. The central arm is BD.
From diagram: \(C_{AB}=5, C_{BC}=5, C_{AD}=5, C_{BD}=5\).

The arm DC contains two capacitors in series: 10 µF and X µF. Let equivalent be \(C_{DC}\).

a. Value of X if balanced:

Condition: \(C_{AB} / C_{AD} = C_{BC} / C_{DC}\) (assuming A and C are input/output).

\[ \frac{5}{5} = \frac{5}{C_{DC}} \Rightarrow C_{DC} = 5 \mu\text{F} \]

Since \(C_{DC}\) is series of 10 and X: \(\frac{10X}{10+X} = 5\).

\[ 10X = 50 + 5X \Rightarrow 5X = 50 \Rightarrow X = 10 \mu\text{F} \]

b. Resultant capacitance between A and C:

If balanced, the central branch (BD) is ignored. The circuit becomes two parallel branches (ABC and ADC).

  • Upper branch (ABC): 5 and 5 in series \(\to 2.5 \mu\text{F}\).
  • Lower branch (ADC): 5 and \(C_{DC}\) (which is 5) in series \(\to 2.5 \mu\text{F}\).

Resultant \(C_{eq} = 2.5 + 2.5 = 5 \mu\text{F}\).

iii. Show that the current flowing through a moving coil galvanometer is directly proportional to the angle of deflection of coil.

Derivation:

  • Deflecting Torque: \(\tau_d = NIAB\) (where N=turns, I=current, A=area, B=magnetic field). The radial field ensures \(\sin\theta=1\).
  • Restoring Torque: \(\tau_r = k\alpha\) (where k=torsional constant, \(\alpha\)=deflection).
  • At equilibrium: \(\tau_d = \tau_r \Rightarrow NIAB = k\alpha\).
  • Therefore, \(I = \left(\frac{k}{NAB}\right)\alpha\).
  • Since k, N, A, B are constants, \(I \propto \alpha\).

Q.6. Attempt any SIX:

v. A red light of wavelength 6400 Å in air has wavelength 4000 Å in glass. If the wavelength of violet light in air is 4400 Å, find its wavelength in glass. (Assume that µr ≈ µv)

Solution:

Refractive index for red: \(\mu_r = \frac{\lambda_{air}}{\lambda_{glass}} = \frac{6400}{4000} = 1.6\).

Given \(\mu_r \approx \mu_v\), so \(\mu_v = 1.6\).

For violet light: \(\lambda_{g,v} = \frac{\lambda_{a,v}}{\mu_v} = \frac{4400}{1.6}\).

\[ \lambda_{g,v} = \frac{44000}{16} = 2750 \text{ Å} \]
vi. The magnetic moment of a magnet of dimensions 5 cm × 2.5 cm × 1.25 cm is 3 Am². Calculate the intensity of magnetization.

Solution:

Volume \(V = 5 \times 2.5 \times 1.25 \text{ cm}^3 = 15.625 \text{ cm}^3 = 15.625 \times 10^{-6} \text{ m}^3\).

Intensity \(I = \frac{M}{V} = \frac{3}{15.625 \times 10^{-6}}\).

\[ I = \frac{3 \times 10^6}{15.625} = 1.92 \times 10^5 \text{ A/m} \]
vii. An A.C. circuit consists of inductor of inductance 125 mH connected in parallel with a capacitor of capacity 50 µF. Determine the resonant frequency.

Solution:

\(L = 125 \text{ mH} = 0.125 \text{ H}\), \(C = 50 \mu\text{F} = 50 \times 10^{-6} \text{ F}\).

Formula: \(f_r = \frac{1}{2\pi\sqrt{LC}}\).

\[ f_r = \frac{1}{2\pi\sqrt{0.125 \times 50 \times 10^{-6}}} \] \[ f_r = \frac{1}{2\pi\sqrt{6.25 \times 10^{-6}}} \] \[ f_r = \frac{1}{2\pi \times 2.5 \times 10^{-3}} \] \[ f_r = \frac{1000}{5\pi} = \frac{200}{\pi} \approx 63.66 \text{ Hz} \]
viii. Calculate the de Broglie wavelength of an electron moving with 1/3rd of the speed of light in vacuum.

Solution:

\(v = \frac{c}{3} = 10^8 \text{ m/s}\). Mass \(m = 9.1 \times 10^{-31} \text{ kg}\). \(h = 6.63 \times 10^{-34}\).

\[ \lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^8} \] \[ \lambda = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-23}} \] \[ \lambda = 0.728 \times 10^{-11} \text{ m} = 0.0728 \text{ Å} \]

Q.7. Select and write the most appropriate answer:

  • i. (B) limit of resolution decreases (since Limit = \(\lambda/2NA\))
  • ii. (B) \(2\pi \times 10^{-5}\) T (Calculated: \(B = \mu_0 n I\), \(n=10\), \(I=5\))
  • iii. (C) energy and charge
  • iv. (C) \(\frac{(E_1 - E_2)\lambda_1\lambda_2}{c(\lambda_2 - \lambda_1)}\)
  • v. (C) the band gap of the material of semiconductor
  • vi. (D) space wave
  • vii. (A) \(\frac{qV}{dm}\) (Acc = \(F/m = qE/m = q(V/d)/m\))

Q.8. Attempt any ONE:

Option 1: Potentiometer Problem... resistance in series to get potential gradient \(10^{-3}\) V/cm.

Solution:

Given: \(L = 4 \text{ m}\), \(R_{wire} = 5 \Omega\). Cell \(E = 2 \text{ V}, r = 1 \Omega\).

Required Gradient \(K = 10^{-3} \text{ V/cm} = 0.1 \text{ V/m}\).

Potential difference across wire \(V_{AB} = K \times L = 0.1 \times 4 = 0.4 \text{ V}\).

Current in circuit \(I = \frac{V_{AB}}{R_{wire}} = \frac{0.4}{5} = 0.08 \text{ A}\).

Also \(I = \frac{E}{R_{wire} + r + R_{ext}}\).

\[ 0.08 = \frac{2}{5 + 1 + R} \] \[ 0.08(6 + R) = 2 \] \[ 6 + R = \frac{2}{0.08} = 25 \] \[ R = 19 \Omega \]

Answer: Series resistance = 19 \(\Omega\).

Option 2: Photoelectric Effect Problem... Threshold 230 nm, Incident 180 nm. Find KEmax.

Solution:

\(\lambda_0 = 230 \text{ nm}\), \(\lambda = 180 \text{ nm}\).

\(KE_{max} = h c \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)\).

Value of \(hc \approx 1240 \text{ eV}\cdot\text{nm}\).

\[ KE_{max} = 1240 \left( \frac{1}{180} - \frac{1}{230} \right) \text{ eV} \] \[ KE_{max} = 1240 \left( \frac{230 - 180}{180 \times 230} \right) \] \[ KE_{max} = 1240 \left( \frac{50}{41400} \right) \] \[ KE_{max} = \frac{62000}{41400} \approx 1.5 \text{ eV} \]

In Joules: \(1.5 \times 1.6 \times 10^{-19} = 2.4 \times 10^{-19} \text{ J}\).

Question Paper Page No. 1 Question Paper Page No. 2 Question Paper Page No. 3 Question Paper Page No. 4 Question Paper Page No. 5
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OCM - March 2025 English Medium Board Paper Solution HSC 12th Standard

Solution: Organisation of Commerce and Management (51)

Board Paper J-278 (2025)
Time: 3 Hrs | Max. Marks: 80 | Date: Day 05

Q. 1. (A) Select the correct options and rewrite the sentences

(1) _____ is regarded as the Father of Scientific Management.
(a) Henry Fayol (b) F.W. Taylor (c) Philip Kotler
Answer: (b) F.W. Taylor
(2) _____ creates time utility.
(a) Warehouse (b) Transport (c) Communication
Answer: (a) Warehouse
(3) Businessmen are _____ of the society.
(a) representatives (b) members (c) trustees
Answer: (c) trustees
(4) The term ‘Market’ is derived from the _____ word ‘Mercatus’.
(a) French (b) Latin (c) Italian
Answer: (b) Latin
(5) Principle of indemnity is not applicable to _____ insurance.
(a) life (b) marine (c) fire
Answer: (a) life

12th OCM Board Papers (March & July 2025)

Q. 1. (B) Give one word /phrase/term

(1) The right person at the right job with right pay.
Answer: Staffing
(2) The first step in online transaction.
Answer: Registration
(3) An activity motivated by profit.
Answer: Economic Activity
(4) One who consumes or uses any commodity or service.
Answer: Consumer
(5) Giving of distinct name to one’s product.
Answer: Branding

12th OCM Board Papers (March & July 2024)

Q. 1. (C) State whether True or False

(1) Every function of management is not based on planning.
Answer: False
Reason: Planning is the primary function and all other functions are based on it.
(2) E-business allows you to work across the globe in any field.
Answer: True
(3) Business ethics is a code of conduct.
Answer: True
(4) Consumer Protection Act provides protection to the producer.
Answer: False
Reason: It provides protection to the consumers.
(5) Air transport is the cheapest mode of transport.
Answer: False
Reason: It is the costliest mode of transport.

12th OCM Board Papers (February & July 2023)

Q. 1. (D) Find the odd one

(1) District Commission, State Commission, NGO, National Commission.
Answer: NGO
(2) NABARD, RBI, SIDBI, EXIM.
Answer: RBI (Central Bank, others are development/specialized banks)
(3) Debit card, Credit card, Aadhar card, ATM card.
Answer: Aadhar card
(4) Writing, Planning, Organising, Staffing.
Answer: Writing
(5) Price, People, Promotion, Product.
Answer: People (Part of 7Ps/Service Mix, others are core 4Ps)

12th OCM Board Papers (2014 - 2022)

  • OCM - March 2022 English Medium: View
  • OCM - March 2022 Marathi Medium: View | Answer Key
  • OCM - July 2022 English Medium: View | Answer Key
  • OCM - March 2020 English Medium: View
  • OCM - March 2020 Marathi Medium: View | Answer Key
  • OCM - March 2019: View
  • OCM - July 2018: View
  • OCM - March 2018: View
  • OCM - July 2017: View
  • OCM - March 2017: View
  • OCM - July 2016: View
  • OCM - March 2016: View
  • OCM - July 2015: View
  • OCM - March 2015: View
  • OCM - October 2014: View
  • OCM - March 2014: View

Q. 2. Explain the following terms / concepts (Any FOUR)

(1) Motion Study: Motion study is the study of the movement of a worker or a machine while completing a specific task. It helps in eliminating unnecessary motions and finding the best method of doing a job to increase efficiency.

(2) Controlling: Controlling is the final function of management. It involves measuring actual performance against standards, finding deviations, and taking corrective actions to ensure organizational goals are met.

(3) Bonded Warehouses: These are warehouses licensed by the government to accept imported goods for storage until customs duty is paid. They work under the control of customs authorities.

(4) Consumer Protection: It refers to the steps taken to protect consumers from unfair trade practices, exploitation, and providing them with their rights and remedies through legal mechanisms like the Consumer Protection Act.

(5) E-mail: Electronic mail (E-mail) is a method of exchanging messages between people using electronic devices. It is a fast and cheap mode of written communication used extensively in business.

(6) Co-ordinating: It is the integration and synchronization of the efforts of group members so as to provide unity of action in the pursuit of common goals. It binds all other functions of management together.

Q. 3. Study the following case / situation and express your opinion (Any TWO)

(1) Mr. Sharad is a businessman. He has his own factories in Pune and Nashik. He lives in Pune with his wife and 2 daughters aged 5 and 8 years old.
(a) Can Mr. Sharad take a life insurance policy for his wife and 2 children?
Yes, Mr. Sharad can take a life insurance policy for his wife and children as he has an insurable interest in their lives.

(b) Can Mr. Sharad take a marine insurance policy for his factories?
No, marine insurance covers risks related to sea transport (cargo/hull). For factories (immovable property), he needs Fire or General insurance.

(c) Which types of insurance should Mr. Sharad take for protecting his factories from loss due to fire?
Mr. Sharad should take a Fire Insurance policy to protect his factories from loss due to fire.
(2) Mr. Suresh made his payment by cheque. At the same time Mr. Saksham made his payment by fund transfer.
(a) Whose payment is faster?
Mr. Saksham's payment (fund transfer) is faster.

(b) Whose payment is related to traditional business?
Mr. Suresh's payment (cheque) is related to traditional business.

(c) Whose payment is related to e-business?
Mr. Saksham's payment (electronic fund transfer) is related to e-business.
(3) An organisation manufacturing paints has been enjoying a prominent market position since many years. It has been dumping its untreated poisonous waste on the river bank, which has created many health problems for the nearby villages.
(a) Which responsibility is neglected by manufacturing organisation?
The organization has neglected its Social Responsibility towards the Society (Protection of Environment).

(b) What kind of pollution they are doing?
They are causing Water Pollution and Soil Pollution.

(c) State any one precautionary measure they need to take.
They should install an Effluent Treatment Plant (ETP) to treat the waste before disposing of it.

Q. 4. Distinguish between the following (Any THREE)

(1) Road Transport vs Water Transport

Point Road Transport Water Transport
Speed It is faster than water transport. It is the slowest mode of transport.
Door-to-door It provides door-to-door service. It does not provide door-to-door service.
Cost Cost is moderate; higher than water transport. It is the cheapest mode of transport.
Suitability Suitable for short distances and perishable goods. Suitable for bulky goods over long distances.

(2) Life Insurance vs Fire Insurance

Point Life Insurance Fire Insurance
Subject Matter Subject matter is Human Life. Subject matter is Property/Assets.
Indemnity Principle of indemnity does not apply. Principle of indemnity applies.
Duration Long-term contract (years). Short-term contract (usually 1 year).
Loss Measurement Loss of life cannot be measured in money. Loss is measurable in money.

(3) Organising vs Staffing

Point Organising Staffing
Meaning It is the process of defining and grouping activities and establishing authority relationships. It involves recruiting, selecting, developing, and compensating the workforce.
Objective To identify and bring together resources. To appoint the right person for the right job.
Factors Deals with internal and external factors. Deals mostly with human factors (internal).
Order It follows the planning function. It usually follows the organizing function.

Q. 5. Answer in brief (Any TWO)

(1) Explain any four principles of management of Henry Fayol.

  • Division of Work: Work should be divided into small tasks and assigned to employees based on their skills to improve efficiency and specialization.
  • Authority and Responsibility: Authority is the right to give orders, and responsibility is the obligation to perform. There must be a balance between the two.
  • Discipline: Employees must obey and respect the rules of the organization. Good discipline is essential for smooth functioning.
  • Unity of Command: Each employee should receive orders from only one superior to avoid confusion and conflict.

(2) Explain any four responsibilities of consumer.

  • Critical Awareness: Consumers should be aware of the price, quality, and quantity of goods before buying.
  • Quality Consciousness: Consumers should look for quality marks like ISI, AGMARK, Hallmark, etc.
  • Ask for Cash Memo: Always insist on a valid proof of purchase (bill/invoice) to file complaints later if needed.
  • Filing Complaints: If cheated, consumers must not remain silent but file a complaint with the appropriate consumer forum.

(3) Explain any four functions of marketing.

  • Marketing Research: Collecting and analyzing information about consumer needs and market trends.
  • Product Development: Designing the product according to consumer requirements.
  • Pricing: Determining the value of the product in monetary terms.
  • Promotion: Communicating with customers to inform and persuade them to buy the product (advertising, sales promotion).

Q. 6. Justify the following statements (Any TWO)

(1) Organising facilitates administration as well as operation of the organisation.

Justification: Organising identifies and groups activities. It assigns roles and responsibilities to specific individuals (Staffing). It clearly defines who is to do what and who is responsible to whom (Hierarchy). This clarity prevents duplication of work and confusion. It ensures resources are used effectively for production (operation) and ensures smooth management (administration). Therefore, organizing bridges the gap between administration and operations.

(2) Consumer organisations and Non-Government organisations play an important role in consumer education.

Justification: Many consumers are unaware of their rights. NGOs and Consumer organizations conduct awareness programs, workshops, and campaigns to educate consumers. They publish magazines and journals (e.g., Grahak Shakti) to spread information. They also provide legal aid and help consumers file complaints in courts. They act as a pressure group on the government to enact consumer-friendly laws. Thus, they play a vital role.

(3) It is easy to set up e-business as compared to traditional business.

Justification: Traditional business requires a physical location (shop/office), which involves high setup costs, registration, and legal formalities. E-business can be started from home with just a computer and internet connection. The procedural requirements are fewer. It does not require maintaining huge inventory initially. The reach is global instantly. Hence, setting up e-business is faster, cheaper, and easier.

Q. 7. Attempt the following (Any TWO)

(1) Explain the functions of an Entrepreneur.

  • Innovation: Introduction of new products, methods, or markets.
  • Determination of Objectives: Setting clear goals for the business.
  • Development of Market: Finding new customers and marketing strategies.
  • New Technology: Adopting latest technology to improve production.
  • Good Relations: Maintaining good relations with employees and stakeholders.

(2) Explain importance of marketing to the consumers.

  • Promotes Product Awareness: Informing consumers about new products.
  • Provides Quality Products: Ensures supply of quality goods.
  • Variety of Products: Offers choices to meet different needs.
  • Consumer Satisfaction: Main aim is to satisfy consumer needs.
  • Regular Supply of Goods: Maintains flow of goods in the market.

(3) Explain nature of principles of management.

  • Universal Application: Applicable to all types of organizations everywhere.
  • General Guidelines: They provide solutions to problems but are not rigid laws.
  • Formed by Practice and Experimentation: Developed over years of research.
  • Flexibility: Can be modified according to changing situations.
  • Behavioral in Nature: Influence human behavior to achieve goals.

Q. 8. Answer the following question in detail (Any ONE)

(1) What is Marketing Mix? Explain 7 Ps of Marketing Mix.

Meaning: Marketing Mix is the combination of different marketing variables that the firm blends and controls to achieve the desired result from the target market. It is a set of marketing tools.

The 7 Ps of Marketing Mix (Extended Mix for Services):

  1. Product: The goods or services offered to the customer to satisfy a need. It includes design, features, quality, and packaging.
  2. Price: The amount of money customers pay to obtain the product. It must be competitive and affordable while covering costs.
  3. Place: The distribution channels used to make the product available to the customer at the right time and location.
  4. Promotion: Activities used to communicate the product's features to the target audience, such as advertising, sales promotion, and personal selling.
  5. People: All people involved in the service delivery (employees, management, and customers). Skilled staff is crucial for service businesses.
  6. Process: The procedures, mechanisms, and flow of activities by which the service is delivered (e.g., the ordering process in a restaurant).
  7. Physical Evidence: The environment in which the service is delivered and any tangible goods that facilitate the performance (e.g., ambiance of a hotel, tickets, brochures).

(2) What is Bank? Explain in detail primary functions of commercial banks.

Meaning: A Bank is a financial institution that deals with money and credit. It accepts deposits from the public and grants loans to those in need.

Primary Functions of Commercial Banks:

A. Accepting Deposits: This is the main function.

  • Demand Deposits:
    • Current Account: Opened by businessmen, unlimited transactions, no interest.
    • Savings Account: For general public to encourage saving, limited withdrawals, low interest.
  • Time Deposits:
    • Fixed Deposit (FD): Money deposited for a fixed period, higher interest rate.
    • Recurring Deposit (RD): Fixed amount deposited monthly for a fixed period.

B. Granting Loans and Advances:

  • Loans:
    • Short/Medium/Long term loans: Given for specific periods (1 year to 5+ years).
  • Advances (Short term):
    • Cash Credit: Overdraft facility against security.
    • Overdraft: Allowed to current account holders to withdraw more than the balance.
    • Discounting of Bills: Bank pays the bill amount before due date after deducting discount charges.
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