Competency-Based Case Study Questions in Class 10 Science

In the reformed CBSE Class 10 Science board examination, Section E tests scientific literacy and competency-based application through three compulsory Case Study questions (carrying 4 marks each, totaling 12 marks). Rather than asking isolated textbook questions, Section E presents real-world experimental scenarios, clinical case reports, industrial processes, or environmental field data.

Students must read the narrative, analyze the data or diagram, and answer structured sub-questions that connect everyday phenomena to foundational chemical equations, physiological mechanisms, or physical laws.

This guide provides the systematic decoding strategy and walks through three full-length solved case study questions across Chemistry, Biology, and Physics.

What You Will Learn

• The 3-tier structure of CBSE Science Section E Case Studies
• Strategic reading framework to extract experimental variables without getting distracted
• Case Study 1 (Chemistry): Industrial Corrosion, Rusting Mechanics, and Sacrificial Protection
• Case Study 2 (Biology): Mendelian Genetics, Monohybrid Inheritance, and Blood Group Transmission
• Case Study 3 (Physics): Domestic Electric Overloading, Short-Circuiting, and Earthing Safety
• Step-by-step scoring templates for 1-mark and 2-mark sub-questions

1. Structure of Science Case Study Questions

Each 4-mark case study question is organized into three distinct components:

• Part (a) [1 Mark]: Foundational definition, chemical formula, or direct scientific term.
• Part (b) [1 Mark]: A logical cause-and-effect relationship or balanced chemical equation.
• Part (c) [2 Marks] (With Internal Choice): An analytical deduction, numerical calculation, or comparative justification.

2. Solved High-Yield Board Exam Case Studies

Case Study 1: The Rusting of Iron and Sacrificial Protection (Chemistry)

Context: An engineering team inspecting a coastal shipping port noticed that the steel pilings supporting the docks were severely corroded. Iron structures exposed to saline coastal air deteriorate much faster than those in dry inland regions. To safeguard the steel pillars, engineers bolted large blocks of zinc metal directly onto the submerged portions of the iron pilings.

                           Coastal Saltwater Air (O₂ + H₂O)
                                        |
       +--------------------------------v--------------------------------+
       | Steel (Iron) Structure                                         |
       |                                                                 |
       |             [ Zinc Block (Sacrificial Anode) ]                  |
       |             Zn oxidizes FIRST, protecting Iron!                 |
       +-----------------------------------------------------------------+

Questions & Step-by-Step Solutions:

(a) Write the chemical formula and chemical name of rust. [1 Mark]

• Answer: Rust is hydrated ferric oxide. $\mathbf{\text{Chemical Formula: } Fe_2O_3 \cdot xH_2O}$

(b) Why does iron corrode significantly faster in coastal areas than in desert areas? [1 Mark]

• Answer: Coastal air contains high levels of moisture (humidity) and dissolved salts (electrolytes). Dissolved salts increase the electrical conductivity of water droplets on the iron surface, accelerating the electrochemical oxidation of iron into rust.

(c) Explain how bolting zinc blocks to the steel pilings prevents rusting. Name this method of protection. [2 Marks]

• Answer:
  1. According to the reactivity series, Zinc ($Zn$) is more reactive than Iron ($Fe$).
  2. Because zinc has a higher tendency to lose electrons, it oxidizes preferentially, sacrificing itself to protect the underlying iron: $Zn \longrightarrow Zn^{2+} + 2e^-$
  3. The iron remains unoxidized as long as zinc is present.
  4. <u>This method of corrosion prevention is called Sacrificial Protection (or Cathodic Protection). (When zinc is coated as a continuous thin layer over iron, it is termed Galvanization).</u>

Case Study 2: Mendelian Inheritance of Plant Height (Biology)

Context: In a classic hybridization experiment, Gregor Mendel crossed pure-breeding tall pea plants ($TT$) with pure-breeding short (dwarf) pea plants ($tt$). All the offspring in the $F_1$ generation were tall. When these $F_1$ tall plants were allowed to self-pollinate, the resulting $F_2$ generation produced a total of $800$ plants, consisting of both tall and dwarf varieties.

    Parents:                   Tall (TT)   ×   Dwarf (tt)
                                     \            /
    F1 Generation:                     All Tall (Tt)
                                             |  (Self-Pollination)
    F2 Generation:             3 Tall (TT, Tt)  :  1 Dwarf (tt)
                                   (75%)               (25%)

Questions & Step-by-Step Solutions:

(a) What was the genotype of the plants in the $F_1$ generation? [1 Mark]

• Answer: The genotype of all $F_1$ plants was heterozygous tall ($Tt$).

(b) Why were there no dwarf plants observed in the $F_1$ generation? [1 Mark]

• Answer: According to Mendel's Law of Dominance, the allele for tallness ($T$) is dominant over the allele for dwarfness ($t$). In the heterozygous condition ($Tt$), the recessive trait of dwarfness remains masked and unexpressed.

(c) Out of the $800$ plants in the $F_2$ generation, calculate the expected number of: (i) Tall plants, and (ii) Dwarf plants. [2 Marks]

• Answer:
  1. In the $F_2$ generation of a monohybrid cross, the phenotypic ratio is strictly $3 \text{ Tall} : 1 \text{ Dwarf}$.
  2. Total plants $N = 800$.
     • (i) Number of Tall Plants: $\text{Tall Plants} = \frac{3}{4} \times 800 = \mathbf{600\text{ plants}}$
     • (ii) Number of Dwarf Plants: $\text{Dwarf Plants} = \frac{1}{4} \times 800 = \mathbf{200\text{ plants}}$
  3. Therefore, <u>there are $600$ tall plants and $200$ dwarf plants</u>.

Case Study 3: Domestic Electrical Safety and Overloading (Physics)

Context: During a winter morning, a family plugged an electric room heater ($2000\text{ W}$), an electric kettle ($1500\text{ W}$), and a microwave oven ($1100\text{ W}$) into a single multi-socket power board connected to a domestic $220\text{ V}$ power supply protected by a $15\text{ A}$ fuse. Within minutes, the circuit wires became hot and electrical power to the room tripped.

Questions & Step-by-Step Solutions:

(a) What is the total power consumed when all three appliances operate simultaneously? [1 Mark]

• Answer: $P_{\text{total}} = 2000\text{ W} + 1500\text{ W} + 1100\text{ W} = \mathbf{4600\text{ Watts}}$

(b) Distinguish between 'Overloading' and 'Short-Circuiting'. [1 Mark]

• Answer:
  • Overloading: Occurs when the total current drawn by multiple appliances exceeds the safe carrying capacity of the circuit wires.
  • Short-Circuiting: Occurs when the Live wire and Neutral wire come into direct physical contact with zero resistance ($R \approx 0$), causing an immediate massive current surge.

(c) Calculate the total current drawn from the $220\text{ V}$ supply. Will the $15\text{ A}$ fuse blow? Justify. [2 Marks]

• Answer:
  1. Formula: $P = V \times I \implies I = \frac{P}{V}$.
  2. Substitute values: $I_{\text{total}} = \frac{4600\text{ W}}{220\text{ V}} = \frac{460}{22} = \mathbf{20.91\text{ Amperes}}$
  3. Evaluation: The total current drawn ($20.91\text{ A}$) is substantially greater than the $15\text{ A}$ safe rating of the fuse.
  4. <u>Yes, the 15 A fuse will blow (melt) immediately due to excessive Joule heating ($H = I^2Rt$), successfully breaking the circuit and preventing an electrical fire!</u>

3. Summary and Examination Tips

Exam Tip: In genetics case studies involving phenotypic vs. genotypic ratios, always double check whether the question asks for the phenotype (tall/dwarf count $= 3:1$) or genotype (pure tall $TT$, hybrid tall $Tt$, dwarf $tt = 1:2:1$)!

Common Mistake: In physics case studies, confusing power with energy. Power is in Watts ($P = VI$), while energy consumed is in Kilowatt-hours ($E = P \times t$)!