Official Syllabus Item: 14, 26, 48
Arithmetic & Applications
Units Conversion, Metric System & Temperature Scales
Master metric ladder conversions (length, mass, area, volume, capacity), hectare/are relations, and Celsius-Fahrenheit-Kelvin temperature scales for RIMC Entrance Exam.
Overview & Fundamental Concept
Units Conversion and Temperature Scales form the foundational dimensional bedrock of the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Serial Numbers 14 ('Conversion of Units'), 26 ('Temperature'), and 48 ('Temperature') of Chapter 8 in the official brochure, questions on this module are staple scoring items in Part A (5 Marks) and essential sub-steps in Part B (10 Marks). The syllabus tests mastery over the decimal metric hierarchy (kilo to milli), dimensional unit transformations across powers of ten (1D length, 2D area, 3D volume/capacity), land measurement units (hectare, are, acre), and inter-conversion between thermometric scales (Celsius, Fahrenheit, and absolute Kelvin). In the descriptive exam booklet, RIMC examiners evaluate candidates on their formal derivation of temperature formulas, explicit square/cube exponent rules during dimensional scaling, step-wise arithmetic reduction, and boxed answers with standard physical symbols.
Official Syllabus Coverage Breakdown
| Syllabus Item | Prescribed Scope | Exam Relevance & Pedagogical Objective |
|---|---|---|
| Item 14 | Conversion of Units: Length, mass, area, volume, capacity, and agricultural land units | Tests multi-dimensional dimensional analysis, square/cube exponent shifts, and real-world metric reductions. |
| Item 26 & 48 | Temperature: Celsius, Fahrenheit, Kelvin scales, boiling/freezing points, and equality derivation | Explicitly duplicated in official brochure syllabus. Assesses thermometric formula manipulation and algebraic equality proofs. |
1. The Universal Decimal Metric Ladder
The International Metric System is a base-$10$ decimal system of measurement. All metric units are built by attaching standardized prefixes to fundamental base units: metre (m) for length, gram (g) for mass, and litre (L) for capacity.
The standard prefix hierarchy (from largest to smallest) is memorized using the classic cadet mnemonic: 'King Henry Died By Drinking Chocolate Milk':
$$\text{Kilo } (10^3) \to \text{Hecto } (10^2) \to \text{Deka } (10^1) \to \mathbf{\text{Base Unit } (10^0)} \to \text{Deci } (10^{-1}) \to \text{Centi } (10^{-2}) \to \text{Milli } (10^{-3})$$
• Moving Down the Ladder (Large to Small): Multiply by $10$ for each step down.
• Moving Up the Ladder (Small to Large): Divide by $10$ for each step up.
Heavy Mass & Commercial Trade Units
- 1 Metric Tonne (t) $= 10\text{ Quintals} = 1,000\text{ kilograms (kg)}$.
- 1 Quintal (q) $= 100\text{ kilograms (kg)}$.
- 1 Kilogram (kg) $= 1,000\text{ grams (g)}$.
- 1 Gram (g) $= 1,000\text{ milligrams (mg)}$.
2. Multi-Dimensional Dimensional Shifts (Area & Volume)
A common pitfall among candidates is applying linear multipliers to area and volume. Linear, areal, and volumetric units scale with powers of the conversion factor:
Agricultural Land Measurement Units
- 1 Are: The area of a square of side $10\text{ m}$:
$$1\text{ are} = 10\text{ m} \times 10\text{ m} = \mathbf{100\text{ m}^2}$$ - 1 Hectare (ha): The area of a square of side $100\text{ m}$:
$$1\text{ hectare} = 100\text{ m} \times 100\text{ m} = \mathbf{10,000\text{ m}^2} = 100\text{ ares}$$ - $$1\text{ km}^2 = 1,000,000\text{ m}^2 = \mathbf{100\text{ hectares}}$$
| Dimension | Linear Base Relation | Exponent Scaling Rule | Standard Exam Conversion |
|---|---|---|---|
| 1D: Length | $1\text{ m} = 100\text{ cm}$ | $10^2$ | $$1\text{ km} = 1,000\text{ m}$$ |
| 2D: Area | $1\text{ m}^2 = (100\text{ cm})^2$ | $(10^2)^2 = 10^4$ | $$1\text{ m}^2 = 10,000\text{ cm}^2$$ $$1\text{ km}^2 = 1,000,000\text{ m}^2$$ |
| 3D: Volume | $1\text{ m}^3 = (100\text{ cm})^3$ | $(10^2)^3 = 10^6$ | $$1\text{ m}^3 = 1,000,000\text{ cm}^3$$ |
| Capacity & Liquid | $1\text{ litre} = 1,000\text{ cm}^3$ | $1\text{ dm}^3 = 1\text{ L}$ | $$1\text{ m}^3 = 1,000\text{ litres} = 1\text{ kL}$$ $$1\text{ cm}^3 = 1\text{ mL}$$ |
3. The Three Primary Temperature Scales & Fundamental Formula
Temperature measures the degree of hotness or coldness of a substance. In science and engineering, three major scales are employed: Celsius ($^\circ\text{C}$), Fahrenheit ($^\circ\text{F}$), and Kelvin (K).
The master linear conversion equation connects the three scales based on the freezing and boiling points of pure water at standard atmospheric pressure:
$$\frac{C - 0}{100 - 0} = \frac{F - 32}{212 - 32} = \frac{K - 273.15}{373.15 - 273.15}$$
Dividing all denominators by $20$ produces the standard RIMC formula:
$$\mathbf{\frac{C}{5} = \frac{F - 32}{9} = \frac{K - 273.15}{5}}$$
Isolated Conversion Formulas
- Fahrenheit to Celsius: $$C = \frac{5}{9}(F - 32)$$
- Celsius to Fahrenheit: $$F = \frac{9}{5}C + 32$$
- Celsius to Kelvin: $$K = C + 273.15 \approx C + 273$$
Cardinal Physical Temperature Benchmarks
- Absolute Zero: $0\text{ K} = -273.15^\circ\text{C} = -459.67^\circ\text{F}$.
- Freezing Point of Water: $0^\circ\text{C} = 32^\circ\text{F} = 273.15\text{ K}$.
- Normal Human Body Temperature: $37.0^\circ\text{C} = 98.6^\circ\text{F} = 310.15\text{ K}$.
- Boiling Point of Water: $100^\circ\text{C} = 212^\circ\text{F} = 373.15\text{ K}$.
4. The -40 Degree Coincidence & Temperature Difference Laws
Examiners frequently ask candidates to prove at what numerical reading the Celsius and Fahrenheit thermometers indicate the exact same value:
$$\text{Let } C = F = x$$
$$\frac{x}{5} = \frac{x - 32}{9} \implies 9x = 5(x - 32) \implies 9x = 5x - 160$$
$$4x = -160 \implies x = -40$$
Thus, $\mathbf{-40^\circ\text{C} = -40^\circ\text{F}}$.
Temperature Differences $(\Delta T)$: A change of $1^\circ\text{C}$ is equivalent to a change of $1.8^\circ\text{F}$ (or $\frac{9}{5}^\circ\text{F}$):
$$\Delta F = \frac{9}{5} \times \Delta C$$
Illustrated Conceptual Diagrams & Visual Schematics
Metric Conversion Staircase & Thermometer Scales
Metric Hierarchy & Temperature Equivalence: To convert to a smaller unit (downstairs), multiply by 10 for each step; to convert to a larger unit (upstairs), divide by 10. Universal temperature relationship: $ rac{C}{5} = rac{F - 32}{9} = rac{K - 273}{5}$. Note that $-40^circ ext{C} = -40^circ ext{F}$.
Topper Strategy & Descriptive Step Breakdown
Mastering RIMC Mathematics Standard
In RIMC, clear step-by-step layout is essential. Write down all given values, declare formulas before substitution, and conclude with full sentences with proper units.
Official Descriptive Marking Scheme & Step Allocation
In RIMC Stage-I, answers without step justification receive only 20-30% of total marks even if the numeric answer is correct. Follow this strict 4-tier layout on every question:
Step 1: Formula Declaration & Given Parameters
1 - 2 Marks
State the temperature or metric conversion formula before substituting numeric values.
Booklet Layout:
Given Fahrenheit reading $F = 95^\circ\text{F}$.\nConversion formula from Fahrenheit to Celsius:\n$$C = \frac{5}{9}(F - 32)$$
Given Fahrenheit reading $F = 95^\circ\text{F}$.\nConversion formula from Fahrenheit to Celsius:\n$$C = \frac{5}{9}(F - 32)$$
Step 2: Numeric Substitution & Bracket Simplification
1 - 2 Marks
Substitute the value, evaluate subtraction inside the parentheses, and show explicit cross-cancellations.
Booklet Layout:
$$C = \frac{5}{9}(95 - 32) = \frac{5}{9} \times 63$$
$$C = \frac{5}{9}(95 - 32) = \frac{5}{9} \times 63$$
Step 3: Stepwise Arithmetic Reduction
2 Marks
Perform clean division ($63 \div 9 = 7$) and multiply to get the final numeric magnitude.
Booklet Layout:
$$C = 5 \times 7 = 35^\circ\text{C}$$
$$C = 5 \times 7 = 35^\circ\text{C}$$
Step 4: Boxed Final Value with Proper Units
1 Mark
State the final equivalence clearly in a complete sentence with degree symbols and unit labels.
Booklet Layout:
$$\mathbf{\text{Answer: } 95^\circ\text{F} = 35^\circ\text{C}}$$
$$\mathbf{\text{Answer: } 95^\circ\text{F} = 35^\circ\text{C}}$$
Examiner Traps & Common Blunders (Marks Penalties)
Squaring / Cubing Omission in Area and Volume Conversions -2 to -3 Marks
Common Mistake: Writing $1\text{ m}^2 = 100\text{ cm}^2$ or $1\text{ m}^3 = 100\text{ cm}^3$.
Cadet Antidote: When converting units with exponents, the conversion factor must be raised to that same exponent: $1\text{ m}^2 = (100)^2\text{ cm}^2 = 10,000\text{ cm}^2$, and $1\text{ m}^3 = (100)^3\text{ cm}^3 = 1,000,000\text{ cm}^3$.
Order of Operations in Temperature Conversion -2 Marks
Common Mistake: Computing $F = \frac{9}{5}C + 32$ by adding 32 first, or in $C = \frac{5}{9}(F - 32)$ multiplying before subtracting 32.
Cadet Antidote: Observe strict BODMAS: when converting $F \to C$, evaluate $(F - 32)$ inside brackets FIRST. When converting $C \to F$, multiply $\frac{9}{5}C$ FIRST, then add $32$.
Confusing Hectares and Ares -2 Marks
Common Mistake: Treating 1 hectare as $100\text{ m}^2$ or 1 are as $10,000\text{ m}^2$.
Cadet Antidote: Anchor: 1 Are $= 10\text{ m} \times 10\text{ m} = 100\text{ m}^2$. 1 Hectare $= 100\text{ m} \times 100\text{ m} = 10,000\text{ m}^2 = 100\text{ ares}$.
Missing Degree Symbols on Celsius/Fahrenheit or Adding Degree to Kelvin -1 Mark
Common Mistake: Writing $35\text{ C}$ without degree sign, or writing $300^\circ\text{K}$ (Kelvin is an absolute scale with no degree symbol).
Cadet Antidote: Use $^\circ\text{C}$, $^\circ\text{F}$, and bare $\text{K}$.
Cadet Topper Mental Verification Shortcuts
The Reverse Operation Check for Temperature
Quickly convert your answer back: if $95^\circ\text{F} \to 35^\circ\text{C}$, check $\frac{9}{5}(35) + 32 = 63 + 32 = 95^\circ\text{F}$ in 5 seconds.
Instant verification eliminates arithmetic slips completely.
The 1 Litre = 1000 cm³ Water Bridge
Remember that $1\text{ litre} = 1000\text{ cm}^3 = 1\text{ dm}^3$ and $1\text{ m}^3 = 1000\text{ litres}$.
A tank of $2\text{ m} \times 3\text{ m} \times 1.5\text{ m} = 9\text{ m}^3 = 9000\text{ litres}$ instantly.
The Metric Ladder Finger Count
Count the steps between prefixes: Kilo(3), Hecto(2), Deka(1), Base(0), Deci(-1), Centi(-2), Milli(-3). The difference in exponents gives the exact number of zeros.
Kilometre to millimetre: $3 - (-3) = 6 \implies 1\text{ km} = 10^6\text{ mm} = 1,000,000\text{ mm}$.
Key Formulas & Conversion Cheat Sheet
Master Temperature Conversion Formula
$$\frac{C}{5} = \frac{F - 32}{9} = \frac{K - 273.15}{5}$$
Universal linear equality connecting Celsius, Fahrenheit, and Kelvin.
Celsius to Fahrenheit
$$F = \frac{9}{5}C + 32$$
Computes Fahrenheit reading from Celsius.
Fahrenheit to Celsius
$$C = \frac{5}{9}(F - 32)$$
Computes Celsius reading from Fahrenheit.
Celsius to Kelvin
$$K = C + 273.15$$
Absolute thermodynamic temperature scale conversion.
Area Unit Conversion
$$1 \text{ ha} = 10,000 \text{ m}^2; \quad 1 \text{ are} = 100 \text{ m}^2; \quad 1 \text{ m}^2 = 10,000 \text{ cm}^2$$
Standard areal relationships in SI and agricultural metrics.
Volume & Capacity Equivalence
$$1 \text{ m}^3 = 1,000 \text{ L}; \quad 1 \text{ L} = 1,000 \text{ cm}^3; \quad 1 \text{ mL} = 1 \text{ cm}^3$$
Volume-capacity bridging relations.
Temperature Differential Relation
$$\Delta F = \frac{9}{5} \times \Delta C = 1.8 \times \Delta C$$
Equivalence between intervals of temperature change.
Heavy Mass Conversions
$$1 \text{ tonne} = 10 \text{ q} = 1,000 \text{ kg}; \quad 1 \text{ q} = 100 \text{ kg}$$
Trade units of heavy commercial mass.
Solved Model Examples (Descriptive Format with Step Mark Distribution)
Model 1: Fahrenheit to Celsius Conversion
5 Marks
Convert a recorded body temperature of $104^\circ\text{ Fahrenheit}$ into Celsius ($^\circ\text{C}$).
Step-by-Step Model Solution:
Step 1: State Formula
1.5 Marks
The formula connecting Fahrenheit and Celsius is:
$$C = \frac{5}{9}(F - 32)$$
$$C = \frac{5}{9}(F - 32)$$
Step 2: Substitute F = 104 and Simplify
2 Marks
$$C = \frac{5}{9}(104 - 32) = \frac{5}{9} \times 72$$
Divide $72$ by $9$: $72 \div 9 = 8$.
$$C = 5 \times 8 = 40^\circ\text{C}$$
Divide $72$ by $9$: $72 \div 9 = 8$.
$$C = 5 \times 8 = 40^\circ\text{C}$$
Step 3: Concluding Statement
1.5 Marks
$$\mathbf{\text{Answer: } 104^\circ\text{F} = 40^\circ\text{C}}$$
Examiner Scoring Key: Award 1.5 marks for formula, 2 marks for evaluation 5/9 * 72 = 40, and 1.5 marks for answer with units.
Model 2: Celsius to Fahrenheit & Kelvin
5 Marks
The temperature of an iron rod is raised by heating to $45^\circ\text{C}$. Express this temperature in:
(a) Fahrenheit ($^\circ\text{F}$)
(b) Kelvin ($\text{K}$)
(a) Fahrenheit ($^\circ\text{F}$)
(b) Kelvin ($\text{K}$)
Step-by-Step Model Solution:
Step 1: Part (a) Conversion to Fahrenheit
2.5 Marks
Formula: $F = \frac{9}{5}C + 32$.
Substitute $C = 45$:
$$F = \frac{9}{5}(45) + 32 = (9 \times 9) + 32 = 81 + 32 = \mathbf{113^\circ\text{F}}$$
Substitute $C = 45$:
$$F = \frac{9}{5}(45) + 32 = (9 \times 9) + 32 = 81 + 32 = \mathbf{113^\circ\text{F}}$$
Step 2: Part (b) Conversion to Kelvin
2.5 Marks
Formula: $K = C + 273.15$.
Substitute $C = 45$:
$$K = 45 + 273.15 = \mathbf{318.15\text{ K}}$$
$$\mathbf{\text{Answer: (a) } 113^\circ\text{F}; \quad \text{(b) } 318.15\text{ K}}$$
Substitute $C = 45$:
$$K = 45 + 273.15 = \mathbf{318.15\text{ K}}$$
$$\mathbf{\text{Answer: (a) } 113^\circ\text{F}; \quad \text{(b) } 318.15\text{ K}}$$
Examiner Scoring Key: Showing F = 81 + 32 = 113°F carries 2.5 marks. K = 318.15 K (without degree symbol) carries 2.5 marks.
Model 3: Derivation of the -40 Coincidence Point
7 Marks
Derive algebraically the exact temperature at which the Celsius and Fahrenheit thermometer scales produce the identical numerical reading.
Step-by-Step Model Solution:
Step 1: State Equivalence Condition
2 Marks
Let the common temperature reading be $x$.
Then, $C = x$ and $F = x$.
Then, $C = x$ and $F = x$.
Step 2: Substitute into Conversion Equation
2.5 Marks
We know that $\frac{C}{5} = \frac{F - 32}{9}$.
Substituting $C = x$ and $F = x$:
$$\frac{x}{5} = \frac{x - 32}{9}$$
Substituting $C = x$ and $F = x$:
$$\frac{x}{5} = \frac{x - 32}{9}$$
Step 3: Cross-Multiply & Solve for x
2.5 Marks
$$9x = 5(x - 32)$$
$$9x = 5x - 160$$
$$9x - 5x = -160 \implies 4x = -160 \implies x = -\frac{160}{4} = -40$$
$$\mathbf{\text{Answer: At } -40^\circ, \text{ both Celsius and Fahrenheit scales read identical values: } -40^\circ\text{C} = -40^\circ\text{F}.}$$
$$9x = 5x - 160$$
$$9x - 5x = -160 \implies 4x = -160 \implies x = -\frac{160}{4} = -40$$
$$\mathbf{\text{Answer: At } -40^\circ, \text{ both Celsius and Fahrenheit scales read identical values: } -40^\circ\text{C} = -40^\circ\text{F}.}$$
Examiner Scoring Key: Full algebraic proof 9x = 5x - 160 -> 4x = -160 -> x = -40 is required. Memorized quote alone gets only 2 marks.
Model 4: Agricultural Land Area Conversions
7 Marks
A military parade ground is rectangular in shape, measuring $250\text{ metres}$ in length and $160\text{ metres}$ in breadth.
(a) Find its area in square metres ($\text{m}^2$).
(b) Express this area in hectares.
(c) Express this area in ares.
(a) Find its area in square metres ($\text{m}^2$).
(b) Express this area in hectares.
(c) Express this area in ares.
Step-by-Step Model Solution:
Step 1: Compute Area in Square Metres (Part a)
2.5 Marks
Length $L = 250\text{ m}$, Breadth $B = 160\text{ m}$.
$$\text{Area} = L \times B = 250 \times 160 = \mathbf{40,000\text{ m}^2}$$
$$\text{Area} = L \times B = 250 \times 160 = \mathbf{40,000\text{ m}^2}$$
Step 2: Convert to Hectares (Part b)
2.5 Marks
We know that $1\text{ hectare} = 10,000\text{ m}^2$.
$$\text{Area in hectares} = \frac{40000}{10000} = \mathbf{4\text{ hectares}}$$
$$\text{Area in hectares} = \frac{40000}{10000} = \mathbf{4\text{ hectares}}$$
Step 3: Convert to Ares (Part c)
2 Marks
We know that $1\text{ are} = 100\text{ m}^2$ (or $1\text{ hectare} = 100\text{ ares}$).
$$\text{Area in ares} = \frac{40000}{100} = \mathbf{400\text{ ares}}$$
$$\mathbf{\text{Answer: (a) } 40,000\text{ m}^2; \quad \text{(b) } 4\text{ hectares}; \quad \text{(c) } 400\text{ ares}}$$
$$\text{Area in ares} = \frac{40000}{100} = \mathbf{400\text{ ares}}$$
$$\mathbf{\text{Answer: (a) } 40,000\text{ m}^2; \quad \text{(b) } 4\text{ hectares}; \quad \text{(c) } 400\text{ ares}}$$
Examiner Scoring Key: Stating conversion factors 1 ha = 10,000 m² and 1 are = 100 m² carries 2.5 marks. Box all three values.
Model 5: Volumetric Reservoir Capacity in Litres & Kilolitres
10 Marks
A water reservoir tank at an army cantonment has length $6\text{ m}$, breadth $4.5\text{ m}$, and depth $3\text{ m}$.
(a) Calculate the volume of the tank in cubic metres ($\text{m}^3$) and cubic centimetres ($\text{cm}^3$).
(b) Find the maximum capacity of the tank in litres and in kilolitres.
(c) If water is pumped out at the rate of $450\text{ litres per minute}$, how many hours will it take to empty the completely filled tank?
(a) Calculate the volume of the tank in cubic metres ($\text{m}^3$) and cubic centimetres ($\text{cm}^3$).
(b) Find the maximum capacity of the tank in litres and in kilolitres.
(c) If water is pumped out at the rate of $450\text{ litres per minute}$, how many hours will it take to empty the completely filled tank?
Step-by-Step Model Solution:
Step 1: Compute Volume in m³ and cm³ (Part a)
3 Marks
$$\text{Volume } V = L \times B \times H = 6 \times 4.5 \times 3 = 27 \times 3 = \mathbf{81\text{ m}^3}$$
Since $1\text{ m}^3 = (100\text{ cm})^3 = 1,000,000\text{ cm}^3$:
$$\text{Volume in cm}^3 = 81 \times 10^6 = \mathbf{81,000,000\text{ cm}^3}$$
Since $1\text{ m}^3 = (100\text{ cm})^3 = 1,000,000\text{ cm}^3$:
$$\text{Volume in cm}^3 = 81 \times 10^6 = \mathbf{81,000,000\text{ cm}^3}$$
Step 2: Compute Capacity in Litres and Kilolitres (Part b)
3 Marks
We know that $1\text{ m}^3 = 1,000\text{ litres} = 1\text{ kL}$.
$$\text{Capacity in Litres} = 81 \times 1000 = \mathbf{81,000\text{ litres}}$$
$$\text{Capacity in Kilolitres} = \mathbf{81\text{ kL}}$$
$$\text{Capacity in Litres} = 81 \times 1000 = \mathbf{81,000\text{ litres}}$$
$$\text{Capacity in Kilolitres} = \mathbf{81\text{ kL}}$$
Step 3: Calculate Emptying Time (Part c)
4 Marks
Pumping rate $= 450\text{ litres/minute}$.
$$\text{Time in minutes} = \frac{81000}{450} = \frac{8100}{45} = 180\text{ minutes}$$
Convert to hours:
$$\text{Time in hours} = \frac{180}{60} = \mathbf{3\text{ hours}}$$
$$\mathbf{\text{Answer: (a) } 81\text{ m}^3 \text{ and } 8.1 \times 10^7\text{ cm}^3; \quad \text{(b) } 81,000\text{ L } (81\text{ kL}); \quad \text{(c) } 3\text{ hours}}$$
$$\text{Time in minutes} = \frac{81000}{450} = \frac{8100}{45} = 180\text{ minutes}$$
Convert to hours:
$$\text{Time in hours} = \frac{180}{60} = \mathbf{3\text{ hours}}$$
$$\mathbf{\text{Answer: (a) } 81\text{ m}^3 \text{ and } 8.1 \times 10^7\text{ cm}^3; \quad \text{(b) } 81,000\text{ L } (81\text{ kL}); \quad \text{(c) } 3\text{ hours}}$$
Examiner Scoring Key: Volume calculation = 81 m³ carries 3 marks. 1 m³ = 1000 L conversion carries 3 marks. Time = 3 hours carries 4 marks.
Model 6: Temperature Difference Across Meteorological Systems
10 Marks
During a severe winter high-altitude exercise in Ladakh, the temperature dropped by $25^\circ\text{ Celsius}$ between noon and midnight.
(a) What is the equivalent temperature drop measured on the Fahrenheit scale?
(b) If the noon temperature was $10^\circ\text{C}$, find the midnight temperature in $^\circ\text{C}$, $^\circ\text{F}$, and Kelvin ($\text{K}$).
(a) What is the equivalent temperature drop measured on the Fahrenheit scale?
(b) If the noon temperature was $10^\circ\text{C}$, find the midnight temperature in $^\circ\text{C}$, $^\circ\text{F}$, and Kelvin ($\text{K}$).
Step-by-Step Model Solution:
Step 1: Compute Temperature Drop in Fahrenheit (Part a)
3 Marks
For a temperature difference $\Delta T$:
$$\Delta F = \frac{9}{5} \times \Delta C$$
Given $\Delta C = 25^\circ\text{C}$:
$$\Delta F = \frac{9}{5} \times 25 = 9 \times 5 = \mathbf{45^\circ\text{F}}$$
The temperature dropped by $45^\circ\text{F}$.
$$\Delta F = \frac{9}{5} \times \Delta C$$
Given $\Delta C = 25^\circ\text{C}$:
$$\Delta F = \frac{9}{5} \times 25 = 9 \times 5 = \mathbf{45^\circ\text{F}}$$
The temperature dropped by $45^\circ\text{F}$.
Step 2: Compute Midnight Temperature in Celsius
2 Marks
$$\text{Midnight Temperature in } ^\circ\text{C} = 10^\circ\text{C} - 25^\circ\text{C} = \mathbf{-15^\circ\text{C}}$$
Step 3: Convert Midnight Temperature to Fahrenheit
2.5 Marks
$$F = \frac{9}{5}C + 32 = \frac{9}{5}(-15) + 32 = 9(-3) + 32 = -27 + 32 = \mathbf{5^\circ\text{F}}$$
Check difference: Noon $F = \frac{9}{5}(10) + 32 = 18 + 32 = 50^\circ\text{F}$. Midnight $= 5^\circ\text{F}$. Drop $= 50 - 5 = 45^\circ\text{F}$ (Matches Part a!).
Check difference: Noon $F = \frac{9}{5}(10) + 32 = 18 + 32 = 50^\circ\text{F}$. Midnight $= 5^\circ\text{F}$. Drop $= 50 - 5 = 45^\circ\text{F}$ (Matches Part a!).
Step 4: Convert Midnight Temperature to Kelvin & Conclude
2.5 Marks
$$K = C + 273.15 = -15 + 273.15 = \mathbf{258.15\text{ K}}$$
$$\mathbf{\text{Answer: (a) Drop = } 45^\circ\text{F}; \quad \text{(b) Midnight = } -15^\circ\text{C} = 5^\circ\text{F} = 258.15\text{ K}}$$
$$\mathbf{\text{Answer: (a) Drop = } 45^\circ\text{F}; \quad \text{(b) Midnight = } -15^\circ\text{C} = 5^\circ\text{F} = 258.15\text{ K}}$$
Examiner Scoring Key: Distinguishing temperature DIFFERENCE (ΔF = 9/5 ΔC) from absolute conversion (F = 9/5 C + 32) is critical and tested rigorously.
RIMC Practice & Drill Exercises
Attempt these questions on paper following proper step-wise conventions, then click Reveal Solution to verify your answers.
Question 1
5 Marks
Convert $40^\circ\text{C}$ into Fahrenheit ($^\circ\text{F}$).
Model Descriptive Solution:
Step 1: Apply Formula
2.5 Marks
$$F = \frac{9}{5}(40) + 32$$
Step 2: Solve
2.5 Marks
$$F = 9(8) + 32 = 72 + 32 = 104^\circ\text{F}$$
Final Answer: $104^\circ\text{F}$
Question 2
5 Marks
Express $5\text{ hectares}$ in square metres ($\text{m}^2$).
Model Descriptive Solution:
Step 1: State Conversion Factor
2.5 Marks
$$1\text{ hectare} = 10,000\text{ m}^2$$
Step 2: Multiply by 5
2.5 Marks
$$5\text{ hectares} = 5 \times 10000 = 50,000\text{ m}^2$$
Final Answer: $50,000\text{ m}^2$
Question 3
5 Marks
Convert $77^\circ\text{F}$ into Celsius ($^\circ\text{C}$).
Model Descriptive Solution:
Step 1: Apply Formula
2.5 Marks
$$C = \frac{5}{9}(77 - 32)$$
Step 2: Solve
2.5 Marks
$$C = \frac{5}{9}(45) = 5 \times 5 = 25^\circ\text{C}$$
Final Answer: $25^\circ\text{C}$
Question 4
7 Marks
Convert $3.75\text{ m}^3$ into litres and cubic centimetres ($\text{cm}^3$).
Model Descriptive Solution:
Step 1: Convert to Litres
3.5 Marks
Since $1\text{ m}^3 = 1000\text{ litres}$:
$$3.75\text{ m}^3 = 3.75 \times 1000 = 3,750\text{ litres}$$
$$3.75\text{ m}^3 = 3.75 \times 1000 = 3,750\text{ litres}$$
Step 2: Convert to cm³
3.5 Marks
Since $1\text{ m}^3 = 1,000,000\text{ cm}^3$:
$$3.75\text{ m}^3 = 3.75 \times 10^6 = 3,750,000\text{ cm}^3$$
$$3.75\text{ m}^3 = 3.75 \times 10^6 = 3,750,000\text{ cm}^3$$
Final Answer: $3,750\text{ litres}$; $3,750,000\text{ cm}^3$
Question 5
7 Marks
The temperature of a patient is $38.5^\circ\text{C}$. Find his temperature on the Fahrenheit scale, and determine by how many degrees Fahrenheit this exceeds the normal human body temperature of $98.6^\circ\text{F}$.
Model Descriptive Solution:
Step 1: Convert 38.5°C to Fahrenheit
3.5 Marks
$$F = \frac{9}{5}(38.5) + 32 = 9(7.7) + 32 = 69.3 + 32 = 101.3^\circ\text{F}$$
Step 2: Compute Difference from Normal Temperature
3.5 Marks
$$\text{Fever Above Normal} = 101.3 - 98.6 = 2.7^\circ\text{F}$$
Final Answer: $101.3^\circ\text{F}$; exceeds normal by $2.7^\circ\text{F}$
Question 6
7 Marks
A square field of side $300\text{ m}$ is to be fenced. If $1\text{ hectare} = 10,000\text{ m}^2$, find the area of the field in hectares, and find the cost of turfing the field at ₹ $250\text{ per are}$.
Model Descriptive Solution:
Step 1: Compute Area in m² and Hectares
3 Marks
$$\text{Area} = 300 \times 300 = 90,000\text{ m}^2 = \frac{90000}{10000} = 9\text{ hectares}$$
Step 2: Convert to Ares and Find Cost
4 Marks
$$\text{Area in ares} = \frac{90000}{100} = 900\text{ ares}$$
$$\text{Cost} = 900 \times 250 = ₹ 2,25,000$$
$$\text{Cost} = 900 \times 250 = ₹ 2,25,000$$
Final Answer: Area $= 9\text{ hectares}$; Cost $= ₹ 2,25,000$
Question 7
7 Marks
Convert $5\text{ quintals and } 45\text{ kg}$ into grams, expressing the answer in standard scientific notation.
Model Descriptive Solution:
Step 1: Convert to Total Kilograms
3 Marks
$5\text{ quintals} = 5 \times 100 = 500\text{ kg}$.
$$\text{Total} = 500 + 45 = 545\text{ kg}$$
$$\text{Total} = 500 + 45 = 545\text{ kg}$$
Step 2: Convert to Grams & Scientific Notation
4 Marks
$$\text{In grams} = 545 \times 1000 = 545,000\text{ g}$$
$$\text{Scientific Notation} = 5.45 \times 10^5\text{ g}$$
$$\text{Scientific Notation} = 5.45 \times 10^5\text{ g}$$
Final Answer: $5.45 \times 10^5\text{ g}$
Question 8
10 Marks
At what temperature is the Fahrenheit scale reading exactly double the Celsius scale reading? Show complete algebraic working.
Model Descriptive Solution:
Step 1: State Condition
3 Marks
Let $C = x$. Then $F = 2x$.
Step 2: Substitute into Conversion Equation
4 Marks
$$\frac{x}{5} = \frac{2x - 32}{9} \implies 9x = 5(2x - 32) \implies 9x = 10x - 160$$
Step 3: Solve for x
3 Marks
$$10x - 9x = 160 \implies x = 160$$
Thus: $C = 160^\circ\text{C}$ and $F = 320^\circ\text{F}$.
Thus: $C = 160^\circ\text{C}$ and $F = 320^\circ\text{F}$.
Final Answer: $160^\circ\text{C}$ (where $F = 320^\circ\text{F}$)
Question 9
10 Marks
A cube of metal with edge $20\text{ cm}$ is melted and recast into smaller cubes of edge $4\text{ cm}$ each.
(a) How many smaller cubes can be made?
(b) Find the ratio of the total surface area of all smaller cubes combined to the surface area of the original large cube.
(a) How many smaller cubes can be made?
(b) Find the ratio of the total surface area of all smaller cubes combined to the surface area of the original large cube.
Model Descriptive Solution:
Step 1: Number of Small Cubes (Part a)
4 Marks
$$V_{\text{large}} = 20^3 = 8000\text{ cm}^3, \quad V_{\text{small}} = 4^3 = 64\text{ cm}^3$$
$$N = \frac{8000}{64} = 125\text{ small cubes}$$
$$N = \frac{8000}{64} = 125\text{ small cubes}$$
Step 2: Surface Area Comparison (Part b)
4 Marks
$$SA_{\text{large}} = 6 \times (20)^2 = 6 \times 400 = 2400\text{ cm}^2$$
$$SA_{\text{one small}} = 6 \times (4)^2 = 6 \times 16 = 96\text{ cm}^2$$
$$SA_{\text{125 small}} = 125 \times 96 = 12,000\text{ cm}^2$$
$$SA_{\text{one small}} = 6 \times (4)^2 = 6 \times 16 = 96\text{ cm}^2$$
$$SA_{\text{125 small}} = 125 \times 96 = 12,000\text{ cm}^2$$
Step 3: Ratio
2 Marks
$$\text{Ratio} = \frac{12000}{2400} = \frac{5}{1} = 5 : 1$$
Final Answer: (a) $125\text{ cubes}$; (b) Ratio $= 5 : 1$
Question 10
10 Marks
Liquid oxygen boils at $-183^\circ\text{C}$ and freezes at $-218.8^\circ\text{C}$ at standard pressure.
(a) Convert both temperatures to Fahrenheit ($^\circ\text{F}$).
(b) Find the liquid range (the difference between boiling and freezing points) in both $^\circ\text{C}$ and $^\circ\text{F}$.
(a) Convert both temperatures to Fahrenheit ($^\circ\text{F}$).
(b) Find the liquid range (the difference between boiling and freezing points) in both $^\circ\text{C}$ and $^\circ\text{F}$.
Model Descriptive Solution:
Step 1: Convert Boiling Point to Fahrenheit
3 Marks
$$F_b = \frac{9}{5}(-183) + 32 = -329.4 + 32 = -297.4^\circ\text{F}$$
Step 2: Convert Freezing Point to Fahrenheit
3 Marks
$$F_f = \frac{9}{5}(-218.8) + 32 = -393.84 + 32 = -361.84^\circ\text{F}$$
Step 3: Compute Liquid Range in Both Scales
4 Marks
$$\text{Range in } ^\circ\text{C} = -183 - (-218.8) = 35.8^\circ\text{C}$$
$$\text{Range in } ^\circ\text{F} = \frac{9}{5}(35.8) = 64.44^\circ\text{F}$$
Check: $-297.4 - (-361.84) = 64.44^\circ\text{F}$.
$$\text{Range in } ^\circ\text{F} = \frac{9}{5}(35.8) = 64.44^\circ\text{F}$$
Check: $-297.4 - (-361.84) = 64.44^\circ\text{F}$.
Final Answer: Boiling: $-297.4^\circ\text{F}$; Freezing: $-361.84^\circ\text{F}$; Range: $35.8^\circ\text{C}$ ($64.44^\circ\text{F}$)
High-Yield FAQs & Revision Notes
What is the mnemonic for the metric ladder?
King Henry Died By Drinking Chocolate Milk (Kilo, Hecto, Deka, Base, Deci, Centi, Milli).
How many square metres are in 1 hectare?
Exactly 10,000 m² (a square measuring 100 m on each side).
What is 1 are in square metres?
1 are = 100 m² (a square measuring 10 m on each side). 1 hectare = 100 ares.
At what temperature are Celsius and Fahrenheit readings equal?
At exactly -40°: -40°C = -40°F.
Why is there no degree symbol on Kelvin?
Kelvin is an absolute thermodynamic temperature scale with zero representing absolute zero, not an arbitrary interval scale.
How do you convert 1 m³ to litres?
1 m³ = 1000 litres = 1 kilolitre (kL).
What is the difference between converting a temperature and a temperature difference?
Converting a temperature uses $F = \frac{9}{5}C + 32$. Converting a temperature interval or change uses $\Delta F = \frac{9}{5}\Delta C = 1.8\Delta C$ (no 32 added).
How many kilograms are in a quintal and a metric tonne?
1 quintal = 100 kg; 1 metric tonne = 10 quintals = 1000 kg.
What is the boiling and freezing point of water on Fahrenheit scale?
Freezing point = 32°F; Boiling point = 212°F.
How does this module map to the official RIMC syllabus?
Directly covers Serial No. 14 ('Conversion of Units') and Serial Nos. 26 and 48 ('Temperature') of Chapter 8 in the official RIMC brochure.
How does this topic relate to the official RIMC syllabus?
This module directly covers Serial No. 14, 26, 48 of the broader syllabus published in Chapter 8 of the official RIMC admission brochure, ensuring 100% syllabus alignment.
Last-Minute Cadet High-Yield Revision Matrix
| Topic / Concept | High-Yield Exam Rules & Memory Anchor |
|---|---|
| Temperature Formula | $\frac{C}{5} = \frac{F - 32}{9} = \frac{K - 273.15}{5}$. |
| -40° Coincidence | $-40^\circ\text{C} = -40^\circ\text{F}$. Prove by equating $x/5 = (x-32)/9$. |
| Hectare & Are | $1\text{ ha} = 10,000\text{ m}^2 = 100\text{ ares}$. $1\text{ are} = 100\text{ m}^2$. |
| Area Scaling | $1\text{ m}^2 = 10,000\text{ cm}^2 = 10^4\text{ cm}^2$. Square the linear factor! |
| Volume & Capacity | $1\text{ m}^3 = 1,000,000\text{ cm}^3 = 1,000\text{ L}$. $1\text{ L} = 1,000\text{ cm}^3$. |
| Temp Intervals | $\Delta F = \frac{9}{5}\Delta C = 1.8\Delta C$. Do not add 32 to temperature changes! |