Unit 1 Scientific Study
Class 10 Science Notes
Table of Contents (Unit 1 Scientific Study)
1. Summary: Scientific Study
1.1 Chapter Overview
This chapter introduces the foundations of Scientific Study, emphasizing that science is a systematic investigation of nature involving facts, theories, models, and experiments. The core focus is on understanding the variables that influence scientific phenomena—distinguishing between causes (independent variables) and effects (dependent variables).
Furthermore, the chapter classifies physical quantities into Fundamental and Derived units, explaining how standard measurements form the basis of scientific communication. It concludes with the concept of Unit Analysis (Dimensional Analysis), a method used to verify the correctness of physics equations by ensuring unit consistency on both sides of an equation.
1.2 Important Terms & Definitions
Variables in Scientific Research
Variable: Any causative factor, characteristic, or trait that has a direct or indirect relationship with an event or phenomenon.
Independent Variable (The Cause)
Definition: The factor that is changed or manipulated by the researcher. It is the “cause” in a cause-and-effect relationship.
Example: In an experiment testing how water affects plant growth, the amount of water is the independent variable.
Dependent Variable (The Effect)
Definition: The factor that is observed or measured in response to changes in the independent variable. It is the “effect” or outcome.
Example: In the same plant experiment, the height of the plant is the dependent variable.
Controlled Variable (Constant)
Definition: Factors that are kept constant throughout the experiment to ensure the test is fair and valid.
Example: The type of soil, pot size, and amount of sunlight must remain the same for all plants.
Units of Measurement
Fundamental Unit: An independent unit that does not depend on other units. There are seven standard fundamental units (e.g., Kilogram, Meter, Second).
Derived Unit: A unit formed by combining two or more fundamental units. Example: Velocity ($m/s$), Force ($N$ or $kg \cdot m/s^2$), Area ($m^2$).
1.3 Important Formulas & Unit Analysis
The Principle of Unit Analysis
For any scientific equation to be valid, the units on the Left-Hand Side (LHS) must be identical to the units on the Right-Hand Side (RHS).
General Rule: $$\text{Unit of LHS} = \text{Unit of RHS}$$
Examples from the Text
1. Valid Equation Example
Equation: $s = v \times t$ (Where $s$ = distance, $v$ = velocity, $t$ = time)
LHS Unit: Distance ($s$) is measured in meters (m).
RHS Unit: Velocity ($v$) $\times$ Time ($t$) = $(m/s) \times s = m$
Conclusion: Since LHS ($m$) = RHS ($m$), the equation is valid.
2. Invalid Equation Example
Equation: $s = \frac{v}{t}$
LHS Unit: Distance ($s$) = m
RHS Unit: Velocity ($v$) / Time ($t$) = $(m/s) / s = m/s^2$
Conclusion: Since LHS ($m$) $\neq$ RHS ($m/s^2$), the equation is invalid.
1.4 Fundamental vs. Derived Units
| Feature | Fundamental Unit | Derived Unit |
|---|---|---|
| Dependency | Does not depend on other units; it is independent. | Depends on fundamental units. |
| Formation | Defined by international standards (SI). | Formed by multiplying or dividing fundamental units. |
| Quantity | There are only seven basic units. | There are many derived units used in science. |
| Examples | Meter ($m$), Kilogram ($kg$), Second ($s$) | Newton ($N$), Pascal ($Pa$), Joule ($J$), Watt ($W$) |
1.5 Quick Review (Key Takeaways)
• Scientific Study involves logical investigation based on measurable evidence.
• Cause & Effect: In any experiment, the Independent Variable is the cause, and the Dependent Variable is the effect.
• Control is Crucial: To prove that ‘A’ causes ‘B’, all other potential causes (Controlled Variables) must be kept distinct and constant.
• Unit consistency: You cannot add apples to oranges. In physics, you cannot equate specific Distance ($m$) to Acceleration ($m/s^2$). Always check units to verify if a formula is physically possible.
• 7 Base Units: Remember that all complex physics units (Force, Energy, Power) are just combinations of the basic 7 units (mostly Mass, Length, and Time).
2.1 Choose the correct options for the following questions:
(a) Which of the following is a fundamental unit?
(b) Which of the following physical quantities has the unit ms⁻¹?
(c) Which of the following is a derived unit?
(d) Which of the following units denote newton?
2.2 Give reason:
(a) Joule is the derived unit of work.
Reason: A derived unit is formed by the combination of two or more fundamental units. Work is defined as the product of force and displacement ($W = F \times d$).
Unit of Force = $kg \cdot m/s^2$ (or Newton)
Unit of Displacement = $m$ (meter)
Therefore, Unit of Work (Joule) = $(kg \cdot m/s^2) \times m = kg \cdot m^2 \cdot s^{-2}$.
Since the unit Joule depends on the fundamental units of mass (kg), length (m), and time (s), it is a derived unit.
(b) Some variables should be controlled while performing an experiment.
Reason: In a scientific experiment, we aim to test the effect of one specific variable (independent variable) on another (dependent variable). If other factors (variables) that could affect the outcome are not controlled (kept constant), we cannot be certain that the results are caused solely by the independent variable. Uncontrolled variables introduce errors and make the test unfair or invalid.
(c) $v^2 = ut$ is not a valid relation.
Reason: We can check validity using unit analysis (dimensional analysis). For an equation to be valid, the units on the LHS must match the units on the RHS.
LHS (Left Hand Side): $v^2$ (velocity squared) = $(m/s)^2 = m^2s^{-2}$
RHS (Right Hand Side): $u \times t$ (velocity × time) = $(m/s) \times s = m$
Since the unit on the LHS ($m^2s^{-2}$) is not equal to the unit on the RHS ($m$), the equation is dimensionally inconsistent and therefore not valid.
2.3 Differentiate between:
(a) Independent variable and dependent variable
| Feature | Independent Variable | Dependent Variable |
|---|---|---|
| Definition | It is the variable that is changed or controlled in a scientific experiment to test the effects on the dependent variable. | It is the variable being tested and measured in a scientific experiment. |
| Role | It acts as the “cause” in the cause-and-effect relationship. | It acts as the “effect” or outcome. |
| Manipulation | The researcher directly manipulates or sets the values of this variable. | The researcher observes and records changes in this variable; they do not change it directly. |
| Graphing | It is typically plotted on the X-axis (horizontal axis) of a graph. | It is typically plotted on the Y-axis (vertical axis) of a graph. |
| Example | In an experiment measuring plant growth over time, “Time” is the independent variable. | In the same experiment, the “Height of the plant” is the dependent variable. |
(b) Fundamental unit and derived unit
| Feature | Fundamental Unit | Derived Unit |
|---|---|---|
| Definition | A unit of measurement that is independent of any other unit is called a fundamental unit. | A unit of measurement that is formed by the combination of two or more fundamental units is called a derived unit. |
| Dependency | They do not depend on other units. | They depend on fundamental units for their definition. |
| Resolution | They cannot be resolved (broken down) into simpler units. | They can be resolved into respective fundamental units. |
| Quantity | There are only 7 fundamental units in the SI system (plus 2 supplementary). | There are a vast number of derived units (e.g., area, volume, force, energy). |
| Examples | Kilogram (kg), Meter (m), Second (s), Kelvin (K). | Newton (N), Pascal (Pa), Joule (J), Watt (W). |
2.4 Answer the following questions:
(a) What is a unit?
Answer: A unit is a standard quantity of measurement against which other quantities of the same kind are compared. For example, a “meter” is a standard unit for measuring length.
(b) Write the SI units of mass, temperature, energy, and density.
Answer:
• Mass: Kilogram (kg)
• Temperature: Kelvin (K)
• Energy: Joule (J)
• Density: Kilogram per cubic meter ($kg/m^3$)
(c) How is the validity of an equation checked? Write an example.
Answer: The validity of an equation is checked using unit analysis (or dimensional analysis). This follows the principle of homogeneity, which states that for an equation to be physically valid, the fundamental units of the terms on the Left Hand Side (LHS) must be identical to the fundamental units of the terms on the Right Hand Side (RHS).
Example: Check if $s = vt$ is valid.
LHS unit (Displacement, $s$) = meter ($m$)
RHS unit (Velocity $v$ $\times$ Time $t$) = $(m/s) \times s = m$
Since LHS = RHS ($m = m$), the equation is valid.
(d) Mention the fundamental units involved in the unit of pressure.
Answer: Pressure ($P$) = Force / Area ($F/A$)
Unit = Newton / $meter^2$ = $(kg \cdot m \cdot s^{-2}) / m^2$
Simplified Unit = $kg \cdot m^{-1} \cdot s^{-2}$
Fundamental units involved: Kilogram (kg), Meter (m), and Second (s).
(e) Find out the fundamental units involved in the given derived unit.
Answer:
i. newton (N): Force = mass × acceleration.
Fundamental units: $kg \cdot m \cdot s^{-2}$
ii. watt (W): Power = Work / time = (Force × distance) / time.
Unit = $(kg \cdot m \cdot s^{-2} \times m) / s$
Fundamental units: $kg \cdot m^2 \cdot s^{-3}$
iii. joule (J): Work/Energy = Force × distance.
Unit = $(kg \cdot m \cdot s^{-2}) \times m$
Fundamental units: $kg \cdot m^2 \cdot s^{-2}$
iv. pascal (Pa): Pressure = Force / Area.
Unit = $(kg \cdot m \cdot s^{-2}) / m^2$
Fundamental units: $kg \cdot m^{-1} \cdot s^{-2}$
(f) Niva claimed that an alternative formula for power is $P = mv^2$ and the formula of pressure is $P = mv/A$. Check the validity of given formulae by the analysis of units.
Answer:
Claim 1: Power ($P$) = $mv^2$
Unit of Power (LHS) = Watt = $kg \cdot m^2 \cdot s^{-3}$
Unit of $mv^2$ (RHS) = mass $\times$ velocity$^2$ = $kg \times (m/s)^2 = kg \cdot m^2 \cdot s^{-2}$
Conclusion: LHS ($kg \cdot m^2 \cdot s^{-3}$) $\neq$ RHS ($kg \cdot m^2 \cdot s^{-2}$). The formula is invalid. (Note: $mv^2$ relates to energy, not power).
Claim 2: Pressure ($P$) = $mv/A$
Unit of Pressure (LHS) = Pascal = $kg \cdot m^{-1} \cdot s^{-2}$
Unit of $mv/A$ (RHS) = $(mass \times velocity) / Area$ = $(kg \times m/s) / m^2 = (kg \cdot m \cdot s^{-1}) / m^2 = kg \cdot m^{-1} \cdot s^{-1}$
Conclusion: LHS ($kg \cdot m^{-1} \cdot s^{-2}$) $\neq$ RHS ($kg \cdot m^{-1} \cdot s^{-1}$). The formula is invalid.
(g) Describe the independent variable, dependent variable, and controlled variable with a suitable example of each.
Answer:
Independent Variable: The variable that is deliberately changed by the experimenter.
Dependent Variable: The variable that changes in response to the independent variable and is measured.
Controlled Variable: The variables that are kept constant to ensure a fair test.
Example: In an experiment to see if fertilizer affects plant height:
• Independent: Amount of fertilizer used.
• Dependent: Height of the plant measured in cm.
• Controlled: Type of plant, amount of water, amount of sunlight, pot size.
(h) Karma’s experiment (Thickness of wire vs Life span of dry cell):
Answer:
Independent Variable: Thickness of the wire. (This is what Karma changes).
Dependent Variable: Life span of the dry cell. (This is the result Karma observes).
Controlled Variables: Length of the wire, type of bulb, type of dry cell (brand/voltage), temperature of the room.
(i) Chandani’s experiment (Substances mixed with soil vs Plant growth):
Answer:
i. Identify variables:
• Independent Variable: The type of substance mixed in the soil (Lime, urea, salt, compost).
• Dependent Variable: The height of the plant (Plant growth).
• Controlled Variables: Amount of soil, size of pots, type of seeds, location (sunny place), amount of water, frequency of watering.
ii. Why did Chandani use 3 pots for each experiment?
Using 3 pots (replication) allows Chandani to calculate an average result for each substance. This reduces the effect of errors or anomalies (e.g., if one seed was bad) and makes the experimental results more reliable and accurate.
(j) Subodh’s experiment (Colour of object vs Heat absorption):
Answer:
Independent Variable: The colour of the enamel on the flask (Black, White, Green, Red).
Dependent Variable: The temperature of the water (ability to hold/absorb heat).
Controlled Variables: Subodh should control:
• The size and shape of the conical flasks.
• The amount (volume) of water in each flask.
• The initial temperature of the water.
• The amount of time they are kept in the sun.
• The intensity of sunlight (all must be in the same spot).
(k) Manisha’s experiment (Eating habits of dog):
Answer:
What is wrong? Manisha is changing two variables at the same time: the amount of food AND the time of giving food. Because both change simultaneously, she will not know which factor caused the change in the dog’s eating speed. This makes it an unfair test.
Correction: She should run two separate experiments (or control one variable).
• Experiment A: Keep the time constant (e.g., always 8:00 AM) and vary the amount of food.
• Experiment B: Keep the amount of food constant (e.g., always 200g) and vary the time of feeding.
(l) Prove that: Unit of electric resistance ohm ($\Omega$) = $kgm^2s^{-3}A^{-2}$
Answer: We know from Ohm’s Law: $V = I \times R$ or $R = V / I$
Where $V$ (Voltage/Potential Difference) = Work Done / Charge = $W / Q$
And Charge ($Q$) = Current ($I$) $\times$ Time ($t$). So, $V = W / (I \times t)$.
Now substitute $V$ back into the resistance formula:
$$R = \frac{W / (I \times t)}{I} = \frac{W}{I^2 \times t}$$
Now, convert to fundamental units:
Unit of Work ($W$, Joule) = Force $\times$ distance = $(kg \cdot m \cdot s^{-2}) \times m = kg \cdot m^2 \cdot s^{-2}$
Unit of Current ($I$) = Ampere ($A$)
Unit of Time ($t$) = Second ($s$)
Substitute these into the equation for $R$:
$$Unit of R (\Omega) = \frac{kg \cdot m^2 \cdot s^{-2}}{A^2 \times s}$$
$$Unit of R (\Omega) = kg \cdot m^2 \cdot s^{-2} \cdot s^{-1} \cdot A^{-2}$$
$$Unit of R (\Omega) = kg \cdot m^2 \cdot s^{-3} \cdot A^{-2}$$
Proved.