Every question in this chapter, answered and explained — step-by-step solutions drawn live from the chapter library across 6 question sections.
Which of the following is a fundamental unit?
Kilogram is one of the 7 SI fundamental units (unit of mass). Newton, pascal, and joule are all derived units formed by combining fundamental units.
Which of the following physical quantities has the unit ms⁻¹?
Velocity = displacement / time = m/s = ms⁻¹. Acceleration = m/s² (ms⁻²), Force = kg·m·s⁻², Density = kg/m³.
Which of the following is a derived unit?
Joule (unit of work/energy) = kg·m²·s⁻², formed by combining fundamental units. Candela, ampere, and kelvin are all fundamental SI units.
Which of the following units denote newton?
By definition, Force = mass × acceleration = kg × (m/s²) = kg·m·s⁻². This combination is given the special name newton (N). So N = kgms⁻².
Joule is the derived unit of work. Give reason.
Work = force × distance. Unit of force is kg·m·s⁻² and unit of distance is m, so unit of work = kg·m·s⁻² × m = kg·m²·s⁻². Since this unit is formed by combining more than one fundamental unit (kg, m, s), it is a derived unit, and this combination is given the special name joule (J).
Some variables should be controlled while performing an experiment. Give reason.
If more than one variable is allowed to change during an experiment, it becomes impossible to know which variable actually caused the observed effect. This makes the result unreliable and the conclusion invalid. So, apart from the independent and dependent variable, all other variables must be kept constant (controlled) to ensure the experiment's findings are valid, reliable, and accurate.
v² = ut is not a valid relation. Give reason.
Check by unit analysis: LHS unit → v² = (m·s⁻¹)² = m²·s⁻². RHS unit → ut = (m·s⁻¹) × s = m. Since LHS unit (m²s⁻²) is not equal to RHS unit (m), the units on both sides are not the same. Therefore, the equation v² = ut is not valid.
Differentiate between independent variable and dependent variable.
| Independent Variable | Dependent Variable |
|---|---|
| Deliberately changed/manipulated by the researcher. | Its value changes as a result of the independent variable; the researcher cannot set it directly. |
| It is the cause in the experiment. | It is the effect/result observed in the experiment. |
| Plotted on the horizontal (x) axis of a graph. | Plotted on the vertical (y) axis of a graph. |
| Example: extension of rubber band in the catapult activity. | Example: distance travelled by the paper bullet in the catapult activity. |
Differentiate between fundamental unit and derived unit.
| Fundamental Unit | Derived Unit |
|---|---|
| Has independent existence; does not depend on other units. | Has no independent existence; formed by combining fundamental units. |
| Cannot be broken down into simpler units. | Can be expressed in terms of two or more fundamental units. |
| There are 7 fundamental units in the SI system. | Many derived units are formed from the 7 fundamental units. |
| Example: kilogram (kg), metre (m), second (s). | Example: newton (N) = kg·m·s⁻², joule (J) = kg·m²·s⁻². |
What is a unit?
A unit is a standard quantity used to measure and express the magnitude of a physical quantity, so that measurements can be communicated clearly and consistently. For example, metre is the unit used to express length.
Write the SI units of mass, temperature, energy, and density.
Mass → kilogram (kg). Temperature → kelvin (K). Energy → joule (J), a derived unit equal to kg·m²·s⁻². Density → kilogram per cubic metre (kg/m³), a derived unit.
How is the validity of an equation checked? Write an example.
The validity of an equation is checked using unit (dimensional) analysis: the fundamental units of the quantity on the left-hand side (LHS) are compared with the fundamental units of the quantity on the right-hand side (RHS). If LHS unit = RHS unit, the equation is valid; otherwise it is invalid. Example: s = v × t. LHS unit = m. RHS unit = (m·s⁻¹) × s = m. Since LHS = RHS, the equation is valid.
Mention the fundamental units involved in the unit of pressure.
Pressure = force / area = (kg·m·s⁻²) / m² = kg·m⁻¹·s⁻². The fundamental units involved are kilogram (kg), metre (m), and second (s).
Find out the fundamental units involved in the given derived units: (i) newton (N) (ii) watt (W) (iii) joule (J) (iv) pascal (Pa).
(i) Newton: N = kg·m·s⁻² → kilogram, metre, second. (ii) Watt: W = kg·m²·s⁻³ → kilogram, metre, second. (iii) Joule: J = kg·m²·s⁻² → kilogram, metre, second. (iv) Pascal: Pa = kg·m⁻¹·s⁻² → kilogram, metre, second.
Niva claimed that an alternative formula for power is P = mv² and the formula of pressure is P = mv/A. Check the validity of the given formulae by the analysis of units.
Correct unit of power = kg·m²·s⁻³. Check P = mv²: unit = kg × (m·s⁻¹)² = kg·m²·s⁻². Since kg·m²·s⁻² ≠ kg·m²·s⁻³, this formula is NOT valid. Correct unit of pressure = kg·m⁻¹·s⁻². Check P = mv/A: unit = kg × (m·s⁻¹) / m² = kg·m⁻¹·s⁻¹. Since kg·m⁻¹·s⁻¹ ≠ kg·m⁻¹·s⁻², this formula is also NOT valid.
Describe the independent variable, dependent variable, and controlled variable with a suitable example of each.
Independent variable: The variable that the researcher deliberately changes by a chosen amount; it is the cause. Example: In a catapult experiment, the extension of the rubber band is the independent variable. Dependent variable: The variable whose value depends on the independent variable and is measured, not set directly. Example: The distance travelled by the paper bullet. Controlled variable: A variable that is kept the same throughout the experiment so it does not affect the result. Example: The thickness of the rubber band and the size of the paper bullet.
Karma connected a dry cell to a bulb using a few pieces of wire and lit the bulb. He was curious to know how the thickness of the used wire affects the life span of a dry cell. In this test, find out the independent variable, dependent variable, and controlled variable.
Independent variable: Thickness of the connecting wire (Karma deliberately changes this). Dependent variable: Life span of the dry cell (this is measured/observed and depends on wire thickness). Controlled variable: Type/rating of the bulb, type and charge of the dry cell, length of the wire, and the material of the wire — these must be kept the same in every trial so only wire thickness affects the result.
Chandani wanted to investigate the effects of substances (lime, urea fertilizer, common salt, and compost manure) mixed with soil on plant growth. She used 12 uniform pots (3 pots per substance), the same soil, same seeds, equal watering, and a sunny location, then measured plant height daily. (i) Identify the independent variable, dependent variable, and controlled variable in Chandani's experiment. (ii) Why did Chandani use 3 pots for each substance?
(i) Independent variable: The type of substance mixed with the soil (lime, urea, salt, or compost). Dependent variable: The height/growth of the plant. Controlled variable: The type and amount of soil, size of the pots, type of seeds, amount of water given, and sunlight exposure — all kept the same across every pot. (ii) She used 3 pots for each substance to repeat the trial (replication). Using only one pot could give an unreliable or accidental result; testing with 3 pots for each condition helps confirm that the observed effect is consistent and not due to chance, making the conclusion more reliable and valid.
Subodh wanted to find out how the colour of an object affects its ability to hold heat. He coated four identical conical flasks with black, white, green, and red enamel, filled them with equal water, sealed them, kept them in the sun, and then measured the water temperature in each. Identify the independent variable and dependent variable in Subodh's experiment. Which variables should be controlled?
Independent variable: The colour of the flask (enamel coating). Dependent variable: The temperature of the water in each flask after being in the sun. Controlled variables: The amount of water in each flask, the size/shape/material of the flasks, the duration of sun exposure, the initial temperature of the water, and the position/intensity of sunlight received by each flask — all must be kept identical so only colour affects the result.
Manisha wanted to test the eating habits of her dog. She decided to study how the amount of food and the time of giving food affect the speed at which the dog ate. What is wrong with the design of Manisha's experiment and how can she correct it?
The problem: Manisha is changing two independent variables at the same time — the amount of food AND the time of feeding. If the dog's eating speed changes, she cannot tell whether it was caused by the amount of food or by the feeding time, since both were varied together. This makes it impossible to draw a valid, reliable conclusion. Correction: She should test only ONE independent variable at a time while keeping the other constant. For example, first she should vary the amount of food while keeping the feeding time fixed (to see the effect of food amount), and in a separate set of trials, vary the feeding time while keeping the amount of food fixed (to see the effect of feeding time).
Prove that: Unit of electric resistance, ohm (Ω) = kg·m²·s⁻³·A⁻².
By Ohm's law, Resistance R = V / I, where V is potential difference (voltage) and I is electric current. Unit of voltage V = Joule/Coulomb = (kg·m²·s⁻²) / (A·s) = kg·m²·s⁻³·A⁻¹ [since 1 Coulomb = 1 Ampere × 1 second, and 1 Joule = kg·m²·s⁻²]. Unit of current I = A (ampere). Therefore, unit of R = V/I = (kg·m²·s⁻³·A⁻¹) / A = kg·m²·s⁻³·A⁻². Hence proved: Ω = kg·m²·s⁻³·A⁻².
A student wants to find out how the length of a simple pendulum affects its time period (the time for one complete swing). Identify the independent variable, dependent variable, and at least two controlled variables in this experiment.
Independent variable: Length of the pendulum (deliberately changed by the student). Dependent variable: Time period of the pendulum (measured, depends on length). Controlled variables: Mass of the bob (should stay the same), the angle/amplitude of swing (should stay small and the same each time), and the location/gravity (kept constant since the experiment is done in the same place).
Prove, using unit analysis, that the formula for kinetic energy, E = ½mv², is dimensionally valid given that the unit of energy is the joule (J) = kg·m²·s⁻².
RHS unit: ½ is just a number (no unit), so we consider mv². Unit of mass (m) = kg. Unit of velocity (v) = m·s⁻¹, so unit of v² = m²·s⁻². Therefore, unit of mv² = kg × m²·s⁻² = kg·m²·s⁻². LHS unit: Energy (E) = joule (J) = kg·m²·s⁻² (given). Since LHS unit (kg·m²·s⁻²) = RHS unit (kg·m²·s⁻²), the formula E = ½mv² is dimensionally valid.
Which of the following is a fundamental unit — ampere, watt, or newton? Explain your answer in one line.
Ampere is the fundamental unit (SI unit of electric current). Watt (kg·m²·s⁻³) and newton (kg·m·s⁻²) are both derived units, formed by combining fundamental units.