Exploration: Entering the World of Secondary Science — CBSE Class 9 Science Important Questions

Answer: (c) mole.

Amount of substance is a fundamental quantity measured in the mole ($\text{mol}$mol). The kilogram measures mass, the candela luminous intensity, and the kelvin temperature.

Answer: (d) Speed.

Speed is derived because $\text{speed}=\dfrac{\text{distance}}{\text{time}}$speed=distance/time, a combination of the fundamental quantities length and time. Length, time and mass are themselves fundamental.

Answer: (c) $2500\ \text{m}$2500 m.

Since $1\ \text{km}=1000\ \text{m}$1 km=1000 m, we have $2.5\ \text{km}=2.5\times1000\ \text{m}=2500\ \text{m}$2.5 km=2.5×1000 m=2500 m.

Answer: (b) Precise but not accurate.

Precision means the readings agree closely with each other; accuracy means they are close to the true value. Here the readings cluster together (precise) but miss the true value (not accurate), suggesting a systematic error such as a zero error.

Answer: (a) Both A and R are true and R is the correct explanation of A. A physical quantity $=\text{numerical value}\times\text{unit}$=numerical value×unit, so a bare number like $5$5 is meaningless without stating whether it is $5\ \text{m}$5 m, $5\ \text{kg}$5 kg, etc.

The main steps are:

1. Observation of a phenomenon.
2. Framing a question.
3. Proposing a hypothesis (a testable guess).
4. Performing a controlled experiment.
5. Analysing the data collected.
6. Drawing a conclusion (accepting or rejecting the hypothesis).

The least count is the smallest measurement (the smallest change in the quantity) that an instrument can read reliably.

An ordinary metre scale has its smallest division of $1\ \text{mm}$1 mm, so its least count is $1\ \text{mm}=0.1\ \text{cm}=0.001\ \text{m}$1 mm=0.1 cm=0.001 m.

Fundamental quantities are basic, independent quantities that are not defined in terms of others. Examples: length (SI unit metre, $\text{m}$m) and time (SI unit second, $\text{s}$s).

Derived quantities are obtained by combining fundamental quantities through multiplication or division. Examples: area $=\text{length}\times\text{breadth}$=length×breadth (SI unit $\text{m}^2$m^2) and speed $=\dfrac{\text{distance}}{\text{time}}$=distance/time (SI unit $\text{m/s}$m/s).

Thus every derived unit is built from the base units, while base units stand on their own.

The SI (International System of Units) is preferred because:

1. It is internationally accepted, so scientists everywhere use the same units and can share results without confusion.
2. It is a decimal (metric) system — conversions use powers of $10$10 (e.g. $1\ \text{m}=100\ \text{cm}$1 m=100 cm), making calculations easy.
3. It is coherent and complete — a small set of base units generates all derived units without arbitrary conversion factors.

Older systems (like the foot-pound-second system) used awkward factors such as $12$12 inches to a foot, which made comparison difficult.

First convert to the same unit. $80\ \text{cm}=0.8\ \text{m}$80 cm=0.8 m.

Area in $\text{m}^2$m^2:
$A=1.2\ \text{m}\times0.8\ \text{m}=0.96\ \text{m}^2$A=1.2 m×0.8 m=0.96 m^2

Area in $\text{cm}^2$cm^2: Since $1\ \text{m}=100\ \text{cm}$1 m=100 cm, $1\ \text{m}^2=10000\ \text{cm}^2$1 m^2=10000 cm^2.
$A=0.96\times10000=9600\ \text{cm}^2$A=0.96×10000=9600 cm^2

If the two sides had been multiplied as $1.2\times80$1.2×80 without converting, the answer $96$96 would be in mixed $\text{m}\cdot\text{cm}$m·cm and physically meaningless. Consistent units are essential for a correct result.

Hypothesis: Plants grow taller when exposed to more sunlight.

Method:

1. Take several identical potted plants of the same species, age and size, planted in the same soil.
2. Place them in locations receiving different amounts of sunlight — for example $2$2, $4$4 and $8$8 hours per day — while keeping water, soil and pot size the same.
3. Keep one plant in a fixed, ordinary condition to act as the control for comparison.
4. Measure each plant's height with a metre scale every day for two weeks, recording data in a table.

Variables:

• Independent variable (the one changed): hours of sunlight.
• Dependent variable (the one measured): height of the plant.
• Controlled variables (kept constant): water, soil, temperature, species.

Analysis and conclusion: Plot height against time for each group. If the plants with more sunlight grow measurably taller while all else is equal, the data support the hypothesis; otherwise it is rejected. Repeating with several plants per group reduces the effect of chance and makes the conclusion reliable.

The seven SI base quantities, their units and symbols are:

1. Length — metre — $\text{m}$m
2. Mass — kilogram — $\text{kg}$kg
3. Time — second — $\text{s}$s
4. Electric current — ampere — $\text{A}$A
5. Temperature — kelvin — $\text{K}$K
6. Amount of substance — mole — $\text{mol}$mol
7. Luminous intensity — candela — $\text{cd}$cd

Building the unit of density:

Density is defined as mass per unit volume:
$\text{density}=\frac{\text{mass}}{\text{volume}}$density=mass/volume

Mass has the base unit $\text{kg}$kg. Volume is a derived quantity, $\text{length}\times\text{breadth}\times\text{height}$length×breadth×height, with unit $\text{m}\times\text{m}\times\text{m}=\text{m}^3$m×m×m=m^3.

Therefore the SI unit of density is:
$\frac{\text{kg}}{\text{m}^3}=\text{kg/m}^3$kg/m^3=kg/m^3

This shows how a physically meaningful derived unit is assembled entirely from base units, which is the strength of the coherent SI system.

(i) It is the displacement method (the water-displacement or 'overflow' method), based on the fact that a submerged solid displaces a volume of water equal to its own volume.

(ii) Volume of stone $=$= final level $-$- initial level:
$V=68\ \text{mL}-50\ \text{mL}=18\ \text{mL}\ (=18\ \text{cm}^3)$V=68 mL-50 mL=18 mL (=18 cm^3)

(iii) The smallest division is $1\ \text{mL}$1 mL, so the least count $=1\ \text{mL}$=1 mL.

(iv) It must be lowered gently so that no water splashes out (which would give a wrong reading) and fully submerged so that the water it displaces exactly equals the stone's whole volume. If part of the stone stuck out, the measured rise would be too small.