Selina Concise Solutions for ICSE Class 7 Physics Chapter 1 Physical Quantities and Measurement

Points to Remember

• The mass of a body is the quantity of matter contained in a body regardless of its volume or any force acting on it.
• The weight of a body is the force with which every body is attracted towards its centre.
• The unit of mass in S.I. system is kilogram (kg). Higher units of mass are quintal and metric tonne.
• The weight of body changes with acceleration due to gravity.
• Weight is zero at the centre of the earth.
• Mass per unit volume of a substance is called density of the body.
• The unit of density in S.I. system is kg m^-3 and g cm^-3 in C.G.S. system.
• The density in S.I. system is 1000 times numerical value in C.G.S. system.
• The density of liquids and gases decreases or increases with the rise or fall in temperature.
• The cycle of upward and downward movements of the fluid form currents in the medium which are known as convectional currents.

Test Yourself

A. Objective Questions

1. Write true or false for each statement

(a) The S.I. unit of volume is litre.
Answer: False.
The S.I. unit of volume is cubic metre.
Volume measures the space occupied by a three-dimensional object in standard units of length. While we use litres for milk or water, the scientific standard is based on the metre.
Teacher's Tip: Remember that "S.I." usually relates to meters, so volume must be m³.
Exam Tip: Always check if the question asks for the "S.I." unit specifically before answering.

(b) A measuring beaker of capacity 200 ml can measure only the volume 200 ml of a liquid.
Answer: True.
A measuring beaker is designed to hold a fixed amount when filled to the brim or the mark. It does not have graduations to measure smaller intermediate volumes like a measuring cylinder does.
Teacher's Tip: Think of a measuring beaker as a fixed-size cup rather than a ruler.
Exam Tip: If a question mentions "fixed volume," a measuring beaker is the correct apparatus.

(c) cm² is a smaller unit of area than m².
Answer: True.
One square metre is much larger because it consists of 10,000 square centimetres. A centimetre is a smaller division of a metre, so its square is also smaller.
Teacher's Tip: Visualize a small postage stamp (cm²) versus a large floor tile (m²).
Exam Tip: Remember that 1 m² = 100 cm times 100 cm = 10,000 cm².

(d) Equal volumes of two different substances have equal masses.
Answer: False.
Equal volumes of two different substances have different masses.
This happens because different materials have different densities, meaning their particles are packed differently. For example, a liter of iron is much heavier than a liter of wood.
Teacher's Tip: Density is the "secret ingredient" that makes same-sized things have different weights.
Exam Tip: Always mention "density" when explaining why masses differ for equal volumes.

(e) The S.I. unit of density is g cm^-3.
Answer: False.
The S.I. unit of density is kg m^-3.
Since the S.I. unit for mass is kg and for volume is m³, the density unit combines them. The g cm^-3 unit belongs to the C.G.S. system used for smaller measurements.
Teacher's Tip: SI units are the "Big Units" (kg, m), while CGS are the "Small Units" (g, cm).
Exam Tip: Pay close attention to units like g/cm³ vs kg/m³ in numerical problems.

(f) 1 g cm^-3 = 1000 kg m^-3.
Answer: True.
This is the conversion factor between the C.G.S. and S.I. units of density. It shows that water, which has a density of 1 g cm^-3, also has a density of 1000 kg m^-3.
Teacher's Tip: Just multiply the "small" density number by 1000 to get the "big" S.I. value.
Exam Tip: Memorize this conversion as it is frequently used to change units in exams.

(g) The density of water is maximum at 4°C.
Answer: True.
Water behaves uniquely and becomes most compact at this specific temperature. Above or below 4°C, the water molecules spread out more, making the density lower.
Teacher's Tip: This "Anomalous Expansion" of water allows fish to survive in frozen lakes.
Exam Tip: 4°C is a magic number for water density; always keep it in mind for multiple-choice questions.

(h) The speed 5 m s^-1 is less than 25 km h^-1.
Answer: True.
To compare them, convert 5 m s^-1 to km h^-1 by multiplying by 3.6, which gives 18 km h^-1. Since 18 is less than 25, the statement is correct.
Teacher's Tip: To quickly change m/s to km/h, multiply by 18 and divide by 5.
Exam Tip: Never compare speeds without converting them to the same unit first.

(i) The S.I. unit of speed is m s^-1.
Answer: True.
Speed is distance divided by time, so its S.I. unit uses the S.I. distance (metre) and S.I. time (second). It represents how many metres an object travels in one single second.
Teacher's Tip: Think of speed as "Metres per Second" to remember the formula.
Exam Tip: Always write the unit as m/s or m s^-1 to avoid losing marks.

2. Fill in the blanks

(a) 1 m³ = 10⁶ cm³
Answer: 10⁶.
Because 1 m = 100 cm, cubing both sides gives 100 times 100 times 100, which is one million. This conversion is essential for changing large volume units into smaller ones.
Teacher's Tip: Count the zeros! 100 has two zeros, so 100³ must have six zeros.
Exam Tip: Write 10⁶ clearly so the exponent is easy to read.

(b) The volume of an irregular solid is determined by the method of displacement of liquid.
Answer: displacement of liquid.
When an object is dipped in water, it pushes aside an amount of water equal to its own volume. We measure how much the water level rises to find the volume of strange shapes.
Teacher's Tip: This is called the "Eureka" method used by Archimedes.
Exam Tip: Mention "displacement" whenever you describe measuring stones or irregular shapes.

(c) Volume of a cube = (one side)³
Answer: (one side)³.
A cube has all sides equal, so length, breadth, and height are the same value. Multiplying the side by itself three times gives the total space inside.
Teacher's Tip: If the side is s, the volume is V = s times s times s.
Exam Tip: Do not confuse side cubed with side multiplied by three.

(d) The area of an irregular lamina is measured by using a grapl paper.
Answer: graph paper.
By tracing the shape on squared paper, we can count the total number of squares it covers. This provides a very good estimate for areas that don't have a standard formula.
Teacher's Tip: Use graph paper with 1 cm squares for the easiest calculations.
Exam Tip: Only count squares that are more than half covered by the shape.

(e) Mass = density times volume.
Answer: volume.
This formula rearranged from Density = Mass / Volume helps calculate how heavy an object is. If you know the material's density and its size, you can find its mass.
Teacher's Tip: Use the formula triangle with Mass on top and Density and Volume at the bottom.
Exam Tip: Ensure density and volume use compatible units (like g and cm³) before multiplying.

(f) The S.I. unit of density is kg m^-3.
Answer: kg m^-3.
Density describes how much mass is packed into a specific volume in the standard metric system. It is the most common way scientists record density worldwide.
Teacher's Tip: The minus sign in m^-3 simply means "per cubic metre".
Exam Tip: Always include the negative sign in the exponent for S.I. density units.

(g) 1 g cm^-3 = 1000 kg m^-3.
Answer: 1000.
This shows that one gram per cubic centimetre is equal to a thousand kilograms per cubic metre. It helps us switch between laboratory scale and industrial scale measurements.
Teacher's Tip: Remember "Gram is small, Kilogram is tall (1000times)".
Exam Tip: Double-check the number of zeros when converting units in word problems.

(h) 36 km h^-1 = 10 m s^-1.
Answer: 10.
By converting 36 km to 36,000 metres and 1 hour to 3,600 seconds, we find the speed in m/s. Dividing 36,000 by 3,600 results exactly in 10.
Teacher's Tip: The shortcut to go from km/h to m/s is to divide by 3.6.
Exam Tip: Show your conversion steps to earn partial marks if the final answer is wrong.

(i) Distance travelled d = speed v times time t.
Answer: speed v.
Distance is found by multiplying how fast you go by how long you travel for. If you walk at 2 m/s for 10 seconds, you cover 20 metres.
Teacher's Tip: Distance is always the product of speed and time.
Exam Tip: Use the notation d = v times t to quickly solve physics problems.

3. Match the following

Column A
(a) Volume of a liquid
(b) Area of a leaf
(c) S.I. unit of volume
(d) S.I. unit of density
(e) S.I. unit of speed

Column B
(i) kg m^-3
(ii) m³
(iii) graph paper
(iv) m s^-1
(v) measuring cylinder

Answer:
(a) Volume of a liquid --- (v) measuring cylinder
(b) Area of a leaf --- (iii) graph paper
(c) S.I. unit of volume --- (ii) m³
(d) S.I. unit of density --- (i) kg m^-3
(e) S.I. unit of speed --- (iv) m s^-1

4. Select the correct alternative

(a) One litre is equal to :
1. 1 cm^-3
2. 1 m³
3. 10^-3 cm³3
4. 10^-3 m³
Answer: 4. 10^-3 m³.
One litre is 1/1000th of a cubic metre because 1 m³ contains 1000 litres. This relationship is used to convert liquid volume into standard cubic units.
Teacher's Tip: Think of 1 m^3 as a giant tank that holds 1000 one-litre bottles.
Exam Tip: 10³ is just a fancy way of writing 1/1000.

(b) A metallic piece displaces water of volume 15 ml. The volume of piece is :
1. 15 cm³
2. 15 m³
3. 15 times 10³ cm³
4. 15 times 10³ cm³
Answer: 1. 15 cm³.
According to the displacement principle, the volume of the object is exactly equal to the volume of the water it moves. Since 1 ml is equal to 1 cm³, 15 ml is 15 cm³.
Teacher's Tip: 1 ml and 1 cm³ are the exact same amount of space.
Exam Tip: Always check if the question provides volume in ml or cm³; they are interchangeable.

(c) A piece of paper of dimensions 1.5 m times 20 cm has area :
1. 30 m²
2. 300 cm²
3. 0.3 m²
4. 3000 m³
Answer: 3. 0.3 m².
First, convert 20 cm to 0.2 m. Multiplying 1.5 m by 0.2 m gives an area of 0.3 square metres.
Teacher's Tip: Never multiply different units; always convert cm to m or vice versa first.
Exam Tip: Area is always squared (m²), never cubed (m³).

(d) The correct relation is :
1. d = M times V
2. M = d times V
3. V = d times M
4. d = M + V
Answer: 2. M = d times V.
Mass is calculated by multiplying the density of a substance by the volume it occupies. This relationship is a fundamental pillar of physics used to find an unknown third value from two known ones.
Teacher's Tip: Remember "My Dear Valentine" as M = D times V.
Exam Tip: Memorize the formula triangle to quickly rearrange this for Mass, Density, or Volume.

(e) The density of alcohol is 0.8 g cm^-3. In S.I. unit, it will be :
1. 0.8 kg m^-3
2. 0.0008 kg m^-3
3. 800 kg m^-3
4. 8 times 10³ kg m^-3
Answer: 3. 800 kg m^-3.
To convert g/cm³ to the S.I. unit kg/m³, you must multiply the value by 1000. Thus, 0.8 times 1000 equals 800.
Teacher's Tip: Multiplying by 1000 just shifts the decimal three places to the right.
Exam Tip: If the answer is in S.I. units, it should generally be a much larger number than the C.G.S. value.

(f) The density of aluminium is 2.7 g cm^-3 and of brass is 8.4 g cm^-3. For the same mass, the volume of:
1. both will be same
2. aluminium will be less than that of brass
3. aluminium will be more than that of brass
4. nothing can be said.
Answer: 3. aluminium will be more than that of brass.
Since aluminium has a lower density, its particles are less tightly packed, requiring more space to reach the same mass as the denser brass. Denser objects take up less room for the same weight.
Teacher's Tip: Low density means "fluffy" or "bulky" for the same weight.
Exam Tip: Volume and Density have an inverse relationship when mass is constant.

(g) A block of wood of density 0.8 g cm^-3 has a volume of 60 cm³. The mass of block will be :
1. 60.8 g
2. 75 g
3. 48 g
4. 0.013 g
Answer: 3. 48 g.
Using the formula Mass = Density times Volume, we multiply 0.8 by 60. The calculation 0.8 times 60 gives exactly 48 grams.
Teacher's Tip: 0.8 times 60 is the same as 8 times 6.
Exam Tip: Always write the unit "g" after your numerical mass answer.

(h) The correct relation for speed is
1. Speed = distance times time
2. speed = distance / time
3. speed = time / distance
4. speed = 1 / distance times time
Answer: 2. speed = distance / time.
Speed is the rate at which distance is covered over a period of time. By dividing the total distance by the time taken, we find how fast the object moved on average.
Teacher's Tip: Think of "Miles per Hour" (Distance per Time) to remember the division.
Exam Tip: Use a division bar instead of the "/" symbol in formal exams to be clearer.

(i) A boy travels a distance 150 m in 1 minute. His speed is
1. 150 m s^-1
2. 2.5 m s^-1
3. 25 m s^-1
4. 9 m s^-1
Answer: 2. 2.5 m s^-1.
First, convert 1 minute to 60 seconds. Then divide the distance (150 m) by time (60 s), which results in 2.5 m/s.
Teacher's Tip: 150 divided by 60 is the same as 15 divided by 6.
Exam Tip: Always convert minutes to seconds before calculating speed in m/s.

B. Short/Long Answer Questions

Question 1: Define the term volume of an object.
Answer: The space occupied by an object is called its volume.
Volume is a measure of how much room an object takes up in three dimensions. Every physical object, whether solid, liquid, or gas, has a volume.
Teacher's Tip: Think of volume as the amount of water an object would push away if dunked in a bucket.
Exam Tip: Use the word "space" to define volume for a perfect answer.

Question 2: State and define the S.I. unit of volume.
Answer: S.I. unit of volume - The S.I. unit of volume is cubic metre. In short form, it is written as m³. One cubic metre is the volume of a cube of each side 1 metre as shown in figure below i.e., 1 m³ = 1 m times 1 m times 1 m.
The cubic metre is the standard international way to quantify large amounts of space. It is defined specifically by the length of a standard meter being cubed.
Teacher's Tip: Imagine a box that is 1 metre tall, 1 metre wide, and 1 metre deep to visualize 1 m³.
Exam Tip: When defining the unit, always mention the side length of the cube (1 m).

Question 3: State two smaller units of volume. How are they related to the S.I. unit?
Answer: A smaller unit of volume is cubic centimetre (symbol cm³) and cubic decimetre (symbol 1 dm³). One cubic centimetre is the volume of a cube of each side 1 centimetre, i.e., 1 cm³ = 1 cm times 1 cm times 1 cm.
Relationship between m³ and cm³
1 m³ = 1 m times 1 m times 1 m
= 100 cm times 100 cm times 100 cm
= 1,000,000 cm³ = 10⁶ cm³.
Relationship between m³ and dm³
1 m³ = 1 m times 1 m times 1 m
= 10 dm times 10 dm times 10 dm
= 1000 dm³
= 10³ dm³
Note 1 m = 10 dm.
Smaller units like cm³ are used for measuring small objects like marbles or medicine doses. These relationships allow scientists to convert between microscopic and macroscopic scales easily.
Teacher's Tip: Decimetre is 1/10th of a metre, and centimetre is 1/100th of a metre.
Exam Tip: Memorize that 1,000,000 cm³ fits into just 1 m³.

Question 4: How will you determine the volume of a cuboid ? Write the formula you will use.
Answer: Volume of a cuboid = length times breadth times height.
A cuboid is a box-like shape where you find the total space by multiplying its three main dimensions. This simple multiplication works for everything from small bricks to large shipping containers.
Teacher's Tip: Use the abbreviation V = l times b times h.
Exam Tip: Ensure all three dimensions are in the same unit before multiplying.

Question 5: Name two devices which are used to measure the volume of an object. Draw their neat diagrams.
Answer: Two devices that are used to measure the volume of an object are:
(i) Measuring cylinder and
(ii) Measuring beaker.
A measuring cylinder has marks to tell you exactly how many millilitres are inside, while a beaker is usually for fixed amounts. Both are essential tools in a science laboratory for handling liquids.
Teacher's Tip: Always read the liquid level at the bottom of the curve (meniscus).
Exam Tip: Practice drawing the vertical lines for the cylinder and the curved spout for the beaker.

Question 6: How can you determine the volume of an irregular solid (say a piece of brass) ? Describe in steps with neat diagrams.
Answer: To measure the volume of a piece of stone.
Take a piece of brass, a measuring cylinder, fine thread of sufficient length and some water.
Place a measuring cylinder on a flat horizontal surface and fill it partially with water. Note the reading of the water level very carefully. Now tie the piece of brass with a thread and dip it completely into water. We see that the level of water rises. Note the reading of the new water level.
The difference in the two levels of water gives the volume of the piece of brass
Initial level of water = 60 ml
Level of water when brass is immersed = 80 ml
therefore Volume of water displaced = 80 ml - 60 ml = 20 ml
therefore Volume of the piece of brass = 20 cm³
Note : 1 ml = 1 cm³.
This method relies on the fact that solid objects occupy space and push water up when submerged. It is the most accurate way to find the volume of shapes that don't have straight edges.
Teacher's Tip: Make sure the object is fully submerged but doesn't touch the bottom or sides.
Exam Tip: Always subtract the initial level from the final level, not the other way around.

Question 7: You are required to take out 200 ml of milk from a bucket full of milk. How will you do it ?
Answer: By using the measuring beaker A measuring beaker is used to measure a fixed volume of liquid from a large volume. Suppose it is required to measure 200 ml of milk from the milk contained in a bucket. For this, take the measuring beaker of capacity 200 ml. Wash it and dry it. Then, immerse the measuring beaker well inside the milk contained in the bucket so that the beaker gets completely filled with the milk. Take out the measuring beaker from the bucket gently so that no milk splashes out and then pour the milk from the measuring beaker into the another empty vessel.
The measuring beaker acts as a standard scoop that holds exactly the volume it was designed for. This is a common practical method used by milkmen and in kitchens.
Teacher's Tip: The beaker must be filled "to the brim" for an accurate measurement.
Exam Tip: Mention that the beaker must be clean and dry before use for hygiene and accuracy.

Question 8: Describe the method in steps to find the area of an irregular lamina using a graph paper.
Answer: Method to find the area of an irregular lamina using a graph paper : First, place the lamina over a graph paper and draw its boundary line on the graph paper with a pencil. Then remove the lamina and count and note the number of complete squares as well as the number of squares more than half within the boundary line (only the squares less than half, are left while counting). The area of lamina is equal to the sum of the area of complete squares and the area of squares more than half. Let n be the total number of complete and more than half or half squares within the boundary of lamina. Since area of one big square is 1 cm times 1 cm = 1 cm², so the area of lamina will be n times 1 cm² or n cm².
This approximation technique turns a complicated shape into a count of simple, equal-sized units. It is widely used for leaves, handprints, or irregular maps.
Teacher's Tip: Mark each counted square with a small dot so you don't count it twice.
Exam Tip: Clearly state the rule: "Count full and more-than-half squares; ignore less-than-half."

Question 9: Define the term density of a substance.
Answer: The density of a substance is defined as the mass of a unit volume of that substance.
Density tells us how "heavy" a material is for its size, helping us distinguish between light things like cork and heavy things like lead. It is a characteristic property, meaning it doesn't change based on how much of the substance you have.
Teacher's Tip: Density is just the "compactness" of a material.
Exam Tip: Always include the phrase "per unit volume" in your definition.

Question 10: State the S.I. and C.G.S. units of density. How are they inte related ?
Answer: The S.I. unit of mass is kilogram (symbol kg) and of volume is cubic metre (symbol m³). Therefore S.I. unit of density is kg/m³ or kg m^-3.
The C.G.S. unit of mass is gram (symbol g) and of volume is cubic centimetre (symbol cm³). Therefore the C.G.S. unit of density is g/cm³ or g cm^-3.
Relationship between kg m^-3 and g cm^-3
1 kg = 1000 g
or 1 g = 1/1000 kg
and 1 m³ = (100 cm)³ = 100 times 100 times 100 cm³ = 10,000,000 cm³
or 1 cm³ = 1/1,000,000 m³
Now 1 g cm^-3 = 1 g/1 cm³ = 11000 kg/11,000,000 m³ = 1,000,000/1000 kg m^-3 = 1,000 kg m^-3
Thus, 1 g cm^-3= 1,000 kg m^-3.
This mathematical derivation shows that the S.I. value is exactly 1000 times larger than the C.G.S. value. This helps scientists communicate measurements across different systems without confusion.
Teacher's Tip: Remember the water rule: density of water is 1 in CGS and 1000 in SI.
Exam Tip: In the derivation, be very careful with the number of zeros in the denominator.

Question 11: 'The density of brass is 8.4 g cm^-3'. What do you mean by the statement ?
Answer: Density of brass is 8.4 g cm^-3. This means that unit volume of brass contain 8.4 g mass.
Specifically, a tiny cube of brass that is 1 cm on each side would weigh exactly 8.4 grams on a scale. It describes the physical property of how concentrated the matter in brass is.
Teacher's Tip: Whenever you see "density is X", it means "one unit of space weighs X".
Exam Tip: For "What do you mean..." questions, always explain it in terms of "mass per 1 cm³".

Question 12: Arrange the following substances in order of their increasing density: (a) iron, (b) cork, (c) brass, (d) water, (e) mercury.
Answer: b < d < a < c < e
(Cork < Water < Iron < Brass < Mercury)
This list goes from the lightest, least compact substance (cork) to the heaviest liquid (mercury). It reflects how tightly packed the atoms are in each material.
Teacher's Tip: Cork floats in water, and iron sinks, which helps you remember their relative densities.
Exam Tip: Use the "<" symbol to clearly show the order from smallest to largest.

Question 13: How does the density of water changes when : (a) it is heated from 0°C to 4°C, (b) it is heated from 4°C to 10°C ?
Answer: (a) Water contracts on heating from 0°C to 4°C and expands on heating above 4°C. (b) The density of water is maximum at 4°C. It decreases when it is cooled from 4°C to 0°C or it is heated above 4°C.
Water is very strange because it gets heavier as it warms up slightly from freezing, reaching peak density at 4°C. After 4°C, it starts acting like a normal liquid and becomes lighter (less dense) as it continues to heat up.
Teacher's Tip: Think of 4°C as the "peak of the mountain" for density.
Exam Tip: Use the term "Anomalous expansion" if you want to impress the examiner.

Question 14: Write the density of water at 4°C.
Answer: The density of water at 4°C is 1.0 g cm^-3, or 1,000 kg m^-3.
These numbers are easy to remember and serve as the baseline for comparing all other substances. If an object's density is higher than 1.0 g/cm³, it will sink in water.
Teacher's Tip: Water's density of "1" makes it the perfect standard for scientists.
Exam Tip: State both the C.G.S. and S.I. values to get full marks.

Question 15: Explain the meaning of the term speed.
Answer: The distance covered or travelled by a body in one second is called the speed of the body, i.e.
Speed = Distance travelled/Timetaken. Speed is usually denoted by the symbol v. If a body travels a distance d in time t, then its speed is given as Speed (v) = fracdt.
Speed tells us how fast an object is moving but does not mention which direction it is going. It is a scalar quantity that measures the rate of motion.
Teacher's Tip: If you travel 100 km in 2 hours, your speed is just 100 divided by 2.
Exam Tip: Always provide the formula v = d/t when defining speed.

Question 16: Write the S.I. unit of speed.
Answer: The S.I. unit of speed is metre/second or metre per second. Its symbol is m s^-1.
This unit combines the S.I. units of distance and time to show change in position every second. It is the global standard for scientific velocity measurements.
Teacher's Tip: m/s and m s^-1 mean the exact same thing.
Exam Tip: Avoid writing units in plural like "metres per seconds"; keep them singular.

Question 17: A car travels with a speed 12 m s^-1, while a scooter travels with a speed 36 km h^-1. Which of the two travels faster ?
Answer: Speed of car = 12 m s^-1. Speed of scooter = 36 km h^-1. Here, 1 km = 1000 m and 1 hr = 3600 sec. therefore Speed of scooter = 36 x 1000/3600 = 10 m s^-1.
therefore Speed of car is more. Car travels faster than scooter.
By converting the scooter's speed into m/s, we found it is only 10 m/s, which is less than the car's 12 m/s. Standardizing units is the only way to compare different speeds fairly.
Teacher's Tip: The car covers 12 metres every second, while the scooter only covers 10.
Exam Tip: Always show the conversion calculation before stating which is faster.

C. Numericals

Question 1: The length, breadth and height of a water tank are 5 m, 2.5 m and 1.25 m respectively. Calculate the capacity of the water tank in (a) m³ (b) litre.
Answer: Given, Length (l) = 5 m, Breadth (b) = 2.5 m, and Height (h) = 1.25 m.
(a) Volume of water tank in m³ = l times b times h = 5 m times 2.5 m times 1.25 m = 15.625 m³.
(b) Volume of water tank in litre = 15.625 times 1000 = 15625 litre.
The tank's volume is found by multiplying its three dimensions together. Since 1 m³ holds 1000 litres, we multiply the cubic result by a thousand to find the total liquid capacity.
Teacher's Tip: "Capacity" is just another word for volume, usually applied to containers.
Exam Tip: Remember 1 m³ = 1000 l for these types of tank problems.

Question 2: A solid silver piece is immersed in water contained in a measuring cylinder. The level of water rises from 50 ml to 62 ml. Find the volume of silver piece.
Answer: Given, initial level of water v₁ = 50 ml. Final level of water v₂ = 62 ml. Volume of silver piece V = v₂ - v₁ = 62 ml - 50 ml = 12 ml or 12 cm³.
The rise in water level accurately represents how much space the silver piece takes up. We use the fact that 1 ml is equivalent to 1 cm³ for the final answer.
Teacher's Tip: This is the displacement method in action.
Exam Tip: Provide the answer in cm³ as it is the standard unit for solid volume.

Question 3: Find the volume of a liquid present in a dish of dimensions 10 cm times 10 cm times 5 cm.
Answer: Volume of water = Length times breadth times height = 10 cm times 10 cm times 5 cm = 500 cm³ or 500 ml.
The dish holds the liquid in a rectangular shape, so we use the cuboid volume formula. The total volume comes out to five hundred cubic centimetres or half a litre.
Teacher's Tip: 10 times 10 is 100, and 100 times 5 is 500.
Exam Tip: Don't forget to mention both cm³ and ml units if possible.

Question 4: A rectangular field is of length 60 m and breadth 35 m. Find the area of the field.
Answer: Length of a rectangular field = 60 m. Breadth of rectangular field = 35 m. therefore Area = 60 m times 35 m = 2100 m².
Area is a 2D measurement found by multiplying length and breadth together. For a large field, we use square metres as the appropriate unit.
Teacher's Tip: Area is "flat space," so it only has two dimensions (l times b).
Exam Tip: Unit of area is always m² (squared), never m or m³.

Question 5: Find the approximate area of an irregular lamina of which boundary line is drawn on the graph paper shown in fig. 1.16. below.
Answer: From figure, the number of complete squares = 11. The number of squares more than half = 9. therefore Total number of squares = 11 + 9 = 20. therefore Area of the 1 square = 1 cm times 1 cm = 1 cm². therefore Area of 20 squares = 20 times 1 cm² = 20 cm². therefore Approximate area of irregular lamina = 20 cm².
By grouping the partial and whole squares, we can estimate the total surface area of the wavy shape. Each square on this specific graph paper represents one square centimetre.
Teacher's Tip: Only count squares that look "mostly full".
Exam Tip: Show the addition (11 + 9) to demonstrate how you arrived at 20.

Question 6: A piece of brass of volume 30 cm³ has a mass of 252 g. Find the density of brass in (i) g cm^-3, (ii) kg m^-3.
Answer: Given, Mass M = 252 g. Volume V = 30 cm³.
(i) Density d = fracMV = frac25230 = 8.4 g cm^-3.
(ii) Since, M = 252 g = 0.252 kg, V = 30 cm³ = 30 times 10^-6 m³.
Density d = frac0.252 kg30 times 10^-6 m³ = 8400 kg m^-3.
First, we find the CGS density by simple division. To get the SI density, we either do complex conversion or use the shortcut of multiplying the CGS result by 1000.
Teacher's Tip: 8.4 times 1000 = 8400. Much easier than converting all units!
Exam Tip: Label your answers clearly as (i) and (ii).

Question 7: The mass of an iron ball is 312 g. The density of iron is 7.8 g cm^-3. Find the volume of the ball.
Answer: Given, Mass M = 312 g. Density d = 7.8 g cm^-3. Since, d = fracMV
V = fracMd. Hence, volume of an iron ball V = frac3127.8 = 40 cm³.
We rearranged the density formula to isolate volume, which is mass divided by density. The result tells us the ball takes up exactly forty cubic centimetres of space.
Teacher's Tip: 312 divided by 7.8 is the same as 3120 divided by 78.
Exam Tip: Always state the formula you are using before plugging in the numbers.

Question 8: A cork has a volume 25 cm³. The density of cork is 0.25 g cm^-3. Find the mass of cork.
Answer: Given, density d = 0.25 g cm^-3, V = 25 cm³. From relation d = M/V
M = d times V = 0.25 times 25 = 6.25 g.
Multiplying the density by the volume gives the total mass of the cork. Since cork is very light (low density), its mass is much smaller than its volume number.
Teacher's Tip: Cork's density is 1/4, so the mass should be 1/4 of the volume.
Exam Tip: The units cm³ in volume and density cancel out, leaving just grams for mass.

Question 9: The mass of 5 litre of water is 5 kg. Find the density of water in g cm^-3.
Answer: Given, Mass M = 5 kg = 5000 g. Volume V = 5 litre = 5000 cm³. Density of water d = M/V = 5000g/5000 cm³ = 1 g cm^-3.
By converting everything to grams and cubic centimetres, we confirm the standard density of water. It perfectly illustrates that 1 kg of water is equal to 1 litre of water.
Teacher's Tip: Water is special: 1 gram = 1 ml = 1 cm³.
Exam Tip: Don't forget to convert kg to g and litres to cm³ first.

Question 10: A cubical tank of side 1 m is filled with 800 kg of a liquid. Find: (i) the volume of tank, (ii) the density of liquid in kg m^-3.
Answer: (i) Volume of a cube = side times side times side. side = 1 m. therefore volume = 1 m times 1 m times 1 m = 1 m³.
(ii) Density of liquid in kg m^-3 = Mass (M)/Volume (V). Mass = 800 kg. Volume = 1 m³. therefore Density = 800/1m³ kg = 800 kg m^-3.
Since the tank is a unit cube, its volume is exactly one cubic metre. The density is simply the mass of the liquid because it's already measured "per cubic metre".
Teacher's Tip: Dividing by 1 is the easiest math in the world!
Exam Tip: Use the exponent ^-3 in the density unit kg m^-3.

Question 11: A block of iron has dimensions 2 m times 0.5 m times 0.25 m. The density of iron is 7.8 g cm^-3. Find the mass of block.
Answer: Given, l = 2 m, b = 0.5 m, h = 0.25 m.
Density of iron = 7.8 g cm^-3 = 7.8 times 1000 kg m^-3 = 7800 kg m^-3.
Volume of block = l times b times h = 2 times 0.5 times 0.25 = 0.25 m³.
From relation d = M/V 
Mass of iron block M = V times d = 0.25 times 7800 kg m^-3= 1950 kg.
We first convert the density into S.I. units to match the dimensions of the block. Multiplying the quarter-cubic-metre volume by the density gives nearly two tonnes of iron.
Teacher's Tip: 0.25 is the same as dividing by 4. 7800 / 4 = 1950.
Exam Tip: Always convert the material's density to match the units of the object's dimensions.

Question 12: The mass of a lead piece is 115 g. When it is immersed into a measuring cylinder, the water level rises from 20 ml mark to 30 ml mark. Find: (i) the volume of the lead piece, (ii) the density of the lead in kg m^-3.
Answer: Ans. Given, M = 115 g. V₁ = 20 ml, V₂ = 30 ml.
(i) Volume of lead piece V = V₂- V₁ = 30 ml - 20 ml = 10 ml or 10 cm³ [because 1 ml = 1 cm³].
(ii) Density of lead piece d = M/V = 115/10cm³ = 11.5 g cm^-3. (since 1 g cm^-3 = 1000 kg m^-3) = 11.5 times 1000 = 11500 kg m^-3.
The volume is determined by the subtraction of the water levels in the cylinder. We then calculate density in CGS and simply multiply by a thousand to reach the S.I. answer.
Teacher's Tip: Lead is very dense, so expect a high number like 11,500.
Exam Tip: State the conversion factor 1 g cm^-3 = 1000 kg m^-3 to show how you got part (ii).

Question 13: The density of copper is 8.9 g cm^-3. What will be its density in kg m^-3 ?
Answer: Density of copper d = 8.9 g cm^-3 = 8.9 times 1000 kg m^-3 = 8900 kg m^-3. [because 1 g cm^-3 = 1000 kg m^-3].
To switch from grams per cubic centimetre to kilograms per cubic metre, you scale up by a factor of one thousand. This converts the laboratory measurement into an industrial standard measurement.
Teacher's Tip: Just move the decimal point three places to the right.
Exam Tip: Don't forget the units (kg m^-3) at the end of your answer.

Question 14: A car travels a distance of 15 km in 20 minute. Find the speed of the car in (i) km h^-1, (ii) m s^-1.
Answer: Distance travelled by car = 15 km. Time taken = 20 minutes.
(i) Speed of car in km h^-1:
Convert 20 minutes to hour: 1 minute = 1/60 hour. therefore 20 minutes = 1 times 20/60 = 1/3 hour.
Speed = Distance/Time = 15 km/1/3 h = 15 times 3 = 45 km h^-1.
(ii) Speed of car in m s^-1:
Convert 15 km into metres: 1 km = 1000 m
 15 km = 15000 m.
Convert minutes into seconds: 1 minute = 60 sec
20 minutes = 60 times 20 = 1200 sec.
Speed = 15000 m1200 sec = 12.5 m s^-1.
We first find the speed using hours as the time unit, which is helpful for driving speeds. Then we convert to the scientific standard of metres per second for technical analysis.
Teacher's Tip: 20 minutes is exactly one-third of an hour.
Exam Tip: Use the fraction 1/3 instead of decimals like 0.33 to get a more accurate final answer.

Question 15: How long a train will take to travel a distance of 200 km with a speed of 60 km h^-1 ?
Answer: Distance covered by train = 200 km. Speed of train = 60 km h^-1.
We know speed = Distance/Time 
60 = 200Time 
Time = 200/60 = 10/3 hours.
= 3 1/3 hours = 3 h + 1/3 hours = 3 h + 1/3 times 60 min = 3 h + 20 min = 3 h 20 min.
By dividing distance by speed, we find the total hours, which we then break down into whole hours and minutes. This tells us exactly how long a passenger would be sitting on the train.
Teacher's Tip: A 1/3 hour is just 20 minutes (since 60/3 = 20).
Exam Tip: Always convert the fractional part of the hour into minutes for a complete answer.

Question 16: A boy travels with a speed of 10 m s^-1 for 30 minute. How much distance does he travel ?
Answer: Speed of boy = 10 m s^-1. Time taken = 30 minutes.
speed = distance travelled/time taken
Distance travelled = Speed times Time taken.
Convert 30 minutes to seconds: 1 minute = 60 sec
30 minute = 60 times 30 = 1800 seconds.
Putting the value of speed and time we get: Distance travelled = 10 ms^-1 times (1800 sec) = 18000 m = 18 km. Ans.
To find distance, we multiply speed by time, but first we must ensure the units are compatible by turning minutes into seconds. The total distance of 18,000 metres is more easily understood as 18 kilometres.
Teacher's Tip: Think: "10 meters every second for 1800 seconds total."
Exam Tip: Express large distance answers in "km" to make them look more professional.

Question 17: Express 36 km h^-1 in m s^-1
Answer: 36 km h^-1 =36 times 1000 m/60 times 60 s = 10 m s^-1.
This common conversion is done by multiplying the kilometres by 1000 and dividing by the 3600 seconds in an hour. It translates a typical car speed into a scientific speed measurement.
Teacher's Tip: Just divide 36 by 3.6 to get 10 instantly.
Exam Tip: Memorize that 36 km/h is exactly 10 m/s; it's a very common exam value.

Question 18: Express 15 m s^-1 in km h^-1.
Answer: 1 metre = 1/1000 km. 15 metre = 15/1000 km. 1 second = 1/3600 hr.
Here, Distance = 151000km. Time taken = 1/3600 hr.
Speed = fracDistance/Time taken = 15/1000 1/3600= 15/1000 times 3600/1= 54 km h^-1.
To reverse the process, we multiply the m/s value by 3.6 to reach the km/h result. It shows that a speed of 15 metres per second is quite fast, over fifty kilometres per hour.
Teacher's Tip: 15 times 3.6 = 54.
Exam Tip: Remember: multiply by 3.6 to go to km/h, divide by 3.6 to go to m/s.