CBSE Class 7 Science Advanced Physics Measurements Short Answer Questions Set 02

Short Answer Questions

Question 1. Density of the material of a paper is 0.4 g cm⁻³. The mass of the paper is 0.5 g and its length and breadth are 10 cm and 4 cm, respectively. Determine the thickness of the paper.
Answer:
Volume of the paper = mass ÷ density = 0.5 ÷ 0.4 = 1.25 cm³.
Volume = length × breadth × thickness, so thickness = 1.25 ÷ (10 × 4) = 0.03125 cm.
Thickness = 0.03125 cm = 0.3125 mm.
In simple words: A sheet is a very flat cuboid, so thickness = volume ÷ area.

Teacher's Note:
1. Get the volume from the mass and density first.
2. Convert cm to mm by multiplying by 10.

Question 2. If the length, breadth and height of a cuboid are all increased to 'n' times their original values, then the ratio of initial and final density of air present inside the cuboid is 8 : 1. Determine the value of 'n'.
Answer:
Original volume = l × b × h = V. New volume = (nl) × (nb) × (nh) = n³V.
The mass of the air inside is the same, so density ∝ 1 ÷ volume.
Initial density : final density = n³V : V = n³ : 1 = 8 : 1, so n³ = 8 and n = 2.
In simple words: If each side doubles, the volume becomes 8 times and the density falls to one-eighth.

Teacher's Note:
1. Volume goes as the cube of the length factor.
2. Check: 2 × 2 × 2 = 8.

Question 3. A wooden block is tied to a metallic rod of mass 100 g and immersed completely in water. When the displaced water is collected in a measuring jar, the volume of the liquid displaced is found to be 400 ml. If the density of the metallic rod is 5 g cm⁻³, and the mass of the wooden block is 300 g, determine the density of the wooden block.
Answer:
Volume of the metallic rod = mass ÷ density = 100 ÷ 5 = 20 cm³.
Total volume displaced = volume of rod + volume of block = 400 cm³ (1 ml = 1 cm³).
Volume of the wooden block = 400 − 20 = 380 cm³.
Density of the wooden block = 300 ÷ 380 = 0.79 g cm⁻³ (about 790 kg m⁻³).
In simple words: Subtract the rod's volume from the total displaced volume, then divide the block's mass by its volume.

Teacher's Note:
1. The tied rod makes the light block sink, so both are fully immersed.
2. Wood is less dense than water, which is why it needs the rod.

Question 4. A student was given a task to find the density of a cube whose mass and side are 100 g and 5 cm, respectively. What is his answer?
Answer:
Volume of the cube = (side)³ = 5³ = 125 cm³.
Density = mass ÷ volume = 100 ÷ 125 = 0.8 g cm⁻³ (800 kg m⁻³).
In simple words: Cube the side to get the volume, then divide the mass by it.

Teacher's Note:
1. Volume of a cube = side × side × side.
2. A density of 0.8 g cm⁻³ is less than water, so the cube would float.

Question 5. A physics student was given a 400 pages book and a metre scale and asked to measure the thickness of the paper in the book. Explain how can he measure it.
Answer:
1. Press the pages of the book together (without the cover) and measure the total thickness T of all the pages with the metre scale.
2. Count the number of sheets (leaves). A sheet has two pages, one on each side, so a 400-page book has 400 ÷ 2 = 200 sheets.
3. Thickness of one sheet of paper = T ÷ 200.
For example, if T = 2 cm, the thickness of a sheet = 2 ÷ 200 = 0.01 cm = 0.1 mm.
In simple words: Measure many sheets together and divide, because one sheet is too thin for a metre scale.

Teacher's Note:
1. Pages are printed on both sides of a sheet, so divide by half the page count.
2. The same idea is used for coins and wires.

Question 6. A wire is wound over a rod and placed over a scale, as shown in figure.
Calculate the radius of the wire.
Answer:
The turns of the wire on the rod run from 1.6 cm to 3.1 cm on the scale, so the length of the coil = 3.1 − 1.6 = 1.5 cm.
Number of turns = 10.
Diameter of the wire = 1.5 cm ÷ 10 = 0.15 cm = 1.5 mm.
Radius of the wire = 1.5 ÷ 2 = 0.75 mm.
In simple words: Length of the coil ÷ number of turns gives the diameter, and half of it is the radius.

Teacher's Note:
1. Count the turns carefully.
2. Radius is half the diameter.

Question 7. A student took a spring balance calibrated in gf. He then suspended an object of mass 1 kg to the spring balance. Determine what would be the reading of the spring balance.
Answer:
Weight of a body of mass 1 kg = 1 kgf.
1 kgf = 1000 gf.
The spring balance reads 1000 gf.
In simple words: A 1 kg mass weighs 1 kgf, and 1 kgf is 1000 gf.

Teacher's Note:
1. A spring balance measures weight, not mass.
2. kgf = 1000 gf.

Question 8. A physics scholar took two spring balances 'S₁' and 'S₂' that are calibrated on the surface of the Moon and the Earth, respectively. He then weighed a body using both S₁ and S₂ at a given place on the surface of the Earth. He determined the ratio of the readings. What is the ratio?
Answer:
A spring balance measures the weight of the body, that is, the pull acting on it, and the same body at the same place has only one weight.
Both balances are marked in the same force unit, so both show the same reading.
Ratio of the readings = 1 : 1.
In simple words: One body at one place has one weight, so both balances show the same value.

Teacher's Note:
1. This is the answer given in the textbook.
2. Only if one balance were marked in mass units for the Moon would the readings differ.

Question 9. A scientist took a physical balance and an object of weight 10 kgwt from the surface of the Earth to the surface of the Moon. There on the Moon, if he placed this 10 kgwt object on one of the pans of the physical balance, find the mass of the object required to balance the physical balance.
Answer:
A 10 kgwt object on the Earth has a mass of 10 kg, and mass does not change with place.
A physical balance compares masses, so the same balance works on the Moon in the same way.
The mass required on the other pan = 10 kg.
In simple words: A physical balance compares masses, and mass is the same on the Moon and Earth.

Teacher's Note:
1. Weight changes on the Moon but mass does not.
2. A spring balance, unlike a physical balance, would show a different reading.

Question 10. The change in volume of a gas with change in temperature is plotted on a graph, as shown in the figure.
Calculate the ratio of density at 50 °C and 100 °C.
Answer:
Volume at 50 °C = 20 cm³ and volume at 100 °C = 30 cm³ (read from the graph).
The mass of the gas does not change, so density = mass ÷ volume, and density ∝ 1 ÷ volume.
Density at 50 °C : density at 100 °C = 30 : 20 = 3 : 2.
In simple words: The gas expands as it gets hotter, so its density at 100 °C is less than at 50 °C.

Teacher's Note:
1. For the same mass, the ratio of densities is the inverse of the ratio of volumes.
2. Read the graph values carefully.

Question 11. Ram took an empty density bottle and a weighing machine. He measured the empty density bottle and found it to be 40 g. He then completely filled it with water and weighed it to be 100 g. With this data can he calculate the capacity of the density bottle? If so, what is its capacity?
Answer:
Yes, he can.
Mass of water = 100 − 40 = 60 g.
Density of water = 1 g cm⁻³, so the volume of water = 60 cm³.
The bottle was completely filled, so the capacity of the density bottle = 60 cm³.
In simple words: The mass of the water that fills the bottle tells us its volume, because 1 g of water takes 1 cm³.

Teacher's Note:
1. Water is the standard liquid for this reason.
2. Subtract the empty bottle's mass first.

Teacher's Note:
1. Mass = density × volume, and the volumes are equal.
2. Do not compare 2, 1 and 3 without looking at the units.

Question 13. Triangulation method is used to determine the distance between two places A and C. A triangle obtained to a reduced scale is as shown below. The ratio of the actual distances from C to B; and A to B is 5. Determine the distance between A and C. The actual distance between B and C is 20 km.

Answer:
Actual distance BC = 20 km and BC ÷ AB = 5, so the actual distance AB = 20 ÷ 5 = 4 km.
On the reduced scale, AB = 5 cm, so 5 cm stands for 4 km and 1 cm stands for 4 ÷ 5 = 0.8 km.
AC on the triangle = 10 cm, so the actual distance AC = 10 × 0.8 = 8 km.
In simple words: First find how many km one cm of the drawing stands for, then multiply by the 10 cm length of AC.

Teacher's Note:
1. Scale = real distance ÷ drawn distance.
2. The drawing is not to scale; only the stated numbers are used.

Question 14. The weight of a body on the earth is 500 gf. The volume of water displaced when it is immersed in water is 250 cm³. Determine the density of the material of the body.
Answer:
Weight of the body = 500 gf, so its mass = 500 g.
The body is completely immersed, so its volume = volume of water displaced = 250 cm³.
Density = mass ÷ volume = 500 ÷ 250 = 2 g cm⁻³ (2000 kg m⁻³).
In simple words: Mass comes from the weight, volume from the displaced water.

Teacher's Note:
1. 1 gf is the weight of a 1 g mass.
2. Density above 1 g cm⁻³ means the body sinks in water.

Question 15. Ram took an empty balloon of mass 5 g and filled it with air. He then measured the mass of the balloon filled with air to be 5.05 g. He considered the density of the air filled inside the balloon as 1.5 kg m⁻³ and found the volume of air present inside the balloon. Find his answer.
Answer:
Mass of the air = 5.05 − 5 = 0.05 g.
Density of air = 1.5 kg m⁻³ = 1.5 × 10⁻³ g cm⁻³ (divide by 1000).
Volume = mass ÷ density = 0.05 ÷ (1.5 × 10⁻³) = 33.3 cm³ (about 33.33 cm³).
In simple words: Find the mass of the air by subtraction and divide by its density in g cm⁻³.

Teacher's Note:
1. Match the units before dividing: grams with g cm⁻³.
2. 1 kg m⁻³ = 10⁻³ g cm⁻³.