Seven.

π x r² = πr²

m/s

a. Volume of a cube = side \(\times\) side \(\times\) side
= 3 cm \(\times\) 3 cm \(\times\) 3 cm = 27 cubic cm or \(\mathrm{cm}^{3}\)
b. Volume of a cylinder \(=\pi \times(\text { radius })^{2} \times \text { height }\)
\(=22 / 7 \times 3 \times 3 \times 7=198 \mathrm{~m}^{3} .\)

Density of gold is 19,300 kg/m³

Area, Volume, Speed, Electric charge and Density.

Comparison of an unknown quantity with a standard quantity is called measurement

Mass is the amount of matter contained in a body. It's unit is kilogram (kg).

Density = Mass / Volume
Hence, Mass = Volume \(\times\) Density
= 125 \(\times\) 7.8 = 975 g.

One light year is defined as the distance travelled by light in vacuum during the period of one year.
1 Light year = 9.46 \(\times\) 10¹⁵m

Physical quantities which can be obtained by multiplying, dividing or by mathematically combining the fundamental quantities are known as derived quantities.
Ex: Area, volume

True.

True.

True.

True.

False.
Correct Statement: One square metre is the area enclosed inside a square of side metre.

False
Correct Statement: A substance which contains less number of molecules per unit volume is said to be rarer.

True

True

cubic meter or m³

m³

(b) Both A and R are true and R is the correct reason.

c. Assertion is true but reason is false.

a. Both assertion and reason are true and reason is the correct explanation of assertion.

Take a leaf from any one of trees in your neighbourhood. Place the leaf on a graph sheet and draw the outline of the leaf with a pencil. Remove the leaf. You can see the outline of the leaf on the graph sheet.
i) Now count the number of whole squares enclosed within the outline of the leaf. Take it to be M
ii) Then count the number of squares that are more than half. Take it as N
iii) Next, Count the number of squares which are half of a whole square. Note it to be P.
iv) Finally, count the number of squares that are less than half. Let it be a Q.
v) Now the approximate area of the leaf can be calculated using the following formula.
Approximate area of the leaf = M+(3/4)N+(1/2)P+(1/4)
This formula can be used to calculate the area of any irregularly shaped plane figures.

i) Find the ratio of masses of spheres A and B.
Sphere A and B are made of the same material
\(\therefore\) The Density of A and B are equal. Take it as D
\(D =\frac{M}{V} \)
\(\therefore M =D \times V\)
Volume of sphere \(A=V_{1}\)
Volume of sphere \(B=V_{2}\)
Mass of sphere \(A=M_{1}\)
Mass of sphere \(\mathrm{B}=\mathrm{M}_{2}\)
\(\therefore \mathrm{M}_{\mathrm{l}} =\mathrm{D} \times \mathrm{V}_{1} \)
\(\mathrm{M}_{2} =\mathrm{D} \times \mathrm{V}_{2}\)
The radius of sphere B is half that of A
So, we can take the radius of sphere A(r)=2
Volume of sphere \(\mathrm{A}\left(\mathrm{V}_{1}\right)=\frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \times \pi \times 2 \times 2 \times 2\)
Volume of sphere \(A\left(V_{1}\right)=\frac{32 \pi{}}{3}\)
We can take the radius of sphere B(r)=1
Volume of sphere \(B\left(V_{2}\right)=\frac{4}{3} \times \pi \times 1 \times 1 \times 1\)
Volume of sphere \(\mathrm{B}\left(\mathrm{V}_{2}\right)=4 / 3 \neq 0\)
Mass of sphere \(\mathrm{A} \mathrm{M}_{1}=\mathrm{D} \times \frac{32 \pi}{3}\)
Mass of sphere \(\mathrm{B} \mathrm{M}=\mathrm{D} \times \frac{4}{3} \pi\)
Ratio of masses of spheres A and B = \(\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\)
\(= \frac{\not D \times \not 32 \times \not  \frac{\pi}{3}}{\not D \times \not  4 \not  \frac{\pi}{3}} \)
\(\mathrm{M}_{1} / \mathrm{M}_{2} =8 / 1 \)
\(\mathrm{M}_{1} \cdot \mathrm{M}_{2} =8: 1\)
Ratio of masses of spheres A and B = 8 : 1
ii) Find the ratio of volume spheres A and B
Volume of sphere \(A\left(V_{1}\right)=\frac{32 \pi}{3}\)
Volume of sphere \(B\left(V_{2}\right)=\frac{4}{2} \pi\)
Ratio of volume of spheres A and B = \(\frac{V_{1}}{V_{2}}\)
\(=\frac{\not  32 \times \not  \frac{\pi}{3}}{\not  4 \not  \frac{\pi}{3}}=8 / 1\)
So Ratio of volume of spheres A and B = 8 : 1
\(V_{1}: V_{2}=8: 1\)
iii) Find the ratio of masses of spheres A and C
i) Spheres A and C have equal radii so the volume of A and C equal and take it as V
ii) Density of A is double that of C now we can take.
Density of A=2 D
Density of C=D
\(\mathrm{M} =\mathrm{D} \times \mathrm{V} \)
\(\text { Mass of sphere } \mathrm{A} =\mathrm{M}_{1} \quad \mathrm{M}_{1}=2 \mathrm{D} \times \mathrm{V} \)
\(\text { Mass of sphere } \mathrm{C} =\mathrm{M}_{3} \quad \mathrm{M}_{3}=\mathrm{D} \times \mathrm{V} \)
\(\mathrm{M}_{1}: \mathrm{M}_{3} =2 \mathrm{D} \times \mathrm{V} / \mathrm{D} \times \mathrm{V}=2 / 1 \)
\(\mathrm{M}_{\mathrm{1}}: \mathrm{M}_{3} =2: \mathrm{1} \)
Ratio of masses of sphere A and C = 2 : 1