Area of a square = side \(\times \) side
= 1 m \(\times \) 1 m
= 1 m² or 1 square metre
Area of 10 squares = 1 square metre \(\times \) 10
= 10 square metre

She moves in a straight line with constant velocity.

Density of gold is 19,300 kg/m³

Astronomical unit and light year.

Area, Volume, Speed, Electric charge and Density.

Density of cylinder \(=\frac { mass \ of \ cylinder }{ volume \ of \ cylinder }\)
\(=\frac{280}{4}=70 \mathrm{~kg} / \mathrm{m}^{3}\)

An object undergoes uniform acceleration when the change (increase or decrease) in its velocity for every unit of time is the same.

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

Aluminium, Iron, Copper, Gold

speed

m³

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

Stability:
Stability is a measure of the body's ability to maintain its original position.
The three types of stability are
(i) Stable equilibrium
(ii) Unstable equilibrium
(iii) Neutral equilibrium
Stable Equilibrium:
i) The frustum can be tilted through quite a big angle without toppling.
ii) Its centre of gravity is raised when it is displaced.
iii) The vertical line through its centre of gravity still falls within its base.
iv) So it can return to its original position.


Unstable Equilibrium:
i) The frustum will topple with the slightest tilting. Its centre of gravity is lowered when it is displaced.
ii) The vertical line through its centre of gravity falls outside its base.

Neutral Equilibrium:
(i) It causes frustum to topple.
(ii) The frustum will rolls about but does not topple.
(iii) Its centre of gravity remains at the same height when it is displaced.
(iv) The body will stay in any position to which it has been displaced.

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.

Final velocity (v) = 20 m/s Initial Velocity (u) = 0 m/s.
Time (t) = 10 s
Acceleration (a) =?
\(\text { Acceleration } =\frac{\text { Change in velocity }}{\text { time }} \)
\(=\frac{\text { Final velocity }(\mathrm{v})-\text { Initial velocity }(\mathrm{u})}{\text { time }}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{20-0}{10} \)
Acceleration of the car = 2 m/s².

Two spheres are same size. So the volume of two spheres are equal and take it as V
Density of copper \(=8900 \mathrm{~kg} / \mathrm{m}^{3}\)
Density of iron \(=7800 \mathrm{~kg} / \mathrm{m}^{3}\)
\(\mathrm{M}=\mathrm{D} \times \mathrm{V}\) mass of copper, take it as \(\mathrm{M}_{1}\)
mass of iron, take it as \(\mathrm{M}_{2}\)
Mass of copper \(\mathrm{M}_{1}=8900 \times \mathrm{V}\)
Mass of iron \(\mathrm{M}_{2}=7800 \times \mathrm{V}\)
Ratio of masses \(\mathrm{M}_{1} / \mathrm{M}_{2}=8900 \times \mathrm{V} / 7800 \times \mathrm{V}\)
\(\mathrm{M}_{1} / \mathrm{M}_{2} =89 / 78 \)
\(\mathrm{M}_{1}: \mathrm{M}_{2} =89: 78\)

Given radius = 10 cm = 0.1m
π = 3.14
Area of a circular disc A = πr² (in m²)
Formula: Area of a circle A = πr²
= 3.14 \(\times\) 10 / 100  \(\times\) 10 / 100
= 0.0314m²
Solution: A = 0.0314m²

Clues - Across
1. KELVIN
2. VOLUME
3. DENSITY
4. CAPACITY
Clues-Down 
a. VELOCITY
b. CUBIC METRE
c. MERCURY
d. LIGHT YEAR