Fill
in the blanks
(a)
The volume of a cube of side 1 cm is equal to.....m3
(b)
The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is
equal to ... (mm)2
(c) A vehicle moving with a speed of 18 km h–1covers....m
in 1 s
(d)
The relative density of lead is 11.3. Its density is ....g cm–3or .
...kg m–3.
Solution
Solution
(a)
Length of edge = 1cm = 1/100 m
Volume of the cube = side3
Plug the value of
side, we get
Volume of the cube =
(1/100 m)3
Answer: volume of a cube of side
1 cm is equal to 10–6 m3.
(b)
Given that
Radius, r = 2.0 cm = 20 mm
(convert cm to mm)
Height, h = 10.0 cm =100 mm
The formula of total surface area
of a cylinder S = 2πr (r + h)
Plug the values in this formula,
we get
Surface area of a cylinder S = 2πr (r + h = 2 × 3.14×20 (20+100)
= 15072 =
1.5 × 104 mm2
Answer
The surface area of a solid cylinder of radius
2.0 cm and height 10.0 cm is equal to 1.5 × 104 (mm)2
(c)
Using the conversion,
Given that
Time, t = 1 sec
speed = 18 km h-1 = 18 km / hour
1 km = 1000 m and
1hour = 3600 sec
Speed = 18 × 1000
/3600 sec = 5 m /sec
Use formula
Speed = distance / time
Cross multiply it, we
get
Distance = Speed × Time = 5 × 1 =
5 m
Answer: A vehicle moving with a speed of 18 km h–1covers
5 m in 1 s
(d)
Formula
Density of lead = Relative
density of lead × Density
of water
Density of water = 1 g/cm3
Plug the values, we
get
Density of lead = 11.3 × 1 g/ cm 3
=
11.3 g cm -3
1 cm = (1/100 m) =10–2
m3
1 g = 1/1000 kg = 10-3 kg
Density of lead = 11.3
g cm-3 = 11.3
Plug the value of 1 cm
and 1 gram
11.3 g/cm3 = 11.3 × 10-3
kg (10-2m)-3 = 11.3 ×10– 3 × 106
kg m-3 =1.13 × 103 kg m –3
Answer:
The relative density of lead is
11.3. Its density is 11.3 g cm -3.g cm–3 or 1.13 × 103 kg m–3.