## 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 | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-2 Units and Measurements

### Q.1:- 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

(a) Length of edge = 1cm = 1/100 m
Volume of the cube = side3
Putting the value of side, we get
Volume of the cube = (1/100 m)3 The volume of a cube of side 1 cm is equal to 10-6 m3

(b) Given,
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)
Putting the values in this formula, we get
Surface area of a cylinder S = 2πr (r + h = 2 x 3.14 x 20 (20+100)
= 15072 = 1.5 × 104 mm2
The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to 1.5 × 104 mm2

(c) Using the conversion,
Given,
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
A vehicle moving with a speed of 18 km h–1covers 5 m in 1 s.

(d) Density of lead = Relative density of lead × Density of water
Density of water = 1 g/cm3
Putting the values, we get
Density of lead = 11.3 × 1 g/ cm3
= 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
Putting 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
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.