### Q.1:- **Fill in the blanks**

(a) The volume of a cube of side 1 cm is equal to…..m^{3}

(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^{–1}covers….m in 1 s

(d) The relative density of lead is 11.3. Its density is ….g cm^{–3}or . …kg m^{–3}

**Answer:-**

(a) Length of edge = 1cm = 1/100 m

Volume of the cube = side^{3}

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} m^{3}

^{
}(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 = 2π*r* (*r* + *h*)

Putting the values in this formula, we get

Surface area of a cylinder S = 2π*r* (*r* + *h* = 2 x 3.14 x 20 (20+100)

= 15072 = 1.5 × 10^{4} mm^{2}

The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to __1.5 × 10__^{4} mm^{2}

(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^{–1}covers __5__ m in 1 s.

(d) Density of lead = Relative density of lead × Density of water

Density of water = 1 g/cm^{3}

Putting 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

Putting the value of 1 cm and 1 gram

11.3 g/cm^{3} = 11.3 × 10^{-3} kg (10^{-2}m)^{-3} = 11.3 ×10^{–3} × 10^{6} kg m^{-3} =1.13 × 10^{3} 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 × 10__^{3} kg m^{–3}.