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Question

Note : In stating numerical answers, take care of significant figures. 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¯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.

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Solution

(a)

Given, the side of the cube is 1cm .

Let a be the side of a cube.

Then,

Volume of a cube = a 3

Substitute the given value in above expression.

Volume of a cube = ( 1cm× 1m 100cm ) 3 = ( 10 2 m ) 3 = 10 6 m 3

So, the volume of the cube of side 1 cm is equal to 10 6 m 3 .

(b)

Given, the radius of a solid cylinder is 2.0cm and height of the cylinder is 10.0cm .

Let r be the radius of the solid cylinder and h be the height of the solid cylinder.

Surface area of a solid cylinder =2πr( h+r )

Substitute the given value in above expression.

Surface area of a solid cylinder =2× 22 7 ×2.0cm×( 10.0cm+2.0cm ) =150.86 cm 2 × 10 2 mm 2 1 cm 2 1.5× 10 4 mm 2

So, 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)

Given, the speed of the vehicle is 18 kmh 1 and the travel time of the vehicle is 1s .

Let v be the speed of the vehicle and t be the travel time.

Then,

Distance covered by the vehicle =v×t

Substitute the given value in above expression.

Distance covered by the vehicle =18km h 1 × 10 3 m 1km × 1h 3600s ×1s =5m

Thus, a vehicle moving with a speed of 18 kmh 1 covers 5m in 1 s .

(d)

Given, the relative density of lead is 11.3 .

Let ρ be the density of lead and ρ w be the density of water. The density of water in g/ cm 3 is 11.3g/ cm 3 . Then,

Relativedensity= ρ ρ w

Substitute the given value in above expression.

11.3= ρ 1g/ cm 3 ρ=11.3g/ cm 3

Density of lead in kg/ m 3 =11.3 g cm 3 × 1kg 1000g × 10 6 cm 3 1 m 3 =1.13× 10 4 kg/ m 3

So, the density of lead is 11.3g/ cm 3 or 1.13× 10 4 kg/ m 3 .


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