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Fill in the blanks <br> (a) The volume of a cube of side 1 cm isk equal to….. m^3 <br> (b) the surface area fo a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to …. (mm)^2 <br> (c ) A vehical moving with a speed of 18km h^(-1) covers ....m in 1s. <br> (d) The relative density of lead is 11.3. its density is ......g cm^(-3) or .....kg m^(-3)

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Solution :
`1 cm =1/100 m` <br> volume of the cube `1 cm^(3)` <br> but , `1cm^(3) =1cm xx1 cm xx1cm =(1/100) m xx(1/100) m xx(1/100) m ` <br> `:. 1 cm^(3)=10^(-6) m^(3)` <br> Hence , the volume of a cube of side 1 cm is equal to `10^(-6) m^(3)` <br> (b) the total surface area of a cylinder of radius r and height h is <br> `S=2pir(r+h)` <br> given that, <br> `r=2 cm =2xx1 cm =2xx10 mm =20 mm` <br> `h=10 cm =10xx10 mm=100 mm` <br> `:. s=2xx3.14 xx20xx(2+100) =15072=1.5xx10^(4) mm^(2)` <br> (c) Using the conservation, <br> `1km//h=5/18 m//s` <br> `18km//h =18xx5/18 =5 m//s` <br> Therefore, distance can be obtaied using the relation <br> Distance =Speed x time =5 x 1 =5 m <br> Hence, the vehicle covers 5 m in 1 s.<br> (d) Relative density of a substance is given by the releation, <br> Relative =density =`("Density of substance")/("Density of water")` <br> Density of water `=1g//cm^(3)` <br> Density of lead=realative density of lead x density of water <br> `=11.3xx1=11.3 g//cm^(3)` <br> Again, `1g =1/1000 kg ` <br> `1 cm^(3) =10^(-6) m^(3)` <br> `1g//cm^(3) =(10^(-3))/(10^(-6)) kg /m^(3) =10^(3) kg //m^(3)` <br> `:. 11.3 g//cm^(3) =11.3 xx10^(3) kg //m^(3)`

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