Question

the external diameter of a hollow steel pipe is 24cm, its length is 35 cm and its thickness is 5cm. what is the weight of the pipe if the steel used in it weight 5 g/cm3​

Answer 1

Answer:

Firstly, find the volume of the pipe.

πh(R+r) (R-r)

35*22/7*(5+3)(5-3)

=5*22*8*2

=1760cm^3

1cm^3 weighs 6 gram

1760cm^3 would weigh=1760*6 =10560gmcm^3

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Answer 2

Answer:

Weight of the is 52250 g.

Step-by-step explanation:

The volume of a cylinder is the density of the cylinder which signifies the amount of material it can carry or how much amount of any material can be immersed in it.

In case of hollow cylinder, we measure two radius, one for inner circle (r2) and one for outer circle(r1) formed by the base of hollow cylinder. Suppose, ‘h’ be the height, then the volume of this cylinder can be written as;

V =  πh($r1^{2} -r2^{2}$)

Let external diameter be d1 and internal be d2

​⇒d1=24cm & d2=24-5-5=14cm

Therefore, r1=12cm & r2=7cm

Volume of hollow cylindrical pipe=π×($r1^{2} -r2^{2}$ )×h

                                                       =$\frac{22}{7}$×($12^{2} -7^{2}$)×35

                                                       =10450

∴Weight of the pipe=5×10450=52250g.

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