Question

25. The total surface area of a hollow cylinder,
which is open from both the sides, is 3575
cm²; area of its base ring is 357.5 cm2 and its
it
height is 14 cm. Find the thickness of the
d cylinder.
​

Answer 1

Step-by-step explanation:

Let the inner radius be r

and outer radius be R

Base Ring =π(R

2

−r

2

)=357.5 cm

2

(R

2

−r

2

)=357.5÷22/7

(R+r)(R−r)=(3575∗7)/220

(R+r)(R−r)=113.75 sq cm............(1)

Total surface area of the cylinder =3575 sq cm

Now, total surface area of a hollow cylinder = outer curved surface + inner curved surface area + 2(Area of the circular base)

=2πRh+2πrh+2π(R

2

−r

2

)

=2πRh+2πrh+2∗357.5=3575

=2πh(R+r)+2×357.5=3575

=2πh(R+r)+751=3575

=2πh(R+r)=3575−751

=2×22/7×14×(R+r)=2824

=(R+r)=

44∗14

2824∗7

=(R+r)=32

Substituting the value of (R+r)=32 in equation (1), we get.

(R+r)(R−r)=113.75

32(R−r)=113.75

R−r=

32

113.75

So, the thickness of the cylinder is 3.55 cm

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