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Ml aggarwal solutions

CHAPTERS

**A rectangular piece of a tin of size 30 cm × 18 cm is rolled in two ways, once along its length (30 cm) and once along with its breadth. Find the ratio of volumes of two cylinders so formed.**

Given,

Size of rectangular tin plate = 30 cm × 18 cm

(i) When rolled along its length (30 cm),

Then, the circumference of the circle so formed = 30 cm

Radius(r_{1}) = C/2π = (30 x 7)/ (2 x 22) = 105/22 cm

And height (h_{1}) = 18 cm

Then, volume = πr_{1}^{2}h_{1} = π x (105/22)^{2} x (18) cm^{3}

If it is rolled along its breadth (18 cm) then,

Circumference = 18 cm

So, radius (r_{2}) = C/2π = (18 x 7)/ (2 x 22) = 63/22 cm

And height (h_{2}) = 30 cm

Then, volume = πr_{2}^{2}h_{2} = π x (63/22)^{2} x (30) cm^{3}

Now, ratio between the two volumes

= π x (105/22)^{2} x (18) : π x (63/22)^{2} x (30)

= (105/22)^{2} x (18) : (63/22)^{2} x (30)

= 5 : 3

Given,

Size of rectangular tin plate = 30 cm × 18 cm

(i) When rolled along its length (30 cm),

Then, the circumference of the circle so formed = 30 cm

Radius(r_{1}) = C/2π = (30 x 7)/ (2 x 22) = 105/22 cm

And height (h_{1}) = 18 cm

Then, volume = πr_{1}^{2}h_{1} = π x (105/22)^{2} x (18) cm^{3}

If it is rolled along its breadth (18 cm) then,

Circumference = 18 cm

So, radius (r_{2}) = C/2π = (18 x 7)/ (2 x 22) = 63/22 cm

And height (h_{2}) = 30 cm

Then, volume = πr_{2}^{2}h_{2} = π x (63/22)^{2} x (30) cm^{3}

Now, ratio between the two volumes

= π x (105/22)^{2} x (18) : π x (63/22)^{2} x (30)

= (105/22)^{2} x (18) : (63/22)^{2} x (30)

= 5 : 3

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