THE VOLUME OF A CYLINDER IS 48.125cm cube , WHICH IS FORMED BY ROLLING A RECTANGULAR PAPER SHEET ALONG THE LENTH OF THE PAPER . IF A CUBOIDAL BOX ( WITHOUT ANY LID i.e , OPEN AT THE TOP ) IS MADE FROM THE SAME SHEET OF PAPER BY CUTTING OUT THE SQURE OF SIDE 0.5 cm FROM EACH OF THE FOUR CORNERS OF PAPER SHEET, THEN WHAT IS THE VOLUME OF THE BOX ?
If the radius of the cylinder is r,
v = πr^2 h, so h = v/(πr^2)
a = 2πrh
now the sheet is w by h, so w = 2πr by v/(πr^2)
the volume of the box is thus
1/2 (w-1)(h-1) = 1/2 (2πr-1)(v/(πr^2) - 1)
since no information was given regarding the relative dimensions of the cylinder, the rectangular sheet could be anything between long and skinny or square.
Steve, don't lose patience.This problem does have answer, only thing is one should think out of the box. Let me tell you how it goes:
Volume of cylinder= (pi) (r)(r)(h)= 48.125
(r)(r)h = 15.3125
15.3125 is not a perfect square. To make it perfect square, h should be 5cm.
Now (r)(r)=15.3125/5 = 3.0625 which makes r = 1.75cm
Therefore, rectangular sheet has length of 2(pi)(r)=11cm and width (h) of 5cm.
After cutting a square of 0.5cm from each corner, the dimension of the cuboid becomes:
Length=11-1=10cm; Breadth=5-1=4cm; Height=0.5cm.
Therefore, Volume of cuboid= 10 x 4 x 0.5 = 20 cubic cm.
Thank you for reading it!
Have a nice day bro!
Jay Bankoti
Let length of paper be l and width be b
so (pi)r^2 b=48.125
r^2 = 48.125/ (pi)*b ..(I)
l = 2(pi)r
so r= l/2(pi).....(II)
from I and II
l^2/4(pi)^2 = 48.125/(pi)b
solving we get
l^2 b = 48.125 *4*(pi)
taking value of
pi as 22/7 we get
l^2*b = 55 * 11
l^2*b = 11*11*5
so l=11
b=5
for cuboid
(l-1)*(b-1)*0.5
=10*4*0.5
=20
To find the volume of the box made from the rectangular paper sheet, we need to calculate the volume of the original cylinder and subtract the volume of the four cut-out corners.
First, let's calculate the volume of the cylinder:
Given: Volume of the cylinder = 48.125 cm³
The volume of a cylinder is given by the formula:
Volume = π * r^2 * h,
Where π is approximately 3.14, r is the radius of the base, and h is the height.
However, in this case, we don't know the height or the radius directly. We need to find a relationship between the radius and the height.
The paper was rolled along the length to form the cylinder, which means the height of the cylinder is equal to the width of the rectangular paper sheet. Let's denote the width as 'w.'
The circumference of the base of the cylinder (formed by rolling the rectangular paper sheet) will be equal to the length of the rectangular paper sheet, which we'll denote as 'l.'
The formula for the circumference of a circle is:
Circumference = 2 * π * r.
In this case, the circumference is 'l,' and the radius is 'w/2.'
So,
l = 2 * π * (w/2),
l = πw.
We know the value of l, which is the circumference of the cylinder, and the value of the volume.
Substituting these values into the volume formula:
Volume = π * r^2 * h,
48.125 = π * (w/2)^2 * w,
48.125 = π * (w^2/4) * w,
48.125 = πw^3/4.
Now, let's simplify the equation to solve for 'w':
48.125 * 4 = πw^3,
192.5 = πw^3,
w^3 = 192.5/π,
w^3 ≈ 61.214,
w ≈ ∛(61.214).
Using a calculator, we find that w ≈ 3.944.
Now that we have the value of 'w,' we can calculate the height of the cylinder, which is equal to its width:
h = w ≈ 3.944.
Next, we'll find the radius of the cylinder (r):
r = w/2 ≈ 3.944/2,
r ≈ 1.972.
Using these values, we can calculate the volume of the cylinder:
Volume = π * r^2 * h,
Volume = π * (1.972)^2 * 3.944.
Using the approximation π ≈ 3.14, we can calculate the volume of the cylinder.
Once we have the volume of the cylinder, we need to subtract the volume of the four cut-out corners.
The cut-out corners are squares with a side length of 0.5 cm.
The volume of each cut-out corner is given by:
Volume = side length^3,
Volume = 0.5^3.
The total volume of the four cut-out corners is:
Total Volume of Cut-outs = 4 * (0.5^3).
Finally, to find the volume of the box, we subtract the total volume of the cut-outs from the volume of the cylinder:
Volume of the Box = Volume of the Cylinder - Total Volume of Cut-outs.
Plug in the values and calculate the volume of the box using these formulas.