The largest possible right circular cylinder | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Edge of the cube $a = 7 \, cm$.
$Quantity \, I$: Volume of the cube left over.
Volume of cube $= a^3 = 7^3 = 343 \, cm^3$.
Volume of the larges

Vedclass · Accepted Answer

Quantity $I < $ Quantity $II$

Vedclass · Answer

(B) Edge of the cube $a = 7 \, cm$.
$Quantity \, I$: Volume of the cube left over.
Volume of cube $= a^3 = 7^3 = 343 \, cm^3$.
Volume of the largest cylinder $= \pi r^2 h = \pi (3.5)^2 (7) = \frac{22}{7} 	imes 12.25 	imes 7 = 269.5 \, cm^3$.
Volume left $= 343 - 269.5 = 73.5 \, cm^3$.
$Quantity \, II$: Surface area of the cube remaining.
Original surface area $= 6a^2 = 6(49) = 294 \, cm^2$.
When the cylinder is cut,two circular faces are removed from the top and bottom faces of the cube,but the curved surface area of the cylinder is exposed.
Remaining surface area $= (6a^2 - 2 \pi r^2) + 2 \pi r h = 294 - 2(\frac{22}{7})(3.5)^2 + 2(\frac{22}{7})(3.5)(7) = 294 - 77 + 154 = 371 \, cm^2$.
Comparing magnitudes: $73.5 < 371$,therefore $Quantity \, I < Quantity \, II$.

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-11x+24=0$
$II.$ $2y^{2}-9y+9=0$

The sum of the roots of the equation,${x^2} + |2x - 3| - 4 = 0$,is

The area of a rectangle is the same as that of a circle of radius $\frac{\sqrt{35}}{\sqrt{11}} \text{ cm}$. If the length of the rectangle exceeds its breadth by $3 \text{ cm}$,find the dimensions of the rectangle.

If $x^{2}+y^{2}+z^{2}=2(x+z-1)$,then find the value of $x^{3}+y^{3}+z^{3}$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{3} - 468 = 1729$
$II.$ $y^{2} - 1733 + 1564 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module