A metal sphere with a radius of 12, cm is mel | vedclass

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

Question

(C) The volume of the metal sphere is given by the formula $V_s = \frac{4}{3} \pi r^3$,where $r = 12\, cm$.$V_s = \frac{4}{3} \pi (12)^3 = \frac{4}{3}

Vedclass · Accepted Answer

$192$

Vedclass · Answer

(C) The volume of the metal sphere is given by the formula $V_s = \frac{4}{3} \pi r^3$,where $r = 12\, cm$.$V_s = \frac{4}{3} \pi (12)^3 = \frac{4}{3} \pi (1728) = 2304 \pi\, cm^3$.The volume of one cone is given by $V_c = \frac{1}{3} \pi r^2 h$,where $r = 3\, cm$ and $h = 4\, cm$.$V_c = \frac{1}{3} \pi (3)^2 (4) = \frac{1}{3} \pi (9)(4) = 12 \pi\, cm^3$.The number of cones obtained is the ratio of the volume of the sphere to the volume of one cone:$N = \frac{V_s}{V_c} = \frac{2304 \pi}{12 \pi} = 192$.Thus,$192$ cones are obtained.

$A$ metal sphere with a radius of $12\, cm$ is melted,and recast into cones with radius $3\, cm$ and height $4\, cm$. Find how many cones are obtained?

Similar Questions

The radius of the base of a cone is $10 \, cm$ and its slant height is $26 \, cm$. Find the volume of the cone. $(\pi = 3.14)$ (in $cm^3$)

The radius of a sphere is increased by $10 \%$. Prove that the volume will be increased by $33.1 \%$ approximately.

State whether the following statement is true or false:
The ratio of the surface areas of two spheres is $81: 49$. Then,the ratio of their volumes is $729: 343$.

The radii of two cylinders are in the ratio $3: 7$ and their heights are in the ratio $28: 15$. Calculate the ratio of their curved surface areas.

The radius of a sphere is $2r$,then its volume will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module