Question
The surface area of a solid metallic sphere is $616\ cm^2$. It is melted and recast into a cone of height $28\ cm$. Find the diameter of the base of the cone so formed $(\text{use}\pi=\frac{22}{7}).$
✓
Answer
The surface area of the metallic sphere is 616 square cm. Let the radius of the metallic sphere is r. Therefore, we have
$\Rightarrow4\pi\text{r}^2=616$
$4\times\frac{22}{7}\times\text{r}^2=616$
$\text{r}^2=\frac{6.16\times7}{22\times4}$
$\Rightarrow\text{r}^2=7\times7$
$\Rightarrow\text{r}=7$
Therefore, the radius of the metallic sphere is 7cm and the volume of the sphere is $\text{V}_1=\frac{4}{3}\pi\times(7)^3\text{cm}^3$
The sphere is melted to recast a cone of height 28cm. Let the radius of the cone is R cm. Therefore, the volume of the cone is $\text{V}_2=\frac{1}{3}\pi\times(\text{R})^2\times28\text{cm}^3$
Since, the volumes of the sphere and the cone are same; we have
$V_1 = V_2$
$\Rightarrow\frac{4}{3}\pi\times(7)^3=\frac{1}{3}\pi\times(\text{R})^2\times28$
$\Rightarrow\text{R}^2=\frac{4\times(7)^3}{28}$
$\Rightarrow\text{R}^2=7^2$
$\Rightarrow\text{R}=7$
Hence, the diameter of the base of the cone so formed is two times its radius, which is $14\ cm$.
$\Rightarrow4\pi\text{r}^2=616$
$4\times\frac{22}{7}\times\text{r}^2=616$
$\text{r}^2=\frac{6.16\times7}{22\times4}$
$\Rightarrow\text{r}^2=7\times7$
$\Rightarrow\text{r}=7$
Therefore, the radius of the metallic sphere is 7cm and the volume of the sphere is $\text{V}_1=\frac{4}{3}\pi\times(7)^3\text{cm}^3$
The sphere is melted to recast a cone of height 28cm. Let the radius of the cone is R cm. Therefore, the volume of the cone is $\text{V}_2=\frac{1}{3}\pi\times(\text{R})^2\times28\text{cm}^3$
Since, the volumes of the sphere and the cone are same; we have
$V_1 = V_2$
$\Rightarrow\frac{4}{3}\pi\times(7)^3=\frac{1}{3}\pi\times(\text{R})^2\times28$
$\Rightarrow\text{R}^2=\frac{4\times(7)^3}{28}$
$\Rightarrow\text{R}^2=7^2$
$\Rightarrow\text{R}=7$
Hence, the diameter of the base of the cone so formed is two times its radius, which is $14\ cm$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Solve the following simultaneous equations using Cramer’s rule.
$2 x+3 y=2 ; x-\frac{y}{2}=\frac{1}{2}$
→$2 x+3 y=2 ; x-\frac{y}{2}=\frac{1}{2}$
Find the ratio in which the point P(m, 6) divides the join of A(-4, 3) and B(2, 8). Also find the value of m.
Find the sum of first $51$ terms of an A.P. whose second and third terms are $14$ and $18 $respectively.
→Are the triangles in figure 1.56 similar? If yes, by which test?
If PA and PB are tangents from an outside point P such that PA = 10cm and $\angle\text{APB}=60^{\circ}.$ Find the length of chord AB.
→In the adjoining figure, seg ML || seg BC, seg NL || seg DC. Prove that AM : AB = AN : AD.
Solve the following examples.
Find the length a diagonal of a rectangle having sides 11 cm and 60cm.
Find the length a diagonal of a rectangle having sides 11 cm and 60cm.
Prove the following.
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$
→$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$
Tiptop Electronics’ supplied an AC of 1.5 ton to a company. Cost of the AC supplied is Rs. 51,200 (with GST). Rate of CGST on AC is 14%. Then find the following amounts as shown in the tax invoice of Tiptop Electronics.
(1) Rate of SGST
(2) Rate of GST on AC
(3) Taxable value of AC
(4) Total amount of GST
(5) Amount of CGST
(6) Amount of SGST
(1) Rate of SGST
(2) Rate of GST on AC
(3) Taxable value of AC
(4) Total amount of GST
(5) Amount of CGST
(6) Amount of SGST
In $\triangle\text{ABC},\text{D}$ is the midpoint of $BC$ and $\text{AE}\perp\text{BC}.$ If $\text{AC}>\text{AB},$ show that.
$\text{AB}^2=\text{AD}^2-\text{BC}.\text{DE}+\frac{1}{4}\text{BC}^2.$
→$\text{AB}^2=\text{AD}^2-\text{BC}.\text{DE}+\frac{1}{4}\text{BC}^2.$