Question

A solid is in the form of a right circular cylinder mounted on a solid hemisphere of radius $14$ cm. The radius of the base of the cylindrical part is $14$ cm and the vertical height of the complete solid is $28$ cm. Find :

(i) The volume of the solid

(ii)The surface area of the solid

(iii) Cost of painting the solid at the rate of Rs.$0.80$ per $c{m}^2$

• A. $14373\frac{1}{3}c{m}^3$, $3080c{m}^2$, Rs. $2464$
• B. $1473\frac{1}{2}c{m}^3$, $3582c{m}^2$, Rs. $2499$
  
• C. $14573\frac{1}{5}c{m}^3$, $3810c{m}^2$, Rs. $2762$
• D. $14370\frac{1}{7}c{m}^3$, $3087c{m}^2$, Rs. $2465$

Answer 1

Answer: (A)

$14373\frac{1}{3}c{m}^3$, $3080c{m}^2$, Rs. $2464$

Given, a solid in form of right circular cylinder mounted on hemisphere of radius $14$ cm. The radius of base part of cylinder is $14$ cm and height of solid is $28$ cm.
Since, height of solid is $28$ cm and height of hemisphere is radius i.e. $14$ cm.
Therefore, height of cylinder $=$ height of solid $-$ radius of hemisphere
$=$ $28$ $-$ $14$
$=$ $14$ cm
(i) Volume of solid $=$ Volume of cylinder $+$ volume of hemisphere
$=\pi r^{2}h+\frac{2}{3}\pi r^3=\frac{22}{7}\times {14}^2\times 14+\frac{2}{3}\times \frac{22}{7}\times {14}^3=8624+\frac{17248}{3}=14373\frac{1}{3}{cm}^3$

(ii) Surface area of solid $=$ curved surface area of cylinder $+$ curved $=2\pi rh+2\pi r^2+\pi r^2$

$=2\times \frac{22}{7}\times 14\times 14+2\times \frac{22}{7}\times {14}^2+\frac{22}{7}\times {14}^2$

$=1232{cm}^2+1232{cm}^2+616$

$=3080{cm}^2$
(iii) Cost of painting at rate Rs. $0.80$ per sq.cm $=$ Surface area of solid $\times$ rate per sq. cm
$=3080\times 0.80$
$=$ Rs. $2464$