A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 3. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm. Find the article. |
OR |
A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap? |
Answer:
Given, Radius (r) of cylinder = Radius of hemisphere = 3.5 cm. Total SA of article = CSA of cylinder \[+2\times \] CSA of hemisphere Height of cylinder, h = 10 cm TSA \[=\text{ }2\pi rh+2\times 2\pi {{r}^{2}}\] \[=2\pi rh+4\pi {{r}^{2}}\] \[=2\pi r\text{ (}h+2r)\] \[=2\times \frac{22}{7}\times 3.5(10+2\times 3.5)\] \[=2\times 22\times 0.5\times (10+7)\] \[=2\times 11\times 17\] \[=374\,\,c{{m}^{2}}\] OR Base diameter of cone = 24 m. \[\therefore \] Radius \[r=12\text{ }m\] Height of cone, \[h=3.5\text{ }m\] Volume of rice in conical heap \[=\frac{1}{3}\pi {{r}^{2}}h\] \[=\frac{1}{3}\times \frac{22}{7}\times 12\times 12\times 3.5\] \[=528\,{{m}^{3}}\] Now, slant height, \[l=\sqrt{{{h}^{2}}+{{r}^{2}}}\] \[=\sqrt{{{(3.5)}^{2}}+{{(12)}^{2}}}\] \[=\sqrt{12.25+144}\] \[=\sqrt{156.25}\] \[=12.5\,m\] Canvas cloth required to just cover the heap = CSA of conical heap \[=\pi rl\] \[=\frac{22}{7}\times 12\times 12.5\] \[=\frac{3300}{7}{{m}^{2}}\] \[471.43\,\,{{m}^{2}}\].
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