Pls solve it fast solids of right cylindrical shape are 49 cm and 35 cm high and their base / - Maths - Mensuration

Question

Pls solve it fast

Rashi Lokwani · Accepted Answer

Dear Student, Answer 1: Diameter of cylinder , $d_{1}$ = 16 cm Height of cylinder, $h_{1}$ = 49 cm Volume of cylinder, $V_{1}$ = $π {r_{1}}^{2} h_{1} = π {(\frac{d_{1}}{2})}^{2} h_{1} = π \frac{{d_{1}}^{2}}{4} h_{1} = \frac{{π d_{1}}^{2} h_{1}}{4} = \frac{{π (16)}^{2} (49)}{4} (i) [∵ r (r a d i u s) = \frac{d (d i a m e t e r)}{2}]$ Now, Diameter of cylinder , $d_{2}$ = 14 cm Height of cylinder, $h_{2}$ = 35 cm Volume of cylinder, $V_{2}$ = $π {r_{2}}^{2} h_{2} = π {(\frac{d_{2}}{2})}^{2} h_{2} = π \frac{{d_{2}}^{2}}{4} h_{2} = \frac{{π d_{2}}^{2} h_{2}}{4} = \frac{{π (14)}^{2} (35)}{4} (i i)$ When both cylinders are melted and moulded into a single cylinder,then the Volume of single cylinder , V = $π r^{2} h = π {(\frac{d}{2})}^{2} h = π \frac{d}{4} h = \frac{{π d}^{2} (56)}{4} (i i i)$ where , height h = 56 cm (given) ⇒Volume of single cylinder , V= $V_{1}$ + $V_{2}$ $\Rightarrow \frac{{π d}^{2} (56)}{4} = \frac{{π (16)}^{2} (49)}{4} + \frac{{π (14)}^{2} (35)}{4} [F r o m (i), (i i) a n d (i i i)] \Rightarrow d^{2} (56) = {(16)}^{2} (49) + {(14)}^{2} (35) \Rightarrow d^{2} = \frac{{(16)}^{2} (49)}{56} + \frac{{(14)}^{2} (35)}{56} \Rightarrow d^{2} = 16 \times 2 \times 7 + \frac{49 \times 5}{2} \Rightarrow d^{2} = 224 + \frac{245}{2} = \frac{448 + 245}{2} = \frac{693}{2} \Rightarrow d^{2} = 346 5 \Rightarrow d = \sqrt{346 5} = 18 6 c m$ Answer 2: Volume of the metallic cylindrical pipe = 748 $c m^{3}$ Length of the cylindrical pipe h =14 cm External radius of the pipe R = 9 cm Let the internal radius of the cylindrical pipe be r cm Therefore the volume of the metallic pipe = $π (R^{2} - r^{2}) h$ Therefore $\frac{22}{7} (9^{2} - r^{2}) * 14 = 748 44 (81 - r^{2}) = 748 81 - r^{2} = \frac{748}{44} = 17 r^{2} = 81 - 17 = 64 r = \sqrt{64} = 8 c m$ Thickness of the metallic pipe = External radius - Internal radius = R - r =9-8=1 cm Please post the remaining queries in separate threads so that we can help you as soon as possible Regards