Match the following columns:

$\begin{matrix} C o l u m n I & C o l u m n I I (a) T h e r a d i i o f c i r c u l a r e n d s o f & (p) 2418 π a b u c k e t i n t h e f o r m o f f r u s t u m o f a c o n e o f h e i g h t 30 c m a r e 20 c m a n d 10 c m r e s p e c t i v e l y . T h e c a p a c i t y o f t h e b u c k e t i s . . . . . c m^{3} . [T a k e π = \frac{22}{7} .] (b) T h e r a d i i o f t h e c i r c u l a r e n d s & (q) 22000 o f a c o n i c a l b u c k e t o f h e i g h t 15 c m a r e 20 c m a n d 12 c m r e s p e c t i v e l y . T h e s l a n t h e i g h t o f t h e b u c k e t i s . . . . c m . (c) T h e r a d i i o f t h e c i r c u l a r e n d s & (r) 12 o f a s o l i d f r u s t u m o f a c o n e a r e 33 c m a n d 27 c m a n d i t s s l a n t h e i g h t i s 10 c m . T h e t o t a l s u r f a c e a r e a o f t h e b u c k e t i s . . . . . c m^{2} . (d) T h r e e s o l i d m e t a l l i c s p h e r e s o f & (s) 17 r a d i i 3 c m, 4 c m a n d 5 c m a r e m e l t e d t o f o r m a s i n g l e s o l i d s p h e r e . T h e d i a m e t e r o f t h e r e s u l t i n g s p h e r e i s . . . . . . c m . \end{matrix}$

The correct answer is

$(a) -_{. . . . . . . .,}$ $(b) -_{. . . . . . . .,}$ $(c) -_{. . . . . . . .,}$ $(d) -_{. . . . . . . . .}$

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Solution

a) Volume of frustum = $\frac{1}{3}$ $π$ $(R^{2} + r^{2} + R r)$ $\times$ h
= $\frac{1}{3}$ $\times$ $\frac{22}{7}$ $\times$ $(20^{2} + 10^{2} + 200)$ $\times$ 30
= 22000 $c m^{3}$
b) Slant height of a conical frustum = $\sqrt{(R - r)^{2} + h^{2}}$
= $\sqrt{(20 - 12)^{2} + 15^{2}}$
= $\sqrt{8^{2} + 15^{2}}$ = $\sqrt{64 + 225}$ = 17 cm
c) Total surface area of cone = $π R^{2}$ + $π r^{2}$ + $π (R + r) l$
= $π$ $[33^{2} + 27^{2} + (33 + 27) 10]$
= $π$ $[1089 + 729 + 600]$
= 2418 $π$
d) Volume of all three spheres = Volume of bigger sphere
Volume of sphere of radius 3 cm = $\frac{4}{3}$ $\times$ $π$ $\times$ $3^{3}$
Volume of sphere of radius 4 cm = $\frac{4}{3}$ $\times$ $π$ $\times$ $4^{3}$
Volume of sphere of radius 5 cm = $\frac{4}{3}$ $\times$ $π$ $\times$ $5^{3}$
Volume of sphere of radius R cm = $\frac{4}{3}$ $\times$ $π$ $\times$ $R^{3}$
$\frac{4}{3}$ $\times$ $π$ $\times$ [ $3^{3}$ + $4^{3}$ + $5^{3}$ ] = $\frac{4}{3}$ $\times$ $π$ $\times$ $R^{3}$
27 + 64 + 125 = $R^{3}$
216 = $R^{3}$
R = 6 cm
So, D = 12 cm

The correct answer is

$(a) - q$ $(b) - s$ $(c) - p$ $(d) - r$

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