Question

Q. Lead spheres of diameter 6cm each are dropped into a cylindrical beaker containing some water and get fully submerged .

If the diameter of the beaker is 18 cm and the water level rises by 40 cm, find the number of lead spheres dropped in the water?

(a) 45
(b) 60
(c) 75
(d) 90​

Answer 1

Answer:

90 lead spheres (Option D)

Given:

Lead sphere diameter=6 cm

Diameter of cylindrical beaker=18 cm and water rises(Height)=40 cm

To find:

Number of lead sphere dropped in the water

Solution:-

Lead sphere diameter=6 cm

$radius = \frac{d}{2}$

$r = 3 \: cm$

$its \: volume = \frac{4}{3}\pi {r}^{3}$

Now

Diameter of cylindrical beaker=18 cm

$r = \frac{18}{2}$

$r = 9 \: cm$

Height of water in beaker=40 cm

$its \: volume = \pi {r}^{2}h$

Let the number of lead sphere be x

Then

$x = \frac{volume \: of \: cylindrical \: beaker}{volume \: of \: lead \: sphere}$

$= \frac{\pi {r}^{2}h }{ \frac{4}{3}\pi {r}^{3} }$

$= \frac{ \frac{22}{7} \times 9 \times 9 \times 40 }{ \frac{4}{3} \times \frac{22}{7} \times 3 \times 3 \times 3 }$

$= 90$

---------------------

Number of lead sphere dropped in the water=90

Answer 2

Solution:

Given:

=> Radius of cylinder (R) = 9 cm

=> Height of cylinder (H) = 40 cm

=> Radius of Sphere (r) = 3 cm

To Find:

=> Number of lead spheres dropped in the water

Formula used:

$\sf{\implies Volume\;of\;cylinder=\pi R^{2}H}$

$\sf{\implies Volume\;of\;sphere=\dfrac{4}{3} \pi r^{3}}$

So,

$\sf{\implies Volume\;of\;lead\;spheres = \dfrac{4}{3} \pi r^{3}}$

$\sf{\implies \dfrac{4}{3} \pi (3)^{3}}$

$\sf{\implies \dfrac{4}{3}\pi \times 27}$

$\large{\boxed{\boxed{\sf{\implies Volume\;of\;lead\;spheres = 36\pi \;cm^{3}}}}}$

Now,

$\sf{\implies Volume\;of\;cylinder=\pi R^{2}H}$

$\sf{\implies 81\times 40 \pi }$

$\large{\boxed{\boxed{\sf{\implies Volume\;of\;cylinder = 3240\pi}}}}$

$\sf{Now,\;Number\;of\;lead\;spheres = \dfrac{Volume\;of\;cylinder}{Volume\;of\;each\;lead\;sphere}}$

$\sf{\implies \dfrac{3240 \pi }{36 \pi}}$

$\large{\boxed{\boxed{\sf{\implies Number\;of\;lead\;spheres = 90}}}}$