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Class 10 Class 12

Surface Areas and Volumes

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have ? Find the surface area of the solid.

Solution not provided.

Ans. 7 cm, 332.5 cm²

127 Views

An iron solid sphere of radius 3 cm is melted and recast into small spherical balls of radius 1 cm each. Assuming that there is no wastage in the process, find the number of small spherical balls made from the given sphere.

Let R be the radius of solid sphere, then R = 3cm
So, Volume (V₁) =

4 over 3 πR cubed

equals space open parentheses 4 over 3 straight x 22 over 7 straight x 3 space straight x space 3 space straight x space 3 close parentheses space cm cubed

Let r be the radius of small spherical ball, then

r = 1 cm

So, Volume (V₂) =

4 over 3 space πr cubed

equals space open parentheses 4 over 3 straight x 22 over 7 straight x space 1 space straight x space 1 straight x space 1 close parentheses space cm cubed

Now, Required number of small spherical balls

equals space fraction numerator Volume space of space solid space sphere space left parenthesis straight V subscript 1 right parenthesis over denominator Volume space of space one space small space spherical space ball space left parenthesis straight V subscript 2 right parenthesis end fraction equals space fraction numerator begin display style 4 over 3 end style space straight x space begin display style 22 over 7 end style straight x space 3 space straight x space 3 space straight x space 3 over denominator begin display style 4 over 3 end style straight x space begin display style 22 over 7 end style straight x space 1 space straight x space 1 space straight x space 1 space end fraction

= 3 x 3 x 3 = 27

1170 Views

Water is flowing at the rate of 3 km/ hr through a circular pipe of 20 cm external diameter into a circular cistern of diameter 10m and depth 2 m. In how much time will the cistern be filled.

Let R cm be the radius and H cm be the height of a circular pipe, then

straight R space equals space 20 over 2 space equals space 10 space cm space equals space 1 over 10 space straight m

H (depth) = 3 km = 3000 m.
Now, Volume of water that flows out of the circular pipe in 1 hour = πR²H

$equals space open parentheses straight pi space straight x space 1 over 10 straight x 1 over 10 straight x 3000 close parentheses space straight m cubed equals space 30 straight pi space straight m cubed$
Let r m be the radius and h m be the depth (height) of a circular cistern, then

$straight r space equals space 10 over 2 equals 5 space straight m space and space straight h space equals space 2 space straight m$

Now, Volume of water that flows in the circular cistern
= π r²h
= π x 5 x 5 x 2 = 50 π m³.
Hence,
Time taken to fill up the cistern

$equals space space space space space space space space space space space space space Volume space of space water space that space flows space space space space space fraction numerator out space the space circular space pipe space in space 1 space hr over denominator Volume space of space water space that space flows space in space the space cistem end fraction equals fraction numerator 50 straight pi over denominator 30 straight pi end fraction equals fraction numerator 5 straight pi over denominator 3 straight pi end fraction equals space 5 over 2 equals 1 space hour space 40 space minutes$

1195 Views

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 6km/h, in how much time will the tank be filled?

Speed of the water
= 6 k.m./hr. = 6000 m/hr.
i.e. length of the water column (h) = 6000 m
And, internal radius of the pipe (r)

$equals space 10 space cm. space space space equals space 1 over 10$
$therefore$ Volume if water that flows in 1 hour = $πr squared straight h$

$equals space open parentheses straight pi space cross times space 1 over 10 cross times 1 over 10 cross times 6000 close parentheses straight m cubed equals 60 space straight pi space straight m cubed$
Now, radius of the base of the tank (R)

$equals 10 over 2 space equals space 5 space straight m$

And, Depth of the tank (h = 2m)
$equals space straight pi space straight R squared space straight x space straight H equals space straight pi space straight x space 5 space straight x space 5 space straight x space 2 space equals space 50 space πm cubed$
Hence, Required time to fill the tank by pipe

$fraction numerator Volume space of space the space tank over denominator Volume space of space water space that space flows space in space 1 space hour end fraction equals space space space fraction numerator 50 straight pi over denominator 60 straight pi end fraction equals 5 over 6 hr.$

401 Views

A spherical copper shell, of external diameter 18 cm, is melted and recast into a solid cone of base radius 14 cm and height $4 3 over 7 space cm.$ Find the inner diameter of the shell.

Let R be the external radius of the copper shell r be the internal radius, h be the height of the conical part and r₁ be its radius.
Now, External d of copper shell = 18 cm.