Question

jaha par 4 number likha h waha se question start ho raha h. If your answer is Wright I will rate it as 5 star. ​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the Concept of Volume of Cuboids has been used . We see that water that can be accumulated in the tank is equal to the volume of the tank . So firstly here, we need to find the volume of the tank . After finding that, we have tk convert its unit from m³ to L .

Let's do it !!

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★ Equations Used :-

$\\\;\boxed{\sf{\pink{Volume\;of\;Cuboid\;=\;\bf{Length\:\times\:Breadth\:\times\:Height}}}}$

$\\\;\boxed{\sf{\pink{1\;m^{3}\;=\;\bf{1000\;\;L}}}}$

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★ Solution :-

Given,

» Length of the cuboidal tank = 8 m

» Breadth of the cuboidal tank = 6 m

» Height of the cuboidal tank = 2.5 m

Firstly let's find the Volume of the tank.

This is given as,

$\\\;\sf{:\rightarrow\;\;Volume\;of\;Cuboid\;=\;\bf{Length\:\times\:Breadth\:\times\:Height}}$

By applying values, we get,

$\\\;\sf{:\Longrightarrow\;\;Volume\;of\;Cuboid_{(Tank)}\;=\;\bf{Length\:\times\:Breadth\:\times\:Height}}$

$\\\;\sf{:\Longrightarrow\;\;Volume\;of\;Cuboid_{(Tank)}\;=\;\bf{8\:\times\:6\:\times\:2.5}}$

$\\\;\bf{:\Longrightarrow\;\;Volume\;of\;Cuboid_{(Tank)}\;=\;\bf{\blue{120\;\;m^{3}}}}$

Now we know that,

$\\\;\sf{\rightarrow\;\;1\;m^{3}\;=\;\bf{1000\;\;L}}$

Then,

$\\\;\sf{\rightarrow\;\;120\;m^{3}\;=\;\bf{120\:\times\:1000\;\;L}}$

$\\\;\sf{\rightarrow\;\;120\;m^{3}\;=\;\bf{\red{120000\;\;L}}}$

$\\\;\underline{\boxed{\tt{Capacity\;\:of\;\:water\;\:tank\;\:can\:\;hold\;=\;\bf{\purple{120000\;\:L}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;Volume\;of\;Cube\;=\;(Side)^{3}}$

$\\\;\sf{\leadsto\;\;Volume\;of\;Cone\;=\;\dfrac{1}{3}\:\pi r^{2}h}$

$\\\;\sf{\leadsto\;\;Volume\;of\;Cylinder\;=\;\pi r^{2}h}$

$\\\;\sf{\leadsto\;\;Volume\;of\;Hemisphere\;=\;\dfrac{2}{3}\:\pi r^{3}}$

$\\\;\sf{\leadsto\;\;Volume\;of\;Sphere\;=\;\dfrac{4}{3}\:\pi r^{3}}$

$\\\;\sf{\leadsto\;\;Volume\;of\;Hollow\;Cylinder\;=\;\pi (R^{2}\:-\;r^{2})h}$

$\\\;\sf{\leadsto\;\;Volume\;of\;Hollow\;Cone\;=\;\dfrac{1}{3}\:\times\:\pi (R^{2}\:-\;r^{2})h}$

Answer 2

Step-by-step explanation:

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$\text{\Large\underline{\red{Question:-}}}$

$\Longrightarrow$ ● A cuboidal tank is 8m long, 6m wide and 2.5m deep. How many liters of water can it hold? [1m³ = 1000liters]

$\text{\Large\underline{\orange{To\ find:-}}}$

$\sf To\ find = \begin{cases} \sf{●\ We\ have\ to\ find\ the\ water\ holding\ capacity\ of\ the\ cuboidal\ tank\ in\ liters.} \end{cases}$

$\text{\Large\underline{\green{Given:-}}}$

$\sf Given = \begin{cases} \sf{●\ Length\ of\ the\ cuboidal\ tank\ =\ 8m.} \\ \\ \sf{●\ Breadth\ of\ the\ cuboidal\ tank\ =\ 6m.} \\ \\ \sf{●\ Height\ of\ the\ cuboidal\ tank\ =\ 2.5m.} \end{cases}$

$\text{\large\underline{\blue{Formula\ to\ be\ used:-}}}$

$\sf{●\ Volume\ of\ cuboid\ =\ Length\ ×\ Breadth\ Height\ ●}$

By putting the values, we get:-

$\leadsto$ $\sf{Volume_{(cuboidal\ tank)}\ =\ Length\ ×\ Breadth\ ×\ Height}$

$\leadsto$ $\sf{Volume_{(cuboidal\ tank)}\ =\ (8\ ×\ 6\ ×\ 2.5)\ cm³}$

$\leadsto$ $\sf{Volume_{(cuboidal\ tank)}\ =\ 120m³}$

$\leadsto$ $\sf{Volume_{(cuboidal\ tank)}\ =\ 120\ ×\ 1m³}$

$\leadsto$ $\sf{Volume_{(cuboidal\ tank)}\ =\ 120\ ×\ 1000\ liters\ .........(1m³\ =\ 1000\ liters)}$

$\leadsto$ $\sf\pink{Volume_{(cuboidal\ tank)}\ =\ 120,000\ liters.}$

Hence:-

● The cuboidal tank can hold 120,000 liters of water.

$\text{\Large\underline{\green{Explore\ more:-}}}$

Volume formulas of some geometric figures:-

● Cube:- = V = a³.

● Cylinder:- = V = πr²h.

● Prism:- = V = B × h.

● Sphere:- = V = (4/3) πr³.

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