Question

find the volume of wood Rekha required to make a top open box of dimension 15 centimetre by 12 CM by its length and thickness of what is meant centimetre

Answer 1

External volune =L×B×H=36×25×16.5=14850 cm

3

Thickness of iron =1.5 cm

Internal l=36−2(1.5)=33 cm

b=25−2(1.5)=22 cm

h=16.5−2(1.5)=13.5 cm

Internal =l×b×h

Volume =33×22×13.5

Volume of iron used =(external -internal) volume

=(14850−9801)cm

3

5049 cm

3

Weight of 1 cm

3

of iron =7.5 gm

Then, weight of the box =5049×7.5

37867.5 gms

37.86 kg.

I hope this helps

Answer 2

Step-by-step explanation:

$Given, Radius of cylindrical container =6cm</p><p></p><p>Given, Radius of cylindrical container =6cmHeight of cylindrical container =15cm</p><p></p><p>Given, Radius of cylindrical container =6cmHeight of cylindrical container =15cmVolume of cylinder ⤵️</p><p></p><p>\pi {r}^{2}hπr2h</p><p></p><p>\pi \times 36 \times 15π×36×15</p><p></p><p>540\pi {cm}^{3}540πcm3</p><p></p><p>Now,as it is divided among 10 children</p><p></p><p>Therefore;Dividing value by 10 ⬇️⬇️⬇️⬇️⬇️</p><p></p><p>\frac{540}{10} = 54 \pi {cm}^{3}10540=54πcm3</p><p></p><p>Volume of cone+Volume of hemispherical top = Volume of ice cream in it .</p><p></p><p>\frac{1}{3}\pi {r}^{2}h + \frac{2}{3} \pi {r}^{3} = \pi {r}^{2}h31πr2h+32πr3=πr2h</p><p></p><p>\frac{1}{3}\pi {r}^{2}(4r) + \frac{2}{3}\pi {r}^{3} = 54\pi31πr2(4r)+32πr3=54π</p><p></p><p>\frac{1}{3}\pi {r}^{3}(4 + 2) = 54\pi31πr3(4+2)=54π</p><p></p><p>2 \pi = 54 = r = 32π=54=r=3</p><p></p><p>Hence,The radius of ice cream c \\ ne = 3cm</p><p></p><p>$