Question

The diameter of a cylindrical pot is 7 decimetre and height is 4 decimetre. The pot contains 108 litres of milk, How many litres of milk have to pour So that the jar will be filled up. (1 litre = 1 c. decimetre)​

Answer 1

Given :-

• Diameter of cylindrical pot = 7 dm

• Height of cylindrical pot = 4 dm

• Amount of milk in cylindrical pot = 108 litres.

To Find :-

• Amount of milk to poured in pot to filled it.

Formula Used : -

$\purple{\boxed{ \bf \:Volume_{(Cylinder)} = \pi \: {r}^{2}}}$

$\purple{\boxed{ \bf \:1 \: litre \: = \: 1000 \: {cm}^{3} = 1 \: {dm}^{3} }}$

Solution :-

Given that,

• Diameter of cylindrical pot, d = 7 dm

We know,

$\purple{\boxed{ \bf \:radius \: = \: \dfrac{1}{2} \: diameter}}$

It implies,

• Radius of cylindrical pot, r = 3.5 dm

Also,

• Height of cylindrical pot, h = 4 dm

So,

Volume of cylindrical pot is

$\rm :\longmapsto\:Volume_{(Cylinder)} = \pi \: {r}^{2}h$

$\rm :\longmapsto\:Volume_{(Cylinder)} = \dfrac{22}{7} \times 3.5 \times 3.5 \times 4$

$\rm :\implies\:Volume_{(Cylinder)} = 154 \: {dm}^{3}$

$\bf\implies \:Volume_{(Cylinder)} = 154 \: litres$

It means,

• Capacity of cylindrical pot = 154 litres

Now,

• Amount of milk in cylindrical pot = 108 litres

Thus,

Amount of milk to poured to filled the cylindrical pot is

$\rm :\longmapsto\:Amount_{(milk \: to \: poured)} \: = 154 - 108$

$\pink{\bf :\longmapsto\:Amount_{(milk \: to \: poured)} \: = 46 \: litres}$

Additional Information :-

$\boxed{ \sf{ \: Volume_{(cuboid)} = lbh}}$

$\boxed{ \sf{ \: Volume_{(cube)} = {(edge)}^{3} }}$

$\boxed{ \sf{ \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2} h}}$

$\boxed{ \sf{ \: Volume_{(sphere)} = \dfrac{4}{3} \pi \: {r}^{3}}}$

$\boxed{ \sf{ \: Volume_{(hemi - sphere)} = \dfrac{2}{3} \pi \: {r}^{3}}}$

$\boxed{ \sf{ \: CSA{(cylinder)} = 2\pi \: rh}}$

$\boxed{ \sf{ \: CSA{(cone)} = \pi \: rl}}$

$\boxed{ \sf{ \: CSA{(cube)} = 4 \times {(edge)}^{2} }}$

$\boxed{ \sf{ \: CSA{(hemi - sphere)} = 2\pi {r}^{2} }}$

$\boxed{ \sf{ \: TSA{(cuboid)} = 2(lb + bh + hl)}}$

$\boxed{ \sf{ \: TSA{(cone)} = {6(edge)}^{2} }}$

$\boxed{ \sf{ \: TSA{(cone)} = \pi \: r(l + r)}}$

$\boxed{ \sf{ \: TSA{(cylinder)} = 2\pi \: r(h + r)}}$