Here r = 1 cm and h = 1 cm.
The diagram is as follows.
Now, Volume of solid = Volume of conical part + Volume of hemispherical part
We know the volume of cone = $\dfrac{1}{3}πr^{2}h$
And,
The volume of the hemisphere = $\dfrac{2}{3}πr^{3}$
So, the volume of the solid will be
=$\dfrac{1}{3}π(1)^{2}[1+2(1)]cm^{3}$
=$\dfrac{1}{3}π1 \times [3]cm^{3}$
= π cm$^{3}$
Given,
Height of cylinder = 12–4 = 8 cm
Radius = 1.5 cm
Height of cone = 2 cm
Now, the total volume of the air contained will be = Volume of cylinder+2×(Volume of the cone)
∴ Total volume = $πr^{2}h+[2×(\dfrac{1}{3} πr^{2}h$ )]
= 18 π+2(1.5 π)
= $66 cm^{3}$.
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It is known that the gulab jamuns are similar to a cylinder with two hemispherical ends.
So, the total height of a gulab jamun = 5 cm.
Diameter = 2.8 cm
So, radius = 1.4 cm
∴ The height of the cylindrical part = 5 cm–(1.4+1.4) cm
=2.2 cm
Now, the total volume of one gulab jamun = Volume of cylinder + Volume of two hemispheres
= $πr^{2}h+(\dfrac{4}{3})πr^{3}$
= $4.312π+(\dfrac{10.976}{3}) π$
= $25.05 cm^{3}$
We know that the volume of sugar syrup = 30% of the total volume
So, the volume of sugar syrup in 45 gulab jamuns = 45×30%(25.05 $cm^{3})$
= 45×7.515 = 338.184 $cm^{3}$
The volume of the cuboid = length x width x height
We know the cuboid’s dimensions as 15 cmx10 cmx3.5 cm
So, the volume of the cuboid = 15x10x3.5 = 525 $cm^{3}$
Here, depressions are like cones, and we know,
Volume of cone = $(\dfrac{1}{3})πr^{2}h$
Given, radius (r) = 0.5 cm and depth (h) = 1.4 cm
∴ Volume of 4 cones = 4x$(\dfrac{1}{3})πr^{2}h$
= 1.46 $cm^{2}$
Now, the volume of wood = Volume of the cuboid – 4 x volume of the cone
= 525-1.46 = 523.54 $cm^{2}$
For the cone,
Radius = 5 cm,
Height = 8 cm
Also,
Radius of sphere = 0.5 cm
The diagram will be like
It is known that,
The volume of cone = volume of water in the cone
= $\dfrac{1}{3}πr^{2}h$ = $(\dfrac{200}{3})π cm^{3}$
Now,
Total volume of water overflown= $(\dfrac{1}{4})×(\dfrac{200}{3}) π $
=($\dfrac{50}{3}$)π
The volume of lead shot
= ($\dfrac{4}{3})πr^{3}$
=$ (\dfrac{1}{6})$ π
Now,
The number of lead shots = $\dfrac{Total\:\: volume\:\: of\:\: water \:\: overflown}{Volume \:\: of\:\: lead \:\: shot}$
= $\dfrac{(\dfrac{50}{3})π}{(\dfrac{1}{6})π}$
= $(\dfrac{50}{3})×6$ = 100
Given the height of the big cylinder (H) = 220 cm
The radius of the base (R) = $\dfrac{24}{2}$ = 12 cm
So, the volume of the big cylinder =$ πR^{2}H$
= $π(12)^{2} × 220 cm^{3}$
= 99565.8 $cm^{3}$
Now, the height of the smaller cylinder (h) = 60 cm
The radius of the base (r) = 8 cm
So, the volume of the smaller cylinder = $πr^{2}h$
= $π(8)^{2}×60 cm^{3}$
= 12068.5 $cm^{3}$
∴ The volume of iron = Volume of the big cylinder + Volume of the small cylinder
= 99565.8 + 12068.5
=111634.5 $cm^{3}$
We know,
Mass = Density x volume
So, the mass of the pole = 8×111634.5
= 893 Kg (approx.)
Here, the volume of water left will be = Volume of the cylinder – Volume of solid
Given,
Radius of cone = 60 cm,
Height of cone = 120 cm
Radius of cylinder = 60 cm
Height of cylinder = 180 cm
Radius of hemisphere = 60 cm
Now,
The total volume of solid = Volume of Cone + Volume of the hemisphere
Volume of cone =$ \dfrac{1}{3}πr^{2}h$
= $\dfrac{1}{3} × π×602×120cm^{3}$
= $144×103π cm^{3}$
Volume of hemisphere = $(\dfrac{2}{3})×π×603 cm^{3} $
= $144×103 π cm^{3}$
So, total volume of solid = $144×103π cm^{3} + 144×103π cm^{3}$
= $288 ×103π cm^{3}$
Volume of cylinder = π×602×180 = 648000 = 648×103 π $cm^{3}$
Now, the volume of water left will be = Volume of the cylinder – Volume of solid
= (648-288) × 103×π = 1.131 $m^{3}$
Given,
For the cylinder part, Height (h) = 8 cm and Radius (R) = $(\dfrac{2}{2})$ cm = 1 cm
For the spherical part, Radius (r) = $(\dfrac{8.5}{2})$ = 4.25 cm
Now, volume of this vessel = Volume of cylinder + Volume of sphere
= π×(1)2×8+$(\dfrac{4}{3})π(4.25)^{3}$
= 346.51 $cm^{3}$
Hence, the child’s calculation is not correct.