102
As a part of school project, Mishika and Sahaj
created a bird-bath from the cylindrical log of
wood by scooping out the hemispherical depression
from one end of the cylinder as shown in the figure
given. Cylinder has a length 2 m out of which 0.6
m is in earth and the diameter is 1.4 m.
On the basis of the above information, answer
the following questions :
(i) Write the radius of the hemispherical
depression.
(ii) Find the volume of water that can be filled in
the hemispherical depression in terms of $\pi$.
(iii) (a) Find the total surface area of log of wood
above the ground after making the bird-bath.
OR
(iii) (b) Compute the volume of log of wood above the ground after
making the bird-bath.
created a bird-bath from the cylindrical log of
wood by scooping out the hemispherical depression
from one end of the cylinder as shown in the figure
given. Cylinder has a length 2 m out of which 0.6
m is in earth and the diameter is 1.4 m.
On the basis of the above information, answer
the following questions :
(i) Write the radius of the hemispherical
depression.
(ii) Find the volume of water that can be filled in
the hemispherical depression in terms of $\pi$.
(iii) (a) Find the total surface area of log of wood
above the ground after making the bird-bath.
OR
(iii) (b) Compute the volume of log of wood above the ground after
making the bird-bath.
Show SolutionHide Solution↓
Ans.
(i) Radius, $r = 0.7$ m (1 Mark)
(ii) Volume of water in hemispherical depression = $\frac{2}{3} \times \pi \times (0.7)^3$ (1/2 Mark)
$= 0.228 \pi$ m$^3$ approx. or $\frac{343}{1500} \pi$ m$^3$ (1/2 Mark)
(iii) (a) Required area = $2 \times \pi \times (0.7) \times (1.4) + 2 \times \pi \times (0.7) \times (0.7)$ (1
frac{1}{2} Mark)
$= 2.94 \pi$ m$^2$ or 9.24 m$^2$ or $\frac{231}{25}$ m$^2$ (1/2 Mark)
OR
(iii) (b) Volume of log of wood = $\pi \times (0.7) \times (0.7) \times (1.4) - \frac{2}{3} \times \pi \times (0.7)^3$ (1
frac{1}{2} Mark)
$= 0.457 \pi$ m$^3$ or 1.437 m$^3$ or $\frac{539}{375}$ m$^3$ (1/2 Mark)
(i) Radius, $r = 0.7$ m (1 Mark)
(ii) Volume of water in hemispherical depression = $\frac{2}{3} \times \pi \times (0.7)^3$ (1/2 Mark)
$= 0.228 \pi$ m$^3$ approx. or $\frac{343}{1500} \pi$ m$^3$ (1/2 Mark)
(iii) (a) Required area = $2 \times \pi \times (0.7) \times (1.4) + 2 \times \pi \times (0.7) \times (0.7)$ (1
frac{1}{2} Mark)
$= 2.94 \pi$ m$^2$ or 9.24 m$^2$ or $\frac{231}{25}$ m$^2$ (1/2 Mark)
OR
(iii) (b) Volume of log of wood = $\pi \times (0.7) \times (0.7) \times (1.4) - \frac{2}{3} \times \pi \times (0.7)^3$ (1
frac{1}{2} Mark)
$= 0.457 \pi$ m$^3$ or 1.437 m$^3$ or $\frac{539}{375}$ m$^3$ (1/2 Mark)