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CBSE Class 10- Mathematics: Chapter β 13 Surface Areas and Volumes Part 1
Problems Based on Conversion of Solids
Question 1:
A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is and the height of the cone is . The solid is placed in a cylindrical tub, full of water, in such a way that the whole solid is submerged in water. If the radius of the cylindrical tub is and its height is find the volume of water left in the cylindrical tub.
Answer:
No. of solid of of hemisphere
On substituting we get,
of
On substituting we get,
Volume of left in the cylinder
Question 2:
A bucket of height and made up of copper sheet is in the form of frustum of right circular cone with radii of its lower and upper ends as and respectively. Calculate
i. the height of the cone of which the bucket is a part
ii. the volume of water which can be filled in the bucket
iii. the area of copper sheer required to make the bucket
Answer:
Let total height be h
β΄ ht. of cone which bucket is a part
Substitute to get Ans. : (ii) (iii)
Question 3:
A sphere and a cube have equal surface areas. Show that the ratio of the volume of the sphere to that of the cube is
Answer:
S. A. of sphere = S. A of cube
ratio of their volume
On simplifying & substituting, we get
Question 4:
A right triangle whose sides are is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.
Answer:
Apply pythagoras theorem to right to get
Vol of double
SA of double cone
Question 5:
Water in a canal wide and deep is flowing with a velocity of . How much area will it irrigate in minutes if of standing water is required for irrigation?
Answer:
Width of canal
Depth of canal
Velocity
Length of water column is formed in
Let of area be irrigated