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2. Heights (in cm) of 25 children are given below:
168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 160,
165, 163, 162, 163, 164, 163, 160, 165, 163, 162
What is the mode of their heights? What do we understand by mode here?

Heights Tally bars Number 160    4 161    1 162    4 163   9 164    3 165    3 168    1 Clearly, 163 occurs 9 times. Therefore,   is the mode of the data.

7. Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Similar to how we find the surface area of the frustum. The volume of the frustum is given by-    =Volume of the bigger cone   -   Volume of the smaller cone                                                    

6. Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

In the case of the frustum, we can consider:- removing a smaller cone (upper part) from a larger cone. So the CSA of frustum becomes:-    CSA of bigger cone   -   CSA of the smaller cone                                                   And the total surface area of the frustum is  =  CSA of frustum +  Area of upper circle and area of lower circle.                                              ...

5.  An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see Fig.).

                                                  

From this, we can write the values of both the radius (upper and lower) and height of frustum. Thus slant height of frustum is :                                                                                                                                                            Now, the area of the tin shed required :    =  Area of frustum  +  Area of the cylinder        

4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Firstly we will calculate the volume of rainfall :  Volume of rainfall   :     And the volume of the three rivers is :     It can be seen that both volumes are approximately equal to each other.  

3. A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?

The total volume of the cistern is :       And the volume to be filled in it is  Now let the number of bricks be n. Then the volume of bricks    :    Further, it is given that brick absorbs one-seventeenth of its own volume of water. Thus water absorbed  :     Hence we write :        Thus the total number of bricks is 1792.

2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of \pi as found appropriate.)

The volume of the double cone will be   =    Volume of cone 1  +  Volume of cone 2.                                                                                                           (Note that sum of heights of both the cone is 5 cm  -  hypotenuse).                                                  Now the surface area of a double cone is :                                              ...

1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.

A number of rounds are calculated by :                                                                                                                      Thus the length of wire in 40 rounds will be    And the volume of wire is: Area of cross-section    Length of wire                                                                                Hence the mass of wire is.                  ...

5.  A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \frac{1}{16}\textup{cm}, find the length of the wire.

The figure for the problem is shown below :                                                     Using geometry we can write :                                                       and             Thus the volume of the frustum is given by :                                                                                             Now, the radius of the wire is :                               ...

4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm 2 . (Take \pi = 3.14)

Firstly we will calculate the slant height of the cone :       Now, the volume of the frustum is :               =   Capacity of the container. Now, the cost of 1-litre milk is  Rs. 20. Then the cost of 10.449-litre milk will be    The metal sheet required for the container is  :                                                                                                           Thus...

3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig.). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

                                                                   

The area of material used is given by :  Area of material   =  Curved surface area of a frustum of cone   +   Area of upper end                                                                                                             

2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

We are given the perimeter of upper and lower ends thus we can find r1 and r2.     And,              Thus curved surface area of the frustum is given by  :                                                                                                                                                                         

1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

The capacity of glass is the same as the volume of glass. Thus the volume of glass  :                 =    Capacity of glass

9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Area of the cross-section of pipe is                                                    Speed of water is given to be   =  3 km/hr Thus, the volume of water flowing through a pipe in 1 min. is              Now let us assume that tank will be completely filled after t minutes. Then we write :     Hence time required for filling of the tank completely is 100 minutes.

8.  Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Speed of water is: 10 Km/hr  And the volume of water flow in 1 minute is   :     Thus the volume of water flow in 30 minutes will be :    Let us assume irrigated area be A. Now we can equation the expression of volumes as the volume will remain the same.                                                                                               Thus the irrigated area is .               ...

7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

According to question volume will remain constant thus we can write : The volume of bucket    =    Volume of heap formed.                          Let the radius of heap be r.     And thus the slant height will be                      Hence the radius of heap made is 36 cm and its slant height is  .

6.  How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Let us assume the number of coins that need to be melted be n. Then we can write :              The volume of n coins   =   Volume of cuboid formed.     Thus the required number of coins is 400.

5. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Let the number of cones that can be filled with ice cream be n. Then we can write :              The volume of a cylinder containing ice cream  =  n ( volume of 1 ice cream cone )                                                  Hence the number of cones that can be filled is 10.

4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

According to the question, the volume is conserved here :  The volume of soil dug out   =  Volume of the embankment made. Let the height of the embankment is h.   Hence the height of the embankment made is 1.125 m.

3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

According to the question, the volume of soil dug will be equal to the volume of the platform created. Thus we can write :   The volume of soil dug = Volume of platform     Thus the height of the platform created is 2.5 m.
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