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14.22 Show that for a particle in linear SHM the average kinetic energy over a period of
oscillation equals the average potential energy over the same period.

Let the equation of oscillation be given by  Velocity would be given as  Kinetic energy at an instant is given by Time Period  is given by The Average Kinetic Energy would be given as follows The potential energy at an instant T is given by  The Average Potential Energy would be given by We can see Kav = Uav

Q 4. 22  \hat i and \hat j are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors \hat i + \hat j, and \hat i - \hat j−? What are the components of a vectorA=2\hat i +3 \hat j along the directions of \hat i + \hat j and \hat i - \hat j? [You may use graphical method]

Let A be a vector such that:-                  Then the magnitude of vector A is given by   :             Now let us assume that the angle made between vector A and x-axis is . Then we have:-                                                                                   Similarly, let B be a vector such that:-        The magnitude of vector B is     :                        Let  be the angle...

Q : 2     The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig \small 13.32. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius \small 1.5\hspace{1mm}cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per \small cm^2 and black paint costs 5 paise per \small cm^2.
 

            

Given, The radius of the wooden spheres =   The surface area of a single sphere =  Again, the Radius of the cylinder support =  Height of the support =   The base area of the cylinder =  Now, Cost of painting silver =   Cost of painting 1 wooden sphere = Cost of painting silver =  Now, Curved surface area of the cylindrical support =  Now, Cost of painting  black =   Cost of painting 1 such...

Q : 3    The diameter of a sphere is decreased by \small 25\%. By what per cent does its curved surface area decrease?
 

Let the radius of the sphere be  Diameter of the sphere =  According to question, Diameter is decreased by  So, the new diameter =  So, the new radius =  New surface area =  Decrease in surface area =   Percentage decrease in the surface area = 

Q : 1    A wooden bookshelf has external dimensions as follows: Height \small =110\hspace{1mm}cm, Depth \small =25\hspace{1mm}cm, Breadth \small =85\hspace{1mm}cm (see Fig. \small 13.31). The thickness of the plank is \small 5\hspace{1mm}cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is \small 20\hspace{1mm} paise per \small cm^2 and the rate of painting is \small 10\hspace{1mm} paise per \small cm^2, find the total expenses required for polishing and painting the surface of the bookshelf.

            

External dimension od the book shelf = (Note: There is no front face) The external surface area of the shelf = We know, each stripe on the front surface is also to be polished. which is 5 cm stretch. Area of front face = Area to be polished = Cost of polishing  area = Cost of polishing  area =  Now, Dimension of inner part =  Area to be painted in 3 rows = Cost of painting  area =...

Q : 10    A capsule of medicine is in the shape of a sphere of diameter \small 3.5\hspace{1mm}mm. How much medicine (in \small mm^3) is needed to fill this capsule?
 

Given, The radius of the spherical capsule = The volume of the capsule =  Therefore,  of medicine is needed to fill the capsule.

Q : 9    Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area \small S'. Find the

            (ii) ratio of S and \small S'.

Given, The radius of a small sphere =  The surface area of a small sphere =  The radius of the bigger sphere =  The surface area of the bigger sphere =  And,  We know, the surface area of a sphere =  The ratio of their surface areas =  Therefore, the required ratio is     

Q : 9    Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area \small S'. Find the

            (i) radius \small r' of the new sphere

Given, The radius of a small sphere =  The radius of the bigger sphere =  The volume of each small sphere=  And, Volume of the big sphere of radius =  According to question,

Q : 8    A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of \small Rs\hspace{1mm}4989.60. If the cost of white-washing is Rs 20 per square metre, find the

(ii) volume of the air inside the dome. 

Let the radius of the hemisphere be  Inside the surface area of the dome =  We know, Surface area of a hemisphere =    The volume of the hemisphere = 

Q : 8    A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of \small Rs\hspace{1mm}4989.60. If the cost of white-washing is Rs 20 per square metre, find the

            (i) inside surface area of the dome

Given,  is the cost of white-washing  of the inside area  is the cost of white-washing  of inside area (i) Therefore, the surface area of the inside of the dome is       

Q : 7     Find the volume of a sphere whose surface area is \small 154\hspace{1mm}cm^2.

Given, The surface area of the sphere =  We know, Surface area of a sphere =    The volume of the sphere = 

Q : 6    A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. 

Given, Inner radius of the hemispherical tank =  Thickness of the tank =   Outer radius = Internal radius + thickness =  We know, Volume of a hemisphere =   Volume of the iron used = Outer volume - Inner volume 

Q : 5    How many litres of milk can a hemispherical bowl of diameter \small 10.5\hspace{1mm}cm hold?
 

The radius of the hemispherical bowl =  We know, Volume of a hemisphere =  The volume of the given hemispherical bowl =  The capacity of the hemispherical bowl =     

Q : 4    The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
 

Given, Let  be the diameters of Earth  The diameter of the Moon =  We know, Volume of a sphere =    The ratio of the volumes =  Therefore, the required ratio of the volume of the moon to the  volume of the earth is 

Q : 3    The diameter of a metallic ball is \small 4.2\hspace{1mm}cm. What is the mass of the ball, if the density of the metal is \small 8.9\hspace{1mm}g per \small cm^3?
 

Given, The radius of the metallic sphere =  We know, Volume of a sphere =   The required volume of the sphere =  Now, the density of the metal is  per ,which means, Mass of  of the metallic sphere =  Mass of  of the metallic sphere = 

Q : 2    Find the amount of water displaced by a solid spherical ball of diameter

            (ii)     \small 0.21\hspace{1mm}m

The solid spherical ball will displace water equal to its volume.  Given, The radius of the sphere =  We know, Volume of a sphere =   The required volume of the sphere =  Therefore, amount of water displaced will be 

Q : 2    Find the amount of water displaced by a solid spherical ball of diameter

            (i) 28 cm 

The solid spherical ball will displace water equal to its volume.  Given, The radius of the sphere =  We know, Volume of a sphere =   The required volume of the sphere =  Therefore, the amount of water displaced will be 

Q : 1    Find the volume of a sphere whose radius is

           (ii) \small 0.63\hspace{1mm}m

Given, The radius of the sphere =  We know, Volume of a sphere =  The required volume of the sphere = 

Q : 1    Find the volume of a sphere whose radius is

            (i) 7 cm

Given, The radius of the sphere =  We know, Volume of a sphere =  The required volume of the sphere = 

Q : 9    A heap of wheat is in the form of a cone whose diameter is \small 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
 

Given, Height of the conical heap =   Base radius of the cone =  We know,  The volume of a cone =  The required volume of the cone formed =  Now, The slant height of the cone =  We know, the curved surface area of a cone =  The required area of the canvas to cover the heap  =   
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