[" 19.A well of diameter "150cm" has a stone parapet around it.If the length of the outer edge of the "],[" parapet is "660cm" ,then find the width of the parapet."],[" 20.An ox in a kolhu (an oil processing apparatus) is tethered to a rope "3m" long.How much "],[" distance does it cover in "14" rounds? "]

A well of diameter 150 cm has a stone parapet around it. If the length of the outer edge of the parapet is 660 cm, then find the width of the parapet.

An ox in a kolhu (an oll processing apparatus) is tethered to a rope 3m long.How much distance does it cover in 14 rounds?

A well of diameter 140 cm has a stone parapet around it. If the length of the outer edge of the parapet is 616 cm, find the width of the parapet.

A car runs 20 times around a rectangular track which is 200-m long and 150-m wide. How much distance does the car cover ?

If the outer and inner diameters of a stone parpapet around a well are 112 cm and 70 cm respectively. Then what is the area of the parapet ?

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

In a rectangular park, there is an elliptical flower bed in the middle and the border. The side lengths of the park are 30m and 20m. Major axis & minor axis are of length 14m and 10m respectively. Around the flower bed in the middle there is a walking path 1.4 m wide and in the middle of each side there is entrance of 2m width to the park upto the walking path. At the border the width of flower bed is Im. Rest of the park is grass area. If the entire walking path is covered by bricks of dimension 10 x 20 cm2, find the approximate number of bricks required. If cost of reading the grass and plating the flowers is Rs.10 and Rs:.50 per square metre and cost of a brick is Rs.2. Find total cost of renovation

Consider a wave sent down a string in the positive direction whose equation is given is y=y_(0)sin[omega(t-x/v)] The wave is propagated along because each string segment pulls upward and downward on the segment adjacent to it a slightly larger value of x and, as a result does work upon the string segment to which wave is travelling. For example, the portion of string at point A is going upward, and will pull the portion at point B upward as well. In fact, at any point along the string, each segment of the string is pulling on the segment just adjacent and to its right, causing the wagve to propagate. It is by this process that the energy is sent along the string. Now we try to calculate how much energy is propagated down the string per second T_(y)=tsintheta~~TtanthetaimpliesT_(y)=-T(dely)/(delx) (The negative sign appears because as shown in the figure II, the slope is negative) the force will act through a distance dy=v_(y)dt=(dely)/(delt) dt Therefore work done by force in time dt is dW=T_(y)dy=-T((dely)/(delx))((dely)/(delt))dtimpliesdW=(omega^(2)y_(0)^(2)T)/vcos^(2)[omega(t-x/V)]dt .....(A) How much average power is transmitted dawn a string having density mu=5xx10^(-4)kg//m and T=5N by a 200Hz vibration of amplitude of 0.20cm

Consider a wave sent down a string in the positive direction whose equation is given is y=y_(0)sin[omega(t-x/v)] The wave is propagated along because each string segment pulls upward and downward on the segment adjacent to it a slightly larger value of x and, as a result does work upon the string segment to which wave is travelling. For example, the portion of string at point A is going upward, and will pull the portion at point B upward as well. In fact, at any point along the string, each segment of the string is pulling on the segment just adjacent and to its right, causing the wagve to propagate. It is by this process that the energy is sent along the string. Now we try to calculate how much energy is propagated down the string per second T_(y)=tsintheta~~TtanthetaimpliesT_(y)=-T(dely)/(delx) (The negative sign appears because as shown in the figure II, the slope is negative) the force will act through a distance dy=v_(y)dt=(dely)/(delt) dt Therefore work done by force in time dt is dW=T_(y)dy=-T((dely)/(delx))((dely)/(delt))dtimpliesdW=(omega^(2)y_(0)^(2)T)/vcos^(2)[omega(t-x/V)]dt .....(A) How much average power is transmitted dawn a string having density mu=5xx10^(-4)kg//m and T=5N by a 200Hz vibration of amplitude of 0.20cm

A room has length 4m, breadth 3m and height 3m. How much paper of width 25cm will be required to cover the four walls of the room