An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression `c=2pir,` where r is the radius of the circle.

Uniform cicular motion is used in physics to describe the motion of an object traveling at a constant speed in a circle. The speed of the object is called tangential velocity and it can be calculated using the formula abovce, where r is the radius of the circle and T is the time is takes for the object to make one complete circle, called a peroid. WHich of the following formulas could be used to find the length of one period if you know the tengential velocity and the radius of the circle ?

The fact tht a changing magnetic flux produces an electric field is basic to the operation of many high energy particle accelerators. Since the principle was first successfully applied to the acceleration of electrons (or beta particles) in a device called the betatron, this method of acceleration is often given that name. The general idea involved is shown in Fig. An electromagnet is used to produce a changing flux through a circular loop defined by the doughnut shaped vacuum chamber. We see that there will be an electric field E along the circular length of the doughnut, i.e. circling the magnet poles, given by 2piaE = d(phi)//dt , where 'a' is the radius of the doughnut. Any charged particle inside the vacuum chamber will experience a force qE and will accelerate. Ordinarily, the charged particle would shoot out the vacuum chamber and becomes lost. However, if the magnetic field at the position of the doughnut is just proper to satisfy the relation, Centripetal force = magnetic force or mv^(2)//a = qvB then the charge will travel in a circle within the doughnut. By proper shaping of the magnet pole piece, this relation can be satisfied. As a result, the charge will move at high speed along the loop within the doughnut. Each time it goes around the loop, it has, in effect, fallen through a potential difference equal to the induced emf, namely epsilon = (d(phi)//dt) . Its energy after 'n' trips around the loop will be q(n(epsilon)) . Variable magnetic flux

The fact tht a changing magnetic flux produces an electric field is basic to the operation of many high energy particle accelerators. Since the principle was first successfully applied to the acceleration of electrons (or beta particles) in a device called the betatron, this method of acceleration is often given that name. The general idea involved is shown in Fig. An electromagnet is used to produce a changing flux through a circular loop defined by the doughnut shaped vacuum chamber. We see that there will be an electric field E along the circular length of the doughnut, i.e. circling the magnet poles, given by 2piaE = d(phi)//dt , where 'a' is the radius of the doughnut. Any charged particle inside the vacuum chamber will experience a force qE and will accelerate. Ordinarily, the charged particle would shoot out the vacuum chamber and becomes lost. However, if the magnetic field at the position of the doughnut is just proper to satisfy the relation, Centripetal force = magnetic force or mv^(2)//a = qvB then the charge will travel in a circle within the doughnut. By proper shaping of the magnet pole piece, this relation can be satisfied. As a result, the charge will move at high speed along the loop within the doughnut. Each time it goes around the loop, it has, in effect, fallen through a potential difference equal to the induced emf, namely epsilon = (d(phi)//dt) . Its energy after 'n' trips around the loop will be q(n(epsilon)) . Working of betatron is not based upon which of the following theories?

The fact tht a changing magnetic flux produces an electric field is basic to the operation of many high energy particle accelerators. Since the principle was first successfully applied to the acceleration of electrons (or beta particles) in a device called the betatron, this method of acceleration is often given that name. The general idea involved is shown in Fig. An electromagnet is used to produce a changing flux through a circular loop defined by the doughnut shaped vacuum chamber. We see that there will be an electric field E along the circular length of the doughnut, i.e. circling the magnet poles, given by 2piaE = d(phi)//dt , where 'a' is the radius of the doughnut. Any charged particle inside the vacuum chamber will experience a force qE and will accelerate. Ordinarily, the charged particle would shoot out the vacuum chamber and becomes lost. However, if the magnetic field at the position of the doughnut is just proper to satisfy the relation, Centripetal force = magnetic force or mv^(2)//a = qvB then the charge will travel in a circle within the doughnut. By proper shaping of the magnet pole piece, this relation can be satisfied. As a result, the charge will move at high speed along the loop within the doughnut. Each time it goes around the loop, it has, in effect, fallen through a potential difference equal to the induced emf, namely epsilon = (d(phi)//dt) . Its energy after 'n' trips around the loop will be q(n(epsilon)) . Magnetic field which keeps the particles in circular path must

Let C_1, C_2, ,C_n be a sequence of concentric circle. The nth circle has the radius n and it has n openings. A points P starts travelling on the smallest circle C_1 and leaves it at an opening along the normal at the point of opening to reach the next circle C_2 . Then it moves on the second circle C_2 and leaves it likewise to reach the third circle C_3 and so on. Find the total number of different path in which the point can come out of nth circle.

Consider a vertical tube open at both ends. The tube consistss of two parts, each of different cross sections and each part having a piston which can move smoothly in respective tubes. The two piston which can move smoothly in respective tube wire. The piston are joined together by an inextensible wire. The combined mass of the two piston is 5 kg and area of cross section of the upper piston is 10 cm^(2) greater than that of the lower piston. Amount of gas enclosed by the pistons is 1 mol . When the gas is heated slowly, pistons move by 50 cm . Find the rise in the temperature of the gas in the form X//R K , where R is universal gas constant. Use g = 10 m//s^(2) and outside pressure = 10^(5) N//m^(2)) .

Consider a vertical tube open at both ends. The tube consistss of two parts, each of different cross sections and each part having a piston which can move smoothly in respective tubes. The two piston which can move smoothly in respective tube wire. The piston are joined together by an inextensible wire. The combined mass of the two piston is 5 kg and area of cross section of the upper piston is 10 cm^(2) greater than that of the lower piston. Amount of gas enclosed by the pistons is 1 mol . When the gas is heated slowly, pistons move by 50 cm . Find the rise in the temperature of the gas in the form X//R K , where R is universal gas constant. Use g = 10 m//s^(2) and outside pressure = 10^(5) N//m^(2)) .

Weight of a body depends directly upon acceleration due to gravity g . Value of g depends upon many factors. It depends upon the shape of earth, rotation earth etc. Weight of a body at a pole is more then that at a place on equator because g is maximum at poles and minimum on equator. Acceleration due to gravity g varies with latitude lambda as per relation given below : g_(rot)=g-Romega^(2)cos^(2)lambda where R is radius of earth and omega is angular velocity of earth. A body of mass m weighs W_(r ) in a train at rest. The train then begins to run with a velocity v around the equator from west to east. It observed that weight W_(m) of the same body in the moving train is different from W_(r ) . Let v_(e ) be the velocity of a point on equator with respect to axis of rotation of earth and R be the radius of the earth. Clearly the relative between earth and trainwill affect the weight of the body. Difference between Weight W_(r ) and the gravitational attraction on the body can be given as

Weight of a body depends directly upon acceleration due to gravity g . Value of g depends upon many factors. It depends upon the shape of earth, rotation earth etc. Weight of a body at a pole is more then that at a place on equator because g is maximum at poles and minimum on equator. Acceleration due to gravity g varies with latitude lambda as per relation given below : g_(rot)=g-Romega^(2)cos^(2)lambda where R is radius of earth and omega is angular velocity of earth. A body of mass m weighs W_(r ) in a train at rest. The train then begins to run with a velocity v around the equator from west to east. It observed that weight W_(m) of the same body in the moving train is different from W_(r ) . Let v_(e ) be the velocity of a point on equator with respect to axis of rotation of earth and R be the radius of the earth. Clearly the relative between earth and trainwill affect the weight of the body. Weight W_(m) of the body can be given as

Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} If buoyant force were also taken into account then value of terminal speed would have