During the medical check-up of 35 students of a class, their weights were recorded as follows: Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula. weight (in kg)=38-40 40-42 42-44 44-46 46-48 48-50 50-52 No. of students=3 2 4 5 14 4 3

The distribution below gives the weight of 30 students in a class. Find the median weight of students: Weight (in kg): 40-45 45-50 50-55 55-60 60-65 65-70 70-75 No. of students: 2 3 8 6 6 3 2

The marks of 10 students of a class in an examination arranged in ascending order are as follows : 13, 35, 43, 46, x, x + 4, 55, 61, 71, 80 If the median marks is 48, find the value of x. Hence find the mode of the given data.

For the following data, draw a histogram and a frequency polygon: Marks: 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 900-100 No. of Students 5 10 4 6 7 3 2 2 3 9

The weight of 12 students (in kg) are : 40,61,54,50,59,37,51,41,48,62,46 and 34. Find the median weight. If the weight of 62 kg is replaced by 35 kg, find the new median weight.

The number of students absent in a class were recorded every day for 120 days and the information is given in the following frequency table. Then mean number of students absent per day is {:("No. of student absent",0,1,2,3,4,5,6,7),("No. of days ",1,4,10,50,34,15,4,2):}

In a mathematics test given to 15 students, the following marks (out of 100) are recorded:41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60 Find the mean, median and mode of this data

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 an object of mass 2 kg and constant b = 4 (N-s)/(m) has terminal speed v_(T) in a liquid then time required to reach 0.63 v_(T) from start of the motion is :

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 on the 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):} Which object would first acquire half of their respective terminal speed in minimum time from start of the motion of all were released simultaneously ?

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):} A small sphere of mass 2.00 g is released from rest in a large vessel filled with oil. The sphere approaches a terminal speed of 10.00 cm/s. Time required to achieve speed 6.32 cm/s from start of the motion is (Take g = 10.00 m//s^(2) ) :

The weights (in kg.) of 15 students of a class are:38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47(i) Find the mode and median of this data.(ii) Is there more than one mode?