How many numbers are there between 200 and 800 which are divisible by both 5 and 7? A. 35 B. 16 C. 17 D. can’t be determined
In the given range, the last number which is divisible by both 5 and 7. i.e., 35 is 210 and the highest number is 770. So the total number of numbers between 200 and 800 which are divisible by both 5 and 7 is (770-210) /35 + 1 = 17 Hence option (c) is correct.
Solid Mechanics: What are the differences between plane stress and plane strain conditions?
In real life situations all the stress and strain associated are 3-dimensional. Many a times we make proper assumptions for simplified solution and analysis of the mathematical model without affecting the solution. In mathematical term a state of plane stress in one in which stress along z-direction is ZERO and a plane strain condition is one in which strain associated along z-direction is ZERO. For physical understanding of the situation let us consider two plates one thick and the other thin. In a thick plate (compared over thin plate)subjected to stresses along x and y-direction ,strains are associated in x and y direction along with strain in z-direction(along thickness),though small, subject to poissons contraction. if we consider a diffrential element on the inside ,this element will resist deformation in thickness direction as more material is present in thickness direction compared over thin plate . so here we can assume that strain along z-direction is zero and this assumption works excellent in this situation. So thick plates are assumed to be plane strain condition. In thin plates subject to biaxial state of stress, plane stress condition is assumed as no stress is acting in z-direction. However subject to biaxial stress poissons contraction will be there but the z-direction stress cannot be sustained.
Two coins are tossed, find the probability that two heads are obtained.
The sample space S is given by. S = {(H,T),(H,H),(T,H),(T,T)} Let E be the event "two heads are obtained". E = {(H,H)} We use the formula of the classical probability. P(E) = n(E) / n(S) = 1 / 4