A convex lens is hold in water. What change do you expect in its focal length?
focal length of the lens increases due to increase in the optical density of the medium i.e water
A research balloon of total mass M is descending vertically with downward acceleration a as shown in below figure. How much ballast must be thrown from the car to give the balloon an upward acceleration a, assuming that the upward lift of the air on the balloon does not change?
Let us consider initially mass of the system (balloon) is, M. So the force (f) acting on the system having downward acceleration, a will be, f = -Ma …… (1) And the weight of the system (W) will be, W = Mg …… (2) Where, g is the free fall acceleration of the system. Therefore the upward force acting on the system (F) will be, F = W + f = Mg+(-Ma) F = Mg-Ma …… (3) Again let us consider m is the mass of single ballast which is thrown from the balloon. Now mass of the system is, M-m. So the force (f1) acting on the system having upward acceleration, a and mass, M-m will be, f1 = (M-m) a …… (4) And the weight of the system (W1) having mass M-m will be, W1 = (M-m) g …… (5) Where, g is the free fall acceleration of the system. To give the balloon an upward acceleration a, the upward force (F) must be equal to the addition of the force (f1) acting on the system having upward acceleration, a and mass, M-m and the weight of the system (W1) having mass M-m will be. So the equation will be, f1+W1 = F …… (6) (M-m) a + (M-m) g = Mg-Ma Ma-ma +Mg-mg = Mg-Ma ma + mg = Ma+ Mg –Mg + Ma m(a+g) = 2Ma m = 2Ma/(a+g) …… (7) From equation (7) we observed that, 2Ma/(a+g) mass of ballast must be thrown from the car to give the balloon an upward acceleration a.
For newtonian fluids, shear stresses are proportional to corresponding shear strain rates. For a simple shear flow, V = u(y)ˆi, show that the shear stress τ is proportional to du/dy.
Consider a small cubical element, at the continuum limit. As a thought experiment, now release the fluid element, with its lower side at y, upper side at y +δy. After a small time δt the element deforms in the xy-plane due to the shear action, see Fig. 1.6. If velocity increases with y, the higher velocity on the upper side (u + δu) means that this side slides further to the right (in x-direction) compared to the lower side, shear angle δθ, sliding length difference δuδt. Since δt is small, δθ ≈ (δu/δy)δt. In the limit δt → 0, dθ/dt = du/dy. Since dθ/dt is the shear strain rate, the shear stress τ is proportional to du/dy (the proportionality constant being the dynamic viscosity µ).