Q1: The acceleration f required to accelerate a rectangular tank containing water horizontally so that the slope of its free surface is 45°, is

ANS:B - g

To determine the acceleration required to accelerate a rectangular tank containing water horizontally so that the slope of its free surface is 45°, we can use principles of physics and trigonometry. When the tank is accelerated horizontally, it induces a pseudo-force in the direction opposite to the acceleration due to inertia. This pseudo-force acts as if the water is being tilted at an angle, resulting in a sloped free surface. Given that the slope of the free surface is 45°, it means that the tangent of the angle is 1. Therefore, the pseudo-force acting on the water must balance the gravitational force acting downward. Let's denote the acceleration as f, and g as the acceleration due to gravity (approximately 9.81 m/s29.81m/s2). The horizontal component of the pseudo-force is f, and the vertical component of the pseudo-force is also f. The gravitational force acting downward is mg, where m is the mass of the water and g is the acceleration due to gravity. Since the tank is rectangular, the gravitational force mg acts through the center of mass, which is at the midpoint of the tank's height. Given that the free surface is at 45°, it means the gravitational force is balanced by the pseudo-force, so we can equate their magnitudes: mg=f To find the value of f, we can express m in terms of the density ρ of water, and the volume V of water: m=ρV The volume �V can be expressed in terms of the dimensions of the tank. Now, the gravitational force mg equals ρVg, and this is equal to f. Therefore, f=ρVg. Now, we need to find V, the volume of water in the tank. Let's assume the dimensions of the tank are l (length), w (width), and ℎh (height). Then, the volume V is: =l×w×h Given that the tank is rectangular and the angle of the free surface is 45°, we can deduce that h=l, the height is equal to the length. Therefore, 2V=l×w×l=lw2. Now, we can express f in terms of ρ, g, l, and w: 2⋅f=ρ⋅lw2⋅g Now, we need to find l and w in terms of f and g. Given that the slope of the free surface is 45°, we can use trigonometric relations: tan⁡(45∘)=tan(45∘)=wl​ 1=wl​ Therefore, l=w. Substituting l=w into 2V=lw2, we get =3V=l3. So, 3V=l3. Now, we can express f as: =3f=ρl3g Now, to find the acceleration f required, we need to compare it with g. f=ρl3g=g 3=1ρl3=1 Given that 3=1l3=1, it implies that =1l=1 (since l is a length, it cannot be negative). Therefore, f=g. Hence, the acceleration required to accelerate the rectangular tank containing water horizontally so that the slope of its free surface is 45° is g.