
To determine the direction of gravitational intensity at the center of a hemispherical shell of uniform mass density, we must consider the symmetrical distribution of mass around the center.
The gravitational intensity (\( \mathbf{g} \)) at a point due to a mass is directed towards the mass since it is an attractive force. In this scenario, the hemispherical shell is symmetric about the center.
For a complete spherical shell, the intensity at the center is zero due to symmetry. However, in the case of a hemispherical shell, the symmetry is broken. The net gravitational field will be directed towards the base of the hemisphere, as the top portion of the sphere, which is missing, does not exert any field.
Thus, for a hemispherical shell, the gravitational intensity at the center is along the axis perpendicular to the base and towards the base of the hemisphere.
Based on the given diagram and the options, the direction of the gravitational intensity at the center (point O) is downward, indicated by option c.
Hence, the correct answer is: c.
A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s–1 ?
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].