Question 91:
easyIf the earth stops rotating about its axis, the acceleration due to gravity will remain unchanged at
The effective acceleration due to gravity at latitude \(\lambda\) is \(g' = g - \omega^2 R \cos^2 \lambda\). At the poles, \(\lambda = 90^\circ\), so \(g' = g\). Thus, rotation has no effect at the poles.
Question 92:
easyOne goes from the centre of the earth to an altitude half the radius of the earth, where will the \(g\) be greatest ?
Inside the earth, \(g\) increases linearly from center to surface: \(g(r) \propto r\). Outside, it decreases: \(g(r) \propto 1/r^2\). Thus, \(g\) is maximum at the surface of the earth.
Question 93:
easyAn iron sphere and an aluminium sphere, both of same radius are dropped from the top of a tower 100m high. At a height 40 m above the ground, both of them will have same.
In free fall under gravity (neglecting air resistance), all bodies experience the same acceleration due to gravity \(g\), regardless of their mass or density.
Question 94:
difficultTwo equal masses m and m are hung from a balance whose scale pans differ in vertical height by ‘h’. The error in weighing in terms of density of the earth \(\rho\) is :
The error in force is \(\Delta F = m(g_1 - g_2) \approx m \frac{2g}{R} h\). Since \(g = \frac{4}{3} \pi G \rho R\), we have \(\frac{2g}{R} = \frac{8}{3} \pi G \rho\). Therefore, \(\Delta F = \frac{8}{3} \pi G \rho m h\).
Question 95:
moderateThe escape velocity for a planet is \(v_e\). A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be
Using conservation of mechanical energy from infinity to the centre: \(0 = \frac{1}{2}mv^2 - \frac{3GmM}{2R}\). Since escape velocity is \(v_e = \sqrt{\frac{2GM}{R}}\), we get \(v^2 = \frac{3GM}{R} = 1.5 v_e^2⇒ v = \sqrt{1.5} v_e\).
Question 96:
moderateThe mass of a spaceship is \(1000\text{ kg}\). It is to be launched from the earth’s surface out into free space. The value of \(‘g’\) and \(‘R’\) (radius of earth) are \(10\text{ m/s}^2\) and \(6400\text{ km}\) respectively. The required energy for this work will be :
The minimum energy required to escape the earth's gravitational pull from the surface is \(E = \frac{GMm}{R} = mgR\). Substituting \(m = 1000\text{ kg}\), \(g = 10\text{ m/s}^2\), and \(R = 6.4 \times 10^6\text{ m}\), we find \(E = 6.4 \times 10^{10}\text{ Joules}\).
Question 97:
easyOne goes from the centre of the earth to an altitude half the radius of the earth, where will the \(g\) be greatest ?
The acceleration due to gravity is zero at the centre, increases linearly inside the earth up to the surface where it is maximum \(g = \frac{GM}{R^2}\), and then decreases as \(1/r^2\) outside the surface. Thus, \(g\) is greatest at the surface.
Question 98:
easyAn iron sphere and an aluminium sphere, both of same radius are dropped from the top of a tower \(100\text{ m}\) high. At a height \(40\text{ m}\) above the ground, both of them will have same:
In the absence of air resistance, all falling bodies have the same acceleration due to gravity \(g\) independent of their mass. Since their masses are different due to different densities, other quantities will differ.
Question 99:
moderateThe acceleration due to gravity increases by \(0.5\%\) when we go from the equator to the poles. What will be the time period of the pendulum at the equator which beats seconds at the poles?
The time period of a seconds pendulum at the poles is \(T_p = 2\text{ s}\). Since \(T propto g^{-1/2}\), we have \(frac{T_e}{T_p} = sqrt{frac{g_p}{g_e}} = sqrt{1 + 0.005} approx 1 + 0.0025\). This gives \(T_e approx 2 times 1.0025 = 2.005\text{ s}\).
Question 100:
moderateTwo equal masses \(m\) and \(m\) are hung from a balance whose scale pans differ in vertical height by \(h\). The error in weighing in terms of density of the earth \(\rho\) is :
The change in gravity over height \(h\) is \(dg = \frac{2g}{R} h\). Using \(g = \frac{4}{3} \pi G \rho R\), we get \(dg = \frac{8}{3} \pi G \rho h\). Thus, the difference in weight (error) is \(m \cdot dg = \frac{8}{3} \pi G \rho m h\).