The acceleration due to gravity at a height \(h\) above the Earth's surface is given by the equation: \[ g_h = g_0 \left( \frac{R}{R + h} \right)^2 \] Where: - \(g_h\) is the acceleration due to gravity at height \(h\), - \(g_0\) is the acceleration due to gravity at the Earth's surface, - \(R\) is the radius of the Earth. At a depth \(d\) inside the Earth, the acceleration due to gravity is given by: \[ g_d = g_0 \left( 1 - \frac{d}{R} \right) \] For the gravity at height \(h = 10 \, km\) and at depth \(d\) to be equal, we set the two equations equal to each other: \[ g_0 \left( \frac{R}{R + h} \right)^2 = g_0 \left( 1 - \frac{d}{R} \right) \] Canceling out \(g_0\) and solving for \(d\): \[ \left( \frac{R}{R + h} \right)^2 = 1 - \frac{d}{R} \] Substitute \(h = 10 \, km\), which corresponds to 10,000 m, and solve for \(d\). After simplifying, we find that \(d = 20 \, km\).
Thus, the depth inside the Earth where the acceleration due to gravity is the same as at a height of 10 km is 20 km.
The acceleration due to gravity \( g \) at a height \( h \) from the Earth's surface and at a depth \( d \) inside the Earth can be expressed using the following relations: 1. At height \( h \) above the Earth's surface, the formula for gravity is: \[ g_h = \frac{g_0}{(1 + \frac{h}{R})^2} \] where \( g_0 \) is the acceleration due to gravity on the Earth's surface and \( R \) is the radius of the Earth. 2. At depth \( d \) inside the Earth, the formula for gravity is: \[ g_d = g_0 \left( 1 - \frac{d}{R} \right) \] where \( d \) is the depth below the surface. We are given that the value of gravity at a height of 10 km is \( x \), so: \[ x = \frac{g_0}{(1 + \frac{10}{R})^2} \] Now, we want to find the depth \( d \) where the value of gravity \( g_d = x \). So we set: \[ x = g_0 \left( 1 - \frac{d}{R} \right) \] Now equating the two expressions for \( x \), we have: \[ \frac{g_0}{(1 + \frac{10}{R})^2} = g_0 \left( 1 - \frac{d}{R} \right) \] Simplifying the equation: \[ \frac{1}{(1 + \frac{10}{R})^2} = 1 - \frac{d}{R} \] Using \( R = 6400 \, \text{km} \), we solve for \( d \) and find that \( d = 20 \, \text{km} \). Thus, the depth inside the Earth where the value of gravity is \( x \) is 20 km.
Thus, the correct answer is (B) 20 km.
Net gravitational force at the center of a square is found to be \( F_1 \) when four particles having masses \( M, 2M, 3M \) and \( 4M \) are placed at the four corners of the square as shown in figure, and it is \( F_2 \) when the positions of \( 3M \) and \( 4M \) are interchanged. The ratio \( \dfrac{F_1}{F_2} = \dfrac{\alpha}{\sqrt{5}} \). The value of \( \alpha \) is 

Match the following:
In the following, \( [x] \) denotes the greatest integer less than or equal to \( x \). 
Choose the correct answer from the options given below:
For x < 0:
f(x) = ex + ax
For x ≥ 0:
f(x) = b(x - 1)2
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].