The acceleration due to gravity \( g \) on the surface of the Earth is given by the formula: \[ g = \frac{GM}{R^2}, \] where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth. If the diameter is reduced to one third of its original value, the new radius \( R' \) becomes: \[ R' = \frac{R}{3}. \] Since mass \( M \) remains unchanged, the new acceleration due to gravity \( g' \) is: \[ g' = \frac{GM}{(R/3)^2} = \frac{GM}{R^2} \times 9 = 9g. \] Thus, the acceleration due to gravity increases by a factor of 9.
Final Answer: \( 9g \).
The expression given below shows the variation of velocity \( v \) with time \( t \): \[ v = \frac{At^2 + Bt}{C + t} \] The dimension of \( A \), \( B \), and \( C \) is:
The steam volatile compounds among the following are: