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: \( 9g \).
A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of:
Match List-I with List-II.
Choose the correct answer from the options given below :
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:
For a given reaction \( R \rightarrow P \), \( t_{1/2} \) is related to \([A_0]\) as given in the table. Given: \( \log 2 = 0.30 \). Which of the following is true?
| \([A]\) (mol/L) | \(t_{1/2}\) (min) |
|---|---|
| 0.100 | 200 |
| 0.025 | 100 |
A. The order of the reaction is \( \frac{1}{2} \).
B. If \( [A_0] \) is 1 M, then \( t_{1/2} \) is \( 200/\sqrt{10} \) min.
C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
D. \( t_{1/2} \) is 800 min for \( [A_0] = 1.6 \) M.