If unit of length and time is doubled the numerical value of g (acceleration due to gravity ) will be

Options

A. doubled
B. halved
C. four time
D. same

Explanation

To solve the problem, we need to analyze how the numerical value of acceleration due to gravity (g) changes when the units of length and time are doubled. Here’s a step-by-step solution:

### Step 1: Understand the given value of g
The standard value of acceleration due to gravity (g) is approximately 9.8 m/s² in SI units.

### Step 2: Identify the changes in units
According to the problem, both the unit of length (meters) and the unit of time (seconds) are doubled.

- New unit of length (m') = 2 meters
- New unit of time (s') = 2 seconds

### Step 3: Express g in terms of the new units
We can express the acceleration due to gravity in the new units as follows:

\[ g = \frac{9.8 \, \text{m}}{\text{s}^2} \]

When we substitute the new units into the equation, we have:

\[ g' = \frac{9.8 \, (2 \, \text{m'})}{(2 \, \text{s'})^2} \]

### Step 4: Simplify the expression
Now, we simplify the expression for g':

\[ g' = \frac{9.8 \times 2 \, \text{m'}}{(2^2) \, \text{s'}^2} \]
\[ g' = \frac{9.8 \times 2 \, \text{m'}}{4 \, \text{s'}^2} \]
\[ g' = \frac{9.8 \times 2}{4} \, \text{m'} \, \text{s'}^{-2} \]

### Step 5: Calculate the new value of g
Now we calculate the numerical value:

\[ g' = \frac{19.6}{4} \, \text{m'} \, \text{s'}^{-2} \]
\[ g' = 4.9 \, \text{m'} \, \text{s'}^{-2} \]

### Conclusion
Thus, the new numerical value of g when the units of length and time are doubled is 4.9 m'/s'². This means that the numerical value of g is halved.

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