A force on an object of mass 100 g is `(10hati+5hatj)N`. The position of that object at t=2s is `(ahati+bhatj)m` after starting from rest. The value of `a/b` will be _________

Explanation

To solve the problem, we need to find the ratio \( \frac{a}{b} \) where the position of an object at time \( t = 2 \) seconds is given as \( (a \hat{i} + b \hat{j}) \) meters. The object has a mass of 100 g and is subjected to a force of \( (10 \hat{i} + 5 \hat{j}) \) N.

### Step-by-Step Solution:

1. **Convert Mass to Kilograms**:
The mass of the object is given as 100 g. We convert this to kilograms:
\[ m = \frac{100 \, \text{g}}{1000} = 0.1 \, \text{kg} \]

2. **Calculate Acceleration**:
Using Newton's second law, \( F = ma \), we can find the acceleration \( a \):
\[ a = \frac{F}{m} = \frac{(10 \hat{i} + 5 \hat{j}) \, \text{N}}{0.1 \, \text{kg}} = (100 \hat{i} + 50 \hat{j}) \, \text{m/s}^2 \]

3. **Use the Equation of Motion**:
The object starts from rest, so the initial velocity \( u = 0 \). The position \( s \) at time \( t \) can be calculated using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting \( u = 0 \), \( a = (100 \hat{i} + 50 \hat{j}) \, \text{m/s}^2 \), and \( t = 2 \, \text{s} \):
\[ s = 0 + \frac{1}{2} (100 \hat{i} + 50 \hat{j}) (2^2) \]
\[ s = \frac{1}{2} (100 \hat{i} + 50 \hat{j}) (4) = (200 \hat{i} + 100 \hat{j}) \, \text{m} \]

4. **Identify Values of \( a \) and \( b \)**:
From the position vector \( s = (200 \hat{i} + 100 \hat{j}) \, \text{m} \), we can identify:
\[ a = 200, \quad b = 100 \]

5. **Calculate the Ratio \( \frac{a}{b} \)**:
Now, we can find the ratio \( \frac{a}{b} \):
\[ \frac{a}{b} = \frac{200}{100} = 2 \]

### Final Answer:
The value of \( \frac{a}{b} \) is \( 2 \).