The mass of a body in a lift at rest is m. If the lift ascends with uniform acceleration a, the effective mass will be

Options

A. `m(1 + g/a)`
B. `m(1+a/g)`
C. `m(1-a/g)`
D. `m(1-g/a)`

Explanation

To find the effective mass of a body in a lift that is ascending with uniform acceleration \( a \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Body
When the lift is at rest, the only force acting on the body is its weight, which is given by:
\[ W = mg \]
where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity.

**Hint:** Identify the forces acting on the body when the lift is at rest.

### Step 2: Analyze the Situation When the Lift Ascends
When the lift ascends with uniform acceleration \( a \), it creates a non-inertial frame of reference. In this case, a pseudo force acts on the body in the opposite direction of the lift's acceleration. The pseudo force \( F_{pseudo} \) can be expressed as:
\[ F_{pseudo} = ma \]

**Hint:** Consider how the acceleration of the lift affects the forces on the body.

### Step 3: Calculate the Effective Weight in the Ascending Lift
The effective weight \( W_{effective} \) of the body in the lift can be calculated by adding the gravitational force and the pseudo force:
\[ W_{effective} = mg + ma \]
This can be factored as:
\[ W_{effective} = m(g + a) \]

**Hint:** Remember that effective weight is the sum of the gravitational force and the pseudo force.

### Step 4: Relate Effective Weight to Effective Mass
The effective mass \( m' \) can be defined in terms of the effective weight:
\[ W_{effective} = m' g \]
Setting the two expressions for effective weight equal gives:
\[ m' g = m(g + a) \]

**Hint:** Use the definition of effective weight to relate it to effective mass.

### Step 5: Solve for Effective Mass
Rearranging the equation to solve for \( m' \):
\[ m' = m \frac{g + a}{g} \]
This can be simplified to:
\[ m' = m \left(1 + \frac{a}{g}\right) \]

**Hint:** Isolate \( m' \) to find the expression for effective mass.

### Final Result
The effective mass of the body when the lift ascends with uniform acceleration \( a \) is:
\[ m' = m \left(1 + \frac{a}{g}\right) \]