A 10 m long iron chain of linear mass density 0.8 kg `m^(-1)` is hanging freely from a rigid support .If g=10 `ms^(-2)` ,then the power required to left the chain upto the point of support in 10 second

Options

A. 10 W
B. 20W
C. 30W
D. 40W

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the mass of the chain
The linear mass density (λ) of the chain is given as 0.8 kg/m, and the length (L) of the chain is 10 m.

\[ \text{Mass of the chain (m)} = \text{Linear mass density} \times \text{Length} = 0.8 \, \text{kg/m} \times 10 \, \text{m} = 8 \, \text{kg} \]

### Step 2: Determine the center of mass of the chain
Since the chain is uniform, the center of mass will be located at the midpoint of the chain. For a 10 m long chain, the center of mass is:

\[ \text{Center of mass} = \frac{L}{2} = \frac{10 \, \text{m}}{2} = 5 \, \text{m} \]

### Step 3: Calculate the potential energy change
When lifting the chain, we need to calculate the change in potential energy (PE). The change in height (h) for the center of mass when lifting the chain to the support point is equal to the distance from the center of mass to the support point, which is 5 m.

The formula for gravitational potential energy is:

\[ \text{Potential Energy (PE)} = m \cdot g \cdot h \]

Substituting the values:

\[ PE = 8 \, \text{kg} \times 10 \, \text{m/s}^2 \times 5 \, \text{m} = 400 \, \text{J} \]

### Step 4: Calculate the power required
Power (P) is defined as the work done (or energy transferred) per unit time. Here, the work done is equal to the change in potential energy we calculated in the previous step. The time taken (t) is given as 10 seconds.

Using the formula for power:

\[ P = \frac{\text{Work Done}}{\text{Time}} = \frac{PE}{t} \]

Substituting the values:

\[ P = \frac{400 \, \text{J}}{10 \, \text{s}} = 40 \, \text{W} \]

### Final Answer
The power required to lift the chain up to the point of support in 10 seconds is **40 Watts**.

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