Statement-1 `:` The work done in bringing a body down from the top to the base along a frictionless incline plane is the same as the work done in brining it down from the vertical side.
STATEMENT-2 `:` Work is defined as `intvec(F) . dvec(r )`

Options

A. Statement-1 is True , Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True , Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True.

Answer

To solve the problem, we need to analyze both statements and verify their correctness step by step.

### Step 1: Understanding Statement 1
Statement 1 claims that the work done in bringing a body down from the top to the base along a frictionless inclined plane is the same as the work done in bringing it down from the vertical side.

### Step 2: Work Done on an Inclined Plane
Let's consider a body of mass \( m \) at a height \( H \) on a frictionless inclined plane. The force acting on the body due to gravity is \( mg \) (where \( g \) is the acceleration due to gravity).

When the body moves down the incline, the height decreases from \( H \) to \( 0 \). The work done by gravity can be calculated using the formula:
\[ W = \int F \cdot dr \]
Here, \( F \) is the gravitational force acting on the body, and \( dr \) is the small displacement along the incline.

### Step 3: Setting Up the Integral
The angle of the incline is \( \theta \). The small displacement \( dr \) can be related to the change in height \( dh \) by the relationship:
\[ dr = \frac{dh}{\cos(\theta)} \]
Thus, the work done can be expressed as:
\[ W = \int_{0}^{H} mg \cdot \frac{dh}{\cos(\theta)} \]
This simplifies to:
\[ W = mg \int_{0}^{H} dh = mgH \]

### Step 4: Work Done on a Vertical Slide
Now, consider the work done when the body is brought down vertically from height \( H \) to \( 0 \). The work done in this case is also:
\[ W = mgH \]

### Step 5: Conclusion for Statement 1
Since the work done in both scenarios (along the inclined plane and vertically) is \( mgH \), we conclude that Statement 1 is true.

### Step 6: Understanding Statement 2
Statement 2 states that work is defined as \( \int \vec{F} \cdot d\vec{r} \). This is indeed the correct definition of work done by a force \( \vec{F} \) along a displacement \( d\vec{r} \).

### Step 7: Conclusion for Statement 2
Since Statement 2 is a correct definition of work, we conclude that Statement 2 is also true.

### Final Conclusion
Both statements are true, and Statement 2 provides a correct explanation for Statement 1.