Question:
Let $F = (2xy + z^{3})i + x^{2}j + 3xz^{2}k$. The work done in moving a particle in this field from (1, -2, 1) to (3, 1, 4) is}
Let $F = (2xy + z^{3})i + x^{2}j + 3xz^{2}k$. The work done in moving a particle in this field from (1, -2, 1) to (3, 1, 4) is}
Show Hint
If the field is conservative, don't bother with the path! Just subtract the potential values at the start and end.
Updated On: Jul 3, 2026
- 72
- 75
- 202
- 81
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Concept
Work done is the line integral $\int F \cdot dr$. If $F$ is conservative ($\nabla \times F = 0$), there exists a potential $\phi$ such that $F = \nabla \phi$.
Step 2: Meaning
Check for exactness: $F = \nabla (x^{2}y + xz^{3})$. Partial derivatives $\frac{\partial}{\partial x}(x^{2}y + xz^{3}) = 2xy + z^{3}$, $\frac{\partial}{\partial y}(x^{2}y + xz^{3}) = x^{2}$, and $\frac{\partial}{\partial z}(x^{2}y + xz^{3}) = 3xz^{2}$ match $F$.
Step 3: Analysis
For a conservative field, work done depends only on endpoints: $\phi(3, 1, 4) - \phi(1, -2, 1)$. $\phi(3, 1, 4) = (3)^{2}(1) + (3)(4)^{3} = 9 + 192 = 201$. $\phi(1, -2, 1) = (1)^{2}(-2) + (1)(1)^{3} = -2 + 1 = -1$.
Step 4: Conclusion
Work done $= 201 - (-1) = 202$. Final Answer: (C)
Work done is the line integral $\int F \cdot dr$. If $F$ is conservative ($\nabla \times F = 0$), there exists a potential $\phi$ such that $F = \nabla \phi$.
Step 2: Meaning
Check for exactness: $F = \nabla (x^{2}y + xz^{3})$. Partial derivatives $\frac{\partial}{\partial x}(x^{2}y + xz^{3}) = 2xy + z^{3}$, $\frac{\partial}{\partial y}(x^{2}y + xz^{3}) = x^{2}$, and $\frac{\partial}{\partial z}(x^{2}y + xz^{3}) = 3xz^{2}$ match $F$.
Step 3: Analysis
For a conservative field, work done depends only on endpoints: $\phi(3, 1, 4) - \phi(1, -2, 1)$. $\phi(3, 1, 4) = (3)^{2}(1) + (3)(4)^{3} = 9 + 192 = 201$. $\phi(1, -2, 1) = (1)^{2}(-2) + (1)(1)^{3} = -2 + 1 = -1$.
Step 4: Conclusion
Work done $= 201 - (-1) = 202$. Final Answer: (C)
Was this answer helpful?
0
0
Top AP ECET BSc Mathematics Mathematics Questions
- The integrating factor of $x \cos x \frac{dy}{dx} + (x \sin x + \cos x) y = 1$ is}
- If $y = y(x)$ is the solution of $\left(\frac{2 + \sin x}{y + 1}\right) \frac{dy}{dx} + \cos x = 0$ with $y(0) = 1$, then $y\left(\frac{\pi}{2}\right) = $
- The solution of $y^{4}dx + 2xy^{3}dy = \frac{ydx - xdy}{x^{3}y^{3}}$ is}
- Which of the following is an integrating factor of $(x^{2} + y^{2} + 2x)dx + 2ydy = 0$?
- The differential equation $\frac{dy}{dx} = \frac{-(x + x^{8} + py^{2})}{y^{8} - y + qxy}$ is exact if}
View More Questions
Top AP ECET BSc Mathematics Vector Calculus Questions
- The velocity vector is given as $\overline{V} = 5xyi + 2y^{2}j + 3yz^{2}k$. The divergence of this vector at (1,1,1) is}
- If $\phi = 2x^{3}y^{2}z^{4}$ then $\nabla^{2}\phi = $
- If $f$ and $F$ are scalar and vector functions respectively, then which of the following is Not Correct?
- For a vector $\overline{A}$, which of the following statement is NOT TRUE?
- If $\overline{a}$ is constant vector then $\nabla \times (\overline{a} \times \overline{r}) = _?$
View More Questions
Top AP ECET BSc Mathematics Questions
- The integrating factor of $x \cos x \frac{dy}{dx} + (x \sin x + \cos x) y = 1$ is}
- If $y = y(x)$ is the solution of $\left(\frac{2 + \sin x}{y + 1}\right) \frac{dy}{dx} + \cos x = 0$ with $y(0) = 1$, then $y\left(\frac{\pi}{2}\right) = $
- The solution of $y^{4}dx + 2xy^{3}dy = \frac{ydx - xdy}{x^{3}y^{3}}$ is}
- Which of the following is an integrating factor of $(x^{2} + y^{2} + 2x)dx + 2ydy = 0$?
- The differential equation $\frac{dy}{dx} = \frac{-(x + x^{8} + py^{2})}{y^{8} - y + qxy}$ is exact if}
View More Questions