Calculus

Detailed SolutionIf $$\overrightarrow {\text{r}} $$ is the position vector of any point on a closed surface S that encloses volume V then $$\iint\limits_{\text{S}} {\overrightarrow {\text{r}} \cdot {\text{d}}\overrightarrow {\text{S}} }$$ is equal to A. $$\frac{1}{2}$$V B. V C. 2V D. 3V

” option2=”\[4{\rm{\hat i}} – {\rm{\hat k}}\]” option3=”\[{\rm{\hat i}} – 4{\rm{\hat j}}\]” option4=”\[{\rm{\hat i}} – 4{\rm{\hat k}}\]” correct=”option4″]

Detailed SolutionVelocity vector of a flow field is given as \[\overrightarrow {\rm{V}} = 2{\rm{xy\hat i}} – {{\rm{x}}^2}{\rm{z\hat j}}{\rm{.}}\] The vorticity vector at (1, 1, 1) is A. \[4{\rm{\hat i}} – {\rm{\hat j}}\] B. \[4{\rm{\hat i}} – {\rm{\hat k}}\] C. \[{\rm{\hat i}} – 4{\rm{\hat j}}\] D. \[{\rm{\hat i}} – 4{\rm{\hat k}}\]

Detailed SolutionIf $$\overrightarrow {\text{a}} $$ and $$\overrightarrow {\text{b}} $$ are two arbitrary vectors with magnitudes a and b, respectively, $${\left| {\overrightarrow {\text{a}} \times \overrightarrow {\text{b}} } \right|^2}$$ will be equal to A. $${{\text{a}}^2}{{\text{b}}^2} – {\left( {\overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} } \right)^2}$$ B. $${\text{ab}} – \overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} $$ C. $${{\text{a}}^2}{{\text{b}}^2} + {\left( {\overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} } \right)^2}$$ D. $${\text{ab}} + \overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} $$

Detailed SolutionThe value of the integral given below is $$\int\limits_0^\pi {{{\text{x}}^2}\cos \,{\text{x dx}}} $$ A. $$ – 2\pi $$ B. $$ – \pi $$ C. $$\pi $$ D. $$2\pi $$

Detailed SolutionThe function f(x) = x2 = x + x …… x times, is defined A. at all real values of x B. only at positive integer values of x C. only at negative integer value of x D. only at rational values of x

Detailed SolutionWhile minimizing the function f(x), necessary and sufficient conditions for a point x0 to be a minima are A. f'(x0) > 0 and f”(x0) = 0 B. f'(x0) < 0 and f''(x0) = 0 C. f'(x0) = 0 and f''(x0) < 0 D. f'(x0) = 0 and f''(x0) > 0

Detailed SolutionThe angle of intersection of the curves x2 = 4y and y2 = 4x at point (0, 0) is A. 0° B. 30° C. 45° D. 90°

Detailed SolutionEquation of the line normal to function $${\text{f}}\left( {\text{x}} \right) = {\left( {{\text{x}} – 8} \right)^{\frac{2}{3}}} + 1$$ at P(0, 5) is A. y = 3x – 5 B. y = 3x + 5 C. 3y = x + 15 D. 3y = x – 15

Detailed SolutionThe direction of vector A is radially outward from the origin, with |A| = krn where r2 = x2 + y2 + z2 and k is a constant. The value of n for which $$\nabla \cdot {\text{A}} = 0$$ is A. -2 B. 2 C. 1 D. 0

Detailed Solution$$\mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{{{\sin }^2}{\text{x}}}}{{\text{x}}}$$ is equal to A. 0 B. $$\infty $$ C. 1 D. -1