The correct option is(C): 498 J.
ΔQ = nCvΔT (Isochoric process)
\(=2×\frac{3R}{2}×20\)
= 498 J
If $10 \sin^4 \theta + 15 \cos^4 \theta = 6$, then the value of $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is:
If the area of the region $\{ (x, y) : |x - 5| \leq y \leq 4\sqrt{x} \}$ is $A$, then $3A$ is equal to
Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to
Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to
Let $C$ be the circle $x^2 + (y - 1)^2 = 2$, $E_1$ and $E_2$ be two ellipses whose centres lie at the origin and major axes lie on the $x$-axis and $y$-axis respectively. Let the straight line $x + y = 3$ touch the curves $C$, $E_1$, and $E_2$ at $P(x_1, y_1)$, $Q(x_2, y_2)$, and $R(x_3, y_3)$ respectively. Given that $P$ is the mid-point of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$, the value of $9(x_1 y_1 + x_2 y_2 + x_3 y_3)$ is equal to
Pressure is defined as the force applied perpendicular to the surface of an object per unit area over which that force is distributed.
When a force of ‘F’ Newton is applied perpendicularly to a surface area ‘A’, then the pressure exerted on the surface by the force is equal to the ratio of F to A. The formula for pressure (P) is:
P = F / A
The SI unit of pressure is the pascal (Pa)
A pascal can be defined as a force of one newton applied over a surface area of a one-meter square.