Q145 / 180 · COMEDK 2024 (Morning Shift) · Physics
Find the value of '$n$' in the given equation $P=\rho^n v^2$ where '$P$' is the pressure, '$\rho$' density and '$v$' velocity.
- A$\mathrm{n}=\dfrac{1}{2}$
- Correct answerB$\mathrm{n}=1$
- C$\mathrm{n}=2$
- D$\mathrm{n}=3$
Solution
The dimensional formula for pressure $P$ is $[M L^{-1} T^{-2}]$.
The dimensional formula for density $\rho$ is $[M L^{-3}]$.
The dimensional formula for velocity $v$ is $[L T^{-1}]$.
Substituting these into the equation $P = \rho^n v^2$, we get:
$[M L^{-1} T^{-2}] = [M L^{-3}]^n [L T^{-1}]^2$
$[M L^{-1} T^{-2}] = [M^n L^{-3n}] [L^2 T^{-2}]$
$[M L^{-1} T^{-2}] = [M^n L^{-3n+2} T^{-2}]$
Comparing the powers of $M$ on both sides, we get $n = 1$.
Checking the powers of $L$: $-1 = -3n + 2 \Rightarrow 3n = 3 \Rightarrow n = 1$.
Checking the powers of $T$: $-2 = -2$, which is consistent.
Answer: $\mathrm{n}=1$
The dimensional formula for density $\rho$ is $[M L^{-3}]$.
The dimensional formula for velocity $v$ is $[L T^{-1}]$.
Substituting these into the equation $P = \rho^n v^2$, we get:
$[M L^{-1} T^{-2}] = [M L^{-3}]^n [L T^{-1}]^2$
$[M L^{-1} T^{-2}] = [M^n L^{-3n}] [L^2 T^{-2}]$
$[M L^{-1} T^{-2}] = [M^n L^{-3n+2} T^{-2}]$
Comparing the powers of $M$ on both sides, we get $n = 1$.
Checking the powers of $L$: $-1 = -3n + 2 \Rightarrow 3n = 3 \Rightarrow n = 1$.
Checking the powers of $T$: $-2 = -2$, which is consistent.
Answer: $\mathrm{n}=1$
Hence, the correct answer is option: 'B'.