Question

A dust particle oscillates in air with a time period which depends on atmospheric pressure P, density of air d and energy of particle E. Time period is proportional to $(P)^a(d)^b(E)^c$. Find the value of (-6a + 2b + 3c).

Answer 1

7

Answer 2

The time period T of oscillation of the dust particle depends on atmospheric pressure P, density of air d, and energy of the particle E. The relationship is given by $T \propto P^a d^b E^c$.

We can write this as $T = k P^a d^b E^c$, where k is a dimensionless constant.

The dimensions of the physical quantities are:

• Time period $[T] = [M^0 L^0 T^1]$
• Pressure $[P] = [ML^{-1}T^{-2}]$
• Density $[d] = [ML^{-3}T^0]$
• Energy $[E] = [ML^2T^{-2}]$

Equating the dimensions on both sides of the relationship $T = k P^a d^b E^c$: $[M^0 L^0 T^1] = [ML^{-1}T^{-2}]^a [ML^{-3}T^0]^b [ML^2T^{-2}]^c$ $[M^0 L^0 T^1] = [M^a L^{-a} T^{-2a}] [M^b L^{-3b} T^{0b}] [M^c L^{2c} T^{-2c}]$ $[M^0 L^0 T^1] = [M^{a+b+c} L^{-a-3b+2c} T^{-2a-2c}]$

Equating the powers of M, L, and T on both sides, we get a system of linear equations:

• For M: $a + b + c = 0$ (1)
• For L: $-a - 3b + 2c = 0$ (2)
• For T: $-2a - 2c = 1$ (3)

From equation (3), we can factor out -2: $-2(a + c) = 1$ $a + c = -1/2$

Substitute this result into equation (1): $(a + c) + b = 0$ $(-1/2) + b = 0$ $b = 1/2$

Now substitute the value of b into equation (2): $-a - 3(1/2) + 2c = 0$ $-a - 3/2 + 2c = 0$ $-a + 2c = 3/2$ $a - 2c = -3/2$

Now we have a system of two equations for a and c: $a + c = -1/2$ $a - 2c = -3/2$

Subtract the second equation from the first: $(a + c) - (a - 2c) = (-1/2) - (-3/2)$ $a + c - a + 2c = -1/2 + 3/2$ $3c = 2/2$ $3c = 1$ $c = 1/3$

Substitute the value of c into the equation $a + c = -1/2$: $a + 1/3 = -1/2$ $a = -1/2 - 1/3$ To subtract the fractions, find a common denominator, which is 6: $a = -3/6 - 2/6$ $a = -5/6$

So, the values of the exponents are $a = -5/6$, $b = 1/2$, and $c = 1/3$.

We are asked to find the value of the expression $(-6a + 2b + 3c)$. Substitute the values of a, b, and c into the expression: $(-6)a + (2)b + (3)c = (-6)(-5/6) + (2)(1/2) + (3)(1/3)$ $= (6 \times 5/6) + (2/2) + (3/3)$ $= 5 + 1 + 1$ $= 7$

The value of $(-6a + 2b + 3c)$ is 7.