A body of mass 2 kg is initially at rest. It starts moving unidirectionally under the influence of a source of constant power P. Its displacement in 4 s is 1 3α2√Pm. The value of α will be _______

JEE Mains - Motion Under Constant Power

Question: A body of mass \( m = 2\,\text{kg} \) is initially at rest. It starts moving unidirectionally under the influence of a source of constant power \( P \). Its displacement in \( 4\,\text{s} \) is given by \[ x(4) = \frac{1}{3}\alpha^2\sqrt{P\,m}. \] Determine the value of \( \alpha \).
[30-Jan-2023 Shift 2]

Select the correct answer:

(A) \(2^{7/4}\)

(B) \(2^{3/2}\)

(C) \(2^{5/4}\)

(D) \(4\)

Solution

Under a constant power source, one common result (derived in many JEE texts) is that the displacement for a body starting from rest is given by \[ x(t) = \frac{4}{3}\sqrt{\frac{P}{m}}\,t^{3/2}. \]

For a mass \( m = 2\,\text{kg} \) and time \( t = 4\,\text{s} \), we have: \[ x(4) = \frac{4}{3}\sqrt{\frac{P}{2}}\,(4)^{3/2}. \] Since \( 4^{3/2} = 8 \), this simplifies to: \[ x(4) = \frac{4}{3}\sqrt{\frac{P}{2}}\,\times 8 = \frac{32}{3}\sqrt{\frac{P}{2}}. \]

The problem also gives the displacement as \[ x(4) = \frac{1}{3}\alpha^2\sqrt{P\,m}. \] With \( m = 2 \), this becomes: \[ x(4) = \frac{1}{3}\alpha^2\sqrt{2P} = \frac{\alpha^2\sqrt{2}}{3}\sqrt{P}. \]

Equate the two expressions for \( x(4) \): \[ \frac{32}{3}\sqrt{\frac{P}{2}} = \frac{\alpha^2\sqrt{2}}{3}\sqrt{P}. \] Multiplying both sides by 3 and writing \(\sqrt{P}\) explicitly: \[ 32\sqrt{\frac{P}{2}} = \alpha^2\sqrt{2}\sqrt{P}. \] Divide both sides by \(\sqrt{P}\) (assuming \(P>0\)): \[ 32\sqrt{\frac{1}{2}} = \alpha^2\sqrt{2}. \] Note that \(\sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}\); hence, \[ 32\left(\frac{1}{\sqrt{2}}\right) = \alpha^2\sqrt{2} \quad \Longrightarrow \quad \frac{32}{\sqrt{2}} = \alpha^2\sqrt{2}. \] Multiply both sides by \(\sqrt{2}\): \[ 32 = \alpha^2 \times 2. \] Thus, \[ \alpha^2 = \frac{32}{2} = 16 \quad \Longrightarrow \quad \alpha = 4. \]

Therefore, the value of \( \alpha \) is 4.