Question

A body of mass \(M\) is suspended from the lower end of an aeroplane accelerating horizontally. If the rope can withstand a maximum tension \(T\), then the maximum acceleration of the aeroplane is:

• A. \sqrt{\frac{T^2}{M^2}-g^2}
• B. \frac{T}{M}-g
• C. \frac{T}{M}+g
• D. \sqrt{\frac{T^2}{M^2}+g^2}

Hint:

For a body suspended in an accelerating frame: \[ T=M\sqrt{g^2+a^2}. \] This relation is frequently used in JEE problems.

Answer 1

The Correct Option is A

Concept: The suspended body experiences: \[ Mg \] vertically downward and an effective horizontal acceleration \[ a. \] The tension balances the resultant of these two perpendicular effects.

Step 1: Draw force balance.
Horizontal component: \[ T\sin\theta=Ma. \] Vertical component: \[ T\cos\theta=Mg. \]

Step 2: Square and add the equations.
\[ T^2\sin^2\theta+T^2\cos^2\theta = M^2a^2+M^2g^2. \] \[ T^2=M^2(a^2+g^2). \]

Step 3: Find maximum acceleration.
\[ a^2 = \frac{T^2}{M^2}-g^2. \] Therefore \[ a = \sqrt{\frac{T^2}{M^2}-g^2}. \] Hence \[ \boxed{ \sqrt{\frac{T^2}{M^2}-g^2} }. \]