Question

If the tension in the cable supporting a lift moving downwards is half the tension when it is moving upwards, the acceleration of the lift is

• A. \( \dfrac{g}{2} \)
• B. \( \dfrac{g}{3} \)
• C. \( \dfrac{g}{4} \)
• D. none of these

Hint:

For lifts: Upward acceleration → \( T = m(g + a) \)
Downward acceleration → \( T = m(g - a) \)

Answer 1

The Correct Option is B

Step 1: Writing expressions for tension.
Let the mass of the lift be \( m \) and acceleration be \( a \).
When the lift moves upwards with acceleration \( a \): \[ T_1 = m(g + a) \] When the lift moves downwards with acceleration \( a \): \[ T_2 = m(g - a) \]
Step 2: Applying the given condition.
It is given that the downward tension is half the upward tension: \[ T_2 = \frac{1}{2}T_1 \] Substituting values: \[ m(g - a) = \frac{1}{2}m(g + a) \]
Step 3: Solving for acceleration.
Canceling \( m \) and simplifying: \[ 2(g - a) = g + a \] \[ 2g - 2a = g + a \] \[ g = 3a \] \[ a = \frac{g}{3} \]
Step 4: Conclusion.
The acceleration of the lift is \( \dfrac{g}{3} \).

Answer 2

Let the lift have mass \( m \) and move with acceleration \( a \) (same magnitude whether going up or down). The cable tension while moving upward is \( T_{up} = m(g+a) \), and while moving downward is \( T_{down} = m(g-a) \); the condition given is \( T_{down} = \tfrac{1}{2}T_{up} \). Each option is substituted back to check consistency.

1. Option \( g/2 \): Substituting \( a = g/2 \), \( T_{up} = m(g + g/2) = 1.5mg \) and \( T_{down} = m(g - g/2) = 0.5mg \); checking the ratio, \( T_{down}/T_{up} = 0.5/1.5 = 1/3 \), not the required \( 1/2 \), so this value does not satisfy the given condition.
2. Option \( g/3 \): Substituting \( a = g/3 \), \( T_{up} = m(g+g/3) = \tfrac{4}{3}mg \) and \( T_{down} = m(g - g/3) = \tfrac{2}{3}mg \); checking the ratio, \( T_{down}/T_{up} = \tfrac{2/3}{4/3} = \tfrac{1}{2} \), exactly matching the required condition that downward tension is half the upward tension.
3. Option \( g/4 \): Substituting \( a = g/4 \), \( T_{up} = m(g+g/4) = 1.25mg \) and \( T_{down} = m(g-g/4) = 0.75mg \); the ratio \( 0.75/1.25 = 0.6 \), not \( 0.5 \), so this does not satisfy the given relationship either.
4. Option "none of these": Since \( a = g/3 \) was shown above to satisfy the given tension ratio exactly, a valid matching value does exist among the choices, so this catch-all option does not apply.

Testing each candidate acceleration against the stated tension ratio shows only one value reproduces the exact \( 1:2 \) relationship between downward and upward tension.

So the correct answer is \( g/3 \).