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Laws of Motion question

2003 · Shift 0 · Q175

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Laws of Motion
2003 · Shift 0 · Q175

Laws of Motion question

2003 · Shift 0 · Q175

JEE MainPhysicsLaws of MotionMCQ+4 / −1

A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads

49N,

when the lift is stationary. If the lift moves downward with an acceleration of

5 m/{s^2}

, the reading of the spring balance will be

A
$24N$
B
$74N$
C
$15N$
D
$49N$

View written solutionFree

Correct answer: A

Reading when the lift is stationary

When the lift is stationary, the spring balance reads the true weight of the bag:

$T = mg = 49\,\text{N}$

So the mass of the bag is

$m = \frac{49}{9.8} = 5\,\text{kg}$

Lift moving downward with acceleration

If the lift accelerates downward with acceleration $a = 5\,\text{m/s}^2$ , then the apparent weight becomes

$T' = m(g-a)$

Substitute the values:

$T' = 5(9.8-5) = 5(4.8) = 24\,\text{N}$

Check the options

A: $24\,\text{N}$ ✅
B: $74\,\text{N}$ ❌
C: $15\,\text{N}$ ❌
D: $49\,\text{N}$ ❌

Therefore, the spring balance reading is

$\boxed{24\,\text{N}}$

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