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The time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration `g/2`, the time period of the pendulum will be ______.

#### Options

`sqrt(3/2)"T"`

`"T"/sqrt3`

`sqrt(2/3)"T"`

`sqrt3"T"`

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#### Solution

`bb(sqrt(2/3)"T")`

**Explanation:**

Using the expression,

T = 2π `sqrt(l/g)`

A pseudo force is exerted downwards when the lift moves with an acceleration of `"g"/2` in the upward direction.

Effective acceleration

`g_"eff" = "g" + "g"/2`

= 3g/2

Therefore, the new time period

T' = `2pisqrt(l/("g"_"eff"))`

= `2pisqrt(l/(3"g"//2))`

= `sqrt(2/3)"T"`.

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