What is the approximate time period of a $3.98 m$ long pendulum?

4 s

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2 s

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8 s

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1 s

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Solution

The correct option is A $4 s$
$T = 2 π \sqrt{\frac{L}{g}} = 2 \times 3.14 \times \sqrt{\frac{3.98}{9.8}} s$

$\Rightarrow 2 \times 3.14 \times \sqrt{0.41} s = (2 \times 3.14 \times 0.64) s = 4.01925 s = 4 s$ (approx.)

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What is the approximate time period of a $3.98 m$ long pendulum?

The correct option is A 4s T=2π√Lg=2×3.14×√3.989.8s⇒2×3.14×√0.41s=(2×3.14×0.64)s=4.01925s=4s (approx.)

The correct option is A $4 s$
$T = 2 π \sqrt{\frac{L}{g}} = 2 \times 3.14 \times \sqrt{\frac{3.98}{9.8}} s$

$\Rightarrow 2 \times 3.14 \times \sqrt{0.41} s = (2 \times 3.14 \times 0.64) s = 4.01925 s = 4 s$ (approx.)