In a vertical U-tube a column of mercury oscillates simple harmonically. If the tube contains $1$ kg of mercury and $1$ cm of mercury column weighs $20$ g, then the period of oscillation is

1

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2

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\sqrt{2}

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Insufficient data

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Solution

The correct option is B $1$ s
$T = 2 π \sqrt{\frac{l}{g}}$
Total length of the tube is $2 l = \frac{1000}{20} = 50 c m = 0.5 m$
$\Rightarrow l = 0.25 m$
Now $T = 2 π \sqrt{\frac{0.25}{9.8}} = 1 s$

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In a vertical U-tube a column of mercury oscillates simple harmonically. If the tube contains 1 kg of mercury and 1 cm of mercury column weighs 20 g, then the period of oscillation is

The correct option is B 1 sT=2π√lgTotal length of the tube is 2l=100020 =50 cm=0.5 m⇒l=0.25 mNow T=2π√0.259.8=1 s

In a vertical U-tube a column of mercury oscillates simple harmonically. If the tube contains $1$ kg of mercury and $1$ cm of mercury column weighs $20$ g, then the period of oscillation is

The correct option is B $1$ s
$T = 2 π \sqrt{\frac{l}{g}}$
Total length of the tube is $2 l = \frac{1000}{20} = 50 c m = 0.5 m$
$\Rightarrow l = 0.25 m$
Now $T = 2 π \sqrt{\frac{0.25}{9.8}} = 1 s$