While measuring the time of fall of a body from a height, a person makes an error or $2 %$ in the value of acceleration due to gravity and an error of $4 %$ in the value of height. The percentage error in the measurement of time of fail is?

Zero

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1 %

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6 %

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3 %

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Solution

The correct option is D $3 %$
Time period of free fall

$t = \sqrt{\frac{2 h}{g}}$

Take log on both side and differentiate.

$\frac{Δ t}{t} = \frac{1}{2} (\frac{Δ h}{h} + \frac{Δ g}{g}) = \frac{1}{2} (0.04 + 0.02) = 0.03$

Hence, error in time is $3 %$

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While measuring the time of fall of a body from a height, a person makes an error or 2% in the value of acceleration due to gravity and an error of 4% in the value of height. The percentage error in the measurement of time of fail is?

The correct option is D 3%Time period of free fall t=√2hg Take log on both side and differentiate. Δtt=12(Δhh+Δgg)=12(0.04+0.02)=0.03 Hence, error in time is 3%

While measuring the time of fall of a body from a height, a person makes an error or $2 %$ in the value of acceleration due to gravity and an error of $4 %$ in the value of height. The percentage error in the measurement of time of fail is?

The correct option is D $3 %$
Time period of free fall

$t = \sqrt{\frac{2 h}{g}}$

Take log on both side and differentiate.

$\frac{Δ t}{t} = \frac{1}{2} (\frac{Δ h}{h} + \frac{Δ g}{g}) = \frac{1}{2} (0.04 + 0.02) = 0.03$

Hence, error in time is $3 %$