Physics

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?


ANSWER

$$3\%$$


SOLUTION

Time period of free fall

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

Take log on both side and differentiate.

$$\dfrac{\Delta t}{t}=\dfrac{1}{2}\left( \dfrac{\Delta h}{h}+\dfrac{\Delta g}{g} \right)=\dfrac{1}{2}\left( 0.04+0.02 \right)=0.03$$

Hence, error in time is $$3\,\%$$ 

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