The circumference of a circle is measured as | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the circumference of a circle $S = 2\pi r = 56 	ext{ cm}$.From this,the radius is $r = \frac{56}{2\pi} = \frac{28}{\pi} 	ext{ cm}$.The err

Vedclass · Accepted Answer

$1/14$

Vedclass · Answer

(C) Given the circumference of a circle $S = 2\pi r = 56 	ext{ cm}$.From this,the radius is $r = \frac{56}{2\pi} = \frac{28}{\pi} 	ext{ cm}$.The error in circumference is given by $\delta S = 2\pi \delta r = 0.02 	ext{ cm}$.Thus,the error in radius is $\delta r = \frac{0.02}{2\pi} 	ext{ cm}$.The area of the circle is $A = \pi r^2$.The relative error in area is given by $\frac{\delta A}{A} = 2 \frac{\delta r}{r}$.Substituting the values,we get $\frac{\delta A}{A} = 2 	imes \frac{0.02 / (2\pi)}{28 / \pi} = 2 	imes \frac{0.02}{2\pi} 	imes \frac{\pi}{28} = \frac{0.02}{28} = \frac{2}{2800} = \frac{1}{1400}$.The percentage error in area is $\frac{\delta A}{A} 	imes 100 = \frac{1}{1400} 	imes 100 = \frac{1}{14} \%$. Therefore,the percentage error is $\frac{1}{14} \%$.

The circumference of a circle is measured as $56 \text{ cm}$ with an error of $0.02 \text{ cm}$. The percentage error in its area is

Similar Questions

Three waves of equal frequency having amplitudes $10\, \mu m, 4\, \mu m$ and $7\, \mu m$ arrive at a given point with successive phase difference of $\pi/2$. The amplitude of the resulting wave in $\mu m$ is given by

Number of moles of $MnO_4^-$ required to oxidize one mole of ferrous oxalate completely in acidic medium will be .............. $moles$.

If $|a|=1, |b|=2$ and the angle between $a$ and $b$ is $120^{\circ}$,then ${(a+3b) \times (3a-b)}^2$ is equal to

Which of the following cranial nerves is present in the rabbit but absent in the frog?

Which kingdoms include decomposers?

Vedclass Test Series

Exam Paper Generator

Online Exam Module