The time period T of a simple pendulum of len | vedclass

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

Question

(B) Given, $T = 2 \pi \sqrt{\frac{l}{g}}$.Taking the natural logarithm on both sides, we get $\ln T = \ln(2 \pi) + \frac{1}{2} \ln l - \frac{1}{2} \ln

Vedclass · Accepted Answer

$1 \%$

Vedclass · Answer

(B) Given, $T = 2 \pi \sqrt{\frac{l}{g}}$.Taking the natural logarithm on both sides, we get $\ln T = \ln(2 \pi) + \frac{1}{2} \ln l - \frac{1}{2} \ln g$.Differentiating both sides, we get $\frac{dT}{T} = \frac{1}{2} \frac{dl}{l}$.Given that the length is increased by $2 \%$, so $\frac{dl}{l} = 0.02$.Substituting this value, we get $\frac{dT}{T} = \frac{1}{2} 	imes 0.02 = 0.01$.Therefore, the percentage change in the time period is $\frac{dT}{T} 	imes 100 = 0.01 	imes 100 = 1 \%$.Hence, the approximate change in the time period is $1 \%$.

The time period $T$ of a simple pendulum of length $l$ is given by $T=2 \pi \sqrt{\frac{l}{g}}$. If the length is increased by $2 \%$, then the approximate change in the time period is:

Similar Questions

The approximate value of $\sqrt{63}$ is

If the percentage error in the radius of a circle is $3\%$,then the percentage error in its area is:

Use differentials to approximate $(25)^{\frac{1}{3}}$.

If the radius of a sphere is measured as $9 \ cm$ with an error of $0.03 \ cm$, then find the approximate error in calculating its surface area. (in $\pi \ cm^2$)

Find the approximate value of $(26)^{\frac{1}{3}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module