A simple pendulum is being used to determine the value of gravitational acceleration (g) at a certain place the length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s , the accuracy in g is x(1.10)% . Then the value of x is ________

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A simple pendulum is being used to determine the value of gravitational acceleration $(g)$ at a certain place the length of the pendulum is $25.0 c m$ and a stop watch with $1 s$ resolution measures the time taken for 40 oscillations to be $50 s$ , the accuracy in $g$ is $x (1.10) %$ . Then the value of x is _________

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 4.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given length of pendulum $l = 25.0 c m$
So, there is an uncertainity of 0.1 cm in measurement of length
Resolution of stop watch is 1 sec, so uncertainity in measurement of time is 1 sec
Now using $T = 2 π \sqrt{\frac{l}{g}}$ (or) $g = 4 π^{2} (\frac{l}{g})$
We have $\frac{Δ g}{g} \times 100 = [\frac{Δ l}{l} + \frac{2 (Δ T)}{T}] \times 100 %$
Accuracy in measurement of g is $\frac{Δ g}{g} \times 100 % = [\frac{0.1}{25} + \frac{2 (1)}{50}] \times 100 % = (0.004 + 0.04) \times 100 % = 4.40 % = 4 (1.10) %; x = 4$