Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

In the experiments, the following observations were recorded: $L=2.820\ m,$ $I=0.087\ cm$ , $M=3\ kg$ . Diameter, $D=0.041\ cm$ . Taking $g=9.81\ \frac{m}{s^2}.$ Using the formula,

$Y=\frac{4MgL}{{\pi D}^2l},$ the allowable error Y is:
Option: 1
$7.96%$

Option: 2
$4.56%$

Option: 3
$6.50%$

Option: 4
$8.42%$

Answers (1)

best_answer

$Y=\frac{4MgL}{{\pi D}^2l}$

So maximum permissible error in $Y=\frac{\Delta Y}{Y}\times 100$

$\left ( \frac{\Delta M}{M} +\frac{\Delta g}{g}+\frac{\Delta L}{L}+\frac{2\Delta D}{D}+\frac{\Delta l}{l}\right )\times 100$

$\left ( \frac{1}{300}+\frac{1}{981}+\frac{1}{2820} +2\times \frac{1}{41}+\frac{1}{87}\right )\times 100$

$0.065\times100=6.5$

Posted by

shivangi.bhatnagar

View full answer

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET UG 2026: Bengaluru Traffic Police fact-check finds late start, longer route behind missed re-exam

anam-khan

UP's NEET aspirant dies by suicide in Maharajganj, family says she was under stress after exam

anam-khan

NEET UG 2026 Re-Exam: Law student held for peddling fake re-NEET question paper created by using AI

Latest Question