AI Mentor

Check Your IQ

Free Expert Demo

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

In the determination of young’s modulus $Y = \frac{4 M L g}{{πld}^{2}}$ by using Searle’s method, a wire of length L = 2m and diameter 0.5mm is used. For a load of M = 2.5 kg an extension of l = 0.25mm in the length of the wire is observed. Quantities ‘d’ and ‘l’ are measured using screw gauge and micrometer respectively. They have the same pitch of 0.5mm.The number of divisions of their circular scale is 100. The contribution to the maximum possible error of the Y measurement is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

due to the error in the measurement of ‘d’ is four times that the due to the error in the measurement of ‘l’

due to errors in the measurements of ‘d’ and ‘l’ are same

due to errors in the measurements of d is twice that due to the error in the measurement of ‘l’

due to the error in the measurement of ‘l’ is twice that due to the error in the measurement of ‘d’

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

errors due to measurement of d $2 \frac{∆ d}{d} = 2 \frac{0.5 \times 10^{- 2}}{0.5} = 2 \times 10^{- 2}$

error due to measurements of l $\frac{∆ l}{l} = \frac{0.5 \times 10^{- 2}}{0.25} = 2 \times 10^{- 2}$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

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

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!