In the experiment of Ohm’s law, a potential difference of 5.0V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00mm. The measured current in the conductor is 200 A. The maximum permissible percentage error in the resistivity of the conductor 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

In the experiment of Ohm’s law, a potential difference of 5.0V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00mm. The measured current in the conductor is 200 A. The maximum permissible percentage error in the resistivity of the conductor 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

8.4

3.9

3.0

7.5

answer is B.

(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

$O h m' s L a w, V = I R$

$\Rightarrow V = I ρ \frac{4 L}{π d^{2}}$

$\Rightarrow ρ = \frac{π d^{2} V}{I 4 L}$

$\Rightarrow \frac{△ ρ}{ρ} \times 100 = 2 (\frac{△ d}{d} \times 100) + (\frac{△ V}{V} \times 100) + (\frac{△ I}{I} \times 100) + (\frac{△ L}{L} \times 100)$

$\Rightarrow \frac{△ ρ}{ρ} \times 100 = 2 (\frac{0.01}{5.00} \times 100) + (\frac{0.1}{5.0} \times 100) + (\frac{1}{200} \times 100) + (\frac{0.1}{10.0} \times 100)$