The length of a germanium rod is 0.58 cm and its area of cross-section is 1 mm². if for Germanium $n_{i} = 2 \cdot 5 \times 10^{19} m^{- 3}, μ_{h} = 0 \cdot 19 m^{2} / V - s$ $μ_{e} = 0 \cdot 9 m^{2} / V - s$ , then the resistance of the rod will be

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

$5 \cdot 0 k Ω$

$2 \cdot 5 kΩ$

$10 k Ω$

$7 \cdot 5 kΩ$

answer is A.

(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

$R = \frac{ρL}{A} = \frac{L}{n_{i} e (μ_{e} + μ_{h}) A} = \frac{0 \cdot 58 \times 10^{- 2}}{(2 \cdot 5 \times 10^{19}) (1 \cdot 6 \times 10^{- 19}) [0 \cdot 19 + 0 \cdot 39] \times (1 \times 10^{- 6})} = 2 \cdot 5 kΩ$

Watch 3-min video & get full concept clarity

hear from our champions

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!