Neural: Numerical Problems Solved Step-by-Step

medium CBSE NEET 2 min read

Question

A nerve impulse travels along a myelinated axon at 120 m/s. If the axon is 1.8 m long (from spinal cord to big toe), how long does the signal take to reach the toe? Compare with an unmyelinated axon conducting at 2 m/s.

Solution — Step by Step

Myelinated speed v1=120v_1 = 120 m/s, unmyelinated v2=2v_2 = 2 m/s, length L=1.8L = 1.8 m. We want time t=L/vt = L/v for both.

t1=1.8120=0.015t_1 = \dfrac{1.8}{120} = 0.015 s = 15 ms. That is faster than a blink.

t2=1.82=0.9t_2 = \dfrac{1.8}{2} = 0.9 s. Nearly one full second — your toe would feel pain way after you stepped on the nail.

t2t1=60\dfrac{t_2}{t_1} = 60. Myelination makes the signal 60× faster here. This is exactly why multiple sclerosis (which damages myelin) causes slow, uncoordinated movements.

Final answer: Myelinated: 15 ms. Unmyelinated: 900 ms. Ratio: 60:1.

Why This Works

Saltatory conduction lets the action potential ‘jump’ from one Node of Ranvier to the next instead of regenerating at every patch of membrane. The myelin sheath is a lipid insulator, so current leaks less and the signal travels further before needing a boost.

Alternative Method

If the question gives you internodal distance and time per node instead of bulk speed, use v=dnode/tnodev = d_{node}/t_{node} and multiply accordingly.

Most neural-system numericals come down to unit hygiene and remembering what each gate does at each phase. Draw the action potential graph before reading the question — it saves time.

Common Mistake

Students often forget to convert units — if length is in cm and speed in m/s, you’ll be off by 100×. Always line up SI units first.

Do not confuse passive channels (follow the gradient) with active pumps (fight the gradient). This single distinction clears half of all neural-system doubts.

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