Question

17-If the distance between the stimulating electrode and the recording electrode of oscilloscope is d= 2cm and the time taken by the nervous message to be recorded is t= 4ms.What would be the speed of propagation of nervous message? a- 5 m/s b-2 m/s c-0.5 m/s d-0.2 m/s e-None of the above

Answer 1

Answer will be (a) 5 m/ s

explanation::

For calculating speed of propagation of nervous impulse it will be distance between the electrodes divided by time taken by nervous message.

Given ,

d = 2 cm

t = 4 ms

Here unit conversion required as the velocity will be in meter/ second.

First we convert d = 2 centimeter into meter...

2 /100 m = .02 m

t = 4 ms into second (milliseconds to seconds)

4/1000 second = .004 sec

So, velocity of nervous impulse = distance / time

velocity =.02 /.004 m/ sec

=2×1000 / 4×100 m/ sec

= 20 / 4 m/ sec

= 5 m / sec

ANSWER-- 5 m/ sec