It is usually assumed that an action potential begins immediately at the cathode. If this were true, both methods for calculating conduction velocity would provide the same answer. However, when a strong stimulus intensity is used, the action potential may begin some distance away from the cathode. Under these conditions, the difference method would be more accurate. Did you observe any important difference between the conduction velocity values?

Provided in the explanation section

Explanation:

Our questions says that:

It is usually assumed that an action potential begins immediately at the cathode. If this were true, both methods for calculating conduction velocity would provide the same answer. However, when a strong stimulus intensity is used, the action potential may begin some distance away from the cathode. Under these conditions, the difference method would be more accurate.Did you observe any important difference between the conduction velocity values?

Answer to this :

By using the difference method, you subtract out any "uncertainties" involved in the measurement of latencies. Say for example, saw we are uncertain as to where the AP's are actually originating within the vicinity of the stimulating electrodes, this "error" will be introduced into both latency measurements, and therefore subtracted out when performing a difference method calculation.

However, the difference method is only experimentally sound when one is dealing with the same population of nerve fibres over the recording electrodes used, which is not the case with the sciatic nerve, as it is a short nerve, and thin at one end.

    The non-uniformity of the nerve, and the difficulty in making accurate measurements of very small distances and latencies are principal points to consider when making conduction velocity measurements. Naturally if the nerve studied were longer and more uniform, we would improve the accuracy of our calculations.

cheers i hope this helped

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current