Question

$\underline{\textbf{solve for the 100 points }}$

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Answer 1

Explanation:-

Young modulus of the wire is given by :-

$y =\dfrac{F}{A}\dfrac{\Delta l }{l}$

Now, we have to differentiate dy /y

$\dfrac{dy}{y} = \dfrac{df}{F}+\dfrac{dA}{A}+\dfrac{dl}{l}+ \dfrac{d(\Delta l)}{\Delta l}$

Now, in this question there is only $\Delta l$ calculation has found to be error in its young modulus.

So,

%error in y = $\dfrac{dy}{y}\times 100$

$\dfrac{d\Delta l}{\Delta l}\times 100$

$d \Delta l = 1 \times 10^{-5}$

$\Delta l = 25 \times 10{-5}$

Now,

$=\dfrac{d \Delta l}{\Delta l }\times 100$

$=\dfrac{1 \times 10^{-5}}{25 \times 10^{-5}}\times 100$

$=\dfrac{1}{25}\times 100$

% error of y $= 4\%$

hence, the percentage errors in young modulus of the wire is 4%.

Note :- It's better to take the help of expert teachers when solving this type of question. Because I m also not able to understand this question as much as possible.

Answer 2

Answer: 4%

Explanation:

We know that, there are two wires in this experiment, first one is reference wire and second is experimental wire. Experimental wires carries  loads while reference wire remains constant. Now, In the first case bubble is perfectly situated in horizontal position. We we install a load on experimental wire, bubble goes up with respect to reference wire. Also when bubble is perfectly set, We note the reading of main scale and spherical scale and reading after installing a load. Thus we find difference of both measurement and load installed by us.

In this experiment we have already measured the cross section area of wire and the length of wire that treated as having  zero error.

Now, As in this question vernier scale lies between same range in both cases(first is when there is load1 and second one is when there is load2).

So reading of main scale didn't change.

Let the least count of spherical scale be LC.

Main Scale Reading in both cases = x

In the first case for length of wire:-

Reading of main scale = x

Reading of spherical scale = 20×LC

Total reading = x + (20×LC)

In the second case length of wire:-

Reading of main scale = x

Reading of spherical scale = 45×LC

Total reading = x + (45×LC)

Change in the length(l) = x + (45×LC) -  {x + (20×LC)}

                                           = 25×LC

Maximum possible error in this measure will be equal to Least Count of Spherical Scale. Hence,

Δl = LC

We know that Young's Modulus,

Y = FL/lA,

In this experiment we are not considering error in  measure of F, L and A are already said.

Hence, % error in  Y,

$\frac{\triangle Y}{Y}\times100=\frac{\triangle l}{l}\times100\\\;\\\frac{\triangle Y}{Y}\times100=\frac{LC}{25\times LC}\times100\\\;\\\frac{\triangle Y}{Y}\times100=\frac{100}{25}\\\;\\\frac{\triangle Y}{Y}\times100=4\;\%$