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Physics: Post your doubts here!

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hahahaha tukka :p

s = ut + (1/2)at²
L = (0)(t) + (1/2)(9.8)(T)²
L = 4.9T²

Now in da same equation, put t = 0.5T
s = ut + (1/2)at²
s = (0)(t) + (1/2)(9.8)(0.5T)²
s = 4.9(0.25T²)
s = 0.25(4.9T²)

Substitute value of L from first equation
s = 0.25L

ohhhhh ._. got it ._. this came in my mid term exam ._. tukka saved me ._.
 
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T
Sorry for the late reply, wasn't online for quite some time.

Let's take this option by option, with the following formula in focus:

ρ = RA/l

Where ρ = resistivity of the material through which current is being passed,
R = resistance of the sample through which current is being passed,
A = Area of the cross section of the sample through which current is being passed (perpendicular to the direction of current), and
l = length of cross section of sample through which current flows (i.e. the distance through which the current flows in the sample used).

A neat way of finding out the area A is by imagining the way the current passes through the sample (say left to right) and imagining using a large blade/ knife to cut it off in between (grotesque, but bear with me for this bit. Alternatively, you can put up a plane perpendicular to the direction of the current and use that). The blade is perpendicular to the current, and the area that the sample occupies on the face of the blade is your area.

So, for option A, let's see what the formula tells us:

If the cube has a one meter side and the current is passed from one face to another, we can say that the current travels 1 meter from one end of the cube to the other; therefore, l = 1 meter. Alternatively, the current first enters the cube on one side, travels 1 meter, and exits the cube, giving us the same value.

Further, the cross section through which the current travels, perpendicular to the direction of current - suppose you place the cube on the table, and cut it parallel to the edges, the area you get after the cut is an area of 1 meter x 1 meter = 1 m^2.
Another way of getting this is to imagine there is no current passing through the cube. You close the circuit, and current starts flowing. Soon, it gets to the beginning of the cube, and starts moving through it. Suppose no charge carrier (electron/proton, either is fine) travels faster than another, you have a "wall" of such charge carriers advancing along the cube. Now ask yourself. How large is that wall? What is the area of that wall? In this case, the area of that wall is 1 m^2, which is our result.

Putting these in the equation, we get

ρ = R * 1 m^2/1 m = R
So ρ = R
Therefore, A is right - the value of ρ is numerically equal to the value of the resistance of the cube in this situation. But for the others, what can we say?

Option B: Doing the maths here again, we can say that the length through which the current passes is l = 1 meter and the cross sectional area through which the current passes is 1 mm^2 = (1/1000 meters)^2 = (10^-3)^2 = 10^-6.
Putting it in the equation,

ρ = R * 10^-6/1 = 10^-6 * R
Which is not right.

Option C: This has already been answered before by usama321 and Asad rehman - the option says that resistivity is dependent on the cross sectional area of the sample used. Even though the formula says that ρ = R/l * A, the wording of the option is something very precise; when it says proportional, what it means is

"...a change in A results in a change in ρ, such that the change in A can be equated to the corresponding change in ρ if a suitable constant k is introduced as a factor of change (i.e. ΔA = k * Δρ)"

This is, of course, false. At a fixed temperature, given no other external electric or magnetic interference (literally and figuratively), the resistivity of a material will not change with any change in area. Instead, the resistance will change to ensure that the value of ρ remains the same. Since resistivity does not depend on the dimensions of a sample and only on the innate nature of the material, C cannot be right.

Changing the cross-sectional area of a sample may change the resistance, but it cannot change the resistivity.

Option D: The same argument as option C can be supplied here, replacing Cross Sectional Area A with length l.

Hope this helped!

Good Luck for all your exams!
(BTW, Thought blocker, thanks for the symbols, bro! Credit to you!)
Look at the line which you said:
Putting these in the equation, we get

ρ = R * 1 m^2/1 m = R
So ρ = R
wont ρ =Rm?
 
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T

Look at the line which you said:
Putting these in the equation, we get

ρ = R * 1 m^2/1 m = R
So ρ = R
wont ρ =Rm?

Yes, ρ = (R Ω) m, but those are a matter of units - the units for the final answer will be in Ωm, which are the units of resistivity (and it should be no other way, to be honest) but that's where the question becomes very specific - it states:

".....resistivity of a material is numerically equal to the resistance in ohms...."

so the units don't matter - the number you get in the end is what matters, i.e. the number you get as a result for the resistance will be the same number for the value of the resistivity. Only the number itself matters, so while the units are important in all situations, the question isn't concerned with the units. So fair play!

Hope this helped!
Good Luck for all your exams!
(and again, credit to Thought blocker for the symbols!)
 
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Yes, ρ = (R Ω) m, but those are a matter of units - the units for the final answer will be in Ωm, which are the units of resistivity (and it should be no other way, to be honest) but that's where the question becomes very specific - it states:

".....resistivity of a material is numerically equal to the resistance in ohms...."

so the units don't matter - the number you get in the end is what matters, i.e. the number you get as a result for the resistance will be the same number for the value of the resistivity. Only the number itself matters, so while the units are important in all situations, the question isn't concerned with the units. So fair play!

Hope this helped!
Good Luck for all your exams!
(and again, credit to Thought blocker for the symbols!)
Fair Play, dude. (y) Bas now no thanking. :)
 
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