ALL CIE PHYSICS DOUBTS HERE !!!

[ashwinkandel]

Hello Sir(astarmathsandphysics) Can you give me explanation for q no. 7. If height falls by h/2 then how loss in GPE is mgh/4

[astarmathsandphysics]

cos only half the mass falls ie moves from one tank to the other

[angel_ak]

answer my question too

[astarmathsandphysics]

average diameter =(1.05+1.04+1.06)/3=1.06mm
r=1.06/2=0.53mm
V=pi r^2 l= 3.14*(0.53*10^-3)^2*0.1=8.82*10^-8 m^3
Since the diameter is +-about 2.5% take only 1 sig fig A

[silvercameron]

Q) Find the diameter of a copper wire which has the same resistance as an aluminium wire of equal length and diameter 1.20 mm. The resistivities of copper and aluminium are 1.7 x 10^-8 and 2.6 x 10^-8 respectively.

[angel_ak]

 Thankew sir, i understood the method..
but m confused why we are considering the ans only in 1 sf? :s

[angel_ak]

and if we consider 1sf the answer should be A ie 9 x 10^-3
but the book says ans is D ie 8.8 x 10^-3

[astarmathsandphysics]

cos the estimates for diameter are 5% apart only take 1 sf

[silvercameron]

anyone? Help with the question i already posted ^ pls.

[astarmathsandphysics]

R=rh*l/A=rho*l/(pi*r^2) so r=sqrt((rho*l)/(R pi)) so if diameter (and so R) increase by a factor (1.7*10^-8)/(2.6*10^-8) and diamneter of copper=(1.7*10^-8)/(2.6*10^-8)*1.2mm

[silvercameron]

thankyou sir. But i dont understand that how could you conclude > "diameter of copper=(1.7*10^-8)/(2.6*10^-8)*1.2mm

I tried this today and did it like this:
resistance of aluminium = resistance of copper.

So :    (rho*l)/A  of aluminium= (rho*l)/ A of copper.

 Radius of aluminium = 1.2/2 = 0.6 mm = 6*10^-4 m
Area of aluminium= pi* r² = pi* (6*10^-4)² = 1.131*10^-6

therfore :     ((2.6*10^-8)*l)/(1.13*10^-6)=((1.7*10^-8)*l)/pi*r²

l and l gets canceled cos they are the same.
(2.6*10^-8)/(1.131*10^-6)= (1.7*10^-8)/(pi*r²)
we cross-multiply, get the radius, multiply it by 2 and finally get diameter= 9.72*10 ^-4 m

^^is my method correct?

[astarmathsandphysics]

Yes. I should have square rooted.

[ashwinkandel]

Questions from Electromagnetism
1.
Figure shows a wire XY which carries a direct current. Plotting compass R, placed alongside the wire, points due north. Compass P is placed below the wire and compass Q is placed above the wire.
a. State the direction of the current in the wire.
b. State in which direction compass P points
c. State in wihch direction compass Q points if the current in the wire is reversed.

2.
At a point on the Earth's surface the horizontal component of the Earth's magnetic field is 1.6*10^-5 T. A piece of wire 3.0 m long and of weight 0.020N lies in the east-west direction on a labortary bench. When a large current flows in the wire, the wire just lifts off the surface of the bench.
i. State the direction of the current in the wire.[1]
ii. Calculate the minimum current needed to lifet the wire from the bench. [3]

[astarmathsandphysics]

a.righthand grip rule current flows from Y yo X
b. same as Q
c.North West
2. by flemings left hand rule current is from west to east
ii. F=BIl
0.02=1.6*10^-5*I*3 so I=0.02/(1.6*10^-5*3)417A

[ashwinkandel]

Can you tell me actually the use of the compass here. What does it indicate or shows?