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Physics: Post your doubts here!

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Total Resistance in the Loop where V1 is connected = 2 ohm
Total Resistance in the loop where v2 is connected = 1 ohm
Now, V1 > v2

Total current in the circuit will be divided equally in the second loop where I2 is labelled so I2 is half the current in the circuit. But I1 must be greater than the current in 6 ohm resistor. 2 (I2) = I1 + Current in 6 ohm resistor. Current in 6 ohm resistor must be less than I2.
Hence I1 > I2
 
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You got me confused by the other answer
For A
Take combined resitance in both loops .
Right loop; Left loop;
1 ohm 2 ohm

V1 + V2 = emf
hence p.d V1 > V2
B and D are incorrect.


Right loop;
Both branches have equal resistances.
Right loop;
more current in 3 ohm resistor than 6 ohm (V=IR)

This gives I1 >I2
 
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Can anyone solve this...
6 A single sheet of aluminium foil is folded twice to produce a stack of four sheets. The total
thickness of the stack of sheets is measured to be (0.80 ± 0.02) mm. This measurement is made
using a digital caliper with a zero error of (−0.20 ± 0.02) mm.
What is the percentage uncertainty in the calculated thickness of a single sheet?
A 1.0% B 2.0% C 4.0% D 6.7%
 
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Solve it please with vivid working!
R is directly proportional to the Length. So 4 times length = 4 times resistance
R is inversely proportional to the diameter square. Diameter is half so resistance is quadrupled. (4 * Resistance)
Combining both = 4 * 4 * R = 192 ohm..... I don't know if it is right :p
 
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What is the order of magnitude of the Young modulus for a metal such as copper?
A: 10^–11 Pa B: 10^–4 Pa C: 10^4 Pa D: 10^11 Pa
9702_w16_qp_11 Q.1
How do we even answer such questions?
 
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84
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Can anyone solve this...
6 A single sheet of aluminium foil is folded twice to produce a stack of four sheets. The total
thickness of the stack of sheets is measured to be (0.80 ± 0.02) mm. This measurement is made
using a digital caliper with a zero error of (−0.20 ± 0.02) mm.
What is the percentage uncertainty in the calculated thickness of a single sheet?
A 1.0% B 2.0% C 4.0% D 6.7%

Answer is 4%?
 
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You got me confused by the other answer
For A
Take combined resitance in both loops .
Right loop; Left loop;
1 ohm 2 ohm

V1 + V2 = emf
hence p.d V1 > V2
B and D are incorrect.


Right loop;
Both branches have equal resistances.
Right loop;
more current in 3 ohm resistor than 6 ohm (V=IR)

This gives I1 >I2
thanks
 
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Total Resistance in the Loop where V1 is connected = 2 ohm
Total Resistance in the loop where v2 is connected = 1 ohm
Now, V1 > v2

Total current in the circuit will be divided equally in the second loop where I2 is labelled so I2 is half the current in the circuit. But I1 must be greater than the current in 6 ohm resistor. 2 (I2) = I1 + Current in 6 ohm resistor. Current in 6 ohm resistor must be less than I2.
Hence I1 > I2
THanks
 
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Thanks a lot for the answer .... but i really dont know how you get there...
I was interested to know whether my answer is right or not before posting the solution.

There's a negative zero error in the device so you need to add the values. (0.8 + 0.2) Uncertainty = (0.02 + 0.02),
Now get the percent uncertainty, (0.04/1) *100
 
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What is the order of magnitude of the Young modulus for a metal such as copper?
A: 10^–11 Pa B: 10^–4 Pa C: 10^4 Pa D: 10^11 Pa
9702_w16_qp_11 Q.1
How do we even answer such questions?
Just remember the YOUNG modulus for any object is 10^11.
learn it.
I found it on Znotes.
 
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R is directly proportional to the Length. So 4 times length = 4 times resistance
R is inversely proportional to the diameter square. Diameter is half so resistance is quadrupled. (4 * Resistance)
Combining both = 4 * 4 * R = 192 ohm..... I don't know if it is right :p
Thanks it make some sense. :cool:
 
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Yeah its right... but why do you add 0.8 and 0.2 rather than subtracting, because the zero error is minus 0.2

In general any zero error is subtracted. In this case (0.8 - (-0.2) ) = 0.8 + 0.2

It's logical. Suppose you're using a micrometer screw gauge and you close it to the end, it doesn't show 0. Instead, it goes beyond 0 which means that it has a negative zero error. For every measurement you take, it will give you a value which is less than the real value and when you add the zero error to the measurement, you will get the real value for it.
 
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