# Physics multiple choice help ✍️

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Hello, I have come across some multiple choice questions when revising. I find that usually that this is one of my weaker areas since I tend to rush through and overlook my answers so I would really appreciate if anyone was able to look over my answers and offer any advice to the methods I used to solve them.

Question A:

Realising that the resistivity of aluminium is 2.7 x 10^-8 Ωm, and that a piece of aluminium wire of length 50 cm has a resistance of 3.7 x10^-3 Ω approximately find the diameter of the cross section of the wire.

1. 21.6 mm

2. 2.16 mm

3. 10.8 mm

4. 1.08 mm

So rearranging the equation R=pL/A so that A=pL/R

A=(2.7 x 10-^8)*0.5/3.7 x10^-3

A=3.64864849 * 10 ^-6

Since A =π r^2

When A = 3.64864849 * 10 ^-6

3.64864849 * 10 ^-6 /π=1.161400936*10^-6

Then r^2= 1.161400936*10^-6

Taking the square root to find r:

r=1.077683133*10^-3 m ~ 1.08 mm ( option 4)

Question B. Various electrical products have a rating of milli-Amp-hours (mAh), which is a rating of:

1. Charge

2. Voltage

3. Power

4. Energy

5. Current

I believe that this is a measure of 1. charge.

Question C. Doubling the diameter of the wire has the following effect on drift velocity:

1. Doubles

2. Halves

3. Quadruples

4. Quarters

I believe, through attempting several arbitrary calculations, that this would divide the drift velocity by four, so I am assuming that the correct answer is 4.

Question D. If electrons generate a current of 0.2 mA through a semiconductor strip that has a cross-sectional area of 1 cm^2, their drift velocity will be of which order of magnitude?

A. ~0.001 ms-1

B. ~0.01 ms-1

C. ~1 ms-1

D. ~100 ms-1

I am uncertain how to answer this question and would appreciate some help as to how I should do so.

Question E. The SI base unit for the coulomb is:

1. A s^-1

2. A s

3. Ohm / Volt

4. W/(Vs)

I think that the SI unit for coulomb is 2 As.

Question F. The number of electrons flowing through a filament lamp in one minute when the current is 800 mA is;

1. 5x10^19

2. 3x10^20

3. 5x10^21

4. 3x10^22

Q=I*T

Q=0.8*60

Q=480 C

Q=n*e

n=Q/e

n=480/(1.6*10^-19)

n=3.0*10^21

But my answer does not correlate to any of the options given. Where have I gone wrong?

Question G. Which is a semiconductor?

1. Glass

2. Germanium

3. Common salt

4. Gold

I think that 2. germanium is the semiconductor.

Question H. At 300 K, Germanium has 2.4 x 10^19 charge carriers per m^3. What current (of electrons) would provide a drift velocity of 0.8 ms-1 when flowing through a strip of Germanium with a rectangular cross-section of 8 mm x 0.4 mm?

A. 9.8 μA

B. 9.8 mA

C. 6.1 μA

D. 6.1 mA

I=nAve

Area=0.008 * 0.0004=3.2*10^-6 m^2

I=2.4 x 10^19*(3.2m*10^-6 )*0.8*(1.6*10^-19)

I=9.8304*10^-6 A

Again I think that my calculations must be incorrect since I have not found a solution to correspond to the given choices.

Question I. The resistivity of copper is 1.72 x 10^-8 Ωm. Connecting two identical 2 m strands of copper wire with a diameter of 4 mm each gives a resistance of:

1. 6.84 x 10^-4 Ω

2. 1.37 x 10^-3 Ω

3. 2.74 x 10^-3 Ω

4. 5.47 x 10^-3 Ω

R=pL/A

R=1.72 x 10^-8*2/π *0.004^2

R=0.00068 ~ 6.84 x 10^-4 Ω (answer 1)

Thank you to anyone who is kind enough to offer their time to help 😁

Question A:

Realising that the resistivity of aluminium is 2.7 x 10^-8 Ωm, and that a piece of aluminium wire of length 50 cm has a resistance of 3.7 x10^-3 Ω approximately find the diameter of the cross section of the wire.

1. 21.6 mm

2. 2.16 mm

3. 10.8 mm

4. 1.08 mm

So rearranging the equation R=pL/A so that A=pL/R

A=(2.7 x 10-^8)*0.5/3.7 x10^-3

A=3.64864849 * 10 ^-6

Since A =π r^2

When A = 3.64864849 * 10 ^-6

3.64864849 * 10 ^-6 /π=1.161400936*10^-6

Then r^2= 1.161400936*10^-6

Taking the square root to find r:

r=1.077683133*10^-3 m ~ 1.08 mm ( option 4)

Question B. Various electrical products have a rating of milli-Amp-hours (mAh), which is a rating of:

1. Charge

2. Voltage

3. Power

4. Energy

5. Current

I believe that this is a measure of 1. charge.

Question C. Doubling the diameter of the wire has the following effect on drift velocity:

1. Doubles

2. Halves

3. Quadruples

4. Quarters

I believe, through attempting several arbitrary calculations, that this would divide the drift velocity by four, so I am assuming that the correct answer is 4.

Question D. If electrons generate a current of 0.2 mA through a semiconductor strip that has a cross-sectional area of 1 cm^2, their drift velocity will be of which order of magnitude?

A. ~0.001 ms-1

B. ~0.01 ms-1

C. ~1 ms-1

D. ~100 ms-1

I am uncertain how to answer this question and would appreciate some help as to how I should do so.

Question E. The SI base unit for the coulomb is:

1. A s^-1

2. A s

3. Ohm / Volt

4. W/(Vs)

I think that the SI unit for coulomb is 2 As.

Question F. The number of electrons flowing through a filament lamp in one minute when the current is 800 mA is;

1. 5x10^19

2. 3x10^20

3. 5x10^21

4. 3x10^22

Q=I*T

Q=0.8*60

Q=480 C

Q=n*e

n=Q/e

n=480/(1.6*10^-19)

n=3.0*10^21

But my answer does not correlate to any of the options given. Where have I gone wrong?

Question G. Which is a semiconductor?

1. Glass

2. Germanium

3. Common salt

4. Gold

I think that 2. germanium is the semiconductor.

Question H. At 300 K, Germanium has 2.4 x 10^19 charge carriers per m^3. What current (of electrons) would provide a drift velocity of 0.8 ms-1 when flowing through a strip of Germanium with a rectangular cross-section of 8 mm x 0.4 mm?

A. 9.8 μA

B. 9.8 mA

C. 6.1 μA

D. 6.1 mA

I=nAve

Area=0.008 * 0.0004=3.2*10^-6 m^2

I=2.4 x 10^19*(3.2m*10^-6 )*0.8*(1.6*10^-19)

I=9.8304*10^-6 A

Again I think that my calculations must be incorrect since I have not found a solution to correspond to the given choices.

Question I. The resistivity of copper is 1.72 x 10^-8 Ωm. Connecting two identical 2 m strands of copper wire with a diameter of 4 mm each gives a resistance of:

1. 6.84 x 10^-4 Ω

2. 1.37 x 10^-3 Ω

3. 2.74 x 10^-3 Ω

4. 5.47 x 10^-3 Ω

R=pL/A

R=1.72 x 10^-8*2/π *0.004^2

R=0.00068 ~ 6.84 x 10^-4 Ω (answer 1)

Thank you to anyone who is kind enough to offer their time to help 😁

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(Original post by

Hello, I have come across some multiple choice questions when revising. I find that usually that this is one of my weaker areas since I tend to rush through and overlook my answers so I would really appreciate if anyone was able to look over my answers and offer any advice to the methods I used to solve them.

Question A:

Realising that the resistivity of aluminium is 2.7 x 10^-8 Ωm, and that a piece of aluminium wire of length 50 cm has a resistance of 3.7 x10^-3 Ω approximately find the diameter of the cross section of the wire.

1. 21.6 mm

2. 2.16 mm

3. 10.8 mm

4. 1.08 mm

So rearranging the equation R=pL/A so that A=pL/R

A=(2.7 x 10-^8)*0.5/3.7 x10^-3

A=3.64864849 * 10 ^-6

Since A =π r^2

When A = 3.64864849 * 10 ^-6

3.64864849 * 10 ^-6 /π=1.161400936*10^-6

Then r^2= 1.161400936*10^-6

Taking the square root to find r:

r=1.077683133*10^-3 m ~ 1.08 mm ( option 4)

Question B. Various electrical products have a rating of milli-Amp-hours (mAh), which is a rating of:

1. Charge

2. Voltage

3. Power

4. Energy

5. Current

I believe that this is a measure of 1. charge.

Question C. Doubling the diameter of the wire has the following effect on drift velocity:

1. Doubles

2. Halves

3. Quadruples

4. Quarters

I believe, through attempting several arbitrary calculations, that this would divide the drift velocity by four, so I am assuming that the correct answer is 4.

Question D. If electrons generate a current of 0.2 mA through a semiconductor strip that has a cross-sectional area of 1 cm^2, their drift velocity will be of which order of magnitude?

A. ~0.001 ms-1

B. ~0.01 ms-1

C. ~1 ms-1

D. ~100 ms-1

I am uncertain how to answer this question and would appreciate some help as to how I should do so.

Question E. The SI base unit for the coulomb is:

1. A s^-1

2. A s

3. Ohm / Volt

4. W/(Vs)

I think that the SI unit for coulomb is 2 As.

Question F. The number of electrons flowing through a filament lamp in one minute when the current is 800 mA is;

1. 5x10^19

2. 3x10^20

3. 5x10^21

4. 3x10^22

Q=I*T

Q=0.8*60

Q=480 C

Q=n*e

n=Q/e

n=480/(1.6*10^-19)

n=3.0*10^21

But my answer does not correlate to any of the options given. Where have I gone wrong?

Question G. Which is a semiconductor?

1. Glass

2. Germanium

3. Common salt

4. Gold

I think that 2. germanium is the semiconductor.

Question H. At 300 K, Germanium has 2.4 x 10^19 charge carriers per m^3. What current (of electrons) would provide a drift velocity of 0.8 ms-1 when flowing through a strip of Germanium with a rectangular cross-section of 8 mm x 0.4 mm?

A. 9.8 μA

B. 9.8 mA

C. 6.1 μA

D. 6.1 mA

I=nAve

Area=0.008 * 0.0004=3.2*10^-6 m^2

I=2.4 x 10^19*(3.2m*10^-6 )*0.8*(1.6*10^-19)

I=9.8304*10^-6 A

Again I think that my calculations must be incorrect since I have not found a solution to correspond to the given choices.

Question I. The resistivity of copper is 1.72 x 10^-8 Ωm. Connecting two identical 2 m strands of copper wire with a diameter of 4 mm each gives a resistance of:

1. 6.84 x 10^-4 Ω

2. 1.37 x 10^-3 Ω

3. 2.74 x 10^-3 Ω

4. 5.47 x 10^-3 Ω

R=pL/A

R=1.72 x 10^-8*2/π *0.004^2

R=0.00068 ~ 6.84 x 10^-4 Ω (answer 1)

Thank you to anyone who is kind enough to offer their time to help 😁

**LukeWatson4590**)Hello, I have come across some multiple choice questions when revising. I find that usually that this is one of my weaker areas since I tend to rush through and overlook my answers so I would really appreciate if anyone was able to look over my answers and offer any advice to the methods I used to solve them.

Question A:

Realising that the resistivity of aluminium is 2.7 x 10^-8 Ωm, and that a piece of aluminium wire of length 50 cm has a resistance of 3.7 x10^-3 Ω approximately find the diameter of the cross section of the wire.

1. 21.6 mm

2. 2.16 mm

3. 10.8 mm

4. 1.08 mm

So rearranging the equation R=pL/A so that A=pL/R

A=(2.7 x 10-^8)*0.5/3.7 x10^-3

A=3.64864849 * 10 ^-6

Since A =π r^2

When A = 3.64864849 * 10 ^-6

3.64864849 * 10 ^-6 /π=1.161400936*10^-6

Then r^2= 1.161400936*10^-6

Taking the square root to find r:

r=1.077683133*10^-3 m ~ 1.08 mm ( option 4)

Question B. Various electrical products have a rating of milli-Amp-hours (mAh), which is a rating of:

1. Charge

2. Voltage

3. Power

4. Energy

5. Current

I believe that this is a measure of 1. charge.

Question C. Doubling the diameter of the wire has the following effect on drift velocity:

1. Doubles

2. Halves

3. Quadruples

4. Quarters

I believe, through attempting several arbitrary calculations, that this would divide the drift velocity by four, so I am assuming that the correct answer is 4.

Question D. If electrons generate a current of 0.2 mA through a semiconductor strip that has a cross-sectional area of 1 cm^2, their drift velocity will be of which order of magnitude?

A. ~0.001 ms-1

B. ~0.01 ms-1

C. ~1 ms-1

D. ~100 ms-1

I am uncertain how to answer this question and would appreciate some help as to how I should do so.

Question E. The SI base unit for the coulomb is:

1. A s^-1

2. A s

3. Ohm / Volt

4. W/(Vs)

I think that the SI unit for coulomb is 2 As.

Question F. The number of electrons flowing through a filament lamp in one minute when the current is 800 mA is;

1. 5x10^19

2. 3x10^20

3. 5x10^21

4. 3x10^22

Q=I*T

Q=0.8*60

Q=480 C

Q=n*e

n=Q/e

n=480/(1.6*10^-19)

n=3.0*10^21

But my answer does not correlate to any of the options given. Where have I gone wrong?

Question G. Which is a semiconductor?

1. Glass

2. Germanium

3. Common salt

4. Gold

I think that 2. germanium is the semiconductor.

Question H. At 300 K, Germanium has 2.4 x 10^19 charge carriers per m^3. What current (of electrons) would provide a drift velocity of 0.8 ms-1 when flowing through a strip of Germanium with a rectangular cross-section of 8 mm x 0.4 mm?

A. 9.8 μA

B. 9.8 mA

C. 6.1 μA

D. 6.1 mA

I=nAve

Area=0.008 * 0.0004=3.2*10^-6 m^2

I=2.4 x 10^19*(3.2m*10^-6 )*0.8*(1.6*10^-19)

I=9.8304*10^-6 A

Again I think that my calculations must be incorrect since I have not found a solution to correspond to the given choices.

Question I. The resistivity of copper is 1.72 x 10^-8 Ωm. Connecting two identical 2 m strands of copper wire with a diameter of 4 mm each gives a resistance of:

1. 6.84 x 10^-4 Ω

2. 1.37 x 10^-3 Ω

3. 2.74 x 10^-3 Ω

4. 5.47 x 10^-3 Ω

R=pL/A

R=1.72 x 10^-8*2/π *0.004^2

R=0.00068 ~ 6.84 x 10^-4 Ω (answer 1)

Thank you to anyone who is kind enough to offer their time to help 😁

Secondly - I haven't had time to look through everything yet. But here are a few thoughts.

A: Careful, you are asked for a diameter, not a radius.

D: The thing that is missing from the equation you would use to work this out is the number density. So you need to know a typical number density for a semiconductor. Not an exact value, but a typical value to the nearest power of 10.

F: you say that Q = 0.8 * 60 = 480 C. This can't possibly be right becasue you are multiplying 60 by a number less than one but end up with a larger answer!

H: Why do you think that your answer is not one of the options?

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(Original post by

Firstly - this would be better asked in the physics section rather than the maths one. You'll get more replies there.

Secondly - I haven't had time to look through everything yet. But here are a few thoughts.

A: Careful, you are asked for a diameter, not a radius.

D: The thing that is missing from the equation you would use to work this out is the number density. So you need to know a typical number density for a semiconductor. Not an exact value, but a typical value to the nearest power of 10.

F: you say that Q = 0.8 * 60 = 480 C. This can't possibly be right because you are multiplying 60 by a number less than one but end up with a larger answer!

H: Why do you think that your answer is not one of the options?

**Pangol**)Firstly - this would be better asked in the physics section rather than the maths one. You'll get more replies there.

Secondly - I haven't had time to look through everything yet. But here are a few thoughts.

A: Careful, you are asked for a diameter, not a radius.

D: The thing that is missing from the equation you would use to work this out is the number density. So you need to know a typical number density for a semiconductor. Not an exact value, but a typical value to the nearest power of 10.

F: you say that Q = 0.8 * 60 = 480 C. This can't possibly be right because you are multiplying 60 by a number less than one but end up with a larger answer!

H: Why do you think that your answer is not one of the options?

Anyhow, yes thank you for spotting that, I think I got carried away and forgot to refer back to the question. In which case the solution to question a would be option 2, since if r=1.08, d=2r; d=2*1.08=2.16 mm.

D: Right, so if the typical value is to the nearest power of 10 would the answer be C. ~1 ms^-1?

F: Sorry, another silly mistake. 0.8*60=48 C

Q=n*e

n=Q/e

n=48/(1.6*10^-19)

n=3.0*10^20 (answer 2)

H: I did not think that my answer corresponded to one of the options since I believe that I calculated a solution in amps, being 9.8304*10^-6 A which is 9.8304*10^-3 mA.

Converting mA to uA: 9.8304*10^-3 mA = 9.8304 uA ~ 9.8 uA (answer A)

I think that I had just forgotten to convert to mA, and uA after calculating in A so I thought that I had arrived at the incorrect answer.

Thank you very much for your help I really appreciate it ✌️

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