You are Here: Home >< Physics

# Multiple Choice Questions help watch

1. Hey guys, I've got some multiple choice questions I'd like some help with please!

With this one I'm confused. I thought surely because the diameter is halved, and this the area is quartered, the youngs modulus of the second sample would be 12E? The three times the length increases the YM by 3, the area is in the denominator so that would increase it by 4, so 12 times the youngs modulus, no??

This one I've just forgotten how to do, I know it's about resolving but I just can't remember, any help would be appreciated.

Honestly kinda a similar thing here too, not sure why it's A, surely others like C make sense too?

Thanks a lot.
2. For 24, EDITED - YM is constant for a given material!!!

For q20, you can use F=ma to find the resultant force.
There are two forces acting on the rocket, its weight and the thrust and these must give a resultant equal to the F=ma you calculated.

For Q18, equilibrium means forces AND moments are balanced. Choose a point in C to take moments around (for example choose the centre) and you'll find there IS a non-zero moment.
3. (Original post by phys981)
For q1, remember that YM depends on cross-sectional area which goes as the square of the diameter. If the diameter changes by factor 0.5, then the area goes by 0.5² or 0.25

If the area is reduced to 1/4 of it's original value, the YM is 4 times greater right? Or am I being confused?
4. Yes! Sorry, and so am I (confised).

YM is a constant for a given material. If we change the lenght and area the extension might also change for a given force but YM remains the same.

5. (Original post by phys981)
Yes! Sorry, and so am I (confised).

YM is a constant for a given material. If we change the lenght and area the extension might also change for a given force but YM remains the same.

So it's kinda a trick question, and the answer is based off of it being the same material and so regardless of other factors, YM is the same?
6. Yes.
7. (Original post by phys981)
Yes.

Any idea on this one? I got it right, it's C, but I just did it by considering the ratios, 2:1 split of the PD and the only one which goes to 4V is the third graph, kinda a BS way of doing it and I don't understand what I've done really
8. You're right, really.

Based on the ratio of resistances, the p.d. across that 10 Ohm resistormust be 4V. So, with the sliding contacts one at either end, they will cover the whole of that 4V. However, as you slide the moveable one towards the right, it covers less and less of the 10Ohm resistor so gets less and less of the 4V until it reaches P and 0V.
9. (Original post by phys981)
You're right, really.

Based on the ratio of resistances, the p.d. across that 10 Ohm resistormust be 4V. So, with the sliding contacts one at either end, they will cover the whole of that 4V. However, as you slide the moveable one towards the right, it covers less and less of the 10Ohm resistor so gets less and less of the 4V until it reaches P and 0V.
Yeah that does make sense, cheers bro! Electricity really isn't one of my strong points, anything you'd recommend doing to get a lot stronger at it? I've always had issues with it really.
10. Just practice really, lots and lots of questions.

If you have a decent textbook with plenty of questions in, it can be useful to practise the bits you find difficult before you attempt the real exam questions.

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 26, 2018
The home of Results and Clearing

### 2,541

people online now

### 1,567,000

students helped last year
Today on TSR

### University open days

1. Keele University
Sun, 19 Aug '18
2. University of Melbourne
Sun, 19 Aug '18
3. Sheffield Hallam University