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AS Physics P1 MCQs Preparation Thread.

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12. You have to consider the angles between T1 and R1 and T2 and R2. T1 cos(angle) gives you R1. What happens to cos(angle) if the angle decreases?
14. At the highest point of a parabolic trajectory, the vertical component of the velocity = o. So the K.E. can come only from the horizontal component, which is constant throughout.
26. Use sin90=nxlambda, then multiply by 2 (because you're taking the maxima on both sides of the middle maximum) and add 1 (the middle maximum).
28. Weight acts downwards so since it's in equilibrium the electric force has to act upwards. Since the plate up is positive and attracting the drop, that drop has to have a negative charge.
 
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12. You have to consider the angles between T1 and R1 and T2 and R2. T1 cos(angle) gives you R1. What happens to cos(angle) if the angle decreases?
14. At the highest point of a parabolic trajectory, the vertical component of the velocity = o. So the K.E. can come only from the horizontal component, which is constant throughout.
26. Use sin90=nxlambda, then multiply by 2 (because you're taking the maxima on both sides of the middle maximum) and add 1 (the middle maximum).
28. Weight acts downwards so since it's in equilibrium the electric force has to act upwards. Since the plate up is positive and attracting the drop, that drop has to have a negative charge.

14. So why is the answer 0.5E instead of E
28. Why g/E instead of E/g
 
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use total momentum before collision=total momentum after collision
therefore, 2*2=1*v
therefore speed of 1 kg trolley is 4..
now u have mass and velocity of both trolleys , right ?
find kinetic energy of both ... and simply add them !
hope that helps ! ;)

BTW bro thankss alott.. this was really very helpful! :)
 
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13. All you gotta do is choose a pivot and work out the moments on either side of it...
14. mgh=1/2m(v^2). mgh/2 = 1/2m(x^2). Eliminate m, g, and 1/2. Cross multiply.
15. Find the angle to the horizontal of the slope. Then find the component of the weight parallel to the slope. Then use F=ma to find the acceleration. Then use a suvat. You can also do it using 1/2m(v^2)=mgh.
34. Resistance = ql/A. A=lxl, so Resistance=ql/lxl. l and l cancel out; R=q/l. l^3=V, so l = cube root of V = V^(1/3), so R=q/V^(1/3).


For Q13. there is already a pivot, so why would we choose one? It doesn't work with the given one.
For Q.14 I have no idea what you did..
I understood Q15. thank you.
And also I do not know what you did in Q34.

Can you explain more please?
 
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For Q13. there is already a pivot, so why would we choose one? It doesn't work with the given one.
For Q.14 I have no idea what you did..
I understood Q15. thank you.
And also I do not know what you did in Q34.

Can you explain more please?

Well, you do use the pivot they've given. (20x60)+(100x10)=50x.
14. The steel sphere is dropped from a height of h. It has g.p.e.=mgh. When it reaches the ground, all its g.p.e. turns into k.e. So mgh=1/2m(v^2). Now it's held at h/2 metres above the ground. So the g.p.e. it has is mgh/2. This too will turn into k.e.
34. I just used the formula for Resistance, Resistivity, Area and Length. R=ρl/A. Now we know that A = l x l. So we can replace A in the equation with l x l. That'll make it R=ρl/l x l. l and l cancels out (from the numerator and denominator) leaving R=ρ/l. Now we know that Volume = l x l x l, so Volume = cube root of l, which is also written as V^(1/3). So we can replace l with V^(1/3). R=ρ/V^(1/3).
 
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hi guys i have done many past papers. so many times i have seen question about bright fringes,dark fringes maximas or minimas
Can anyone explain me fully that how to find maximas minimas or greatest no of maximas in diffraction grating and one thing more no of lines per mm is D or 1/N ??? in this case wavelength and no of lines per mm is given
i think this type of question also came in oct /nov 2012 variant 12 que 30 plzz try it
 
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