- Messages
- 576
- Reaction score
- 308
- Points
- 73
i didn't get 31, B and C means almost same i guess, B is energy dissipated, nd C is energy dissipated per unit time. how does dat change the P.D description :/3. Intensity = energy/(area * time) (also given in the question)
= kgm²s¯² / s¯1 * m2
= kgs¯3
Hence the correct answer is D.
8. You should know how to “convert” displacement/acceleration/velocity graphs to one another. Anyway, here is how this should be:
At the first stage, the speed is constant. This means that the object’s displacement is increasing at a uniform rate. At the second stage, the speed is decreasing, this logically means that the displacement is increasing, but at a decreasing rate. This means that eventually, at S there will be a straight horizontal line. The only graph that shows these 2 stages correctly is C.
28.
y = λd / a = 6.0 * 10-9 / a (this is for the original wavelength of 600nm)
ay = 6.0 * 10-9
Now we want to find the new value of d,
Y = (4.0 * 10-9 * d) / a
Substitute a from the old equation, a and y will cancel out and the value of d turns out to be 1.5m.
31. The best way to solve this is to see each option, and find out which one is correct.
A is wrong because the word force shouldn't be there, it should be replaced with energy.
B is wrong because the word energy should be replaced with power.
Now from C and D, use the equation P = I2R.
P = IV
V = P/I
This is the ratio of power dissipated to current, so C is correct.
34. C. You have to learn these graphics (diode, filament lamp, ohmic conductor and thermistor/semi-conductor type material). Google them if you don't already know them.
37. The easiest way to solve these questions is by using some fake values of “I” entering the parallel junction. You know that the current is divided into the inverse proportional of the resistances in parallel, so that means if say, you have 6 A entering the junction, 4 A would go the 2 Ohm resistor and 2 A to the 6 Ohm resistor in the 1st parallel combination (lets call this “block 1”).
In “block 2”, 6 A is also entering the junction, and 3 A is shared equally between the resistors. I1 is 4 A and I2 is 2 A, and I1 > I2. Since you also have the currents entering the resistors, you can easily find out the values of V too and compare them.
