1) Given we effectively have a flat rotating cylinder than it behooves us to make use of cylindrical coordinates. The divergence is:
Ñ = ^h ¶ / ¶r + ^m/ r (¶ / ¶f) + ^k ¶ / ¶z
And we have: r = r^h + z^k
(z = constant)
The velocity is:
v = dr/ dt = r’ ^h + r f’ ^m = r w ^m
Ñ x v = (^h x ^m) w + ^m/ r (¶ / ¶f) [r w ^m]
+ ^k x ^m ¶ (r w)/ ¶z
+ ^k x ^m ¶ (r w)/ ¶z
But: (¶ / ¶f) ^m = - ^h
So that:
Ñ x v = 2 (^h x ^m) w = 2 w (B)
2) Based on the presentation in (1) it must be in the +z direction (D)
3) curl ix2 + jy2 + k (x2 – y2)
= Ñ x i x2 + j y2 + k (x2 – y2)
= (i x j) (¶ / ¶x) y2 + (i x k) ¶ / ¶x (x2 + y2 )
+ (j x i) ¶ / ¶y (x2 ) + (j x k) ¶ / ¶y (x2 - y2 )
+ (k x i) ¶ / ¶z (x2 ) + (k x j) ¶ / ¶z (x2 - y2 )
= -2 ( iy + j x)
4) First, construct a sketch of the string, e.g. in u(= y) , x coordinates as shown in order to do a brief analysis as shown:
We then have in terms of the string tension T:
T u = T sin q
As usual per these approaches, assume q is very small, in which limit, sin q » tan q
Then:
T sin q = T tan q = T (¶ u/ ¶x)
Take the force difference:
[T u ] x + dx - [T u ] x = ¶ / ¶x (T (¶ u/ ¶x) dx
To ensure no net horizontal forces due to tension, we limit the situation to small slopes. Also, neglect the possibility of a vertical force per unit length, so:
r dx (¶ 2 u/ ¶ t2) = ¶ / ¶x (T (¶ u/ ¶x) dx
Which simplifies to:
r (¶ 2 u/ ¶ t2) = T (¶ 2 u/ ¶x2 )
Now, transform to standard wave equation in 1-dimension:
¶2 u/ ¶x2 - 1/ c2 (¶ 2 u/ ¶ t2) = 0
From this we can solve for the velocity c:
c= (T/ r ) ½
The boundary conditions can be written:
u(x, t) = u(L, t) = 0
and:
u(x,t) = X(x) Y(t)
Then, on forming the particular 2nd order differential equation:
c2/ X (d 2 X/ dt 2 ) = 1/ Y (d 2 Y/ dt 2 )
From which:
1/ Y (d 2 Y/ dt 2 ) = - w2
And: c2/ X (d 2 X/ dt 2 ) = - w2
The solutions are of the form:
X = C cos wx/ c + D sin wx/ c
Check boundary conditions: X(0) = C = 0
And X(L) = D sin wL/ c = 0
Then w will assume values:
w = n p c/ L = n p / L [T/ r ] ½
The frequency is f = w/ 2p = n p / L [T/ r ] ½ w/ 2p
Or f = n/ 2L [T/ r ] ½
Note: Though in the actual GRE there will be multiple choice options, this is generally the type of problem you want to avoid because – in the absence of memory (say already knowing the wave equation for a string) and the useful DEs, the derivation is excessively long. Bear in mind this is a timed test. So you have 3 hours (180 mins.) to get through 100 problems. That works out to roughly 1 m 48s per problem. That means if you need to send something like 8 minutes deriving the answer for one problem, you will lose valuable time for 4-5 others (even given 2 may be easy) . If, however, you can make an educated guess, say based on a knowledge of dimensional analysis, then that might be worth a try!
5) (E)
A rigid body may move in any of three directions. It may also rotate about any of three axes. Therefore, it must have six degrees of freedom.
6) (A)
If a rigid body is constrained to rotate about a given axis it has only one rotational degree of freedom. It cannot have any translational freedom since this would change the axis of rotation.
7) (B)
It has one degree of rotational freedom since the axis of rotation must be parallel to the plane. It can only have one degree of translational freedom to keep the disk perpendicular to the plane. Hence the total is 1 R + 1T = 2 DF.
8) We have: I = K x
(1….0…..0)
(0….1…..1)
(0….1… ..1)
We write out the determinant with eigenvalue l:
(1 - l….0…..0)
(0….1 - l ..1)
(0….1… ..1 - l)
Leading to the characteristic equation:
(1 - l)3 – (1 - l) = 0
(1 - l) [ ((1 - l)2 – 1] = 0
Or:
(1 - l) (l2 – 2l) = 0
Yielding eigenvalues: l= 0, l = 2
Then:
T = Kl, so T1 = 0, T2 = K, and T3 =2K
So (B) is the answer.
9)We will have to take: (I – T1)C
So that:
K [(1….0…..0)
[(0….1…..1)
[(0….1… ..1)
-
K (1….0…..0)](x)
(0….1….. 0)] (y)
(0….0… ..1)] (z)
=
(0….0…..0) (x)
(0….0…..1) (y)
(0….1… ..0) (z) = 0
So that: 0 =
(0)
(z)
(y)
Therefore the ans. is (A) or i
10) By the analog of the parallel axis theorem:
I o = IG - M(R2 I – RR)
D I = I o - IG = M(R2 I – RR)
RR = r o 2 x
(0….0………0)
(0….1/2…..1/2)
(0….1/2… ..1/2)
D I = M r o 2 x
[(1….0…..0)
[(0….1…..0)
[(0….0… ..1)
-
(0….0……..0)]
(0….1/2.. 1/2)]
(0….1/2… ..1/2)]
= M r o 2 x
(1….0………..0)
(0….1/2…..-1/2)
(0….-1/2… ..1/2)
11) For a circular loop: B = mo I / 2r at the center of the loop, where mo is the magnetic permeability of free space.
We have: r = 0.05 m, B = (0.70) (10 -4) Wb/m2
(Remember the conversion factor: 1G = 10 -4 Wb/m2 )
Then, solving for I: I = 2 Br/ mo
where mo = 4p x 10 -7 H/m
I= 2 (0.070) 10 -4 A/ (4p x 10 -7 H/m) = 5.6 A
12) The work done in bringing the charge is equal to the change in the electrostatic energy of the system. Let W1 and W2 be the electrostatic energy of the system without and with the presence of the dielectric, respectively.
Then:
W1 = 1/8 p ∫ 0 ¥ D ×E ( 4p r2 ) dr =
1/8 p ∫ 0 b (q/ r2)2 ( 4p r2 ) dr +
1/8 p ∫ a b (q/ r2)2 ( 4p r2 ) dr +
1/8 p ∫ a ¥ (q/ r2)2 ( 4p r2 ) dr
And:
W2 = 1/8 p ∫ 0 b (q/ r2)2 ( 4p r2 ) dr +
1/8 p ∫ b a (kq/ r2)2 (q/ r2) ( 4p r2 ) dr +
1/8 p ∫ a ¥ (q/ r2)2 ( 4p r2 ) dr
Work done = W2 – W1 =
1/8 p ∫ b a (kq/ r2) (q/ r2) ( 4p r2 ) dr
- 1/8 p ∫ b a (q/ r2)2 ( 4p r2 ) dr
= (k – 1) q 2 [1/b - 1/a]
13) We have an ideal monatomic gas so:
U (internal energy) = 3/2 [n RT/2]
The change in internal energy is:
DU = 3 nR [ 373 K - 273K] / 2
= 3 (8.32 J/K) (100K) /2= 1250 J
14) The change in entropy of the gas, given
dS = dQ/ T , is:
dS = dQ/ T , is:
D S = ∫ 273 373 nR dT/ T
= 5/2 nR ln (373 – 273) = 6.56 J/K
15) The f electron has ℓ =3 so that the total angular momentum quantum number possibilities are:
j = ℓ + ½, ℓ - ½
Then: j = 7/2, 5/2
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