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Category: Physics Olympiad
Q. An insulating rod running from (z=-a to z=a) carries indicated line charges. Determine the leading term in the multipole expansion of the potential is given by $$ \lambda=kcos(\frac{\pi z}{2a})$$ where k is a constant.
Solution:
Potential due to multipole expansion is given by $$ V(\vec{r})=\frac{1}{4\pi\epsilon_0}\sum_{0}^{\inf} \frac{P_n(cos\theta)}{r^{n+1}}I_n$$ where (I_n=\int_{-a}^{+a}z^n\lambda(z)dz)
Inserting the value of charge density in the integral to obtain: $$I_0=k\int_{-a}^{a} cos(\frac{\pi z}{2a})dz=\frac{2ak}{\pi}sin(\frac{\pi z}{2a}$$
Applying limits (z=-a to z=a) $$ I_0= \frac{2ak}{\pi}[sin (\pi/2)+sin(\pi/2)]=\frac{4ak}{\pi}$$
The potential $$ V(r,\theta)=\frac{1}{4\pi\epsilon_0}(4ak/pi)$$
.
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Inductance of a Toroid
Show that the inductance of a toroid of rectangular cross-section is given by $$ L=\frac{\mu_0N^2Hln(b/a)}{2\pi}$$ where (N) is the total number of turns, (a) is the inner radius, (b) is the outside radius and (H) is the height of the toroid.
Solution:
ToroidUsing the definition of the self inductance of a solenoid, we express (L) in terms of flux (\phi), (N) and (I):
$$ L=\frac{N\phi}{I}$$
We apply Ampere’s law to a closed path of radius (a<r<b):
$$ \oint \vec{B}.\vec{dl}=B(2\pi r)$$ $$=\mu_0NI$$ $$ \Rightarrow B=\frac{\mu_0NI}{2\pi r}$$ We express the flux in a strip of height (H) and width (dr):
$$ d\phi=BHdr=\frac{\mu_0NIH}{2\pi}\int_{a}^{b}\frac{dr}{r}$$ $$ =\frac{\mu_0NIH}{2\pi}ln(\frac{b}{a})$$
Substitute for flux (\phi) in the equation (1) we obtain the expression for (L)
$$ L=\frac{\mu_0N^2Hln(b/a)}{2\pi}$$
When a gas expands adiabatically, its volume is doubled while its absolute temperature is decreased by a factor (1.32). Compute the number of degrees of freedom for the gas molecules.
Solution:
The number of degrees can be found from the relation $$ f=\frac{2}{\gamma-1}$$
We can find (\gamma) from the adiabatic relation,$$ T_2V_2^{\gamma-1}= T_1V_1^{\gamma-1} $$
$$( \frac{V_2}{V_1})^{\gamma-1}=\frac{T_1}{T_2}=1.32$$
$$ 2^{\gamma-1}=1.32$$
where $$ \gamma=1+\frac{log 1.32}{log2}=1.4$$
The number of degrees of freedom $$ f=\frac{2}{1.4-1}=5$$
Categories
Efficiency of Heat Engine
Here is a problem where we need to calculate the efficiency of the heat engine and work done by it. Let’s see how we can solve it.
The problem:
A heat engine absorbs heat of (10^5Kcal) from a source, which is at (127^\circ) and rejects a part of heat to sink at (27^\circ). Calculate the efficiency of the engine and the work done by it.
Solution:
The efficiency of the engine is $$ \eta=1-\frac{T_2}{T_1}$$ $$=1-\frac{300}{400}$$
$$ =0.25$$ that is, (25\% )
Work done by the engine $$ W= \eta\times Q$$ $$=0.25\times 10^8 Cal$$
$$ =0.25\times 10^8\times4.81J$$
$$ =1.05\times10^8J$$
A ring of radius (R) carries a linear charge density (\lambda). It is rotating with angular speed (\omega). What is the magnetic field at the centre?
Discussion:
Linear charge density $$ \lambda=\frac{Q}{2\pi R}
$$
When the ring is rotated about the axis, the motion of the electrons in a circular orbit is equivalent to a current carrying loop.
Current $$ I=\frac{Q}{T}=\frac{Q\omega}{2\pi}$$
since Time period (T=2\pi/\omega).
Now, magnetic field around the centre of a current carrying loop is given by $$ B=\mu_0I/2R$$
Putting the value of (I) in the above equation, we get
$$ B=\frac{\mu_0\omega}{2}.\frac{Q}{2\pi R}
$$$$ \Rightarrow B=\frac{\mu_0\lambda\omega}{2}
$$
A rigid triangular molecule consists of three non-collinear atoms joined by rigid rods. The constant pressure molar specific heat (C_p) of an ideal gas consisting of such molecules is
(a) (6R)
(b) (5R)
(c) (4R)
(d) (3R)
Degrees of freedom are the number of independent parameters that define its configuration
If (N) be the number of particles in a system and (k) be the number of constraints between the number of degrees of freedom is given by $$ f=3N-k$$ $$f=(3*3)-3$$ $$=6$$
Relation between (f) and (C_p) $$ C_p=(f/2+1)R$$ $$ \Rightarrow C_p=(6/2+1)R$$ $$C_p=4R$$
A cylinder contains (16g) of (O_2). The work done when the gas is compressed to (75\%) of the original volume at constant temperature of (27^\circ) is ________.
Discussion:
Given mass of (O_2), m=(16g)
Number of moles of (O_2), $$n=\frac{m}{M}$$ where (M)=molecular weight of (O_2)=(32g)
$$n=\frac{16}{32}=\frac{1}{2}$$
Temperature (T=27^\circ=300K)
If (V_1) be the original volume and (V_2) be the final volume
Work done by the gas in the isothermal process $$ W_0=nRTlog(V_2/V_1)$$$$=0.58.31ln(3/4)$$$$=1508.31ln(3/4)$$$$=-358.56J$$
Categories
A Problem on Doppler Effect
Try this Problem based on Doppler Effect where we find the tone of the whistle and speed of the Train. First, do it yourself and then read the solution.
The Problem: Doppler Effect
A train passes through a station with constant speed. A stationary observer at the station platform measures the tone of the train whistle as (484Hz) when it approaches the station and (442Hz) when it leaves the station. If the sound velocity is (330m/s), then the tone of the whistle and the speed of the train are
(a) (462hz, 54km/h)
(b) (463Hz, 52Km/h)
(c) (463Hz, 56Km/h)
(d) (464Hz, 52Knm/h)
Solution:
When train approaches the station, the frequency heard by the observer
$$ n_1=n\frac{v}{v-v_s}=n(\frac{330}{330-v_s})$$
Here, $$ v=330m/s$$
n is the actual frequency of the whistle
$$ 484 =n(330/330-v_s)$$….. (i)
When the train leaves the station $$ n_2=n\frac{v}{v+v_s}=n(\frac{330}{330+v_s}) $$
$$ 442=n(\frac{330}{330+v_s})$$…. (ii)
Divide Eqs (i) by (ii), we get
$$ \frac{484}{442}=330+v_s/330-v_s$$
$$ 1.09=(330+v_s)/(330-v_s)$$
$$ 330+v_s=1.09(330-v_s)$$
$$v_s=\frac{31.35}{2.09}$$$$=15m/s$$
Substituting (v_s) in Eqn (i) gives $$ 484=n(330/330-15)$$ $$=n(330/315)$$ $$n=\frac{484*21}{22}$$
$$=462Hz$$
Categories
Variation of Specific Heat
In this post, let’s learn about variation of specific heat by finding out the difference between mean specific heat and specific heat at midpoint.
The Problem:
The variation of the specific heat of a substance is given by the expression $$ C=A+BT^2$$ where (A) and (B) are constants and (T) is the Celsius temperature. Find the difference between the mean specific heat and specific heat at midpoint.
Discussion:
The variation of the specific heat of a substance is given by the expression $$ C=A+BT^2$$ where (A) and (B) are constants and (T) is the Celsius temperature.
Mean specific heat
$$ \bar{C}=\frac{\int C dT}{dT}=\frac{\int_{0}^{T}(A+BT^2)dT}{T}$$ $$= \frac{AT+BT^3/3}{T}$$ $$=A+BT^2/2$$
C(midpoint)$$ = A+B(T/2)^2$$ $$=A+\frac{BT^2}{4}$$
Hence, the difference between mean specific heat and specific heat at midpoint $$= \bar{C}-C(midpoint)$$ $$=A+BT^2/3-(A+BT^2/4)$$ $$=\frac{BT^2}{12}$$