Categories

## Magnetic Field at the Centre of a Ring

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
$$B=\frac{\mu_0\omega}{2}.\frac{Q}{2\pi R}$$$$\Rightarrow B=\frac{\mu_0\lambda\omega}{2}$$

Categories

## Specific Heat of a Rigid Triangular Molecule

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 $C_p$ $$C_p=(f/2+1)R$$ $$\Rightarrow C_p=(6/2+1)R$$ $$C_p=4R$$

Categories

## Work Done on Compression of Gas

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 $$W_0=nRTlog(V_2/V_1)$$$$=0.5*8.31*ln(3/4)$$$$=150*8.31*ln(3/4)$$$$=-358.56J$$

Categories

## A Problem on 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

## Light through Prisms (KVPY ’10)

White light is split into a spectrum by a prism and it is seen on a screen. If we put another identical inverted prism behind in contact, what will be seen on the screen?

(A) Violet will appear where red was

(B) The spectrum will remain the same

(C) There will be no spectrum but only the original light with no deviation

(D) There will be no spectrum, but the original will be laterally displaced

Discussions:

The system will behave as a slab since an inverted prism is put behind in contact with the first prism. Hence, there will be no spectrum, but only original light with no deviation.