NEB Class 12 Science Physics (NEB Released Items) Question Paper 2081 (Set B) Nepal

a) When the electron moves making an angle of $0^\circ$ with the electric field (i.e. along the field line), the electric force acts along the same line as its motion. The electron is only accelerated/decelerated along that line, so its path remains a straight line (it is not deflected sideways). [1]

b) The acceleration of the electron is

with $e = 1.6\times10^{-19}\ \text{C}$, $E = 1.5\times10^3\ \text{N/C}$ and $m = 9.1\times10^{-31}\ \text{kg}$:

[2]

c) For the beam to pass undeviated, the magnetic force must balance the electric force (velocity selector condition):

[2]

a) The circuit is a series resonance (series RLC) circuit. Reason: the current is shown to be maximum at a particular (resonant) frequency; at resonance the inductive and capacitive reactances cancel ($X_L = X_C$), so the impedance is minimum and the current peaks. [2]

b) When current is replaced by impedance $Z$, the plot is inverted: instead of a peak, the impedance-frequency graph becomes U-shaped with a minimum value at the resonant frequency. (Impedance is minimum exactly where current is maximum, so the new curve is opposite to the current curve.) [2]

  Z |  \           /
    |   \         /
    |    \       /
    |     \____/      <-- minimum at resonant frequency f0
    +------------------- f

c) Practical application: tuning circuits in a radio, TV or telephone/mobile phone (used to select a desired frequency/station). [1]

a) Closed organ pipe (one end closed, one end open), length $L$:

A closed pipe has a node at the closed end and an antinode at the open end, so only odd harmonics are present. If $v$ is the speed of sound:

• Fundamental (1st harmonic): $L = \dfrac{\lambda_1}{4} \Rightarrow \lambda_1 = 4L$, so $f_1 = \dfrac{v}{4L}$. (One node and one antinode.)
• First overtone (3rd harmonic): $L = \dfrac{3\lambda_3}{4} \Rightarrow \lambda_3 = \dfrac{4L}{3}$, so $f_3 = \dfrac{3v}{4L} = 3f_1$. (Two nodes, two antinodes.)
• Second overtone (5th harmonic): $L = \dfrac{5\lambda_5}{4} \Rightarrow \lambda_5 = \dfrac{4L}{5}$, so $f_5 = \dfrac{5v}{4L} = 5f_1$. (Three nodes, three antinodes.)

Thus the frequencies are in the ratio $f_1 : f_3 : f_5 = 1 : 3 : 5$. (Wave patterns to be sketched accordingly.) [3]

b) Apparent frequency — source moving, listener stationary (Doppler effect):

Let $v$ = speed of sound, $v_s$ = speed of source, $f$ = real frequency emitted, and $\lambda$ the emitted wavelength.

In one time period $T = 1/f$, the source moves a distance $v_s T$ towards the listener, so the $f$ waves emitted in that time are compressed into a length $(v - v_s)T$. The apparent wavelength is

(i) Source moving towards the listener:

Since $v - v_s < v$, $f' > f$ (apparent frequency increases). [1]

(ii) Source moving away from the listener:

Now the waves are spread over a length $(v + v_s)T$, so

Since $v + v_s > v$, $f' < f$ (apparent frequency decreases). [1]

c) Why the whistle of an approaching train is shriller:

As the train (source) moves towards the stationary observer, by the Doppler effect the apparent frequency $f' = \dfrac{fv}{v - v_s}$ is greater than the real frequency $f$. A higher frequency is perceived as a higher (shriller) pitch, so the whistle of an approaching train sounds shriller. [1]