Question

A train of mass $100$ tons ($1$ton  $ = 1000\,\,Kg$) runs on a meter gauge track (distance between the two rails is  $1\,\,m$.  The coefficient of friction between the rails and the train is  $0.045$.  The train is powered by an electric engine of $90\,\% $  efficiency. The train is moving with uniform speed of  $72\,\,\,Kmph$  at its highest speed limit. Horizontal and vertical component of earth's magnetic field are  $BH = {10^{ - 5}}\,T$ and  $BV = 2 \times {10^{ - 5}}\,T$.  Assume the body of the train and rails to be perfectly conducting. If now a resistor of ${10^{ - 3}}\,\,\Omega $ is attached of between the two rails, the extra units of energy (electricity) consumed during a  $324\,\,Km$  run of the train is ( $1$  unit of power $ = 1\,\,KWh$)  (assume the speed of train to remain unchanged)
(A) $8 \times {10^{ - 4}}\,\,KWh$ (B) $8 \times {10^{ - 5}}\,\,KWh$ (C) $8 \times {10^{ - 6}}\,\,KWh$ (D) $8 \times {10^{ - 7}}\,\,KWh$

Answer 1

Hint: The above given problem can be solved using the formula derived for finding the energy consumed by an object which includes the power by which the object drives and the time taken by the object to complete the certain task.

Complete step by step answer:
The data given in the problem are;
Distance that the train travelled is, $L = 324\,\,Km$.
Speed by which the train is travelling is, $s = 72\,\,\,Kmph$.
Mass of the train is, $m = 1000\,\,Kg$.
The vertical component of earth's magnetic field is, $BV = 2 \times {10^{ - 5}}\,T$.
The vertical component of earth's magnetic field is, $BH = {10^{ - 5}}\,T$.
Distance between the two rails is, $l = 1\,\,m$.

The time taken by the train is given as;
$t = \dfrac{L}{s}$
Where, $t$ denotes the time taken by the train, $L$ denotes the distance travelled by the train, $s$ speed of the train.
$
  t = \dfrac{{324}}{{72}} \\
  t = \dfrac{9}{2}\,\,h \\
 $

Extra power in the engine = power displaced in the resistor.
$\varepsilon = \left( {{B_V}} \right) \times V \times l$
Where, ${B_V}$ denotes the vertical component of earth's magnetic field, $l$ denotes the distance between the two rails.
Substitute the values of for the above formula;
$
  \varepsilon = \left( {2 \times {{10}^{ - 5}}} \right) \times 20 \times 1 \\
  \varepsilon = 40 \times {10^{ - 5}} \\
 $

$P = \dfrac{{{\varepsilon ^2}}}{R}$
Where, $P$ denotes the power displaced in the track resistor, $R$ denotes the resistance of the track.
Substitute the values of the resistor of the track in the above formula;
$
  P = \dfrac{{{{\left( {40 \times {{10}^{ - 5}}} \right)}^2}}}{{{{10}^3}}} \\
  P = \dfrac{{16 \times {{10}^{ - 8}}}}{{{{10}^{ - 3}}}} \\
  P = 16 \times {10^{ - 5}}\,\,W \\
 $

The formula for energy consumed by the rails is given by;
${E_C} = P \times t$
Substitute the values for power displaced in the track resistor and the time taken by the train in the above formula;
$
  {E_C} = \dfrac{{16 \times {{10}^{ - 5}}}}{{90\,\% }} \times \dfrac{9}{2} \\
  {E_C} = \dfrac{{16 \times {{10}^{ - 5}}}}{{\dfrac{{90}}{{100}}}} \times \dfrac{9}{2} \\
  {E_C} = 8 \times {10^{ - 7}}\,\,KWh \\
 $

Therefore, the energy consumed by the rails is ${E_C} = 8 \times {10^{ - 7}}\,\,KWh$.

Hence, the option (D) ${E_C} = 8 \times {10^{ - 7}}\,\,KWh$ is the correct answer.

Note: Energy consumption is the result of power or energy usage of a system by making use of the energy supply. The consumption is done in GigaJoule per year, kilograms of oil equivalent per year, and in Watts.