Angular speed of rotation of the earth is ω0. A train is running along the equator at a speed v from west to east. A very sensitive balance inside the train shows the weight of an object as W1. During the return journey when the train is running at same speed from east to west the balance shows the weight of the object to be W2. Weight of the object when the train is at rest was shown to be W0 by the balance. Then (W2 – W1) is equal to

When the train is at restW0=mg−mV02R V0=ω0R,ω0= angular speed of the earth]When the train is moving from West to East V1 = V0 + v∴W1=mg−mV0+v2RFor train running due westV2 = V0 – vW2=mg−mV0−v2R∴W2−W1=mRV0+v2−V0−v2=mR4V0v=4mω0RvR=4mω0v=4W0gω0v

Angular speed of rotation of the earth is ω0. A train is running along the equator at a speed v from west to east. A very sensitive balance inside the train shows the weight of an object as W1. During the return journey when the train is running at same speed from east to west the balance shows the weight of the object to be W2. Weight of the object when the train is at rest was shown to be W0 by the balance. Then (W2 – W1) is equal to

When the train is at restW0=mg−mV02R V0=ω0R,ω0= angular speed of the earth]When the train is moving from West to East V1 = V0 + v∴W1=mg−mV0+v2RFor train running due westV2 = V0 – vW2=mg−mV0−v2R∴W2−W1=mRV0+v2−V0−v2=mR4V0v=4mω0RvR=4mω0v=4W0gω0v

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Angular speed of rotation of the earth is $ω_{0}$ . A train is running along the equator at a speed v from west to east. A very sensitive balance inside the train shows the weight of an object as W₁. During the return journey when the train is running at same speed from east to west the balance shows the weight of the object to be W₂. Weight of the object when the train is at rest was shown to be W₀ by the balance. Then (W₂ – $W_{1})$ is equal to

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$\frac{W_{0} ω_{0} v}{4 g}$

$\frac{W_{0} ω_{0} v}{2 g}$

$\frac{2 W_{0} ω_{0} v}{g}$

$\frac{4 W_{0} ω_{0} v}{g}$

answer is D.

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Detailed Solution

When the train is at rest
$W_{0} = mg - \frac{{mV}_{0}^{2}}{R} [V_{0} = ω_{0} R, ω_{0}$ = angular speed of the earth]
When the train is moving from West to East V₁ = V₀ + v
$∴ W_{1} = mg - \frac{m {(V_{0} + v)}^{2}}{R}$
For train running due west
V₂ = V₀ – v
$\begin{array}{l} W_{2} = mg - \frac{m {(V_{0} - v)}^{2}}{R} \\ ∴ W_{2} - W_{1} = \frac{m}{R} [{(V_{0} + v)}^{2} - {(V_{0} - v)}^{2}] \\ = \frac{m}{R} [4 V_{0} v] = \frac{4 m (ω_{0} R) v}{R} = 4 {mω}_{0} v \\ = 4 \frac{W_{0}}{g} ω_{0} v \end{array}$