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

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{4 W_{0} ω_{0} v}{g}$

$\frac{2 W_{0} ω_{0} v}{g}$

$\frac{W_{0} ω_{0} v}{2 g}$

$\frac{W_{0} ω_{0} v}{4 g}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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
$\begin{array}{l} W_{2} = mg - \frac{m {(V_{0} - v)}^{2}}{R} \\ ∴ W_{2} - W_{1} = \frac{m}{R} [{(V_{0} + v)}^{2} - {(V_{0} - v)}^{2}] \\ = \frac{m}{R} [4 V_{0} v] = \frac{4 m (ω_{0} R) v}{R} = 4 {mω}_{0} v \\ = 4 \frac{W_{0}}{g} ω_{0} v \end{array}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th