Warmup Test(2)!!

Level pending

If Earth suddenly stops rotating about its own axis, the increase in its temperature will be

Details:

1) $R$ is the radius of Earth.

2) $J$ is the mechanical equivalent of heat.

3) $\omega$ is the angular velocity of Earth.

4) $s$ is the average specific heat of Earth.

None of these

\dfrac {R^2\omega^2}{3Js}

\dfrac {R^2\omega^2}{Js}

\dfrac {R^2\omega^2}{5Js}

1 solution

A Former Brilliant Member
Nov 26, 2018

We know that,
Work done = $J\times$ Heat

$W= J\times H = J\times (M_es\Delta \theta)$

$\Delta \theta$ is the rise in temperature and $M_e$ is the mass of Earth.

The Moment of Inertia of Earth = $I = \dfrac 25 M_e \times R^2$

Now, $W= J\times H = \dfrac 12 \times I\times \omega^2$

$\Rightarrow J\times (M_es\Delta \theta) = \dfrac 12 (\dfrac 25 M_e R^2) \times \omega^2$

$\Rightarrow \boxed{\Delta \theta = {\dfrac {R^2\omega^2}{5Js}}}$