If the earth shrinks such that its mass does not change but radius decreases to one quarter of its original value then one complete day will take

Question

If the earth shrinks such that its mass does not change but radius decreases to one quarter of its original value then one complete day will take (A) 96 h (B) 48 h (C) 6 h (D) 1.5 h

Tardigrade Admin · Accepted Answer

We know that angular momentum of spin =I ω By the conservation of angular momentum (2/5) M R2 ⋅ (2 π/T)=(2/5) M((R/4))2 ⋅ (2 π/T') T'=(T/16)=(24/16)=1.5 h

Q. If the earth shrinks such that its mass does not change but radius decreases to one quarter of its original value then one complete day will take

96 h

48 h

6 h

1.5 h

We know that angular momentum of spin $= I ω$
By the conservation of angular momentum
$\frac{2}{5} M R^{2} \cdot \frac{2 π}{T} = \frac{2}{5} M (\frac{R}{4})^{2} \cdot \frac{2 π}{T ^{'}}$
$T^{'} = \frac{T}{16} = \frac{24}{16} = 1.5 h$

Q. If the earth shrinks such that its mass does not change but radius decreases to one quarter of its original value then one complete day will take

96 h

48 h

6 h

1.5 h

We know that angular momentum of spin =Iω By the conservation of angular momentum 52​MR2⋅T2π​=52​M(4R​)2⋅T′2π​ T′=16T​=1624​=1.5h

We know that angular momentum of spin $= I ω$
By the conservation of angular momentum
$\frac{2}{5} M R^{2} \cdot \frac{2 π}{T} = \frac{2}{5} M (\frac{R}{4})^{2} \cdot \frac{2 π}{T ^{'}}$
$T^{'} = \frac{T}{16} = \frac{24}{16} = 1.5 h$