
A raw egg and hard boiled are made to spin on a table with the same angular speed about the same axis. The ratio of the time taken by the two to stop is
(A) Equal to one
(B) Less than one
(C) More than one
(D) Equal to two
Answer
244.2k+ views
Hint: To solve this question, we need to use the first rotational equation of motion for calculating the time required for the eggs to stop. Then using the relation between the angular acceleration and moment of inertia we can obtain the ratio of the time in the form of the ratio of moments of inertia. Finally comparing the moments of inertia for the two eggs, we will get the final answer.
Complete step-by-step solution:
Let the time required for the raw egg and the hard boiled egg to stop be ${t_1}$ and ${t_2}$ respectively.
Also, let their respective moments of inertia be ${I_1}$ and ${I_2}$.
Now, a raw egg has a liquid called yolk inside itself. While spinning, this yolk will experience centrifugal force due to which it will get attached to the inner circumference of the raw egg.
But a hard boiled egg is solid from the inside.
So the mass of a raw egg is separated at a larger distance from the axis of rotation. But since the mass of a hard boiled egg is distributed throughout its volume, so its effective radius of gyration is less than that of the raw egg. We know that the moment of inertia is proportional to the square of the moment of inertia. So the moment of inertia of hard boiled egg is less than that of the raw egg. This means that
${I_1} > {I_2}$.........................(1)
Since both the eggs are rotating on the same table, they will experience the same frictional torque which stops them. Let the torque be $\tau $. So their angular accelerations are given by
\[{\alpha _1} = - \dfrac{{{\tau _{}}}}{{{I_1}}}\].....(2)
\[{\alpha _2} = - \dfrac{\tau }{{{I_2}}}\]...................(3)
(Negative signs are due to the fact that the torque is opposing the motion)
We know that
${\omega _f} - {\omega _i} = \alpha t$
Since both of them are stopping so ${\omega _f} = 0$. Let the initial angular speed of both the eggs be \[{\omega _i} = \omega \]. Substituting these above we get
$0 - \omega = \alpha t$
$ \Rightarrow t = - \dfrac{\omega }{\alpha }$
For the raw egg we have
${t_1} = - \dfrac{\omega }{{{\alpha _1}}}$
Putting (2) above, we get
\[{t_1} = \dfrac{{{I_1}\omega }}{\tau }\] …………….(4)
Similarly for the hard boiled egg we get
\[{t_2} = \dfrac{{{I_2}\omega }}{\tau }\]...........(5)
Dividing (4) by (5) we get
$\dfrac{{{t_1}}}{{{t_2}}} = \dfrac{{{I_1}}}{{{I_2}}}$
From (1) we can say that
$\dfrac{{{t_1}}}{{{t_2}}} > 1$
Thus, the required ratio is more than one.
Hence, the correct answer is option C.
Note: An egg gets solid from inside after boiling due to the heat energy breaking its proteins and linking them with the other. But we do not need to know all this. The phrase “hard boiled egg” is itself suggesting that the egg is hard or solid from the inside.
Complete step-by-step solution:
Let the time required for the raw egg and the hard boiled egg to stop be ${t_1}$ and ${t_2}$ respectively.
Also, let their respective moments of inertia be ${I_1}$ and ${I_2}$.
Now, a raw egg has a liquid called yolk inside itself. While spinning, this yolk will experience centrifugal force due to which it will get attached to the inner circumference of the raw egg.
But a hard boiled egg is solid from the inside.
So the mass of a raw egg is separated at a larger distance from the axis of rotation. But since the mass of a hard boiled egg is distributed throughout its volume, so its effective radius of gyration is less than that of the raw egg. We know that the moment of inertia is proportional to the square of the moment of inertia. So the moment of inertia of hard boiled egg is less than that of the raw egg. This means that
${I_1} > {I_2}$.........................(1)
Since both the eggs are rotating on the same table, they will experience the same frictional torque which stops them. Let the torque be $\tau $. So their angular accelerations are given by
\[{\alpha _1} = - \dfrac{{{\tau _{}}}}{{{I_1}}}\].....(2)
\[{\alpha _2} = - \dfrac{\tau }{{{I_2}}}\]...................(3)
(Negative signs are due to the fact that the torque is opposing the motion)
We know that
${\omega _f} - {\omega _i} = \alpha t$
Since both of them are stopping so ${\omega _f} = 0$. Let the initial angular speed of both the eggs be \[{\omega _i} = \omega \]. Substituting these above we get
$0 - \omega = \alpha t$
$ \Rightarrow t = - \dfrac{\omega }{\alpha }$
For the raw egg we have
${t_1} = - \dfrac{\omega }{{{\alpha _1}}}$
Putting (2) above, we get
\[{t_1} = \dfrac{{{I_1}\omega }}{\tau }\] …………….(4)
Similarly for the hard boiled egg we get
\[{t_2} = \dfrac{{{I_2}\omega }}{\tau }\]...........(5)
Dividing (4) by (5) we get
$\dfrac{{{t_1}}}{{{t_2}}} = \dfrac{{{I_1}}}{{{I_2}}}$
From (1) we can say that
$\dfrac{{{t_1}}}{{{t_2}}} > 1$
Thus, the required ratio is more than one.
Hence, the correct answer is option C.
Note: An egg gets solid from inside after boiling due to the heat energy breaking its proteins and linking them with the other. But we do not need to know all this. The phrase “hard boiled egg” is itself suggesting that the egg is hard or solid from the inside.
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