Question 541: Centre of Mass and Rotational Motion

Question

Two circular loops P and Q of radii ' r ' and ' nr ' are made respectively from a uniform wire. Moment of inertia of loop Q about its axis is four times that of loop P about its axis. The value of ' $n$ ' is

Accepted Answer

The moment of inertia of a circular loop about its axis is given by the formula:
$$ I = m r^2 $$
where $ m $ is the mass of the loop and $ r $ is the radius.
Since the loops are made from the same uniform wire, the mass of the wire is proportional to its length. Therefore, the mass of each loop is proportional to its circumference:
$$ m_P \propto 2 \pi r $$
$$ m_Q \propto 2 \pi (nr) $$
The ratio of their masses is:
$$ \frac{m_Q}{m_P} = \frac{nr}{r} = n $$
Thus, the mass of loop $ Q $, $ m_Q = n \cdot m_P $.
Using the moment of inertia formula, the moments of inertia for $ P $ and $ Q $ are:
$$ I_P = m_P r^2 $$
$$ I_Q = m_Q (nr)^2 = n \cdot m_P (nr)^2 = n \cdot m_P \cdot n^2 \cdot r^2 = n^3 m_P r^2 $$
Since it is given that $ I_Q = 4 I_P $:
$$ n^3 m_P r^2 = 4 m_P r^2 $$
Cancelling $ m_P r^2 $ from both sides, we get:
$$ n^3 = 4 $$
Taking the cube root of both sides:
$$ n = \sqrt[3]{4} = 2^{2/3} $$
Thus, the value of $ n $ is option B: $ (2)^{2/3} $.

Answer

$(2)^{1 / 3}$

Answer

$(2)^{2 / 3}$

Answer

$(2)^{3 / 4}$

Answer

$(2)^{1 / 4}$

Centre of Mass and Rotational Motion

Two circular loops P and Q of radii ' r ' and ' nr ' are made respectively from a uniform wire. Moment of inertia of loop Q about its axis is four times that of loop P about its axis. The value of ' n ' is

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