To find the ratio of velocities of the two particles having the same mass, we need to consider the condition of constant centripetal force. The formula for centripetal force (\(F_c\)) is given by:
\(F_c = \frac{mv^2}{r}\)
where:
Since the mass of the two particles is the same and the centripetal force is constant, we can equate the expressions for the two particles and consider their ratio:
\(\frac{mv_1^2}{r_1} = \frac{mv_2^2}{r_2}\).
Simplifying, we get:
\(\frac{v_1^2}{r_1} = \frac{v_2^2}{r_2}\).
Rearranging gives the relation between their velocities and radii:
\(v_1^2 \cdot r_2 = v_2^2 \cdot r_1\).
We know the radii are in the ratio \(r_1 : r_2 = 3:4\). Substituting these values, we have:
\(v_1^2 \cdot 4 = v_2^2 \cdot 3\).
Solve for \(\frac{v_1}{v_2}\):
\(\frac{v_1^2}{v_2^2} = \frac{3}{4}\).
Taking the square root of both sides, we get:
\(\frac{v_1}{v_2} = \frac{\sqrt{3}}{2}\).
Hence, the velocities are in the ratio \(\sqrt{3} : 2\).
Step 1: Given Data: - Masses \(m_1 = m_2\) - Radius ratio \(\frac{r_1}{r_2} = \frac{3}{4}\)
Step 2: Use the Centripetal Force Formula: - Centripetal force \(F = \frac{mv^2}{r}\). - Since the centripetal force is constant, \(F_1 = F_2\): \[ \frac{m_1 v_1^2}{r_1} = \frac{m_2 v_2^2}{r_2} \]
Step 3: Simplify the Equation: - With \(m_1 = m_2\), we get: \[ \frac{v_1^2}{r_1} = \frac{v_2^2}{r_2} \]
\[ \Rightarrow \frac{v_1}{v_2} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]
So, the correct answer is: \(\sqrt{3} : 2\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)