Given:
- The particles have equal charges \( q \).
- They are accelerated through the same potential difference \( V \).
- Radii of circular paths in a magnetic field are \( R_1 \) for particle \( X \) and \( R_2 \) for particle \( Y \).
Step 1. Relate radius to mass and velocity:
For a particle moving in a circular path in a magnetic field, the radius \( R \) is given by:
\(R = \frac{mv}{qB}\)
where:
- \( m \) is the mass of the particle,
- \( v \) is the velocity after acceleration,
- \( q \) is the charge, and
- \( B \) is the magnetic field strength.
Step 2. Express \( v \) in terms of \( V \):
Since each particle is accelerated through the same potential difference \( V \), the kinetic energy gained by each particle is:
\(\frac{1}{2} mv^2 = qV\)
Solving for \( v \), we get:
\(v = \sqrt{\frac{2qV}{m}}\)
Step 3. Substitute \( v \) into the radius formula:
\( R = \frac{m}{qB} \cdot \sqrt{\frac{2qV}{m}} = \frac{\sqrt{2m \cdot qV}}{qB}\)
Step 4. Determine the ratio of radii for particles \( X \) and \( Y \):
\(\frac{R_1}{R_2} = \frac{\sqrt{2m_X \cdot qV}}{\sqrt{2m_Y \cdot qV}} = \sqrt{\frac{m_X}{m_Y}}\)
Step 5. Solve for the mass ratio:
Squaring both sides, we get:
\(\frac{m_X}{m_Y} = \left(\frac{R_1}{R_2}\right)^2\)
Thus, the mass ratio \( \frac{m_X}{m_Y} \) is \( \left(\frac{R_1}{R_2}\right)^2 \).
The Correct Answer is: \( \left( \frac{R_1}{R_2} \right)^2 \)
The wire loop shown in the figure carries a steady current \( I \). Each straight section of the loop has length \( d \). A part of the loop lies in the \( xy \)-plane and the other part is tilted at \( 30^\circ \) with respect to the \( xz \)-plane. The magnitude of the magnetic dipole moment of the loop (in appropriate units) is:
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
The standard enthalpy and standard entropy of decomposition of \( N_2O_4 \) to \( NO_2 \) are 55.0 kJ mol\(^{-1}\) and 175.0 J/mol respectively. The standard free energy change for this reaction at 25°C in J mol\(^{-1}\) is (Nearest integer)