In a magnetic field as shown, in figure two horizontal wires of same mass and lengths l1 and l2 are free to slide on different vertical rails with velocities v1 and v2 respectively. If the resistances of two circuits are same and a1 and a2 are accelerations of two horizontal wires respectively the condition for a1> a2. If two falling conductors attain velocities v1 and v2 respectively after falling through same height h, ratio of energy dissipated as heat to the energy dissipated in resistor per unit time for two conductors is same. Find which of the following conditions will be satisfied

mgh-12mv12mgv1=mgh-12mv22mgv2 ghv2-12v12v2=ghv1-12v22v1 ghv1-v2=12v1v2v2-v1 gh+12v1v2=0

In a magnetic field as shown, in figure two horizontal wires of same mass and lengths l1 and l2 are free to slide on different vertical rails with velocities v1 and v2 respectively. If the resistances of two circuits are same and a1 and a2 are accelerations of two horizontal wires respectively the condition for a1> a2. If two falling conductors attain velocities v1 and v2 respectively after falling through same height h, ratio of energy dissipated as heat to the energy dissipated in resistor per unit time for two conductors is same. Find which of the following conditions will be satisfied

mgh-12mv12mgv1=mgh-12mv22mgv2 ghv2-12v12v2=ghv1-12v22v1 ghv1-v2=12v1v2v2-v1 gh+12v1v2=0

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In a magnetic field as shown, in figure two horizontal wires of same mass and lengths l₁ and l₂ are free to slide on different vertical rails with velocities v₁ and v₂ respectively. If the resistances of two circuits are same and a₁ and a₂ are accelerations of two horizontal wires respectively the condition for a₁> a_2. If two falling conductors attain velocities v₁ and v₂ respectively after falling through same height h, ratio of energy dissipated as heat to the energy dissipated in resistor per unit time for two conductors is same. Find which of the following conditions will be satisfied

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$\frac{1}{2} gh + v_{1} v_{2} = 0$

$gh + \frac{1}{2} v_{1} v_{2} = 0$

$gh + \frac{v_{1}^{2} v_{2}}{v_{1} + v_{2}} = 0$

$gh + v_{1} v_{2} = 0$

answer is A.

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Detailed Solution

$\frac{mgh - \frac{1}{2} {mv}_{1}^{2}}{{mgv}_{1}} = \frac{mgh - \frac{1}{2} {mv}_{2}^{2}}{{mgv}_{2}} {ghv}_{2} - \frac{1}{2} v_{1}^{2} v_{2} = {ghv}_{1} - \frac{1}{2} v_{2}^{2} v_{1} gh (v_{1} - v_{2}) = \frac{1}{2} v_{1} v_{2} (v_{2} - v_{1}) gh + \frac{1}{2} v_{1} v_{2} = 0$