JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 19)
Two particles of the same mass m are moving in circular orbits because of force, given by $$F\left( r \right) = {{ - 16} \over r} - {r^3}$$
The first particle is at a distance r = 1, and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to :
The first particle is at a distance r = 1, and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to :
$$6 \times {10^{ - 2}}$$
$$3 \times {10^{ - 3}}$$
$${10^{ - 1}}$$
$$6 \times {10^{ 2}}$$
Explanation
In circular motion the force required
$$\left| F \right| = {{m{v^2}} \over r}$$
$$\therefore\,\,\,$$ $${{m{v^2}} \over r} = {{16} \over r} + {r^3}$$
$$ \Rightarrow $$ mv2 = 16 + r4
$$\therefore\,\,\,$$ kinetic energy (K) = $${1 \over 2}$$ mv2 = $${1 \over 2}$$ [ 16 + r4]
$$\therefore\,\,\,$$ Kinetic energy of first particle (K1) = $${1 \over 2}$$ [16 + 1]
Kinetic energy of second particle (K2) = $${1 \over 2}$$ [16 + 44]
$$\therefore\,\,\,\,$$ $${{{K_1}} \over {{K_2}}}$$ = $${{{{16 + 1} \over 2}} \over {{{16 + 256} \over 2}}}$$ = $${{17} \over {272}}$$
$$ \Rightarrow $$ $$\,\,\,$$ $${{{K_1}} \over {{K_2}}} = 6 \times {10^{ - 2}}$$
$$\left| F \right| = {{m{v^2}} \over r}$$
$$\therefore\,\,\,$$ $${{m{v^2}} \over r} = {{16} \over r} + {r^3}$$
$$ \Rightarrow $$ mv2 = 16 + r4
$$\therefore\,\,\,$$ kinetic energy (K) = $${1 \over 2}$$ mv2 = $${1 \over 2}$$ [ 16 + r4]
$$\therefore\,\,\,$$ Kinetic energy of first particle (K1) = $${1 \over 2}$$ [16 + 1]
Kinetic energy of second particle (K2) = $${1 \over 2}$$ [16 + 44]
$$\therefore\,\,\,\,$$ $${{{K_1}} \over {{K_2}}}$$ = $${{{{16 + 1} \over 2}} \over {{{16 + 256} \over 2}}}$$ = $${{17} \over {272}}$$
$$ \Rightarrow $$ $$\,\,\,$$ $${{{K_1}} \over {{K_2}}} = 6 \times {10^{ - 2}}$$
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