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In a satellite if the time of revolution is `T`, then kinetic energy is proportional toA. `1/T`B. `1/(T^(2))`C. `1/(T^(3))`D. `T^(2//3)` |
Answer» Correct Answer - B `K=1/2mv^(2)=1/2m omega^(2) R^(2)=1/2m((2pi)/T)^(2) R^(2)` `=(2pi^(2)mR^(2))/(T^(2))` `K prop 1/(T^(2))` |
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