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If momentum of an object is increased by 10%, then is kinetic energy will increase byA. `40%`B. `20%`C. `10%`D. `21%` |
Answer» Correct Answer - D `(K.E._(2))/(K.E._(1))=(L_(2)^(2))/(L_(1)^(2))` `= (R_(2)^(2)omega_(2)^(2))/(R_(1)^(2)omega_(1)^(2))=(1.1)^(2)=1.21` `therefore (K.E._(2))/(K_(.E._(1))-1=1.21-1 therefore K.E._(1)=21%` `therefore K.E._(2)` increased by 21%. |
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