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In Arrangement Given In Figure, If The Block Of Mass M Is Displaced, The Frequency Is Given By
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Two Springs Have Spring Constants KA And KB And KA > KB . The Work Required To Stretch Them By Same Extension Will Be
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A Mass M Attached To A Spring Oscillates Every 2 Sec. If The Mass Is Increased By 2 Kg, Then Timeperiod Increases By 1 Sec. The Initial Mass Is
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Onefourth Length Of A Spring Of Force Constant K Is Cut Away. The Force Constant Of The Remaining Spring Will Be
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Two Springs Of Force Constants K And 2K Are Connected To A Mass As Shown Below. The Frequency Of Oscillation Of The Mass Is
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A Spring Has Length L And Spring Constant K. If Spring Is Divided In Two Equal Parts Then Spring Constant Is
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If The Length Of Secondâ€™s Pendulum Is Decreased By 2%, How Many Seconds It Will Lose Per Day
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If The Length Of Simple Pendulum Is Increased By 300%, Then The Time Period Will Be Increased By
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If The Length Of Simple Pendulum Is Increased By 300%, Then The Time Period Will Be Increased By
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The Period Of Oscillation Of A Simple Pendulum Of Constant Length At Earth Surface Is T. Its Period Inside A Mine Is
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