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If the length of a clock pendulum increa...

If the length of a clock pendulum increases by 0.2% due to atmospheric temperature rise, then the loss in time of clock per day is

A

86.4s

B

43.2s

C

72.5s

D

32.5s

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The correct Answer is:
To solve the problem, we need to determine the loss in time of a clock per day due to an increase in the length of the pendulum. Here's a step-by-step solution: ### Step 1: Understand the relationship between pendulum length and time period The time period \( T \) of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. ### Step 2: Calculate the change in length Given that the length of the pendulum increases by 0.2%, we can express this mathematically. If the original length is \( L \), then the new length \( L' \) is: \[ L' = L + 0.002L = 1.002L \] ### Step 3: Calculate the new time period Now, we can find the new time period \( T' \) using the new length: \[ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{1.002L}{g}} = 2\pi \sqrt{\frac{L}{g}} \sqrt{1.002} \] Using the approximation \( \sqrt{1+x} \approx 1 + \frac{x}{2} \) for small \( x \), we get: \[ T' \approx T \left(1 + \frac{0.002}{2}\right) = T(1 + 0.001) = 1.001T \] ### Step 4: Calculate the change in time period The change in time period \( \Delta T \) is: \[ \Delta T = T' - T \approx 1.001T - T = 0.001T \] ### Step 5: Determine the loss in time over one day The clock is supposed to keep time for one day (86400 seconds). The loss in time per day can be calculated as: \[ \text{Loss in time} = \Delta T \cdot \frac{86400 \text{ seconds}}{T} \] Since \( T \) for a clock pendulum is approximately 2 seconds (for a simple pendulum of reasonable length), we can substitute: \[ \text{Loss in time} = 0.001 \cdot 86400 \text{ seconds} = 86.4 \text{ seconds} \] ### Final Answer The loss in time of the clock per day is approximately **86.4 seconds**. ---
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AAKASH INSTITUTE-OSCILLATIONS-Assignment (Section - A) (OBJECTIVE TYPE QUESTIONS)
  1. Two pendulums of length 1.21 m and 1.0 m starts vibrationg. At some in...

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  2. The time period of oscillations of a simple pendulum is 1 minute. If i...

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  3. If the length of a clock pendulum increases by 0.2% due to atmospheric...

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  4. A simple pendulum is oscillating in a trolley moving on a horizontal s...

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  5. The time period of oscillation of a simple pendulum is sqrt(2)s. If it...

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  6. The graph between time period (T) and length (l) of a simple pendulum ...

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  7. A hollow sphere is filled with water through a small hole in it. It is...

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  8. A uniform rod of mass m and length l is suspended about its end. Time ...

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  9. A uniform disc of mass m and radius r is suspended through a wire atta...

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  10. A solid cylinder of denisty rho(0), cross-section area A and length l ...

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  11. A block of mass m hangs from three springs having same spring constant...

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  12. Two masses m(1) = 1kg and m(2) = 0.5 kg are suspended together by a ma...

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  13. A mass m is attached to two springs of same force constant K, as shown...

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  14. A clock S is based on oscillations of a spring and clock P is based on...

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  15. A 100 g mass stretches a particular spring by 9.8 cm, when suspended v...

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  16. An assembly of identicl spring mass system is placed on a smooth horiz...

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  17. The time period of a mass suspended from a spring is T. If the spring ...

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  18. When a block of mass m is suspended separately by two different spring...

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  19. In damped oscillations damping froce is directly proportional to speed...

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  20. In forced oscillations , a particle oscillates simple harmonically wit...

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