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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Thermal Properties of Matter

NTA Abhyas 2020 · Physics · Thermal Properties of Matter

A seconds pendulum clock having steel wire is calibrated at 20 ° C...

A seconds pendulum clock having steel wire is calibrated at 20°C20 ° C. When temperature is increased to 30°C30 ° C, then calculate how much time does the clock lose or gain in one week~[αsteel=1.2×10−5∘C−1]\left[{α}_{steel} = 1\text{.}2 \times 1{0}^{- 5}{}^{∘}C^{-1}\right]

Options

  1. A0.3628s\text{0.3628}s
  2. B3.626s\text{3.626}s
  3. C362.8s\text{362.8}s
  4. D36.28s\text{36.28}s

Correct answer

D. 36.28s\text{36.28}s

Step-by-step solution

The time period of the second's pendulum = 2s2s

Change in the time period, ΔT=12TαΔθ\Delta \text{T} = \dfrac{1}{2} \text{T} \alpha \Delta \theta

\Delta \text{T} = \left(\dfrac{1}{2}\right) \left(2\right) & \left(1 \text{.}2 \times1 {0}^{- 5}\right)\left(3 {0}^{∘} - 2 {0}^{∘}\right)

=1.2×10−4s<br/><br/>Newtimeperiod,= 1 \text{.}2 \times1 {0}^{- 4} \text{s}<br /><br />New time period, T^{′} = T+\text{T} +\Delta T\text{T}

T′=2.00012s<br/><br/>Timelostinaweek,<br/><br/>Δt=ΔTT′×tT^{′} = 2 \text{.}0 0 0 1 2 \text{s}<br /><br />Time lost in a week,<br /><br /> \Delta \text{t} = \dfrac{\Delta \text{T}}{T^{′}} \times \text{t}

=1.2×10−42.00012×(7×24×3600)= \dfrac{1 \text{.} 2 \times 1 {0}^{- 4}}{2 \text{.} 0 0 0 1 2} \times\left(7 \times2 4 \times3 6 0 0\right)

=36.= 3 6 \text{.}}28s\text{s}
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