A seconds pendulum clock has a steel wire. The clock shows correc… - Vidyadip

MCQ

A seconds pendulum clock has a steel wire. The clock shows correct time at $25^{\circ} C$. .......... $s$ time does the clock lose or gain, in one week, when the temperature is increased to $35^{\circ} C$ ? $\left(\alpha_{\text {toel }}=1.2 \times 10^{-5} /{ }^{\circ} C \right)$

A
$321.5$
B
$3.828$
C
$82.35$
✓
$36.28$

✓

Answer

Correct option: D.

$36.28$

d
(d)

$\frac{\Delta T}{T}=\frac{1}{2} \alpha \theta$

$=\frac{1}{2} \times 1.2 \times 10^{-5} \times 10$

$\frac{\Delta T}{T}=6.0 \times 10^{-5}$

$\text { Hence time lost in } 1 \text { week }=6.0 \times 10^{-5} \times T$

$=6.0 \times 10^{-5} \times 7 \times 24 \times 3600$

$=36.28 \,s$

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