A seconds pendulum clock has a steel wire. The clock shows correct time at 25° C. How much time does the clock lose or gain, in one week, when the temperature is increased to 35° C ? (α text steel =1.2 × 10-5 / ° C )

Question

A seconds pendulum clock has a steel wire. The clock shows correct time at 25° C. How much time does the clock lose or gain, in one week, when the temperature is increased to 35° C ? (α text steel =1.2 × 10-5 / ° C ) (A) 321.5 s (B) 3.828 s (C) 82.35 s (D) 36.28 s

Tardigrade Admin · Accepted Answer

(Δ T/T)=(1/2) α θ =(1/2) × 1.2 × 10-5 × 10 (Δ T/T)=6.0 × 10-5 Hence time lost in 1 week =6.0 × 10-5 × T =6.0 × 10-5 × 7 × 24 × 3600 =36.28 s

Q. A seconds pendulum clock has a steel wire. The clock shows correct time at $2 5^{\circ} C$ . How much time does the clock lose or gain, in one week, when the temperature is increased to $3 5^{\circ} C$ ?
$(α_{steel} = 1.2 \times 1 0^{- 5} /^{\circ} C)$

321.5 s

3.828 s

82.35 s

36.28 s

$\frac{Δ T}{T} = \frac{1}{2} α θ$
$= \frac{1}{2} \times 1.2 \times 1 0^{- 5} \times 10$
$\frac{Δ T}{T} = 6.0 \times 1 0^{- 5}$
Hence time lost in $1$ week $= 6.0 \times 1 0^{- 5} \times T$
$= 6.0 \times 1 0^{- 5} \times 7 \times 24 \times 3600$
$= 36.28 s$

Q. A seconds pendulum clock has a steel wire. The clock shows correct time at 25∘C. How much time does the clock lose or gain, in one week, when the temperature is increased to 35∘C ? (αsteel ​=1.2×10−5/∘C)