A clock pendulum made of invar has a period of 2 s at 20° C. If the clock is used in a climate where average temperature is 40° C, what correction (in seconds) may be necessary at the end of 10 days to the time given by clock? (α text invar =7 × 10-7 ° C -1, 1 text day =8.64 × 104 s )

Question

A clock pendulum made of invar has a period of 2 s at 20° C. If the clock is used in a climate where average temperature is 40° C, what correction (in seconds) may be necessary at the end of 10 days to the time given by clock? (α text invar =7 × 10-7 ° C -1, 1 text day =8.64 × 104 s ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

T=2 π √(l/g) ⇒ (Δ T/T)=(1/2) (Δ l/l)=(1/2) α Δ θ ⇒ Δ T=(T/2)(α Δ θ) =((10 × 8.64 × 104/2))(7 × 10-7)(20)=6 s

Q. A clock pendulum made of invar has a period of 2s at 20∘C. If the clock is used in a climate where average temperature is 40∘C, what correction (in seconds) may be necessary at the end of 10 days to the time given by clock? (αinvar ​=7×10−7∘C−1,1 day =8.64×104s)

T=2πgl​​⇒TΔT​=21​lΔl​=21​αΔθ ⇒ΔT=2T​(αΔθ) =(210×8.64×104​)(7×10−7)(20)=6s

$T = 2 π \frac{l}{g} \Rightarrow \frac{Δ T}{T} = \frac{1}{2} \frac{Δ l}{l} = \frac{1}{2} α Δ θ$
$\Rightarrow Δ T = \frac{T}{2} (α Δ θ)$
$= (\frac{10 \times 8.64 \times 1 0 ^{4}}{2}) (7 \times 1 0^{- 7}) (20) = 6 s$