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# a pendulum clock becomes 5sec fast each day at temp at 15 c and becomes 10 sec slow each day at temp 30 c .the temp at which the clock gives correct time is

arun
123 Points
5 years ago
in this we have to use mainly three concepts which are  L2=L1(1+(alpha)(change in temprature))
another is binomial theorem which is (1 + x )n= 1 + nx         for x
and the last one which is $T = 2\pi\sqrt{\frac{l}{g}}$
now it is given that clock is 5 seconds fast each day i.e. 1.16 $\times$ 10-4 seconds
i.e. time period at that temprature is (2 – 1.16 $\times$ 10-4)
and let at $\theta$ temprature it shows correct time i.e. it’s time period is 2 seconds
now, $\frac{T_{1}}{T_{o}}=\sqrt{\frac{L_{1}}{L_{o}}}$
and L1 = Lo(1 + $\alpha$(15  – $\theta$))
$\Rightarrow \frac{T_{1}}{T_{o}}=\sqrt{\frac{L_{o}(1 + \alpha (15 - \theta ))}{L_{o}}}$
using binomial theorem
$\Rightarrow \frac{T_{1}}{T_{o}}=1 - \frac{1}{2}\alpha(15 - \theta )$
$\frac{2 - 1.16 \times 10^{-4}}{2} = 1 + \frac{1}{2}\alpha (\theta - 15)$
$\theta - 15 = \frac{1.16 \times 10^{-4}}{\alpha}$        …......................1
similarly
$30 - \theta = \frac{2.32 \times 10^{-4}}{\alpha}$   …............................2
now subtracting adding equation 1 and 2 we get
$15 = \frac{3.48 \times 10^{-4}}{\alpha}$
$\Rightarrow \alpha = \frac{3.48 \times 10^{-4}}{15}$
substituting it in equation 1 we get
$\theta - 15 = 5$
$\Rightarrow \theta = 20$
so clock will give correct time at 20 oC
Sandesh
13 Points
3 years ago
A simple method to solve this is the following:Time lost or gained= aplha∆t/2 X given time periodCase 1: Time gained is 5s so the condition is like this: 5=alpha(t-15)/2 where its (t-15) as its gained so t is at more temp.Case 2: Time lost is 10s so the condition is like this: 10=alpha(30-t)/2 where its (30-t) as its lost so t is at a less temp.Dividing these two equations we get the temp at which clock gives correct time.10/5=(30-t)/2 X 2/(t-15)2(t-15)=30-t2t-30=30-t3t=60T=20°
Harsh Jaiswal
19 Points
3 years ago
We know that: T = 2π √(l/g)
dT/T= 1/2∝ ∆T
And we can say that temp., would be between 15oC and 30oC
5/T=1/2∝(θ- 15) ------1nd
10/T= 1/2  ∝(30- θ) ------2nd
Diviving 1st and 2nd eqn
we get,
5/10= (θ-15)/(30- θ)
⸫ θ = 20oC
Yash Chourasiya
11 months ago
Dear Student

Please see the solution in the attachment.