Question

A pendulum clock keeps correct time at 20°C. The
correction to be made during summer per day
where the average temperature is 40°C,​

Answer 1

answer : time loss , ∆t = $10\alpha$ × 86400 sec

you didn't mention about coefficient of linear thermal expansion of pendulum. Let $\alpha$ is coefficient of linear thermal expansion of pendulum.

initially, length of pendulum is l

and then time period , T = 2π√{l/g}....(1)

after increase temperature 20°C to 40°C , length of pendulum increases, $l'=l(1+\alpha(40-20)=l(1+20\alpha)$

here it is clear that, $l(1+20\alpha)$ > l

so, T' = 2π√{l'/g} > T

so, time period increases during summer day (when temperature is 40°C), due to increase in time period, clock becomes slow.

∆T can be written as ∆T = T' -T

and then, ∆T/T = (T' - T)/T

= (T'/T) - 1

= $\sqrt{\frac{l(1+20\alpha)}{l}}-1$

from binomial expansion,

if (1 + a)ⁿ , where a <<< 1 then, (1 + a)ⁿ = (1 + na)

so, here $(1+20\alpha)^{1/2}=1+\frac{20}{2}\alpha$

= $(1+\frac{20}{2}\alpha)-1$

= $10\alpha$

and we also know, ∆T/T = ∆t/t

or, ∆t/t = $10\alpha$

here t is time taken in a day,

e.g., t = 24 × 60 × 60 = 86400 sec

then, time loss , ∆t = $10\alpha$ × 86400 sec

now you can get answer by putting value of $\alpha$

Answer 2

Explanation:

☢️PLEASE MARK AS BRAINLIEST☢️