Question

A clock with an iron pendulam keeps correct time at 20°c . How much will it make the time loss or gain per day if temperature change to 40°c ? Co-efficient of cubical expansion of iron is 36×10^(-6)/°c

Answer 1

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A)

We know that for an ideal pendulum time period T is given by the expression

T=2π√Lg

where L is the length of the pendulum and g accelaration due to gravity.

Given that the pendulum keeps correct time at 20

∴Tcorrect=2π√L20g .......(1)

Now we need to find time period at 40∘C

We know that with increase in temperature length L0 of metallic rod expands and the expression is

LT=L0(1+αΔT) .......(2)

where α is the coefficient of linear expansion and ΔT is the change in temperature.

It can be shown that the coefficient of volumetric expansion

γ≈3α

As such from (2) we get

L40=L20(1+36×10−63×(40−20))

L40=1.00024L20 .....(3)

We see that at 40∘C length of the pendulum is greater than at 20∘C.

As the time period is directly proportional to the square root of length, increase in length implies increase in time period. Which amounts to clock loosing time.

From (1)

T40=2π√L40g Using (3)

T40=2π√1.00024L20g

⇒T40=√1.00024×Tcorrect

To calculate loss per day we insert seconds in 1 day (24 hours) as the correct time and deduct seconds equal to one day. We get

ΔTime=√1.00024×86400−86400

=10.4s, rounded to one decimal place.

Answer 2

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We know that for an ideal pendulum time period T is given by the expression

T=2π√Lg

where L is the length of the pendulum and g accelaration due to gravity.

GIVEN:-

pendulum keeps correct time at 20

So,Tcorrect=2π√L20g _____(1)

Finding time period at 40°C➡

We know that with increase in temperature length L0 of metallic rod expands and the expression is

LT=L0(1+αΔT) ______(2)

where α is the coefficient of linear expansion and ΔT is the change in temperature.

Coefficient of volumetric expansion

γ≈3α

So from (2) we get

L40=L20(1+36×10−63×(40−20))

L40=L20(1+36×10−63×(40−20))L40=1.00024L20 ______(3)

As the time period is directly proportional to the square root of length, increase in length implies increase in time period. Which amounts to clock loosing time.

Therefore , From (1)

From (1)T40=2π√L40g Using (3)

T40=2π√1.00024L20g

T40=√1.00024×Tcorrect

For calculating loss per day we insert seconds in 24 hours as the correct time and deduct seconds equal to one day.

We get,

ΔTime=√1.00024×86400−86400

=10.4s ( by rounding to one decimal place)