JEE Main 26-Aug-2021 Shift 2 Solved Paper

Increase in length of pendulum is 0.1%.
i.e.

ΔL

L

×100=0.1
Time period of pendulum is given by
T=2π√

L

g

Here, 2 π and g are constant.

ΔT

T

×100=

1

2

ΔL

L

×100
⇒

ΔT

T

×100=

1

2

×0.1
ΔT=0.05×

T

100

In one single day, the time in seconds is
T=24×60×60=86400s
∴ΔT=0.05×

86400

100

=43.2s
Thus, the error in time per day is 43.2 s.