JEE Main 26-Aug-2021 Shift 2 Solved Paper

Given, accuracy of acceleration due to gravity = 4%
Accuracy of time period = 3%
Energy stored in pendulum at any instant is given as
E=

mgL‌θ2

2

...(i)
Time period of pendulum is given by expression
T=2π√

L

g

Rearrange the above expression for L, we get
L=g(

T

2π

)2 ...(ii)
Substituting the value of L in Eq. (i), we get
E=

mg2θ2

2

(T2π)2=

mg2θ2T2

8π2

Now, the accuracy in measurement of energy can be calculated as

ΔE

E

×100=2

Δg

g

×100+2

ΔT

T

×100 (∵ m,θ and π are constant)
⇒

ΔE

E

×100=2×4%+2×3%=14%
Thus, the accuracy to which E is known is 14%.