If the length of a simple pendulum is increased by 2 %, then the time period

Question

If the length of a simple pendulum is increased by 2 %, then the time period (A) increases by 1% (B) decreases by 1% (C) increases by 2 % (D) decreases by 2 %

Tardigrade Admin · Accepted Answer

l2=1.02 l1. Time period (T)=2 π × √(l/g) propto √l Therefore (T2/T1)=√(l2/l1)=√(1.02 l1/l1)=1.01. Thus time period increased by 1 %.

Q. If the length of a simple pendulum is increased by $2%$ , then the time period

increases by 1%

decreases by 1%

increases by 2 %

decreases by 2 %

$l_{2} = 1.02 l_{1}$ .
Time period $(T) = 2 π \times \frac{l}{g} \propto l$
Therefore $\frac{T _{2}}{T _{1}} = \frac{l _{2}}{l _{1}} = \frac{1.02 l _{1}}{l _{1}} = 1.01$ .
Thus time period increased by $1%$ .

Q. If the length of a simple pendulum is increased by 2%, then the time period