Subject: General Aptitude · Chapter: Ratio and Proportion · Exam: · Marks: · Difficulty:
The time of oscillation of a pendulum varies as the square root of its length. If a pendulum of length 40 cm oscillates once in a second, find the length of the pendulum oscillating once in 2.5 seconds.
Answer
Answer (as printed):
Explanation
$T \propto \sqrt{l} \Rightarrow T=k\sqrt{l}$ for some constant $k$. When $l=40$ cm, $T=1$ sec, $T=k\sqrt{l} \Rightarrow 1=k\sqrt{40} \Rightarrow k=\frac{1}{\sqrt{40}}$. Let the required length be $x$ cm. Then, $2.5=\frac{1}{\sqrt{40}}\cdot\sqrt{x} \Rightarrow \sqrt{x}=2.5\times\sqrt{40} \Rightarrow x=(2.5\times\sqrt{40})^2=6.25\times 40=250$. Hence, required length = 250 cm.
Explanation as extracted from the printed page; notation may be imperfect.