Question

The length of a seconds pendulum is 1m. The decrease in its length to decrease its time period of oscillation by 1 second is:

• A. $0.75m$
• B. $0.5m$
• C. $0.414m$
• D. $0.25m$

Answer 1

The correct option is A $0.75m$

$T=2\pi \sqrt{\frac{e}{g}}$

$T_1=2\pi \sqrt{\frac{e_1}{g}}$

$T_2=2\pi \sqrt{\frac{e_2}{g}}$

$T_1-T_2=1$

$2\pi \sqrt{\frac{e_1}{g}}-\sqrt{\frac{e_2}{g}}=1$

$\sqrt{e_1}-\sqrt{e_2}=\frac{1}{2}$

Taking ${\pi }^2=g$

$1-\frac{1}{2}=\sqrt{e_2}$

$e_2^2=\frac{1}{4}$

$\therefore L_1-L_2=1-\frac{1}{4}$$=\frac{3}{4}=0.75m$