A simple pendulum oscillating in air has a period of s. If it is completely immersed in non-viscous liquid, having density of the material of the bob, the new period will be [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A simple pendulum oscillating in air has a period of $\sqrt{3}$ s. If it is completely immersed in non-viscous liquid, having density ${(\frac{1}{4})}^{th}$ of the material of the bob, the new period will be [2023]

1 $2 \sqrt{3}$ s

2 $\frac{2}{\sqrt{3}}$ s

3 $2$ s

4 $\frac{\sqrt{3}}{2}$ s

Ans.
(3)

For simple pendulum, $T = 2 π \sqrt{\frac{L}{g_{eff}}}$

According to the question, $g_{eff} = g (1 - \frac{ρ_{l}}{ρ_{0}}) = g (1 - \frac{1}{4}) = \frac{3}{4} g$

So, $\frac{T'}{T} = \frac{2 π \sqrt{\frac{L}{3 g / 4}}}{2 π \sqrt{\frac{L}{g}}} \Rightarrow T' = \frac{2}{\sqrt{3}}$

As, $T = \sqrt{3}$ $s$ so $T' = \frac{2}{\sqrt{3}} \times \sqrt{3} = 2 s$

Want Us to Email you About Special Offers & Updates?