Two identical simple pendulums each of length | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $L = 5 \, cm = 0.05 \, m$ be the length of the string and $d = 6 \, cm = 0.06 \, m$ be the distance between the bobs. At equilibrium, the bob

Vedclass · Accepted Answer

$0.12$

Vedclass · Answer

(A) Let $L = 5 \, cm = 0.05 \, m$ be the length of the string and $d = 6 \, cm = 0.06 \, m$ be the distance between the bobs. At equilibrium, the bob is acted upon by three forces: tension $T$, gravitational force $mg$, and electrostatic force $F_e = \frac{kq^2}{d^2}$. Let $	heta$ be the angle the string makes with the vertical. From the geometry, $\sin 	heta = \frac{d/2}{L} = \frac{3 \, cm}{5 \, cm} = 0.6$. Thus, $\cos 	heta = \sqrt{1 - \sin^2 	heta} = 0.8$. In equilibrium, $	an 	heta = \frac{F_e}{mg}$. Substituting the values: $	an 	heta = \frac{0.6}{0.8} = 0.75$. $F_e = \frac{(9 	imes 10^9) 	imes (2 	imes 10^{-6})^2}{(0.06)^2} = \frac{9 	imes 10^9 	imes 4 	imes 10^{-12}}{0.0036} = \frac{36 	imes 10^{-3}}{36 	imes 10^{-4}} = 10 \, N$. Now, $mg = \frac{F_e}{	an 	heta} = \frac{10}{0.75} = \frac{10}{3/4} = \frac{40}{3} \, N$. $m = \frac{40}{3 	imes 10} = \frac{4}{3} \, kg \approx 1.33 \, kg$. (Note: Given the options provided in the original prompt were incomplete, the calculated value is $4/3 \, kg$).

Similar Questions

Two identical charges repel each other with a force equal to $10 \text{ mg}$ wt when they are $0.6 \text{ m}$ apart in air $(g = 10 \text{ ms}^{-2})$. The value of each charge is:

Two identical pendulums $A$ and $B$ are suspended from the same point. The bobs are given positive charges,with $A$ having more charge than $B$. They diverge and reach equilibrium,with $A$ and $B$ making angles $\theta_1$ and $\theta_2$ with the vertical,respectively. Then:

Two point charges $+3 \mu C$ and $+8 \mu C$ repel each other with a force of $40 \ N$. If a charge of $-5 \mu C$ is added to each of them,then the force between them will become: (in $N$)

Four charges equal to $-Q$ are placed at the four corners of a square and a charge $q$ is at its centre. If the system is in equilibrium, the value of $q$ is

Two electrons are moving towards each other,each with a velocity of $10^6 \, m/s$. What will be the closest distance of approach between them in $m$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module