A wire fixed at the upper end stretches by le | vedclass

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

Question

(D) The force applied to the wire increases linearly from $0$ to $F$ as it stretches from $0$ to $l$.The average force applied during the stretching p

Vedclass · Accepted Answer

$\frac{Fl}{2}$

Vedclass · Answer

(D) The force applied to the wire increases linearly from $0$ to $F$ as it stretches from $0$ to $l$.The average force applied during the stretching process is given by $F_{av} = \frac{0 + F}{2} = \frac{F}{2}$.The work done $(W)$ in stretching the wire is equal to the product of the average force and the displacement (extension).$W = F_{av} 	imes l = \left( \frac{F}{2} ight) 	imes l = \frac{Fl}{2}$.Therefore,the work done is $\frac{Fl}{2}$.

$A$ wire fixed at the upper end stretches by length $l$ by applying a force $F$. The work done in stretching is

Similar Questions

Which kingdom consists of organisms that are multicellular,aquatic or terrestrial,heterotrophic,and eukaryotic?

Which one of the following elements on doping with germanium,makes it a $p$-type semiconductor?

During the electrolytic reduction of alumina,the reaction at cathode is

If $x-y+1=0$ meets the circle $x^2+y^2+y-1=0$ at $A$ and $B$,then the equation of the circle with $AB$ as diameter is

An electron accelerated through a potential difference $V$ passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2V$,the electron in the same magnetic field will experience a force:

Vedclass Test Series

Exam Paper Generator

Online Exam Module