The cross-sectional area of all wires is 10^{ | vedclass

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

Question

(B) The displacement of point $C$ is the sum of the elongations of wire $AB$ and wire $BC$.Elongation of wire $AB$ $(\Delta L_{AB})$ = $\frac{Mg L_{AB

Vedclass · Accepted Answer

$9 	imes 10^{-6} \, m$

Vedclass · Answer

(B) The displacement of point $C$ is the sum of the elongations of wire $AB$ and wire $BC$.Elongation of wire $AB$ $(\Delta L_{AB})$ = $\frac{Mg L_{AB}}{A Y_{AB}} = \frac{10 	imes 10 	imes 0.1}{10^{-4} 	imes 2.5 	imes 10^{10}} = \frac{1}{2.5} 	imes 10^{-5} = 0.4 	imes 10^{-5} = 4 	imes 10^{-6} \, m$.Elongation of wire $BC$ $(\Delta L_{BC})$ = $\frac{Mg L_{BC}}{A Y_{BC}} = \frac{10 	imes 10 	imes 0.2}{10^{-4} 	imes 4 	imes 10^{10}} = \frac{2}{4} 	imes 10^{-5} = 0.5 	imes 10^{-5} = 5 	imes 10^{-6} \, m$.Total displacement of point $C$ = $\Delta L_{AB} + \Delta L_{BC} = 4 	imes 10^{-6} + 5 	imes 10^{-6} = 9 	imes 10^{-6} \, m$.

The cross-sectional area of all wires is $10^{-4} \, m^2$. What is the displacement of point $C$?

Similar Questions

$A$ neutron makes a head-on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is

Glycogen is a polymer of

How many types of sporophylls are found in Pteridophytes?

Two soap bubbles combine to form a single bubble. In this process,the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure,and $T$ is the surface tension of the soap solution,the following relation is true :

An insulated container contains $4 \, moles$ of an ideal diatomic gas at temperature $T$. Heat $Q$ is supplied to this gas,due to which $2 \, moles$ of the gas are dissociated into atoms,but the temperature of the gas remains constant. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module