Question

A wire of steel of radius $r$ and length $L$ breaks due to weight $W$. What will be the breaking weight of another steel wire of radius $2r$ and length $2L$ ?
A. $4W$
B. $8W$
C. $\dfrac{W}{4}$
D. $\dfrac{W}{8}$

Answer 1

In order to solve this question we need to understand mass density of material. Mass Density of material is defined as mass of particle distributed per unit volume of material. It gives real information and can be useful in case of buoyancy force. Also density can be of various types like mass density (mass per unit volume), number density of electrons in solid state (number of electrons per unit volume), and density of energy states (number of particles per energy state).

Complete step by step answer:
Let the density of steel be $\rho$. Since the wire is of cylindrical shape, so the wire radius $R = r$ and length be $H = L$. So volume of cylinder is $V = \pi {R^2}H$
Putting values we get, $V = \pi {r^2}L$
Weight of material given $W$
So using formula of density we get, $\rho = \dfrac{W}{V}$
$\therefore W = \rho V$
Putting values we get, $W = \rho \pi {r^2}L \to (i)$

For another wire of steel, density is $\rho$
Radius is $R = 2r$
Height of cylinder $H = 2L$
Let new weight be $W'$
So new volume of wire is $V' = \pi {R^2}H$
Putting values we get, $V' = \pi {(2r)^2}(2L)$
$V' = 8\pi {r^2}L$
So using formula of density we get, $\rho = \dfrac{{W'}}{{V'}}$
$\therefore W' = \rho V'$
Putting values we get, $W' = 8\rho \pi {r^2}L$
From equation $(i)$ we get, $W' = 8W$

The correct option is B.

Note: It should be remembered that here the mass density in both cases is same because wire in both cases is of the same material which is steel. Also Weight is maximum hold capacity a steel have, more than this weight steel forces could not balance gravity and broke or deformation in steel exist. Also steel has high holding capacity because its atoms are more closely packed and bond energy is large enough.