A wire can be broken by applying a load of 200 N. The force required to break another wire of the same length and same material, but double in diameter, is

Young’s modulus $Y = \frac{{FL}}{{Al}}\,or\,F = \frac{{YAl}}{L}$ or $F \propto A\;{\rm{or\;}}F \propto {r^2}\;{\rm{or\;}}F \propto {d^2}$ $\therefore \,\frac{{{F_1}}}{{{F_2}}} = \frac{{d_1^2}}{{d_2^2}}$ Given, ${d_1} = d,\;\;{d_2} = 2d,\;{F_1} = 200{\rm{N}}$ , $\therefore \,\frac{{200}}{{{F_2}}} = \frac{{{{\left( d \right)}^2}}}{{{{\left( {2d} \right)}^2}}} = \frac{1}{4}$ or ${F_2} = 4 \times 200 = 800{\rm{N}}$

When a weight of $10\,kg$ is suspended from a copper wire of length 3 metres and diameter $0.4\,mm$, its length increases by $2.4\,cm$. If the diameter of the wire is doubled, then the extension in its length will be

$l \propto \frac{{FL}}{{\pi {r^2}Y}} \Rightarrow l \propto \frac{L}{{{r^2}}}$ [F and Y are constant] $\frac{{{l_1}}}{{{l_2}}} = \frac{{{L_1}}}{{{L_2}}}{\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2} = \frac{1}{2}{\left( {\sqrt 2 } \right)^2}\;\therefore \frac{{{l_1}}}{{{l_2}}} = 1:1$

The correct Option is "$0.6\,cm$"

The length of an elastic string is a metre when the tension is 44 N, and b metre when the tension is 5 N. The length in metre when the tension is 9 N, is

${T_1} = K\left( {l - {l_1}} \right)\,,\,{T_2} = K\left( {l - {l_2}} \right)\,So,\,\frac{{{T_1}}}{{{T_2}}} = \;\frac{{l - {l_1}}}{{\left( {l - {l_2}} \right)}}\;$ $\therefore \;\;{T_1}l - {T_1}{l_2} = {T_2}l - {T_2}{l_1}$ , $\left( {{T_1} - {T_2}} \right)l = {T_1}{l_2} - {T_2}{l_1}\,,\,l = \;\frac{{{T_1}{l_2} - {T_2}{l_1}}}{{\left( {{T_1} - {T_2}} \right)}}\;$ , $l = \left( {5a - 4b} \right)...\left( i \right)\,,\,k = \frac{1}{{b - a\;}}....\left( {ii} \right)$ So, length of wire when tension is 9 N. $9 = kl'\;(l' = \,change{\rm{ }}in{\rm{ }}length)$ , $9 = \frac{1}{{\left( {b - a} \right)}}\; \times l' \Rightarrow l' = \;9b - 9a$ Hence, final length $ = l + l'\,,\, = 5a - 4a + 9a - 9a\,,\,{l_0} = 5b - 4a$

The correct Option is "$5b - 4a$"

The following four wires of length $L$ and radius $r$ are made of the same material. Which of these will have the largest extension, when the same tension is applied?

Young’s modulus $\;Y = \frac{F}{A} \times \frac{L}{l}$ , $\;Y = \frac{F}{{\pi {r^2}}} \times \frac{L}{l}\,,\,Y \propto \frac{L}{{{r^2}}}$ Option (d) has the largest extension when the same tension is applied.

The correct Option is "L = 100 cm, r = 0.2 mm"

The mass and length of a wire are $M\,and\,L$ respectively. The density of the material of the wire is $d$. On applying the force $F$ on the wire, the increase in length is $l$, then the Young’s modulus of the material of the wire will be

$Y = \frac{F}{A}\frac{L}{l} = \frac{{Fd{L^2}}}{{Ml}}$ As M= Volume $ \times $ density $ = A \times L \times d\therefore A = \frac{M}{{Ld}}$

The correct Option is "$\frac{{Fd{L^2}}}{{Ml}}$"

A wire of area of cross-section ${10^{ - 6}}{m^2}$ is increased in length by $0.1\% $. The tension produced is 1000 N. The Young’s modulus of wire is

$Y = \frac{{FL}}{{Al}} = \frac{{1000 \times 100}}{{{{10}^{ - 6}} \times 0.1}} = {10^{12}}N/{m^2}$

The correct Option is "${10^{12}}N/{m^2}$"

A steel wire of 1m long and $1m{m^2}$ cross section area is hang from rigid end. When weight of $1kg$ is hung from it then change in length will be (given $Y = 2 \times {10^{11}}N/{m^2}$)

$l = \frac{{MgL}}{{YA}} = \frac{{1 \times 10 \times 1}}{{2 \times {{10}^{11}} \times {{10}^{ - 6}}}} = 0.05\;mm$

The correct Option is "$0.05\,mm$"

For a constant hydraulic stress on an object, the fractional change in the object’s volume $\left( {\frac{{\Delta V}}{V}} \right)$ and its bulk modulus ($B$) are related as

$B = \frac{{\Delta P}}{{\Delta V/V}} \Rightarrow \frac{1}{B} \propto \frac{{\Delta V}}{V}\;\;\;\;[\Delta p = constant]$

$\frac{{\Delta V}}{V} \propto B$

The correct Option is "$\frac{{\Delta V}}{V} \propto \frac{1}{B}$"

$\frac{{\Delta V}}{V} \propto {B^2}$

$\frac{{\Delta V}}{V} \propto {B^{ - 2}}$

The strain-stress curves of three wires of different materials are shown in the figure. $P,Q\,and\,R$ are the elastic limits of the wires. The figure shows that.

As stress is shown on x-axis and strain on y-axis So we can say that $Y = \cot \theta = \frac{1}{{\tan \theta }} = \frac{1}{{{\rm{slope}}}}$ So elasticity of wire $P$ is minimum and of wire $R$ is maximum

Elasticity of wire $P$ is maximum

Elasticity of wire $Q$ is maximum

Tensile strength of $R$ is maximum

The correct Option is "None of the above is true"

Wires A and B are made from the same material. A has twice the diameter and three times the length of B. If the elastic limits are not reached, when each is stretched by the same tension, the ratio of energy stored in A to that in B is

$U = \frac{1}{2}Fl = \frac{{{F^2}L}}{{2AY}}.\;U \propto \frac{L}{{{r^2}}}\;[F$ and Y are constant] $\therefore \frac{{{U_A}}}{{{U_B}}} = \left( {\frac{{{L_A}}}{{{L_B}}}} \right) \times {\left( {\frac{{{r_B}}}{{{r_A}}}} \right)^2} = \left( 3 \right) \times {\left( {\frac{1}{2}} \right)^2} = \frac{3}{4}$