Mechanical Properties of Solids MCQs for NEET — Physics Questions with Answers

(d)The object is spherical and the bulk modulus is represented by B. It is the ratio of normal stress to the volumetric strain. 

Hence             $B=\frac{F/A}{∆V/V}$

$\Rightarrow \frac{∆V}{V}=\frac{P}{B}\Rightarrow \left|\frac{∆V}{V}\right|=\frac{P}{B}$

Hence p is applied pressure on the object and $\frac{∆V}{V}$ is

volume strain 

Fractional decrease in volume 

$\Rightarrow \frac{∆V}{V}=3\frac{∆R}{R} \left[So, V=\frac{4}{3}{\mathrm{\pi R}}^3\right]$

Volume of the sphere decreases due to the decrease in its radius.

Hence $\frac{∆V}{V}=\frac{3∆R}{R}=\frac{P}{B}\Rightarrow \frac{∆R}{R}=\frac{P}{3B}$

Given, Y_steel=2Y_brass and L_s=L_b and A_s=A_b

such that ΔL_s=ΔL_b

As we know,Young's modulus

Y=stress/strain=W/A/ΔL/L



i.e. W_s/W_b=Y_s/Y_b=2Y_b/Y_b=2/1

=> 2:1

Thus, weight added to the steel and brass wires must be in the ratio of 2:1

For Hooke's law, Young Modulus 

y=F/A/Δl/l

=F x l/Δl x A ...(i)

V = A x l=constant ...(ii)

From Eqs. (i) and (ii),

y=F x lxl / Δl x V

Δl=F/V x y x l²  

=>Δl∝l²  

(a) ΔL=FL/AY or ΔL∝L/d²  [∴ A=πd²/4]

Therefore,ΔL will be maximum for that wire for which L/A is maximum.

(a) Young's modulus of wire does not varies with dimension of wire. It is the property of given material.

(c)

      $$\begin{gathered} l=\frac{FL}{AY}\Rightarrow l\propto \frac{1}{r^2}(F,L and Y are cons\tan t) \\ \frac{l_2}{l_1}={\left(\frac{r_1}{r_2}\right)}^2={\left(2\right)}^2=4\Rightarrow l_2=4l_1=4cm \end{gathered}$$

(c) $l=\frac{mgL}{AY}=\frac{1\times 10\times 1.1}{1.1\times {10}^{11}\times {10}^{-6}}m=0.1mm$

(b) To double the length of wire,

Stress = Young's modulus

$\therefore \frac{F}{A}=2\times {10}^{12}\frac{dyne}{c{m}^2}$

If A=$1$ then F=$2 \times {10}^{12}dyne$

(d) When a elastic body is stretched and applied deforming force is removed then the body is expected to return to its original configuration instantaneously. But sometimes some materials take some time to return to its original configuration. This temporary delay in achieving its original configuration is termed as elastic after effect. This elastic after effect is very short for quartz fiber. But elastic after effect is more for glass. 

(c) Breaking Force = µ X Area of cross section of wire ($\mathrm{\pi }$r²)

If radius of wire is double then breaking force will become four times.