Past NEET papers

Hint

(c) Stress \(=\frac{\text { force }}{\text { cross-section area }}=\frac{M g}{A}\)

Strain \(=\frac{\text { change in length }}{\text { original length }}=\frac{\Delta L}{L}=\frac{L_1-L}{L}\)

Young’s modulus, \(Y=\frac{\text { stress }}{\text { strain }}=\frac{M g L}{A\left(L_1-L\right)}\)

Hint

(c) Fracture point and ultimate strength point is close for material \(X\), hence \(X\) is brittle in nature and both points are far apart for material \(Y\) hence it is ductile.

Hint

\(
\Delta \ell=\left(\frac{F}{A Y}\right) 3 \ell \dots(i)
\)
\(
\Delta \ell=\left(\frac{F^{\prime}}{3 A Y}\right) \ell \dots(ii)
\)
From equation (i) \& (ii),
\(
\Delta \ell=\left(\frac{F}{A Y}\right) 3 \ell=\left(\frac{F^{\prime}}{3 A Y}\right) \ell \text { or, } F^{\prime}=9 F
\)

Hint

(a) Young’s modulus \(\mathrm{Y}=\frac{\mathrm{W}}{\mathrm{A}} \cdot \frac{\mathrm{l}}{\Delta l}\)
\(
\frac{W_1}{Y_1}=\frac{W_2}{Y_2}
\)
\([\because \mathrm{A}, l, \Delta l\) same for both brass and steel]
\(
\frac{W_1}{W_2}=\frac{Y_1}{Y_2}=2 \quad\left[Y_{\text {steel }} / Y_{\text {brass }}=2 \text { given }\right]
\)

Hint

(b) As \(\mathrm{Y}=\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{\Delta l}{l}} \Rightarrow \Delta l=\frac{\mathrm{Fl}}{\mathrm{AY}}\)
But \(\mathrm{V}=\mathrm{A} l\) so \(\mathrm{A}=\frac{\mathrm{V}}{l}\)
Therefore \(\Delta l=\frac{F l^2}{\mathrm{VY}} \propto l^2\)
Hence graph of \(\Delta l\) versus \(l^2\) will give a straight line.

Hint

(d) \(\mathrm{F}=\frac{\mathrm{YA}}{\mathrm{L}} \times \Delta \mathrm{L}\)
So, extension, \(\Delta \mathrm{L} \propto \frac{\mathrm{L}}{\mathrm{A}} \propto \frac{\mathrm{L}}{\mathrm{D}^2}\)
\([\because\) F and \(Y\) are constant \(]\)
\(
\begin{aligned}
& \Delta \mathrm{L}_1 \propto \frac{100}{1^2} \propto 100 \text { and } \Delta \mathrm{L}_2 \propto \frac{200}{2^2} \propto 50 \\
& \mathrm{~L}_3 \propto \frac{300}{3^2} \propto \frac{100}{3} \text { and } \mathrm{L}_4 \propto \frac{50}{\frac{1}{4}} \propto 200
\end{aligned}
\)
The ratio of \(\frac{L}{\mathrm{D}^2}\) is maximum for case (d).

Hint

(c) From formula,
Increase in length \(\quad \Delta L=\frac{F L}{A Y}=\frac{4 F L}{\pi D^2 Y}\)
\(
\begin{aligned}
\frac{\Delta L_S}{\Delta L_C} & =\frac{F_S}{F_C}\left(\frac{D_C}{D_S}\right)^2 \frac{Y_C}{Y_S} \frac{L_S}{L_C}=\frac{7}{5} \times\left(\frac{1}{p}\right)^2\left(\frac{1}{s}\right) q \\
& =\frac{7 q}{\left(5 s p^2\right)}
\end{aligned}
\)

Note:

If a wire of length \(l\) is suspended from a rigid support and \(\mathrm{A}\) is the area of cross-section of the wire, then mass of the wire, \(\mathrm{m}=\mathrm{A} / \mathrm{p}\). Then extension in the wire due to its own weight can be find as follows.
Young Modulus, \(Y=\frac{m g}{A} \times \frac{(l / 2)}{\Delta l}\)
Here, \(\mathrm{L}=1 / 2\) because the weight of the wire acts at the midpoint of the wire.
\(\therefore\) Extension in the wire due to its own weight
\(\Delta l-\frac{l^2 p g}{2 Y}\)

Hint

(c) We know that Young’s modulus
\(
\mathrm{Y}=\frac{\mathrm{F}}{\pi \mathrm{r}^2} \times \frac{\mathrm{L}}{\ell}
\)
Since Y, F are same for both the wires, we have,
\(
\ell \propto \frac{\mathrm{L}}{\mathrm{r}^2}
\)
or, \(\frac{\ell_1}{\ell_2}=\frac{r_2^2 \times L_1}{r_1^2 \times L_2}=\frac{\left(D_2 / 2\right)^2 \times L_1}{\left(D_1 / 2\right)^2 \times L_2}\)
or, \(\frac{\ell_1}{\ell_2}=\frac{\mathrm{D}_2^2 \times \mathrm{L}_1}{\mathrm{D}_1^2 \times \mathrm{L}_2}=\frac{\mathrm{D}_2^2}{\left(2 \mathrm{D}_2\right)^2} \times \frac{\mathrm{L}_2}{2 \mathrm{~L}_2}=\frac{1}{8}\)
So, \(\ell_1: \ell_2=1: 8\)

Hint

In stretching a wire, the work done against internal restoring force is stored as elastic potential energy in wire and given by

\(
\begin{aligned}
U=W & =\frac{1}{2} \times \text { Force }(F) \times \text { Elongation }(l) \\
& =\frac{1}{2} F l=\frac{1}{2} \times M g \times l=\frac{1}{2} M g l
\end{aligned}
\)

Hint

(c) Bulk modulus is given by
\(
\begin{aligned}
& \mathrm{B}=\frac{\mathrm{p}}{\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)} \text { or } \frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\mathrm{p}}{\mathrm{B}} \\
& 3 \frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\mathrm{p}}{\mathrm{B}} \text { (here, } \frac{\Delta R}{R}=\text { fractional decreases in radius} \\
\end{aligned}
\)
\(
\Rightarrow \frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\mathrm{p}}{3 \mathrm{~B}}
\)

Note:
If the object of mass \(m\) and radius \(r\) is sorrounded by a liquid in a cylindrical container
\(
\frac{\Delta R}{R}=\frac{1}{3} \frac{\Delta V}{V} \dots(i)
\)
Bulk modulus, \(B=-V \frac{\Delta P}{\Delta V}\)
\(
\begin{aligned}
& \therefore \frac{\Delta V}{V}=\frac{\Delta P}{B}=\frac{m g}{A B} \quad\left(\because \Delta P=\frac{m g}{A}\right) \\
& \therefore \frac{\Delta R}{R}=\frac{1}{3} \frac{m g}{A B} \quad \text { (Here } \mathrm{A}=\text { area of object) }
\end{aligned}
\)

Hint

(c) Energy stored per unit volume
\(
\begin{aligned}
& =\frac{1}{2} \times \text { stress } \times \text { strain } \\
& =\frac{1}{2} \times \text { stress } \times(\text { stress } / \text { Young’s modulus }) \\
& =\frac{1}{2} \times(\text { stress })^2 /(\text { Young’s modulus })=\frac{S^2}{2 \mathrm{Y}}
\end{aligned}
\)

Hint

We know, velocity of transverse wave
\(
\begin{aligned}
& v=\sqrt{\frac{T}{\mu}} \\
& \therefore \quad v_i=\sqrt{\frac{T}{\mu}} \text { and } v_f=\sqrt{\frac{2 T}{\mu}} \\
& \therefore \quad \frac{v_i}{v_f}=\frac{1}{\sqrt{2}}
\end{aligned}
\)

Hint

\(
\begin{aligned}
F_{v p} & =20000+3000 \\
& =23000 \mathrm{~N}
\end{aligned}
\)
Minimum power \(P_{\min }=\vec{F} \cdot \vec{v}\)
\(
\begin{aligned}
P_{\min } & =F V=23000 \times \frac{3}{2} \\
& =34500 \mathrm{~W}
\end{aligned}
\)

Hint

It is true that stretching of spring is determined by shear modulus of the spring as when coil spring is stretched neither its length nor its volume changes, there is only change in its shape.
Tensile strength of steel is more than that of copper.
Hence Assertion is true and reason is false.

Hint

\(
\begin{aligned}
& \text { Longitudinal stress }=\frac{\text { Internal restoring force }}{\text { Area }}=\frac{F_{\text {ext }}}{\text { Area }} \\
& \text { Stress }=\frac{W}{A}
\end{aligned}
\)