Mechanical Properties of Solids

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As ivory ball is more elastic than wet clay ball. So ivory ball tries to regain its original shape quickly, hence more energy and momentum transferred to ivory than clay ball. So ivory ball rises high

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Length of steel cable l=9.1m

Radius r= 5mm= $5*{10}^{-3}$

Tension in the cable F =800N

Young's modulus of steel Y= 2 $*\frac{{10}^{11}N}{m^2}$

So $?l=\frac{F}{\pi r^2}*\frac{l}{Y}=\frac{800}{3.14*{(5*{10}^{-3})}^{-2}}*\frac{9.1}{2*{10}^{11}}=4.64*{10}^{-4}m$

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Bulk modulus of rubber is 9.8*10⁸ N/m²

density of sea water is 10³ kg m^-3

percentage decrease in volume $\frac{?V}{V}*100=0.1$

$\frac{?V}{V}=\frac{1}{1000}$

Let the rubber ball be taken up to depth h.

Change in pressure =h $\rho$ g

Bulk modulus K= $\frac{p}{\frac{?V}{V}}=\frac{h\rho g}{\frac{?V}{V}}$

So h = $\frac{k*?V/V}{\rho g}=\frac{9.8*{10}^8*1/1000}{{10}^3*9.8}=100m$

This is a short answer type question as classified in NCERT Exemplar

Young's modulus of steel Y= 2 $*\frac{{10}^{11}N}{m^2}$

Coefficient of thermal expansion $\alpha$ = 10^-5

Length =1m

Area = 10^-4m²

Increase in temperature $?$ t= 200⁰C

Tension produced in steel rod F = YA $\alpha ?t$ =2 $*{10}^{11}*{10}^{-4}*{10}^{-5}*200$

= 4 $*{10}^{4}N$

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Young's modulus Y= $\frac{F}{A}*\frac{L}{l}$

For first wire Y= $\frac{f}{\pi r^2}*\frac{L}{l}$

For second wire Y= $\frac{2f}{\pi ({2r)}^2}*\frac{2L}{l^{\text{'}}}$

From above equations $\frac{f}{\pi r^2}*\frac{L}{l}=\frac{f}{\pi r^2}*\frac{L}{l\text{'}}$

So l=l' hence Y will also same.

This is a short answer type question as classified in NCERT Exemplar

Young's modulus Y= $\frac{F}{A}*\frac{l}{?l}$

For a perfectly rigid body change in length $?l=0$

Y= $\infty$ so young's modulus for perfectly rigid body is infinite

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Work done =1/2F $*?$ l

As spring of both are equally compressed

So Young's modulus Y= $\frac{F}{A}*\frac{l}{?l}$    as $?l$ is inversely proportional to young's modulus

And also we know W is inversely proportional to young's modulus

So $\frac{W_{steel}}{W_{copper}}=\frac{Y_{copper}}{Y_{steel}}1$

W_steelcopper 

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stress= $\frac{magnitudeofinternalreactionforce}{areaofcrosssection}$

Therefore stress is scalar quantity not vector quantity.

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Young's modulus Y= stress/longitudinal strain

For same longitudinal strain, $\frac{Y_{steel}}{Y_{rubber}}=\frac{{stress}_{steel}}{{stress}_{rubber}}$

Y_steel>Y_rubber

so we can say that  stress_steel > stress_rubber