Physics NCERT Exemplar Solutions Class 11th Chapter Nine

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(d) Y young's modulus is inversely proportional to temperature, so if we increase temperature , young's modulus decreases.

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(d) breaking stress = breaking force/area of cross section

Breaking force will not depend upon length.

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(b) No frictional force exists in case of ideal fluid. Hence tangential forces are zero.

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

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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.

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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