physics ncert solutions class 11th

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Y= $\frac{stress}{strain}=\frac{\frac{F}{A}}{\frac{?L}{L}}=\frac{F}{\pi ({D/2)}^2}*\frac{L}{?L}$

D²= $\frac{4FL}{\pi ?LY}$

D= $\sqrt[]{\frac{4FL}{\pi ?LY}}$

$\frac{D_{copper}}{D_{iron}}=\sqrt[]{\frac{Y_{iron}}{Y_{copper}}}$

This is a multiple choice answer as classified in NCERT Exemplar

(c) Consider the diagram where a spring is stretched by applying a load to its free end. Clearly the length and shape of the spring changes. The change in length corresponds to longitudinal strain and change in shape corresponds to shearing strain.

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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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Answer-d

Explanation- for small objects  sizes less than 100 m centre of mass very close with the centre of gravity of the body. But when the size of object increases it weight changes and its CM and CG become far from each other.

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Answer-a, c

Explanation- a geostationary satellite is having same sense of rotation as that of earth i.e west east direction.

A polar satellite goes around earth's pole in north south direction.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a,c,d

Explanation- F₁=-F₂=- $\frac{r_{12}}{r_{12}^3}G{M}_o^2\left(\frac{m_1{m}_2}{M_o^2}\right)$ ⁿ

r₁₂=r₁-r₂

acceleration due to gravity ,g = F/mass= $\frac{G{M}_o^2({m_1{m}_2)}^n}{r_{12}^2{\left(M_o\right)}^{2n}}*\frac{1}{mass}$

as g is constant hence constant of proportionality will not be constant in kepler's third law.

Hence kepler's law will not be valid.

For negative n, g= $\frac{G{M}_o^2{\left(m_1{m}_2\right)}^{-n}}{r_{12}^2{\left(M_o\right)}^{-2n}}*\frac{1}{mass}$

= $\frac{G{M}_o^{2(1+n)}}{r_{12}^2}\frac{{{(m}_1{m}_2)}^{-n}}{mass}$

M_o>m₁ or m₂ so objects lighter than water will sink