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Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

Consider a long steel bar under a tensile stress due to forces F acting at the edges along the length of the bar (Fig. 9.5). Consider a plane making an angle θ with the length. What are the tensile and shearing stresses on this plane?

(a) For what angle is the tensile stress a maximum?

(b) For what angle is the shearing stress a maximum?

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Payal Gupta

Contributor-Level 10

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

Let the cross sectional area A . consider the equilibrium of the plane aa'. A force F must be acting on this plane making an angle of $\frac{π}{2} - θ$ with the normal ON. Resolving F into components along the plane (FP) and normal to the plane.

By resolving into components

We get F_p= Fcos $θ$

And F_N= Fsin $θ$

Let area of the face aa' be A' then

A/A'=sin $θ$ so A=A'sin $θ$

The tensile stress = normal force/area=Fsin $θ / A'$

= $\frac{F s i n θ}{A / s i n θ} = \frac{F}{A}$ sin^{2
$θ$}

Shearing stress = parallel foce/Area

= $\frac{F c o s θ}{A / s i n θ} = \frac{F}{A} s i n θ c o s θ$

$\frac{F}{2 A} (2 s i n θ c o s θ = \frac{F}{2 A} s i n 2 θ$

a) For stress to be maximum , sin^{2
$θ$}=1

So $θ$ = $\frac{π}{2}$

b) Shearing stress to be maximum

sin2 $θ = 1$

So $θ$ = $\frac{π}{4}$

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

A mass m is placed at P a distance h along the normal through the centre O of a thin circular ring of mass M and radius r. If the mass is removed further away such that OP becomes 2h, by what factor the force of gravitation will decrease, if h = r ?

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Gravitational force at h height F = $\frac{G M m h}{{{(r}^{2} + h^{2})}^{3 / 2}}$

When mass is displaced upto distance 2h then

F'= $\frac{G M m 2 h}{{{[r}^{2} + {(2 h)]}^{2}]}^{\frac{3}{2}}}$

= $\frac{2 G M m h}{{(r^{2} + 4 h^{2})}^{\frac{3}{2}}}$

When h = r then F= $\frac{G M m h}{[{r^{2} + {(r)}^{2}]}^{3 / 2}}$

So F = $\frac{G M m}{2 \sqrt[]{2 r^{2}}}$

F'= $\frac{2 G M m r}{{(r^{2} + 4 r^{2})}^{\frac{3}{2}}} = \frac{2 G M m}{5 \sqrt[]{5 r^{2}}}$

$\frac{F^{'}}{F} = \frac{4 \sqrt[]{2}}{5 \sqrt[]{5}}$

F' = $\frac{4 \sqrt[]{2}}{5 \sqrt[]{5}} F$

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance R to 2R from the centre of the earth. What is the gain in its potential energy?

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation-potential energy of the objects at the surface of the earth is =- $\frac{G M m}{R}$

PE of the object at a height equal to radius of earth = $\frac{G M m}{2 R}$

Gain in potential energy = $\frac{G M m}{2 R}$ - (- $\frac{G M m}{R}$ )= $\frac{G M m}{2 R}$

= $\frac{g R^{2} m}{2 R} = \frac{1}{2} m g R$ (GM=gR²)

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Shown are several curves (Fig. 8.2). Explain with reason, which ones amongst them can be possible trajectories traced by a projectile (neglect air friction).

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation-The trajectory of a particle under gravitational force of the earth will be in conic section with the centre of the earth as a focus. Only c meets this requirements

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Show the nature of the following graph for a satellite orbiting the earth.

(a) KE vs orbital radius R

(b) PE vs orbital radius R

(c) TE vs orbital radius R.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Orbital speed of the satellite v_o= $\sqrt[]{\frac{G M}{R}}$

KE of the satellite of mass m, E_k = 1/2mv_o²= Gmm/2R

So kinetic energy is inversely proportional to distance.

b)potential energy of a satellite E_p=-GMm/R

so it is also inversely proportional to R

c)total energy of the satellite E=E_k+E_p= $\frac{G M m}{2 R} - \frac{G M m}{R}$ = - $\frac{G M m}{2 R}$

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the midpoint of the line joining their centres be in stable equilibrium or unstable equilibrium? Give reason for your answer.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Let M be the mass and R be the radius of sphere . an object of mass m be placed at the mid point P of the line joining their centres

Force F₁=F₂=GMm/ (5R)²

If we displace it through x distance

New force towards sphere A, F₁'= GMm/ (5R-x)²

Sphere B F₂' = GMm/ (5R+x)²

As F₁'>F₂' therefore a resultant force (F₁'-F₂') acts on the objects towards sphere A therefore objects start to move towards A hence equilibrium is unstable.

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Mean solar day is the time interval between two successive noon when sun passes through zenith point (meridian). Sidereal day is the time interval between two successive transit of a distant star through the zenith point (meridian). By drawing appropriate diagram showing earth's spin and orbital motion, show that mean solar day is four minutes longer than the sidereal day. In other words, distant stars would rise 4 minutes early every successive day. (Hint: you may assume circular orbit for the earth).

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation-Every day the earth advances in the orbit by approximately 1⁰. Then it will have to rotate by 361⁰ to have the sun at zenith point again. So 361⁰ corresponds to 24h

So 1^o corresponds to 24/361= 0.066h=3.99 min =4min

Hence distant stars would rise 4 min early every successive day.

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

What is the angle between the equatorial plane and the orbital plane of (a) Polar satellite? (b) Geostationary satellite

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Angle between the equatorial plane and orbital plane of a polar satellite is 90⁰ and angle between equatorial plane and orbital plane of a geostationary satellite is 0⁰.

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Out of aphelion and perihelion, where is the speed of the earth more and why ?

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Phelion is the location of the earth where it is the greatest distance from the sun and perihelion is the location of the earth where it is the nearest to the sun

The areal velocity (1/2 rv) of the earth around the sun is constant . so speed of earth is more at perihelion.

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

The gravitational force between a hollow spherical shell (of radius R and uniform density) and a point mass is F. Show the nature of F vs r graph where r is the distance of the point from the centre of the hollow spherical shell of uniform density.

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