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physics ncert solutions class 11th

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4 months ago

physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Different points in earth are at slightly different distances from the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the c.m. (centre of mass) causing translation and a net torque at the c.m. causing rotation around an axis through the c.m. For the earth-sun system (approximating the earth as a uniform density sphere)

(a) The torque is zero.

(b) The torque causes the earth to spin.

(c) The rigid body result is not applicable since the earth is not even approximately a rigid body.

(d) T

...more

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a

Explanation- As the earth is revolving around the sun in a circular motion due to gravitational attraction. The force of attraction will be of radial nature and angle between them is zero so torque is also zero.

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physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Nine

A steel rod of length 2l, cross sectional area A and mass M is set rotating in a horizontal plane about an axis passing through the centre. If Y is the Young's modulus for steel, find the extension in the length of the rod. (Assume the rod is uniform.)

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Payal Gupta

Contributor-Level 10

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

Consider an element of width dr at r

Let T(r) and T(r+dr) be the tensions at r+dr respectively

So net centrifugal force = w²rdm

= w²r $μ d r$

T(r)-T(r+dr)= $μ$ w²rdr

-dT= $μ$ w²rdr

- $\int_{T = 0}^{T} d T = \int_{r = l}^{r = r} μ w^{2} r d r$

T(r)= $\frac{μ w^{2}}{2} (l^{2} - r^{2})$

Let the increase in length of the element dr be $? r$

So Young's modulus Y= stress/strain= $\frac{\frac{T (r)}{A}}{\frac{? r}{d r}}$

$\frac{? r}{d r} = \frac{T (r)}{A} = \frac{μ w^{2}}{2 Y A}$ (l²-r²)

$? r = \frac{1}{Y A} \frac{μ w^{2}}{2} (l^{2} - r^{2}) d r$

Change in length in right part = $\frac{1}{Y A} \frac{μ w^{2}}{2} \int_{0}^{l} (l^{2} - r^{2}) d r$

= $\frac{1}{Y A} \frac{μ w^{2}}{2} [l^{3} - \frac{l^{3}}{3}] = \frac{1}{3 Y A} μ w^{2} L^{2}$

Total change in length = $\frac{2 μ w^{2} l^{2}}{3 Y A}$

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physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Eight

As observed from earth, the sun appears to move in an approximate circular orbit. For the motion of another planet like mercury as observed from earth, this would

(a) Be similarly true.

(b) Not be true because the force between earth and mercury is not inverse square law

(c) Not be true because the major gravitational force on mercury is due to sun.

(d) Not be true because mercury is influenced by forces other than gravitational forces.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-c

Explanation-The gravitational force between sun and earth follows inverse square law. Due to relative motion between earth and mercury, the orbit of mercury observed from the earth will not be approximately circular since the major gravitational force on mercury is due to the sun.

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physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Nine

A steel wire of mass µ per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m^–3 (Young's modules Y=2*10¹¹ Nm^–2). (b) If the yield strength of steel is 2.5*10⁸ Nm^–2, what is the maximum weight that can be hung at the lower end of the wire?

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Payal Gupta

Contributor-Level 10

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

When a small element of length dx is considered at x from the load x=0

(a) letT(x) and T(x+dx) are tensions on the two cross sections a distance dx apart then

T(x+dx)+T(x)=dmg= $μ$ dxg

dT= $μ$ gx+C

at x=0 T(0)=mg

C=mg

T(x)= $μ$ gx+Mg

Let length dx at x increases by dr then

Young's modulus Y= stress/strain

$\frac{\frac{T (x)}{A}}{\frac{d r}{d x}} = Y$

$\frac{d r}{d x} = \frac{T (x)}{Y A}$

r= $\frac{1}{Y A} \int_{0}^{L} (μ g x + M g) d x$

= $\frac{1}{Y A} [\frac{μ g x^{2}}{2} + M g x]$ ₀^L

r= $\frac{1}{2 * 10^{11} * π * 10^{- 6}} [\frac{π * 786 * 10^{- 3} * 10 * 10}{2} + 25 * 10 * 10]$

so r = 4 $* 10^{- 3} m$

(b) tension will be maximum at x=L

T= $μ g L$ +Mg=(m+M)g

The force = (yield strength) area= $250 * π N$

(m+M)g= 250 $π$

Mg $\approx 250 π$

M= 25 $π \approx 75 k g$

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4 months ago

physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Eight

The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity

(a) Will be directed towards the centre but not the same everywhere.

(b) Will have the same value everywhere but not directed towards the centre.

(c) Will be same everywhere in magnitude directed towards the centre.

(d) Cannot be zero at any point.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-d

Explanation- We treated that the mass of earth is at centre . in this case a=g=0 at centre but if earth is not uniform then value of g is different at different point.

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4 months ago

physics ncert solutions class 11th 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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physics ncert solutions class 11th 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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4 months ago

physics ncert solutions class 11th 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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physics ncert solutions class 11th 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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physics ncert solutions class 11th 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.