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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod. It is heated in such a way that temperature of each rod increases by ?T. Find change in the angle ABC. [Coeff. of linear expansion for Cu is α₁ ,Coeff. of linear expansion for Al is α₂ ]

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Payal Gupta

Contributor-Level 10

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

l₁=AB ,l₂=AC ,l₃=BC

Cos $θ$ = $\frac{l_{3}^{2} + l_{1}^{2} - l_{2}^{2}}{2 l_{3} l_{1}}$

2l₃l₁cos $θ$ = $l_{3}^{2} + l_{1}^{2} - l_{2}^{2}$

Differentiating 2( $l_{3} d l_{1} + l_{1} d l_{3}$ )cos $θ$ -2 $l_{1} l_{3} s i n θ d θ$

= 2 $l_{3} d l_{1} + 2 l_{1} d l_{1} - 2 l_{2} d l_{2}$

= d $l_{1} = l_{1} α_{1} ? t$

=d $l_{2} =$ $l_{2} α_{1} ? t$

=d $l_{3} = l_{3} α_{2} ? t$

$l_{1} = l_{2} = l_{3} = l$

( $l^{2} α_{1} ? t$ + $l^{2} α_{1} ? t$ )cos $θ$ + $l^{2} s i n θ d θ$ = $l^{2} α_{1} ? t$ + $l^{2} α_{1} ? t$ - $l^{2} α_{1} ? t$

sin $θ d θ = 2 α_{1} ? t$ (1-cos $θ$ )- $α_{2} ? t$

$θ = 60$

d $θ * s i n 60 = 2 α_{1} ? t (1 - c o s 60) - α_{2} ? t$

= 2 $α_{1} ? t * \frac{1}{2} - α_{2} ? t = (α_{1} - α_{2}) ? t$

d $θ$ = change in the angle ABC

\frac{(α_{1} - α_{2}) ? t}{s i n 60} = \frac{2 (α_{1} - α_{2}) ? t}{3}

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Both earth and moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon

(a) Will be elliptical.

(b) Will not be strictly elliptical because the total gravitational force on it is not central.

(c) Is not elliptical but will necessarily be a closed curve.

(d) Deviates considerably from being elliptical due to influence of planets other than earth.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b

Explanation- as observed from the sun two types of forces are acting on the moon one is due to gravitational attraction between the sun and the moon and the other is due to gravitational attraction between the earth and the moon hence total force is not normal.

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Satellites orbiting the earth have finite life and sometimes debris of satellites fall to the earth. This is because,

(a) The solar cells and batteries in satellites run out.

(b) The laws of gravitation predict a trajectory spiralling inwards.

(c) Of viscous forces causing the speed of satellite and hence height to gradually decrease.

(d) Of collisions with other satellites.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer-c

Explanation- As the total energy of the earth satellite bounded system is negative . due to the viscous force acting on the satellite, energy decreases continuously and radius also decreases.

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New answer posted

a year ago

Physics 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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New answer posted

a year ago

Physics 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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New answer posted

a year ago

Physics 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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New answer posted

a year ago

Physics 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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New answer posted

a year ago

Physics 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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New answer posted

a year ago

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?

0 Follower 21 Views

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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New answer posted

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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