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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Particles of masses 2M, m and M are respectively at points A, B and C with AB = ½ (BC). m is much-much smaller than M and at time t = 0, they are all at rest (Fig. 8.1). At subsequent times before any collision takes place:

(a) m will remain at rest.

(b) m will move towards M.

(c) m will move towards 2M.

(d) m will have oscillatory motion.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-c

Explanation- force on B due to A = F_BA= $\frac{G (2 M m)}{{A B}^{2}}$

force on B due to C = F_BC= $\frac{G (M m)}{{B C}^{2}}$

BC = 2AB

F_BC= $\frac{G (M m)}{(2 {A B)}^{2}} = \frac{G (M m)}{4 {A B}^{2}}$ BA

So it move towards BA

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by $\frac{Y π r^{4}}{4 R}$ . Y is the Young's modulus, r is the radius of the trunk and R is the radius of curvature of the bent surface along the height of t

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Payal Gupta

Contributor-Level 10

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

Consider the diagram according, the bending torque on the trunk of radius r of the tree = $\frac{Y π r^{4}}{4 R}$

When the tree is about to buckle Wd= $\frac{Y π r^{4}}{4 R}$

If R>>h, then the centre of gravity is at a height l $\approx \frac{h}{2} f r o m t h e g r o u n d$

From $? A B C R^{2} \approx (R - d)$ ²+ (h/2)²

If d <2+

So d = h²/8R

If w_o is the weight /volume

$\frac{Y π r^{4}}{4 R} = w_{o} (π r^{2} h) \frac{h^{2}}{8 R}$

h= ( $\frac{2 Y}{w_{o}}$ )^1/3r^2/3

critical height = h= ( $\frac{2 Y}{w_{o}}$ )^1/3r^2/3

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Choose the wrong option.

(a) Inertial mass is a measure of difficulty of accelerating a body by an external force whereas the gravitational mass is relevant in determining the gravitational force on it by an external mass.

(b) That the gravitational mass and inertial mass are equal is an experimental result.

(c) That the acceleration due to gravity on earth is the same for all bodies is due to the equality of gravitational mass and inertial mass.

(d) Gravitational mass of a particle like proton can depend on the presence of neighbouring heavy objects but the inertial mass cannot.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-d

Explanation- gravitational mass of proton is equivalent to its inertial mass and independent of presence neighbouring heavy objects.

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

In our solar system, the inter-planetary region has chunks of matter (much smaller in size compared to planets) called asteroids. They

(a) Will not move around the sun since they have very small masses compared to sun.

(b) Will move in an irregular way because of their small masses and will drift away into outer space.

(c) Will move around the sun in closed orbits but not obey Kepler's laws.

(d) Will move in orbits like planets and obey Kepler's laws.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-d

Explanation- asteroids are also being acted upon by central gravitational forces, hence they are moving in circular orbits like planets and obey Kepler's law.

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10 months 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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10 months 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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10 months 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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10 months 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

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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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10 months 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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10 months 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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