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7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Nine

Is stress a vector quantity?

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Payal Gupta

Contributor-Level 10

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

stress= $\frac{m a g n i t u d e o f i n t e r n a l r e a c t i o n f o r c e}{a r e a o f c r o s s s e c t i o n}$

Therefore stress is scalar quantity not vector quantity.

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7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Nine

The Young's modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?

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Payal Gupta

Contributor-Level 10

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

Young's modulus Y= stress/longitudinal strain

For same longitudinal strain, $\frac{Y_{s t e e l}}{Y_{r u b b e r}} = \frac{{s t r e s s}_{s t e e l}}{{s t r e s s}_{r u b b e r}}$

Y_steel>Y_rubber

so we can say that stress_steel> stress_rubber

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7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Eight

If the law of gravitation, instead of being inverse-square law, becomes an inverse-cube law-

(a) Planets will not have elliptic orbits.

(b) Circular orbits of planets is not possible.

(c) Projectile motion of a stone thrown by hand on the surface of the earth will be approximately parabolic.

(d) There will be no gravitational force inside a spherical shell of uniform density.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a, c

Explanation- as we know F= $\frac{G M m}{a^{2}} = \frac{m v^{2}}{a}$

V= orbital speed= $\frac{\sqrt[]{G m}}{a}$

Time period of revolution of planet T = $\frac{2 π a}{v} = \frac{2 π a}{\frac{\sqrt[]{G m}}{a}}$ = $\frac{2 π a^{2}}{\sqrt[]{G M}}$

T² $\propto a^{4}$ ……….1

Hence orbit will be elliptical.

F= $\frac{G M}{a^{3}} m$ =g'm

It is clear from equation 1 that its path is parabola.

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7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Nine

A stone of mass m is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P.

(a) Find the distance y from the top when the mass comes to rest for an instant, for the first time.

(b) What is the maximum velocity attained by the stone in this drop?

(c) What shall be the nature of the motion after the stone has reached its lowest point?

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Payal Gupta

Contributor-Level 10

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

Till the stone drops through a length L it will be in free fall. After that the elasticity of the spring will force it to a SHM. Let the stone come to rest instantaneously at y.

The loss in PE of the stone is the PE stored in the stretched string .

Mgy=1/2 k(y-L)²

Mgy = $\frac{1}{2} k y^{2} - k y L + \frac{1}{2} k L^{2}$

= $\frac{1}{2} k y^{2} - (k L + m g) y + \frac{1}{2} K L^{2} = 0$

Y= $\frac{(K L - m g) \pm \sqrt[]{{(k L + m g)}^{2} - K^{2} L^{2}}}{k} = \frac{(k L + m g) \pm \sqrt[]{2 m g k L + m^{2} g^{2}}}{k}$

b)in SHM the maximum velocity is attained when the body passes through the equilibrium position i.e when instantaneous acceleration is zero. That is mg-kx=0

so mg=kx

from the conservation of energy

$\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = m g (L + x)$

$\frac{1}{2} m v^{2} = m g (L + x) - \frac{1}{2} k x^{2}$

mg=kx

x=mg/k

$\frac{1}{2} m v^{2} = m g (L + \frac{m g}{K}) - \frac{1}{2} k \frac{m^{2} g^{2}}{k^{2}}$

$\frac{1}{2} m v^{2} = m g L + \frac{1}{2} \frac{m^{2} g^{2}}{k}$

v²=2gL+mg²/K

v= (2gL+mg²/K)^1/2

c)when stone is at lowest position i.e at instantaneo

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

Till the stone drops through a length L it will be in free fall. After that the elasticity of the spring will force it to a SHM. Let the stone come to rest instantaneously at y.

The loss in PE of the stone is the PE stored in the stretched string .

Mgy=1/2 k(y-L)²

Mgy = $\frac{1}{2} k y^{2} - k y L + \frac{1}{2} k L^{2}$

= $\frac{1}{2} k y^{2} - (k L + m g) y + \frac{1}{2} K L^{2} = 0$

Y= $\frac{(K L - m g) \pm \sqrt[]{{(k L + m g)}^{2} - K^{2} L^{2}}}{k} = \frac{(k L + m g) \pm \sqrt[]{2 m g k L + m^{2} g^{2}}}{k}$

b)in SHM the maximum velocity is attained when the body passes through the equilibrium position i.e when instantaneous acceleration is zero. That is mg-kx=0

so mg=kx

from the conservation of energy

$\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = m g (L + x)$

$\frac{1}{2} m v^{2} = m g (L + x) - \frac{1}{2} k x^{2}$

mg=kx

x=mg/k

$\frac{1}{2} m v^{2} = m g (L + \frac{m g}{K}) - \frac{1}{2} k \frac{m^{2} g^{2}}{k^{2}}$

$\frac{1}{2} m v^{2} = m g L + \frac{1}{2} \frac{m^{2} g^{2}}{k}$

v²=2gL+mg²/K

v= (2gL+mg²/K)^1/2

c)when stone is at lowest position i.e at instantaneous distance Y from p, then equation of motion of the stone is

$\frac{m d^{2} y}{d t^{2}} = m g - k (y - L)$

$\frac{d^{2} y}{d t^{2}} + \frac{k}{m} (y - L) - g = 0$

Make a transformation of variables z= $\frac{k}{m} (y - L) - g$

It is differential equation of second order which represents SHM

Comparing with equation $\frac{d^{2} z}{d t^{2}} + w^{2} z = 0$

Anglular frequency of harmonic motion w = $\sqrt[]{\frac{k}{m}}$

The solution of above equation will be of the type z = Acos (wt+ $\emptyset$ ) where w = $\sqrt[]{\frac{k}{m}}$

Y= (L+mg/k)+A'cos(wt+ $\emptyset$ )

Thus , the stone will perform SHM with angular frequency w= $\sqrt[]{\frac{k}{m}}$ about a point

Y_o=L+mg/k

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New answer posted

7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Which of the following options are correct?

(a) Acceleration due to gravity decreases with increasing altitude.

(b) Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).

(c) Acceleration due to gravity increases with increasing latitude.

(d) Acceleration due to gravity is independent of the mass of the earth.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a, c, d

Explanation- acceleration due to gravity at altitude g'= $\frac{g}{{(1 + \frac{h}{R})}^{2}} \approx g (1 - \frac{2 h}{R})$

At depth d, g'=g (1-d/R)

In both cases value of g decreases

But in case of latitude the value of g increases when we increase $\emptyset$

$g_{\emptyset} = g - w^{2} R {c o s}^{2} \emptyset$

also we conclude from the formula that it is independent upon mass.

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7 months ago

Class 11th 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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7 months ago

Class 11th 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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7 months ago

Class 11th 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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7 months ago

Class 11th 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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7 months ago

Class 11th 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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