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Physics NCERT Exemplar Solutions Class 11th Chapter Nine

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

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

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

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

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

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

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 Exemplar Solutions Class 11th Chapter Nine Class 11th

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

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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