12gm of gas occupy a volume of 4 × 10-3 m3 at a temperature of 7°C. After the gas is heated at constant pressure its density becomes 6 × 10-4gm/cm3. What is the temperature to which the gas was heated (in kelvin)?

P = (RT/M? )ρ ⇒ P/ρT = constant ⇒ (12)/ (4 × 10? ³ × 10? ) × 280 = 6 × 10? × T ⇒ T = 1400K

In a young's double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved 8 cm towards the slits, the change in fringe width is 4 × 10-5 m. If the separation between the slits is 1 mm, find the wavelength of light used (in nm ).

β = (λD/d) ⇒ Δβ = (ΔDλ/d) ⇒ 4 × 10? = (λ/10? ³) × 8 × 10? ² ⇒ λ = 5 × 10? m = 500 nm

A bucket of height 20 cm is placed under a tap such that water flows into the bucket at a constant rate of 100 cm3/s. There is a small hole in the bucket of cross-sectional area of 1 cm2 at height of 3 cm from the bottom. Area of hole is very small as compared to area of the bucket. Find the maximum height (in cm ) above the bottom upto which the water is filled in the bucket. (Take g = 10 m/s2 )

Q = a√2gh h = Q²/ (a² × 2g) = (10? )²/ (10? × 2 × 10) = 5 cm Height above the ground = 3 cm + 5 cm = 8 cm

A wooden wheel having radius of 1 m is to be fitted with an iron ring around it. The diameter of the ring is 6 mm smaller than the wooden wheel. If the coefficient of linear expansion of iron is 1.2 × 10-5/ °C, then find the minimum increase in temperature (in °C) to fit the ring on the wheel.

r = r? (1 + αΔT) ⇒ r – r? = r? αΔT ⇒ 3 × 10? ³ = 1 × 1.2 × 10? ΔT ⇒ ΔT = 250°C

A capacitor of capacitance 3µF is charged to a potential difference of 12V. Another identical capacitor, initially uncharged is filled with a dielectric having dielectric constant K and it is connected to the charged capacitor. The final common potential difference across both capacitors is found to be 2V. Find the value of k.

V_common = (C? V? + C? V? )/ (C? + C? )? 2 = (3? F)*12)/ (3? F)+ (3? F)*K)? 2 = 12/ (K+1)? K = 5

Magnetic field exists in the region between lines PQ and RS as shown in the figure. A particle having charge to mass ratio of a enters the magnetic field as shown in the figure with a speed v. Lines PQ and RS are parallel to each other. Find the value of d such that the charged particle exits the magnetic field region with velocity perpendicular to line RS.

(√3v)/(2aB)

v/(2aB)

v/(aB)

The molar specific heat C v of a diatomic gas is (5/2)R when we treat its molecules to be rigid having no vibrations. But this assumption is not always valid especially at high temperatures. In that case, molecules are treated as non-rigid having vibrational energy also. C v for diatomic gas with non-rigid gas molecules will be:

(7/2)R

(3/2)R

3R

If the monochromatic source in Young's double slit experiment is replaced by white light, then

There will be a central bright white fringe surrounded by a few coloured fringes

Interference pattern will disappear

There will be a central dark fringe surrounded by a few coloured fringes

All bright fringes will be of equal width

Two particles are executing SHM which are described by the equations x 1 = asin(ωt + π/6) and x 2 = asin(ωt + π/4). Find the minimum time interval between the particles to cross their mean position.

π/(12ω)

π/(6ω)

π/(4ω)

π/(3ω)

A particle is moving in a circle of radius 1m with angular velocity (ω) given as a function of angular displacement ( θ ) by the relation ω = θ² + 2θ. The total acceleration of the particle when θ = 1rad is:

15 m/s²

9 m/s²

12 m/s²

An inductor and a resistor are connected to an ac supply of 120 V, 50 Hz. The resistance of the resistor is 20Ω. If the voltages leads the current by a phase difference of 45°, find the inductance of the inductor.

63.7mH

40mH

66.7mH

A vernier calliper whose 10 vernier scale divisions coincide with 9 main scale divisions is used to take a reading. Each main scale division is 1 mm. Before taking the reading it is observed that the 4 th division of the vernier scale coincides with the main scale and zero of main scale is behind zero of vernier scale. When the reading is taken using this vernier calliper, the main scale reads 2.5 cm and the 7 th division of the vernier scale coincides with the main scale. Find the final value of the measurement.

2.53 cm

2.54 cm

2.57 cm

2.61 cm

Three rods of equal mass having length L, L and √2L are joined to form a triangular frame as shown in the figure. The system is rotating about an axis passing through the vertex of the triangle and perpendicular to the plane of the triangle with an angular velocity ω. Find the angular momentum of the system.

mL²ω

(1/3)mL²ω

(2/3)mL²ω

A fish moving up in water (μ = 4/3) with a speed of 4 cm/s observes a bird coming directly towards itself with a speed of 12 cm/s. Find the actual speed of the bird with respect to a stationary observer on the ground.

6 cm/s

5 cm/s

8 cm/s

9 cm/s

An inverted cone is rotating about its vertical axis. A particle is kept on the inner surface of cone and it is at rest relative to the cone at a height of 0.4 m above its vertex. The coefficient of friction between the surface of cone and the particle is 0.6 and the apex angle of cone is 90°. The maximum angular velocity of revolution of the cone can be: (take g = 10/s 2 )

10rad/s

12.5rad/s

7.5rad/s

The rest mass of the deuteron, proton and neutron is equivalent to energy of 1877MeV, 939MeV, 940MeV respectively. A deuteron may disintegrate to a proton and neutron if it.

captures a γ - ray photon of energy 2MeV

emits a γ - ray photon of energy 2MeV

emits a γ - ray photon of energy 3MeV

captures a γ - ray photon of energy 3MeV

In an imaginary electric field, the electric potential energy of an electron at a distance r from the centre of force field is given by U = kr 2 , where k is a positive constant of appropriate dimensions. If the electron is moving in a circular orbit of radius r about the centre, then the orbital time period of the electron is proportional to:

r³/²

r¹/²

The property which is not of an electromagnetic wave travelling in free space is that:

They originate from charges moving with uniform speed

The energy density in electric field is equal to energy density in magnetic field

They travel with a speed equal to 1 μ 0 ε 0

An electron moving with a velocity of 2 m/s along +ve x-axis at a point in a magnetic field experiences a force of 2N along -ve y-axis. If the electron moves with of a velocity of 2 m/s along +ve y-axis at the same point, it experiences a force of 2N along +ve x-axis. Find force that the electron would experience if it were moving with a velocity of 2m/s along z-axis.

2N + ve along y-axis

2N along -ve x-axis

Varying velocity and varying acceleration

Constant velocity but varying acceleration

A satellite is moving around a planet of radius R in a circular orbit of radius r. The speed of the satellite is v. Find the acceleration due to gravity on the surface of the planet.

(v²r)/R²

Physics NCERT Exemplar Solutions Class 11th Chapter Nine: Overview, Questions, Preparation

Q: 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?

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 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 Fp= Fcos θ And FN= 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 ' = F s i n θ A / s i n θ = F A sin2 θ Shearing stress = parallel foce/Area = F c o s θ A / s i n θ = F A s i n θ c o s θ F 2 A ( 2 s i n θ c o s θ = F 2 A s i n 2 θ a) For stress to be maximum , sin2 θ =1 So θ = π 2 b) Shearing stress to be maximum sin2 θ = 1 So θ = π 4

Q: 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?

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 T x A d r d x = Y d r d x = T x Y A r= 1 Y A ∫ 0 L ( μ g x + M g ) d x = 1 Y A μ g x 2 2 + M g x 0L r= 1 2 × 10 11 × π × 10 - 6 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 ≈ 250 π M= 25 π ≈ 75 k g

Q: 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.)

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 - ∫ T = 0 T d T = ∫ r = l r = r μ w 2 r d r T(r)= 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= TrA?rdr ?rdr=T(r)A=μw22YA (l2-r2) ?r=1YAμw22(l2-r2)dr Change in length in right part = 1YAμw22∫0ll2-r2dr = 1YAμw22l3-l33=13YAμw2L2 Total change in length = 2μw2l23YA

Updated on Apr 24, 2025 14:35 IST

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Mechanical Properties of Solids Questions and Answers
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Jee Mains 2020

Mechanical Properties of Solids Questions and Answers

1. 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?

Explanation- 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 $θ$ = $\frac{π}{4}$

2. 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?

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

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

b) tension will be maximum at x=L

T= $μ g L$ +Mg=(m+M)g

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

(m+M)g= 250 $π$

Mg $\approx 250 π$

M= 25 $π \approx 75 k g$

3. 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.)

Explanation- 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 = $\frac{2 μ w^{2} l^{2}}{3 Y A}$

4. 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 ]

Explanation- 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 $θ \times s i n 60 = 2 α_{1} ∆ t (1 - c o s 60) - α_{2} ∆ t$

= 2 $α_{1} ∆ t \times \frac{1}{2} - α_{2} ∆ t = (α_{1} - α_{2}) ∆ t$

d $θ$ = change in the angle ABC

\frac{(α_{1} - α_{2}) ∆ t}{s i n 60} = \frac{2 (α_{1} - α_{2}) ∆ t}{\sqrt[]{3}}

Commonly asked questions

To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. (The bulk modulus of rubber is 9.8×10⁸ N/m² ; and the density of sea water is 10³ kg m^-3.)

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

Bulk modulus of rubber is 9.8×10⁸ N/m²

density of sea water is 10³ kg m^-3

percentage decrease in volume $\frac{? V}{V} \times 100 = 0.1$

$\frac{? V}{V} = \frac{1}{1000}$

Let the rubber ball be taken up to depth h.

Change in pressure =h $ρ$ g

Bulk modulus K= $\frac{p}{\frac{? V}{V}} = \frac{h ρ g}{\frac{? V}{V}}$

So h = $\frac{k \times ? V / V}{ρ g} = \frac{9.8 \times 10^{8} \times 1 / 1000}{10^{3} \times 9.8} = 100 m$

A copper and a steel wire of the same diameter are connected end to end. A deforming force F is applied to this composite wire which causes a total elongation of 1cm. The two wires will have

(a) The same stress.

(b) Different stress.

(c) The same strain.

(d) Different strain.

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) for steel wire Y_steel= stress/strain= $\frac{f / A}{s t a r i n}$

When F and A are same for both the wires . hence stress will be same for both the wire

(Strain)_steel= stress/Y_steel and strain_copper=stress/Y_copper

Y_{steel $\neq$}Y_copper

hence they both have different starin

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?

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 $θ$ = $\frac{π}{4}$

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?

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

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

(b) tension will be maximum at x=L

T= $μ g L$ +Mg=(m+M)g

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

(m+M)g= 250 $π$

Mg $\approx 250 π$

M= 25 $π \approx 75 k g$

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.)

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 = $\frac{2 μ w^{2} l^{2}}{3 Y A}$

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 ]

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

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

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}

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 the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.

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

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?

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

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?

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

Is stress a vector quantity?

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.

Identical springs of steel and copper are equally stretched. On which, more work will have to be done?

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

Work done =1/2F $\times ?$ l

As spring of both are equally compressed

So Young’s modulus Y= $\frac{F}{A} \times \frac{l}{? l}$ as $? l$ is inversely proportional to young’s modulus

And also we know W is inversely proportional to young’s modulus

So $\frac{W_{s t e e l}}{W_{c o p p e r}} = \frac{Y_{c o p p e r}}{Y_{s t e e l}} 1$

W_steelcopper

What is the Young’s modulus for a perfect rigid body ? 9.18 What is the Bulk modulus for a perfect rigid body?

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

Young’s modulus Y= $\frac{F}{A} \times \frac{l}{? l}$

For a perfectly rigid body change in length $? l = 0$

Y= $\infty$ so young’s modulus for perfectly rigid body is infinite

A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by l. Another wire of the same material of length 2L and radius 2r, is pulled by a force 2f. Find the increase in length of this wire.

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

Young’s modulus Y= $\frac{F}{A} \times \frac{L}{l}$

For first wire Y= $\frac{f}{π r^{2}} \times \frac{L}{l}$

For second wire Y= $\frac{2 f}{π ({2 r)}^{2}} \times \frac{2 L}{l^{'}}$

From above equations $\frac{f}{π r^{2}} \times \frac{L}{l} = \frac{f}{π r^{2}} \times \frac{L}{l'}$

So l=l’ hence Y will also same.

A steel rod (Y = 2.0 × 10¹¹ Nm^-2; and α = 10^-50 C^-1) of length 1 m and area of cross-section 1 cm² is heated from 0° C to 200° C, without being allowed to extend or bend. What is the tension produced in the rod?

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

Young’s modulus of steel Y= 2 $\times \frac{10^{11} N}{m^{2}}$

Coefficient of thermal expansion $α$ = 10^-5

Length =1m

Area = 10^-4m²

Increase in temperature $?$ t= 200⁰C

Tension produced in steel rod F = YA $α ? t$ =2 $\times 10^{11} \times 10^{- 4} \times 10^{- 5} \times 200$

= 4 $\times 10^{4} N$

A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 5 mm. When the car just begins to move, the tension in the cable is 800 N. How much has the cable stretched? (Young’s modulus for steel is 2 × 10¹¹ Nm^–2.)

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

Length of steel cable l=9.1m

Radius r= 5mm= $5 \times 10^{- 3}$

Tension in the cable F =800N

Young’s modulus of steel Y= 2 $\times \frac{10^{11} N}{m^{2}}$

So $? l = \frac{F}{π r^{2}} \times \frac{l}{Y} = \frac{800}{3.14 \times {(5 \times 10^{- 3})}^{- 2}} \times \frac{9.1}{2 \times 10^{11}} = 4.64 \times 10^{- 4} m$

Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the floor and why?

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

As ivory ball is more elastic than wet clay ball. So ivory ball tries to regain its original shape quickly, hence more energy and momentum transferred to ivory than clay ball. So ivory ball rises high

Modulus of rigidity of ideal liquids is

(a) Infinity

(b) Zero

(c) Unity

(d) Some finite small non-zero constant value

This is a multiple choice answer as classified in NCERT Exemplar

(b) No frictional force exists in case of ideal fluid. Hence tangential forces are zero.

The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will

(a) Be double

(b) Be half

(c) Be four times

(d) Remain same

This is a multiple choice answer as classified in NCERT Exemplar

(d) breaking stress = breaking force/area of cross section

Breaking force will not depend upon length.

The temperature of a wire is doubled. The Young’s modulus of elasticity

(a) Will also double

(b) Will become four times

(c) Will remain same

(d) Will decrease

This is a multiple choice answer as classified in NCERT Exemplar

(d) Y young’s modulus is inversely proportional to temperature, so if we increase temperature , young’s modulus decreases.

A spring is stretched by applying a load to its free end. The strain produced in the spring is

(a) Volumetric

(b) Shear

(c) Longitudinal and shear

(d) Longitudinal

This is a multiple choice answer as classified in NCERT Exemplar

(c) Consider the diagram where a spring is stretched by applying a load to its free end. Clearly the length and shape of the spring changes. The change in length corresponds to longitudinal strain and change in shape corresponds to shearing strain.

A rigid bar of mass M is supported symmetrically by three wires each of length l . Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to

(a) Y_copper /Y_iron

(b) $\sqrt[]{\frac{Y_{i r o n}}{Y_{c o p p e r}}}$

(c) $\frac{Y_{i r o n}^{2}}{Y_{c o p p e r}^{2}}$

(d) Y_iron/Y_copper .

This is a multiple choice answer as classified in NCERT Exemplar

Y= $\frac{s t r e s s}{s t r a i n} = \frac{\frac{F}{A}}{\frac{? L}{L}} = \frac{F}{π ({D / 2)}^{2}} \times \frac{L}{? L}$

D²= $\frac{4 F L}{π ? L Y}$

D= $\sqrt[]{\frac{4 F L}{π ? L Y}}$

$\frac{D_{c o p p e r}}{D_{i r o n}} = \sqrt[]{\frac{Y_{i r o n}}{Y_{c o p p e r}}}$

A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars (Fig. 9.1). A mass m is suspended from the midpoint of the wire. Strain in the wire is

(a)

(b) x/L

(c) x²/2L

(d)

This is a multiple choice answer as classified in NCERT Exemplar

(a)

BO+OC- (BD+DC)

= 2BO -2BD

= 2 (BO-BD)

=2 [ (x²+L²)^1/2-L]

=2L [ (1+ $\frac{x^{2}}{L^{2}}$ )^1/2-L]

= 2L [ (1+ $\frac{1}{2} \frac{x^{2}}{L^{2}} - 1$ ]= $\frac{x^{2}}{L}$

Strain = $\frac{? L}{2 L} = \frac{\frac{x^{2}}{L}}{2 L} = \frac{x^{2}}{2 L^{2}}$

A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports (Fig. 9.2). It can be done in one of the following three ways The tension in the strings will be

(a) The same in all cases

(b) Least in (a)

(c) Least in (b)

(d) Least in (c)

This is a multiple choice answer as classified in NCERT Exemplar

2Tsin $θ$ =mg

Total horizontal forces = Tcos $θ - T c o s θ = 0$

T=mg/2sin $θ$

As mg is constant T $\propto \frac{1}{s i n θ}$

T_max= mg/sin $θ$ _min

Sin $θ$ _min=0, $θ$ _min= 0

T_min=mg/2sin $θ$ _max

_{$s i n θ$ max}= 1, $θ$ =90⁰

Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass M is attached to each of the free ends at the centre of the rods.

(a) Both the rods will elongate but there shall be no perceptible change in shape.

(b) The steel rod will elongate and change shape but the rubber rod will only elongate.

(c) The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse.

(d) The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre

This is a multiple choice answer as classified in NCERT Exemplar

(d) a mass M is attached at the centre. As the mass is attached to both the rods, both rod will be elongated, but due to different elastic properties of material rubber changes shape also.

The stress-strain graphs for two materials are shown in Fig.9.3 (assume same scale).

(a) Material (ii) is more elastic than material (i) and hence material (ii) is more brittle.

(b) Material (i) and (ii) have the same elasticity and the same brittleness.

(c) Material (ii) is elastic over a larger region of strain as compared to (i).

(d) Material (ii) is more brittle than material (i).

This is a multiple choice answer as classified in NCERT Exemplar

(c), (d) The ultimate tensile strength for material ii is greater hence material ii is elastic over larger region as compared to material (i) for material (ii) fracture point is nearer, hence it is more brittle.

A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight.

(a) Tensile stress at any cross section A of the wire is F/A

(b) Tensile stress at any cross section is zero

(c) Tensile stress at any cross section A of the wire is 2F/A

(d) Tension at any cross section A of the wire is F

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) Forces at cross section is F.

Now applying formula . stress = tension/area=F/A

Tension = applied force =F

A rod of length l and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths (Fig. 9.4). The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. Y_Al=70 $\times 10^{9} N m^{- 2}$ and Y_steel= 200 $\times 10^{9} N m^{- 2}$

(a) Mass m should be suspended close to wire A to have equal stresses in both the wires

(b) Mass m should be suspended close to B to have equal stresses in both the wires

(c) Mass m should be suspended at the middle of the wires to have equal stresses in both the wires

(d) Mass m should be suspended close to wire A to have equal strain in both wires

This is a multiple choice answer as classified in NCERT Exemplar

(b), (d) Let mass m is placed at x from the end B respectively.

T_A and T_B be the tensions in wire A and wire B respectively.

For the rotational equilibrium of the system,

\sum τ = 0

T_Bx-T_A(l-x)=0

$\frac{T_{B}}{T_{A}}$ = $\frac{l - x}{x}$

Stress in wire A = S_A= $\frac{T_{A}}{a_{A}}$

Stress in wire B = S_B= $\frac{T_{B}}{a_{B}}$ where a are the area of wire

We know that a_B=2a_A

Now for equal stress

S_A=S_B

$\frac{T_{A}}{a_{A}} = \frac{T_{B}}{a_{B}}$

$\frac{T_{B}}{T_{A}} = 2$

$\frac{l - x}{x} = 2$

So x =l/3 and l-x= 2l/3

Hence mass m should placed to B.

For equal strain

Strain_A= Strain_B

$\frac{Y_{A}}{S_{A}} = \frac{Y_{B}}{S_{B}}$

$\frac{Y_{s t e e l}}{Y_{A l}} = \frac{T_{A}}{T_{B}} \times \frac{a_{B}}{a_{A}} = \frac{x}{l - x} (\frac{2 a_{A}}{a_{A}})$

$\frac{200 \times 10^{9}}{70 \times 10^{9}} = \frac{x}{l - x}$

After solving we get x= x= 10l/17

l-x=l=10l/17=7l/17

For an ideal liquid

(a) The bulk modulus is infinite.

(b) The bulk modulus is zero.

(c) The shear modulus is infinite.

(d) The shear modulus is zero.

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) An ideal liquid is not compressible

Hence $?$ V=0

Bulk modulus B= strss /volumetric strain= $\frac{\frac{F}{A}}{\frac{? V}{V}} = \infty$

Compressibility K= 1/B=1/ $\infty = 0$

As there is no tangential force exists. So shear strain =0

Physics NCERT Exemplar Solutions Class 11th Chapter Nine Exam

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