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The Indus valley was known for its high quality cotton.

Correct option is: (a) cotton

Correct option is: (a) Mohen-jo-daro

The archaeologists made various excavations in the north western region of India which led to the discovery of the mins of the pre-historic cities of Harappan in West Punjab and Mohenjo-daro in Sind. The sites of early excavations were found on the river Indus and its tributaries. Therefore, Indus Valley Civilization came to be known as Harappan Civilization.

The elaborate social structure and standard of living confirmed the presence of Trade in the Indus Valley Civilisation, e.g., the cities like Mohenjo-daro, Harappa and Lothal were important centers of metallurgy. Balakot and Chanhudaro were centres for shell-working and bangle-making.

1. The ruins of the sites reveal that the Harappan people were primarily urban and their cities were designed skillfully. 

2. The unique features of the city was its elaborate drainage system. A brick-lined drainage channel flowed alongside every street. 

3. The Great Bath was also unearthed. The pool was filled with water taken from a well nearby.The walls of the pool were made watertight using specially-made bricks and gypsum mortar. 

4. The Great Granary consisted of two blocks with an aisle between them. Each block had six halls with corridors. It was used for storing food grains.

Correct Option  (d)  -v²/R cosθ i+ v²/R sinθ j

(d) Momentum 

λ = h/p, when wavelength λ is same, momentump is also same.

This is because the capillary rise is inversely proportional to the radius of the capillary tube.

Raheem: 

∠AOD = 140°, ∠BOC = 40° 

they are not adjacent angles and not linear pair. 

So, Raheem is not correct. 

Mary: 

∠COD = 140°, ∠BOC = 40° 

they are adjacent angles, but they are not linear pair. 

So, Mary answer is not correct. 

Roshitha: 

∠XOZ = 140°, ∠ZOY = 40° they are adjacent angles and linear pair. 

So, Roshitha answer is correct.

The ratio of increase in their lengths is (1 : 8)

Elastic limit is the value of applied force beyond which the material (body) does not come back to its original shape when the applied force is reduced.

Young's modulus (Y) of a material is defined as the ratio of normal stress to the longitudinal strain, i.e.,

Y = normal stress/longitudinal strain

Within elastic limit, the ratio of lateral strain to the linear strain is called the Poisson‘s ratio.

Elastic Iimit is the maximum stress on whose removal, the bodies regain their original dimensions.

The theoretical value of Poisson's ratio is 0.5

A force that produces a change in the shape or size of a body is called a deforming force.

(a) the same stress.

(d) different strain.

1. The relative displacement of the bottom face and the top face of the cube is called shearing strain.

2. Shearing strain = \(\frac{\triangle\,l}{L}\) = tan θ

where, ∆l = displaced length,

L = Original length.

3. When ∆l is very small,

tan θ ≈ θ and shearing strain = θ

Restoring force

A heavy wire (even when no weight is attached to it) is under stress, when it is suspended from a roof. It is because the weight of the heavy wire acts as the deforming force.

Correct answer is (C) hardness

It is the property of elasticity.

The physical quantities having different values in different directions are called tensor physical quantities. For example - stress.

As the separation tends to decrease, further, the force between molecules becomes strong repulsive, which prevents the molecules of the given substance from collapsing to the condensed state.

It is because of the loss of the strength of the material due to repeated alternating strains, to which the wire is subjected.

The work done in stretching the wire is stored in it in the form of the elastic potential energy.

When a wire is stretched, interatomic forces come into play and these forces oppose the increase in length of the wire. Therefore, in order to stretch the wire, work has to be done against the interatomic forces.

Given :

Breaking stress = 7.5 × 10⁸ dyne cm^-2,

Ρ = 2.7 g cm^-3.

Suppose l be the greatest length of the wire that can be hang vertically without breaking.

Mass of wire,

w = Area of cross-section × Length × Density = Alρ

Hence, weigh of wire,

w = mg = Alρg

This is equal to the maximum force that the wire can withstand.

So, breaking stress = \(\frac{W}{A}\) = \(\frac{Alρg}{A}\) = lρg

or 7.5 × 10⁸ = l × 2.7 × 980

So, l = \(\frac{7.5 × 10^8}{2.7×980}\)

= 2.834 × 10⁵ cm

= 2.834 km.

The angular tilt between different layers of a body due to tangential force (Shear Force) is called Shear Strain.

Given : 

Young’s modulus Y for the wire = 12.5 x 10¹¹ dyne cm^-2 

=12.5 x 10¹⁰Nm^-2

Diameter =D=2.5mm=2.5 x 10^-3 m 

Force=F=100kg f 

=100x 9.8N 

=980N

Formula to be used :ΔL/Lx100 

A=πr²=π(1.25 x 10^-3)³ m²  

Use the formula:Y=FL/A Δ L

Percentage inrease in length =ΔL/L x100 

=(F/AY )x100 

=(F/π r²Y )x100 

=[980/3.142 x (1.25 x 10^-3)² x 12.5 x 10¹⁰ ] x 100

=15.96 x 10^-2=0.16 %

 ∴The percentage increase in the length of a wire is 0.16 %

Given :

Y = 12.5 × 10¹¹ dyne cm^-2

= 12.5 × 10¹⁰ Nm^-2

Diameter, D = 2.5 mm = 2.5 × 10^-3 m.

∴ radius, r = \(\frac{D}{2}\) = 1.25 × 10^-3 m.

F = 100 kgf = 100 × 9.8 N = 980 N. \(\frac{∆L}{L}\) × 100 = ?

A = πr² = π(1.25 × 10^-3)² m².

From the relation, Y = \(\frac{FL}{A∆L},\)

We get % increase in length = \(\frac{∆L}{L}\) × 100

= \(\frac{F}{AY}\) × 100 = \(\frac{F}{πr^2Y}\) × 100

= \(\frac{980}{3.142×(1.25×10^−3)×12.5×10^{10}}\) × 100

= 0.16 %

Correct option: 3. Cl > F > Br > l

Explanation:

The E.A of   F₂ is less them E.A Cl₂

(a) 2F

(b) FF

The solids in which atoms and molecules are not arranged in definite and regular manner are called glassy or amorphous solids, e.g. glass, rubber, sulphur etc.

If one end of a cylinder of length l is clamped and other is given a twist ϕ by an application of a couple, then the relation between angle of ϕ and shear θ is

θ = xϕ/1 (where x is distance from axis)

Shear at surface (x = r) is maximum and given by

θ_max = rϕ/l

Torsional rigidity of hollow cylinder of internal and external radii r₁ and r₂.

C_h = {πη(r₂⁴ - r₁⁴)}/{2l}

Restoring couple of solid cylinder

T_s = C_sϕ = πηr₄ϕ/2l

and the restoring couple of hollow cylinder

T_h = C_hϕ = {(πr₂⁴ - r₁⁴)ϕ}/{2l}

For same mass, material and length, the hollow cylinder is stronger than the solid cylinder.

When a beam is bent, the strain produced is longitudinal and so elastic modulus involved is Young's modulus.

The bending moment is the algebraic sum of moments of all restoring forces developed in the filaments of the bent beam about a neutral axis. If Y is Young's modulus, R radius of curvature of neutral filament and I, the geometrical moment of inertia, then longitudinal strain at a distance Z from neutral filament = Z/R

Bending moment = YI/R

For a beam of circular cross-section of radius r,

I = πr⁴/4

For a beam of rectangular cross-section

I = bd³/12

where b is the breadth and d its depth.

For a beam supported at ends loaded in the middle by a load W = Mg, the depression at the centre is given by

δ = Wl³/48YI

For a beam of rectangular cross-section

I = bd³/12 and δ = Wl³/4Ybd³

F = 100 x 9.8 N

Y = 12.5 x 10¹⁰ Nm^-2

A = (π/4)d² = (π/4) x (2.2 x 10^-3)² = 3.8 x 10^-6 m³

Y = FL/Al

⇒ l/L = F/AY = {100 x 9.8}/{(3.8 x 10^-6) x (12.5 x 10¹⁰)}

= 0.0021

% increase in length = (l/L) x 100 = 0.021 x 100 = 0.21%

3.  Cl₂  > Br₂   > F₂  > I₂

The ratio of the forces required is 1 : 1

Bulk modulus, K = Δp.V/ΔV

K = 0.80 x 10¹⁰ Nm^-2 = {0.80 x 10¹⁰}/{10⁴} Ncm^-2

Δp = 20,000 N cm^-2

⇒ ΔV/V = Δp/B = {20000 x 10⁴}/{0.80 x 10¹⁰} = 1/40

or, ΔV = V/40

Now, volume V' = V - V/40 = 39V/40

Since the mass of lead will remain the same, we have Vρ = V'ρ'

New density,

ρ' = {V x 11.4 x 40}/{39V} = 11.7 g cm^-3

Interatomic forces of attractive nature are required to form a solid.

We prefer steel to copper in the manufacture of springs because steel is more elastic than copper.

3. OF₂  < H₂O < OCl₂  < ClO₂

Solids are of two types:

(i) Crystalline solids

(ii) Glassy or amorphous solids.

Bulk modulus.

(Take Bulk modulus of elasticity of water = 2.2 x 10⁹ Nm^-2)

Given:

V = 1 litre = 10^-3 m³

ΔV/V = 0.10/100 = 10^-3

B = pV/ΔV

or, p = BΔV/V = (2.2 x 10⁹) x 10^-3 = 2.2 x 10⁶ Pa

1. HClO₄ < HNO₃ < H ₂CO₃ < B (OH)₃

Correct option: 2 Cl₂ , KClO₃

Explanation:

3Cl₂ + 6KOH → 5KCl + KClO₃ + 3H₂O

If groups of atoms are not given a chance to arrange themselves in proper order say be checking their mobility at low temperature, amorphous solids are formed, e.g., amorphous carbon.