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The change in the speed of the ball is due to force of gravity.

Mass of the body = 10kg 

Initial velocity, u = 4 m/s 

Final velocity, v = 8 m/s 

Time, t = 2s 

a) Momentum before the force acts, p₁ = mu = 40 kg.m/s 

b) Momentum after the force acts, p₂ = mv = 80 kg.m/s 

c) Gain in momentum for 2s = p₂ – p₁ = 40 kg.m/s

Gain in momentum per second = 40/2 = 10 kg.m/s 

d) Acceleration, a = (v-u)/t = 2 m/s_² 

Force = ma = 20N

A gas thermometer.

Mass of the body = 2.5kg 

Initial velocity, u = 20 m/s 

Final velocity, v = 0 m/s 

Force, F = 10N 

Acceleration, a = F/m = 10/2.5 = 4 m/s²

Since, v < u, negative sign is used = -4 m/s² 

Time, t = (v-u)/a = 5s

Do the values of coefficients of expansion differ, when the lengths are measured in C.G.S. system or in S.I No.

Here, l = 1.5 m, r = 1.0 cm = 1 x 10^-2 m

Mass issuing out per second

 m = 4 x 10^-3 kg s^-1

ρ = 1.3 x 10³ kg m^-3

η = 0.83 Pa s

Now volume issuing out per second

V = m/ρ = {4 x 10^-3}/{1.3 x 10³}= 3.08 x 10^-6 m³ s^-1

Using Poiseulli's relation

V = π/8 x pr⁴/ηl

⇒ p = 8 V ηl/πr⁴ = {8 x 3.08 x 10^-6 x 0.83 x 1.5}/{3.14 x (1 x 10^-2)⁴}

= 976.97 Pa

Mass of the body = 20 kg 

Initial velocity, u = 0 m/s 

Final velocity, v = 100 m/s 

Force, F = 100 N 

Acceleration, a = F/m = 100/20 = 5 m/s² 

Time, t = (v-u)/a = 20s

Mass of the car = 2400 kg 

Initial velocity, u = 20 m/s 

Final velocity, v = 0 m/s 

Time, t = 10s 

Retardation, a = (v-u)/t = 2 m/s² 

Force = ma = 4800N

Given polynomial

(i) f(x) = 0 is a constant polynomial as 0 is a constant.

(ii) g(x) = 2x³ -7x +4 is a cubic polynomial as degree of the polynomial is 3. 

(iii) h(x) = -3x + 1/2 is a linear polynomial as degree of the polynomial is 1.

(iv) p(x) = 2x² -x +4 is a quadratic as the degree of the polynomial is 2. 

(v) q(x) = 4x +3 is a linear polynomial as the degree of the polynomial is 1.

(vi) r(x) = 3x³ +4x² +5x -7 is a cubic polynomial as the degree is 3.

(a) The body with large reflectivity would be poor c of absorber of radiations. poor absorbers are poor radiators. Hence, a  body with large reflectivity is a poor emitter.

(b) Brass is a good Conductor of heat. When we touch brass tumbler with our fingers, our body heat is quickly conducted to the brass tumbler and hence temperature of the finger tips is reduced. Thus the brass-tumbler feels colder.

On the Other hand, wood is a bad conductor, hence our body heat is not conducted to the eooden tray.

(c) The temperature of red hot iron in the Oven is given by E₁ = σ. T4 When iron is taken out in the open temperature (T₀), its radiation energy is given by, E_s = σ. (T⁴ - T₀⁴). Thus, the pyrometer  measures low Values is the hot iron in the open.

(d) The atmosphere server purpose of a blanket over the earth and it dose not allow earth's heat to be radiated during night.

(e) This is due to the fact that steam contains more heat in the form of latent heat (540 calories/gram) than water.

1. We know that a + r + 1 = 1, where ‘a’ is absorbance, ‘r’ is reflectance and ‘t’ is transmittance or emittance. According to Kirchhoff’s law, emittance ∝ absorbance, i.e., good absorbers are also good emitters and so are poor reflectors. Therefore, if reflectivity is large (‘r’ is large) then ‘a’ is small and hence emittance is smaller (the object behaves as a poor emitter).

2. The coefficient of thermal conductivity of brass is higher than that of wood. When a brass tumbler is touched, heat quickly flows from human body to the tumbler and so the tumbler appears colder. But in the case of the wooden tray heat does not flow from the human body to the wooden tray, and so it feels relatively hotter.

It has positive value for metals and alloys. For semiconductors and insulators, the value of α (temperature coefficient) is negative.

Correct option is: (C) 0°C and 100°C

Boyle's temperature : Temperature at which a real gas obeys ideal gas equation over an appreciable range of pressure

Answer is (c) z = 0

In all three figures, the liquid is at the same. Temperature is also the same for each case. Hence, the surface tension in figure (b) and (c) is the same as in figure.

Since the length of the film in all the cases is 40 cm, the weight supported in each case is 4.5 × 10^-2 N.

The weight supported by (b) and (c) are same as that in (a) and is equal to 4.5 × 10^–2 N. The weight supported = 2 σ l, where σ is surface tension and l is the length which is same in all the three cases, hence weight supported is same.

Take case (a):

The length of the liquid film supported by the weight, l = 40 cm = 0.4 cm The weight supported by the film, W = 4.5 × 10^–2 N 

A liquid film has two free surfaces. 

Surface tension = W/2l = 4.5 x 10^-2/2 x 0.4 = 5.625 x 10^-2 Nm^-1 

In all the three figures, the liquid is the same. Temperature is also the same for each case. Hence, the surface tension in figure (b) and figure (c) is the same as in figure (a), i.e., 5.625 × 10^–2 N m^–1 .

Since the length of the film in all the cases is 40 cm, the weight supported in each case is 4.5 × 10^–2 N.

1. When balloon is blown, air that is blown inside makes the balloon expand. 

2. A given size of balloon can expand upto certain limit. 

3. Once that limit is reached and air is still blown inside the balloon, balloon cannot expand further. 

4. As a result, air causes additional pressure on inner surface of balloon. 

5. Since, pressure inside balloon is now greater than pressure outside balloon, balloon bursts equalizing the two pressures.

Because of high value of specific heat of water, the wet clothes can take away a large amount of heat from the body. Due to this, the temperature of the body may fall much below the normal temperature.

1. In cold climate, temperature of water in ponds and lakes starts falling. 

2. On getting colder, water contracts. As a result, density of water increases and it goes down. To replace it, warmer water from below rises up. This process continues till temperature of water at the bottom of pond becomes 4°C. 

3. Water, due to its anomalous behaviour possesses maximum density at 4°C. 

4. If the temperature lowers further, ice is formed at the surface of pond with water below it. 

5. Ice being poor conductor of heat blocks the further heat exchange between atmosphere and water in the pond and maintains water below surface in liquid state.

As shown in the fig.(a)

Length, l = 40 cm = 40 x 10^-2 m

Also, Force, F = 4.5 x 10^-2 N

surface tension, T = F/2l = {4.5 x 10^-2}/{2 x 40 x 10^-2} = 5.625 x 10^-2 Nm^-1

In case of figures (b) and (c), length is again 40 x 10^-2 m

Small weight that the film can support is

F = T x (2l) = 5.625 x 10^-2 x 2 x 40 x 10^-2

= 4.5 x 10^-2 N

'J' represents Joule's mechanical equivalent of heat. It is not a physical constant. its a conversion factor, i.e., 1 calorie = 4.18 Joules. It means 4.18 joules of energy and 1 calorie of heat are equivalent. Thus, amount of work (in J) is obtained from the quantity of heat (in cals) by multiplying the latter by 4018 and vice- versa.

On heating, the physical changes such as expansion, change of state, change of colour, change of electrical properties (resistance, conductivity), etc., may take place.

The cause of hotness of a body is the kinetic energy of the molecules constituting the body.

Lord Kelvin suggested that on a scale of temperature, whose zero coincides with -273. 15°C, the temperature of a body will always be positive. Such a scale of temperature. is called absolute scale of temperature and -273. 15°C as the zero of this new scale, called absolute zero.

The solid expands on heating. A small is left between the rails so as to allow the expansion of the rails summer. In case, the space is not left, the rails will bend.

Given as function is f(x) = 2x² – 5x + 3 on [1, 3]

Here, the given function f is a polynomial. Therefore, it is continuous and differentiable everywhere.

Then, let us find the values of function at the extreme values.

⇒ f (1) = 2(1)² – 5(1) + 3

⇒ f (1) = 2 – 5 + 3

⇒ f (1) = 0…… (1)

⇒ f (3) = 2(3)² – 5(3) + 3

⇒ f (3) = 2(9) – 15 + 3

⇒ f (3) = 18 – 12

⇒ f (3) = 6…… (2)

From the equation (1) and (2), we can say that, f(1) ≠ f (3)

Therefore, Rolle’s Theorem is not applicable for the function f in interval [1, 3].

Given as the function is f (x) = 2sinx + sin2x on [0, π]

As we know that sine function continuous and differentiable over R.

Let us check the values of function f at the extremes

⇒ f (0) = 2sin(0) + sin2(0)

⇒ f (0) = 2(0) + 0

⇒ f (0) = 0

⇒ f (π) = 2sin(π) + sin2(π)

⇒ f (π) = 2(0) + 0

⇒ f (π) = 0

f(0) = f(π), therefore there exist a c belongs to (0, π) such that f’(c) = 0.

Let us find the derivative of function f.

f'(x) = d(2sin x + sin 2x)/dx

f'(x) = 2cos x + cos 2x(d(2x)/dx)

⇒ f’(x) = 2cosx + 2cos2x

⇒ f’(x) = 2cosx + 2(2cos²x – 1)

⇒ f’(x) = 4 cos²x + 2 cos x – 2

f'(c) = 0, from definition

⇒ 4cos²c + 2 cos c – 2 = 0

⇒ 2cos²c + cos c – 1 = 0

⇒ 2cos²c + 2 cos c – cos c – 1 = 0

⇒ 2 cos c (cos c + 1) – 1 (cos c + 1) = 0

⇒ (2cos c – 1) (cos c + 1) = 0

cos c = 1/2 or cos c = -1

c = (π/3) ∈ (0,π)

Thus, Rolle's theorem is verified. 

(i) 90° – θ (ii) 90° – θ (iii) 180° – 2θ

Correct option is (C) intersecting lines

By comparing with standard form of system of linear equations, we get

\(a_1=3,b_1=2,c_1=-80\) and

\(a_2=4,b_2=3,c_2=-110\)

\(\because\) \(\frac{a_1}{a_2}=\frac34,\) \(\frac{b_1}{b_2}=\frac23\)

\(\because\) \(\frac34\neq\frac23\)

\(\therefore\) \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\)

\(\therefore\) Both gives lines are intersecting lines.

Correct option is C) intersecting lines

The given system of equations is: 

2x + y – 5 = 0 

4x + 2y – 10 = 0 

The above equations are of the form 

a₁ x + b₁ y − c₁ = 0 

a₂ x + b₂ y − c₂ = 0 

Here, a₁ = 2, b₁ = 1, c₁ = −5 

a₂ = 4, b₂ = 2, c₂ = −10

So according to the question, we get 

\(\frac{a_1}{a_2}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\) 

\(\frac{b_1}{ b_2}\) = \(\frac{1}{2}\) and, 

\(\frac{c_1}{c_2}\) = \(\frac{−5}{−10}\) = \(\frac{1}{2}\) 

⇒ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{ b_2}\) = \(\frac{c_1}{c_2}\) 

Hence, we can conclude that the given system of equation has infinity many solutions.

The product of the force & the time for which it acts on a body is called impulse of a force. The force acting on a body for short interval of time is called impulsive force.

Initial momentum = \(\vec p\)

Final momentum = - \(\vec p\)

Change in momentum = Δ p 

= - \(\vec p\)- \(\vec p\)= -2 \(\vec p\).

When we are on the floor, exerts reaction on us. When we jump the reaction on us is zero. Therefore while jumping on a cement Floor we weigh less than on cement floor.

F = 50N t = 0 :1 s

impulse = 50 N × 0.1s = 5 Ns.

Given: m = 1.0 g = 1.0 × 10^-3 kg,

h = 1 km = 10³ m, v = 50 ms^-1

To find: Work done (W_f)

Formula: W_f = ∆ K.E – W_g

Calculation:

i. The change in kinetic energy of the drop

∆ K.E = (K.E.)_final (K.E.)_initial

∴ ∆ K.E. = \(\frac{1}{2}\)mv² − 0

= \(\frac{1}{2}\) × 1.0 × 10^-3 × (50)²

∴ ∆ K.E.= 1.25 J

ii. Work done by the gravitational force is,

W_g = mgh = 1.0 × 10^-3 × 9.8 × 10³ = 9.8 J

∴ W_g = 9.8J

From formula,

W_f = ∆K.E. – Wg = 1.25 – 9.8

W_f = -8.55 J

Work done by the unknown force is – 8.55 J.

Conservative forces are those forces against which the work done doesn’t depend on the path followed but depends only on the initial and final positions. Non-conservative forces are those in which the work done depends on the path taken.

(b) along the radius

Correct option is: (C) frictional

Friction forces are non-conservative force because its depend on the path.

Correct option is: (C) frictional

Correct option is: (B) zero

1. Consider an object dropped from some point at height h.

2. While coming down its potential energy decreases.

∴ Work done = decrease in P.E of the body.

3. But, in this case, some part of the energy is used in overcoming the air resistance. This part of energy appears in some other forms such as heat, sound, etc. Thus, the work is not entirely converted into kinetic energy.

In this case, the work-energy theorem can mathematically be written as,

∴ ∆ PE = ∆ K.E. + W_{air resistance}

∴ Decrease in the gravitational P.E. = Increase in the kinetic energy + work done against non-conservative forces.

(A) Real and pseudo forces,

(B) Conservative and non-conservative forces,

(C) Contact and non-contact forces,

(D) Inertial and non-inertial frames of reference.

Correct Answer is : (a) Gujarat

As per census 2011, the calculated population of tribals in India is 182.81 Lakh.

Correct Answer is : (d) All these

In five districts of Rajasthan, namely – Dungarpur, Banswara, Pratapgarh, Udaipur and Sirohi (19 Tehsils and 23 panchyat samitis), the Tribal Development Program is operational.

Two objectives of tribal Development Program are: 

1. To reduce the difference in level of development between tribals and other areas. 

2. Improving standard of life of Tribals.

This program has been devised for those areas where population of tribal people is 50% or more.

1. Major tribal areas: 9 states and 2 union territories are identified as home of such tribal areas which are mainly in Madhya Pradesh, Chhattisgarh, Odisha, Maharashtra, Jharkhand, Gujarat, Andhra Pradesh and Rajasthan (19 tehsils and 23 panchayat samitis).

2. Objectives of this project: The projects that are created for tribal areas have the following objectives : 

• To reduce the difference between tribal and non-tribal development rate. 
• To raise the standard of living.

The programmes created for tribal areas have some of the following reforms: Agricultural and horticulture, animal rearing, forestry, small and cottage industries, reforms in marketing. These include education as well as potable water, sufficient housing, medical and nutrition facilities also.

A tribal area is divided into following three levels under Tribal Development Program: 

1. Macro level – Tribal level 

2. Medium Level – Tehsil

3. Small – Scale Development Unit.