k₁ = 1 N m−¹, k₂ = 2 N m⁻¹
k_p = k₁ + k₂ N m⁻¹
k_p = 1 + 2 = 3 N m⁻¹
k_p > k₁ and k_p > k₂
Therefore, the effective spring constant is greater than both k₁ and k₂.

When the water is heated from 30°C to 60°C, there is only a slight change in its volume. So we can treat this process as isochoric. In an isochoric process the work done by the system is zero. The given heat supplied is used to increase only the internal energy.
ΔU = Q = ms_v ΔT
The mass of water = 500 g = 0.5 kg
The change in temperature = 30K
The heat Q = 0.5\(\times\)4184\(\times\)30 = 62.76 kJ

The excess of pressure inside the soap bubble is
\(\Delta P={ P }_{ 2 }-{ P }_{ 1 }=\frac { 4T }{ R } \)
But \(\Delta P={ P }_{ 2 }-{ P }_{ 1 }=\rho gh\Rightarrow \rho gh=\frac { 4T }{ R } \)
\(\Rightarrow\) Surface tension,
\(T=\frac { \rho ghR }{ 4 } =\frac { (800)(9.8)(4\times { 10 }^{ -3 })\left( 2\times { 10 }^{ -2 } \right) }{ 4 } \)=T= 15.68\(\times\)10^-2 N m^-1

Using the equation,
\(|\vec A-\vec B|=\sqrt{5^2+7^2-2\times 5\times 7\cos60^o}=\sqrt{25+49-35}=\sqrt{39}\) units
The angle that \(\vec A-\vec B\) makes with the vector \(\vec A\) is given by
\(\tan\alpha_2=\frac{7\sin60^o}{5-7\cos60^o}=\frac{7\sqrt{3}}{10-7}=\frac{7}{\sqrt{3}}\)= 4.041
\(\alpha_2=tan^{-1}(4.041)=76^o\)

Impulse: When a large force works on a body for very small time interval, it is called impulsive force. An impulsive force does not remain constant, but changes first from zero to maximum and then from maximum to zero. In such case we measure the total effect of force.
\(\vec{F} =\frac{d \vec{p}}{d t} \text { or }  \)
\(\int_{t_{1}}^{t_{2}} \vec{F} d t =\int_{p_{1}}^{p_{2}} d \vec{p} \Rightarrow \vec{I}=\overrightarrow{p_{2}}-\overrightarrow{p_{1}}=\overrightarrow{\Delta p}\)
\(\therefore\) the impulse of a force is equal to the change in momentum. This statement is known as Impulse momentum theorem. If no external force acts on a system (called isolated) of constant mass, the total momentum of the system remains constant with time. According to this law for a system of particles \(\vec{F}=0 \ then \ \vec{p}=\) constant.
Examples: Hitting, kicking, catching, jumping, diving, collision etc.

The answer is wrong.
Actually, the seasons in the Earth arise due to the rotation of Earth around the sun with 23.5^o tilt.

The spring constant, k = 0.1 N m^-1
The displacement, x = 25 cm = 0.25 m
(a) The potential energy stored in the spring is given by
\(U=\frac { 1 }{ 2 } { kx }^{ 2 }=\frac { 1 }{ 2 } \times 0.1\times { (0.25) }^{ 2 }=0.0031J\)
(b) The work done W_s by the spring force _{\(\bar { F } \)} is given by,
\({ W }_{ s }=\int _{ 0 }^{ x }{ \overrightarrow { { F }_{ s } } .\overrightarrow { dr } } =\int _{ 0 }^{ x }{ (-k\ x\hat { i } ).(dx\hat { i } ) } \)
The spring force \(\overrightarrow { { F }_{ s } } \)  acts in the negative x direction while elongation acts in the positive x direction.
\({ W }_{ s }=\int _{ 0 }^{ x }{ (-kx)dx=-\frac { 1 }{ 2 } { kx }^{ 2 } } \)
\({ W }_{ s }=-\frac { 1 }{ 2 } \times 0.1\times { (0.25) }^{ 2 }=-0.0031\ J\)
Note that the potential energy is defined through the work done by the external agency. The positive sign in the potential energy implies that the energy is transferred from the agency to the object. But the work done by the restoring force in this case is negative since restoring force is in the opposite direction to the displacement direction.
(c) During compression also the potential energy stored in the object is the same.
\(U=\frac { 1 }{ 2 } { kx }^{ 2 }=0.0031\ J\)
Work done by the restoring spring force during compression is given by
\({ W }_{ s }=\int _{ 0 }^{ x }{ { \overrightarrow { F } }_{ s } } \overrightarrow { dr } =\int _{ 0 }^{ x }{ (kx\hat { i } ).(-dx\hat { i } ) } \)
In the case of compression, the restoring spring force acts towards positive x-axis and displacement is along negative x direction.
\({ W }_{ s }=\int _{ 0 }^{ x }{ (-kx)dx=-\frac { 1 }{ 2 } { kx }^{ 2 } } =-0.0031J\)

By the principle of homogeneity, a / V² is of the dimensions of pressure and b is of the dimensions of volume.
[a] = [pressure] [V²] = [ML⁻¹T⁻²] [L⁶]
= [ML⁵T^-2]
[b] = [V] = L³

In general our interest our interest is to find how many gas molecules have the range of speed from v to v + dv. This is given by Maxwell's speed distribution function.
\({ N }_{ v }=4\pi N{ \left( \frac { m }{ 2\pi KT } \right) }^{ \frac { 3 }{ 2 } }{ v }^{ 2 }{ e }^{ \frac { { mv }^{ 2 } }{ 2KT } }\)   ....(1)
The above expression is graphically shown as follows

From the figure it is clear that, for a given temperature the number of molecules having lower speed increases parabolically but decreases exponentially after reaching most probable speed. The rms speed, average speed and most probable speed are indicated in the figure. It can be seen that the rms speed is greatest among the three. To Know the number of molecules in the range of speed between \(50 \mathrm{~m} \mathrm{~s}^{-1} \ and \ 60 \mathrm{~m}\mathrm{s}^{-1}\), we need to integrate \(\int_{50}^{60} 4 \pi N\left(\frac{m}{2 \pi k T}\right)^{\frac{3}{2}} v^{2} e^{\frac{m v^{2}}{2 k T}} d v=\mathrm{N}\left(50\right.\ to \ \left.60 \mathrm{~ms}^{-1}\right)\). In general the number of molecules within the range of speed v and v + dv is given by
\(\int_{v}^{v+d v} 4 \pi N\left(\frac{m}{2 \pi k T}\right)^{\frac{3}{2}} v^{2} e^{\frac{m v^{2}}{2 k T}} d v=N(v \text { to } v+d v) \)
The exact integration is beyond the scope of the book. But we can infer the behaviour of gas molecules from the graph.
(i) The area under the graph will give the total number of gas molecules in the system.
(ii) Figure shows the speed distribution graph for two different temperatures. As temperature increases, the peak of the curve is shifted to the right. It implies that the average speed of each molecule will increase. But the area under each graph is same since it represents the total number of gas molecules.