From the shadows it is revealed that Earth has spherical shape with bulging along equator and flat at poles.

F = k_Ax_A = k_Bx_B
\({ x }_{ A }=\frac { F }{ { k }_{ A } } ,{ x }_{ B }=\frac { F }{ { k }_{ B } } \)
The work done on the springs are stored as potential energy in the springs.
\({ U }_{ A }=\frac { 1 }{ 2 } { k }_{ A }{ x }_{ A }^{ 2 };\quad { U }_{ B }=\frac { 1 }{ 2 } { k }_{ B }{ x }_{ B }^{ 2 }\)
\(\frac { { U }_{ A } }{ { U }_{ B } } =\frac { { k }_{ A }{ x }_{ A }^{ 2 } }{ { k }_{ B }{ x }_{ B }^{ 2 } } =\frac { { { k }_{ A }\left( \frac { F }{ { k }_{ A } } \right) }^{ 2 } }{ { { k }_{ B }\left( \frac { F }{ { k }_{ B } } \right) }^{ 2 } } =\frac { \frac { 1 }{ { k }_{ A } } }{ \frac { 1 }{ { k }_{ B } } } \)
\(\frac { { U }_{ A } }{ { U }_{ B } } =\frac { { k }_{ B } }{ { k }_{ A } } \)
k_A > k_B implies that U_B > U_A.Thus, more work is done on B than A.

n₁ = 72 kmh^-1 n₂ =? ms^-1
L₁ = 1 km       L₂ = 1m
T₁=1h             T₂=1s
\(n_2=n_1[\frac{L_1}{L_2}]^a[\frac{T_1}{T_2}]^b\)
The dimensional formula for velocity is [LT^-1]
a = 1 b =-1
\(n_2=72[\frac{1\ km}{1\ m}][\frac{1\ h}{1\ s}]^{-1}\)
\(=72[\frac{1000\ m}{1\ m}]^1[\frac{3600\ s}{1\ s}]^{-1}=72\times1000\times\frac{1}{3600}=20ms^{-1}\)
72km h^-1 = 20ms^-1

Sum of the clockwise moments is equal to sum of the anticlockwise moments when a body is in rotational equilibrium or algebraic sum of moments at any point is zero.

Kinetic energy is KE\(=\frac { 1 }{ 2 } { mv }_{ x }^{ 2 }\)
Multiply numerator and denominator by m
\(KE=\frac { 1 }{ 2m } { m^{ 2 }v }_{ x }^{ 2 }=\frac { 1 }{ 2m } \left( { mv }_{ x } \right) ^{ 2 }=\frac { 1 }{ 2m } { P }_{ x }^{ 2 }\)
where, P_x is the linear momentum of the particle executing simple harmonic motion.
Total energy can be written as sum of kinetic energy and potential energy, therefore, from equation (10.73) and also from equation (10.75), we get
E = KE + U(x) \(=\frac { 1 }{ 2m } { P }_{ x }^{ 2 }+\frac { 1 }{ 2m } { m\omega ^{ 2 }v }_{ x }^{ 2 }\) = constant

The steam engine is not a Carnot engine, because all the process involved in the steam engine are not perfectly reversible. But we can calculate the maximum possible efficiency of the steam engine by considering it as a Carnot engine.
\(\eta =1-\frac { { T }_{ L } }{ { T }_{ H } } =1-\frac { 300K }{ 523K } =0.43\)
The steam engine can have maximum possible 43% of efficiency, implying this steam engine can convert 43% of input heat into useful work and remaining 57% is ejected as heat. In practice the efficiency is even less than 43%.

Capillary descent, cos140 = cos(90+50)–sin50 = –0.7660
\(h=\frac { 2Tcos\theta }{ r\rho g } =\frac { 2\times ({ 0.465N \ m }^{ -1 })({ cos \ 140 }^{ 0 }) }{ \left( 2\times { 10 }^{ -3 }m \right) \left( 13.6\times { 10 }^{ 3 } \right) \left( { 9.8 }ms^{ -2 } \right) } \)
\(=\frac{2 \times 0.456 \times(-0.7660)}{2 \times 13.6 \times 9.8} \)
\(=\frac{-0.6986}{266.56}=-2.62 \times 10^{-3} \mathrm{~m}\)
where, negative sign indicates that there is fall of mercury (mercury is depressed) in glass tube.

Relative velocity of B with respect to A
 \({ v }_{ BA }={ v }_{ B }-{ v }_{ A }\)
= 50 km h^-1 + (-50) km h^-1 
= 0 kmh^-1
Similarly, relative velocity of A with respect to B i.e., v_AB is also zero.
Thus each train will appear to be at rest with respect to the other.

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.