Mass of the earth M = 6.024 5 10²⁴ kg
Radius of the earth R = 6.4\(\times\)10⁶ m
Gravitational constant G' = 6.67\(\times\)10¹¹
Gravitational constant G = 6.67\(\times\)10¹¹
Acceleration due to gravity g' =?
g' = 
g' = 9.8\(\times\)10²² m/s²
g = 9.8 m/s²
∴ g' = g\(\times\)10²² (or) 10²²
g = g' m/s²
Weight W = mg
W' = mg'
 = 10²² .W
W = 10²²

Final momentum on each object Δp = FΔt = 100\(\times\)5 = 500 kg ms^-1
Final speed on the object of mass 5 kg = 500/5 =100 m s^-1
Final speed on the object of mass 20 kg = 500/20 =25 m s^-1
Note that momentum on each object is the same after 5 seconds but speed is not the same after 5 seconds. The heavier mass acquires lesser speed than the one with lower mass.

Gravitation force of sun.

Less reaction force.

The average acceleration is defined as the ratio of change in velocity over the time interval \(a_{avg}=\frac{\triangle\vec V}{\triangle t}\) . It is a vector quantity.

Due to conservation of angular momentum.

Angular momentum of planet.

No, moment of inertia and radius of gyration depends on axis of rotation and also on the distribution of mass of the body about its axis.

\(\vec { \tau } =\vec { r } \times \vec { F } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 5 \\ 3 & -2 & 4 \end{matrix} \right| \)
\(\vec { \tau } =\left( 12-\left( 10 \right) \right) \hat { i } +\left( 15-8 \right) \hat { j } +\left( -4-9 \right) \hat { k } \)
\(\vec { \tau } =22\hat { i } +7\hat { j } -13\hat { k } \)

four

one

The percentage error in V is 5% and in I it is 2%
The total percentage error in R is given by
\({\triangle R \over R}\times 100={\triangle V\over V}\times 100 +{\triangle I\over I}\times 100=5\%+2\%=7\%\)

For symmetrically shaped bodies of uniform composition (Eg: spheres, cylinders, rectangular solids), the centre of mass is located at the geometrical centre of the body.

When a cat falls to ground from a height, it stretches its body along with the tail so that its moment of inertia becomes high. Since angular momentum (L = Iω) remain constant, the value of angular speed ω decreases and therefore the cat is able to land on the ground gently.

The heavy boxes should be loaded first so that the CG of the loaded cart remains in the lowest position. This ensures stability of equilibrium.

Since fmal and initial positions are same hence displacement of the boy will be, \(\triangle r=r-r=0\).

When bodies have the same mass i.e., m₁ = m₂ and second body (usually called target) is at rest (u₂ = 0), By substituting m₁ = m₂ and u₂ = 0 in equations
\({ v }_{ 1 }=\left( \frac { { m }_{ 1 }-{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { 2m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\)...(1)
and equations \({ v }_{ 2 }=\left( \frac { { 2m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { m }_{ 2 }-{ m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\) , we get...(2)
from equations (1) => v₁ = 0
from equations (2) => v₂ = u₁

Work is said to be done by the force when the force applied on a body displaces it. Eg: Horse pulls a cart, engine pulls a train.

P.E of the spring when stretched through a distance x,
\(U={1\over2}{kx}^{2}=10J\)
When x becomes lx, the potential energy will be
\(u'={1\over2}k(2x)^2=4\times{1\over2}{kx}^{2}=4\times10=40J\)
\(\therefore\) Workdone = u' - u = 40 - 10 = 30 J

When an apple falls towards the Earth, the direction of motion (direction of velocity) of the apple and that of force are in the same downward direction.

The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.

A coordinate system and the position of an object is described relative to it, then such a coordinate system is called frame of reference.

At highest point of the vertical loop weight of the pilot provides the necessary centripetal force.
\([mg=\frac{mv^{2}}{r}]\)

A clock is used to measure, the time interval. An atomic standard of time, is based on the periodic vibration produced in a Cesium atom. Some of the clocks developed later are electric oscillators, electronic oscillators, solar clock, quartz crystal, clock, atomic clock, decay of elementary particles, radioactive dating.

Range of masses: from heavenly bodies to electron, 10⁵⁵ kg (mass of known observable universe) to 10^-31 kg (mass of an electron) [the actual mass of an electron is 9.11\(\times\)10^-31 Kg].

The acceleration of a particle at any instant 'r' is equal to the rate of change of velocity.
Acceleration \(\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}\). It is a vector quantity.

At Earth's surface \(\mathrm{KE}=\frac{1}{2} \mathrm{M} v_{\mathrm{e}}^{2}, \mathrm{P} . \mathrm{E}=-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\)
At infinity K.E = \(\frac{1}{2} M V_{\infty}^{2}, \mathrm{P} . \mathrm{E}=\mathrm{O}\)
By law of conservation of energy,
\(\frac{1}{2} \mathrm{Mv}_{\mathrm{e}}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}} =\frac{1}{2} \mathrm{M} v_{\infty}^{2}  \)
\(v_{\mathrm{e}}^{2} =v_{\infty}^{2}+\frac{2 \mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}  \)
\(\text {But } \mathrm{GM}_{\mathrm{E}} =\mathrm{qR_{ \textrm {E } } ^ { 2 }}  \)
\(\therefore v_{\mathrm{e}}^{2} =v_{\infty}^{2}+2 \mathrm{gR}_{\mathrm{E}}  \)
\(v_{\mathrm{c}} =\sqrt{\mathrm{v}_{\infty}^{2}+2 \mathrm{gR}}\)

The value of semi major axis is directly proportional to twice the square of time period of a planet.
i.e, a \(\propto 2 \mathrm{~T}^{2}\)
It is given that for planet Kurinji,
\(\mathrm{T}_{1}=2, \quad \mathrm{a}_{1}=8=2 \times 2^{2} \Rightarrow 2 \mathrm{~T}_{1}{ }^{2}\)
For planet Mullai \(\quad \mathrm{T}_{2}=3, \quad \mathrm{a}_{2}=18=2 \times 3^{2} \Rightarrow 2 \mathrm{~T}_{2}{ }^{2}\)
For planet Marutham \(\quad \mathrm{T}_{3}=4, \quad \mathrm{a}_{3}=32=2 \times 4^{2} \Rightarrow 2 \mathrm{~T}_{3}{ }^{2}\)
For planet Neithal \(\quad \mathrm{T}_{4}=5, \quad \mathrm{a}_{4}=50=2 \times 5^{2} \Rightarrow 2 \mathrm{~T}_{4}{ }^{2}\)
For planet Paalai \(\quad \mathrm{T}_{5}=6, \quad \mathrm{a}_{5}=72=2 \times 6^{2} \Rightarrow 2 \mathrm{~T}_{5}^{2}\)
\(\therefore \alpha \propto 2 \mathrm{~T}^{2}\)

Initial momentum of the object P_i = 10 \(\times\)15 = 150 kgm s^-1
Final momentum of the object p_f = 0
Δp = 150 - 0 = 150 kg ms^-1
(a) Impulse J = Δp = 150 N s and Average force \({ F }_{ avg }=\frac { \triangle p }{ \triangle t } =\frac { 150 }{ 0.03 } \)= 5000N
b) Impulse J = Δp = 150 N s and Average force F_avg =\(\frac { 150 }{ 10 } \)= 15N

Circular wheels roll on the road and rolling friction comes into play during the motion of automobile.
Rolling friction is less than the sliding friction. It is due to this reason that wheels of automobiles are made circular.

The statement is wrong. The earth revolves around the sun under the force of attraction of the sun. This force (centripetal) is always perpendicular to the motion of the earth. Therefore, \(\theta\) = 90° and W = FS cos 90° = 0.Hence, sun does to work on the earth.

Yes, there is ail internal energy in a body due to the thermal agitation of the particles of the body, while the vector sum of the momenta of the moving particles may be zero.