Answers (a), (b) and (c) are all anticlockwiseand answer (d) alone is in the clockwise direction

Unit of a (Planck's constant) - Js
Unit of c (Velocity of light) - ms^-1
Unit of G (Gravitational Constant) - \(\frac{\mathrm{Nm}^{2}}{\mathrm{Kg}^{2}}\)
\(\therefore \text { Unit of } \frac{h c}{G} \text { is }=\frac{J s \times m s^{-1}}{N m^{2} / k g^{2}} \)
\(=\frac{N m s \times m s^{-1} \times k g^{2}}{N m^{2}}[J=N m] =\mathrm{kg}^{2}\)

Let the particle of mass m cross a horizontal distance x in time t.

Angular momentum \(\overrightarrow{L}=\int{\overrightarrow{\tau}}{dt}\)
But \(\overrightarrow{\tau}=\overrightarrow{r}\times \overrightarrow{F}\)
\(\overrightarrow{r}=x\hat { i } +y\hat { j} \) and \(\hat { F }=-mg \hat { j } \)
\(\therefore \overrightarrow { \tau }=(x\hat { i }+y\hat { j } ) \times (-mg \hat { j } )\)
\(\overrightarrow { \tau } =-mgx(\hat { i } \times \hat { j } )=-mgx\hat { k } \)
\(\overrightarrow { L } = -mg \int{(xdt)}\hat {k} =-gv\ cos\theta (\int t dt)\hat { k } \)
Let initial time t = 0 and final time t = t_f
\(\overrightarrow {L}=-mg \cos \theta \left( ^{ t }\int _{ 0 }^{ f }{ tdt } \right) \hat k = -\frac{1}{2}mgv\ cos \theta\ t^{2}_{f} \hat{k}\)
Negative sign indicates, \(\overrightarrow {L}\) point inwards. 

We use p = mv
For the mass of 10 g, m = 0.01 kg
p = 0.01\(\times\)10 = 0.1 kg m s^-1
For the mass of 1 kg
p = 1\(\times\)10 = 0.1 kg m s^-1
Thus even though both the masses have the same speed, the momentum of the heavier mass is. 100 times greater than that of the lighter mass.

Consider a particle moving from a point P₁ having position vector \(\vec { { r }_{ 1 } } ={ x }_{ 1 }\vec { i } +{ y }_{ 1 }\vec { j } +{ z }_{ 1 }\vec { k } \) to a point P₂ where its position vector is \(\vec { { r }_{ 2 } } ={ x }_{ 2 }\vec { i } +{ y }_{ 2 }\vec { j } +{ z }_{ 2 }\vec { k } \)
The displacement vector is given b y   \(\Delta\vec{r}=\vec{{r}_{2}}-\vec{{r}_{1}}\)
 = \((x_2-x_1)\vec{i}+(y_2-y_1)\vec{j}+(z_2-z_1)\vec{k}\)

(i) Two spring balances are attached.
(ii) One end is fixed with rigid support and the other free end pulled by the force.
(iii) Pull one end with same force and note the readings on both the balances.
(iv) Repeat it. We can notice same reading in both balances.

Unit of a physical quantity is defined as the established standard used for comparison of the given physical quantity. It is classified into fundamental and derived unit.

Least count of screw gauge
\(={{Pitch}\over{No. of\ divisions}}\)
\(={{1}\over{100}}=0.01\ mm\) (or) 0.001 cm
Least count of vernier calipers
\(=1MSD-1-1VSD=(1-19/20)MSD\)
\(={{1}\over{20}}\) = 0.05 cm.
So screw gauge is more precise than vernier.

It plays the same role in rotatory motion as the mass does in translatory motion.

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

Formula:
(i) Impulse = Area ABC
= \(\frac { 1 }{ 2 } \) \(\times\)18000\(\times\)(2.5 -1)
=1.35\(\times\)104 kg ms^-1
(ii) Force = \(\frac { impulsw }{ time } =\frac { 1.35\times { 10 }^{ 4 } }{ (2.5-1) } \) = 9000 N
(iii) Maximum force = 18000 N

1. Greater the mass concentrated away from the axis, greater the moment of inertia.
2. Moment of inertia of a body about an axis of rotation resists a change in it. In rotational motion, moment of inertia increases with increase in torque to change it's rotation.

(i) Consider a force \(\vec {(F)}\) acting on a body which moves displacement in some direction\(​​\vec{(dr)}\) as shown in figure.
work done = force\(\times\)displacement, W = Fs
(ii) When force and displacement are inclined in each other, the expression for work done (W) by the force on the body is mathematically written as, \(W=\vec{F}=\vec{dr}\)
Here, the product \(\vec{F}.\vec{dr}\) is a scalar product (or dot product). Thus, work done is a scalar quantity.
w=F dr cosθ
where, θ is the angle between applied force and the displacement of the body.
The work done by the force depends on the force (F), displacement (dr) and the angle (θ) between them.

(a) v_x - remains constant
(b) v_y - decreases and increases
(c) a - remains downward
(d) r - varies

Let us consider a rigid body rotating with angular velocity \(\omega\) about an axis as shown in figure. Every particle of the body will have the same angular velocity \(\omega\) and different tangential velocities v based on its positions from the axis of rotation.
Let us choose a particle of mass mi situated at distance r_i from the axis of rotation. It has a tangential velocity v_i given by the relation, v_i = r_i \(\omega\). The kinetic energy KE_i of the particle is,
KE_i =  \(\frac { 1 }{ 2 } { m }_{ i }{ v }_{ i }^{ 2 }\)
Writing the expression with the angular velocity,
\(KE=\frac { 1 }{ 2 } { m }_{ i }\left( { r }_{ i }\omega \right) ^{ 2 }=\frac { 1 }{ 2 } \left( { m }_{ i }{ r }_{ i }^{ 2 } \right) \omega ^{ 2 }\)

For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with summation as,
\(KE=\frac { 1 }{ 2 } \left( \sum { { m }_{ i }{ r }_{ i }^{ 2 } } \right) \omega ^{ 2 }\)
where, the term \(\sum { { m }_{ i }{ r }_{ i }^{ 2 } } \) is the moment of interiaI of the whole body. \(\sum { { m }_{ i }{ r }_{ i }^{ 2 } } \)
Hence, the expression for KE of the rigid body in rotational motion is,
KE = \(\frac{1}{2}\)I\(\omega^2\)
This is analogous to the expression for kinetic energy in translational motion.
KE = \(\frac{1}{2}\)Mv²
Relation between rotational kinetic energy and angular momentum
Let a rigid body of moment of inertia \(\omega\) rotate with angular velocity \(\omega\).
The angular momentum of a rigid body is, L = I\(\omega\)
The rotational kinetic energy of the rigid body is, KE = \(\frac{1}{2}I\omega^2\)
By multiplying the numerator and denominator of the above equation with I, we get a relation between Land KE as,
KE = \(\frac{1}{2}\)\(\frac { I^{ 2 }\omega ^{ 2 } }{ I } =\frac { 1 }{ 2 } \frac { \left( I\omega \right) ^{ 2 } }{ I } \)
\(KE=\frac { { L }^{ 2 } }{ 2I } \)

Angle \(\theta\) = 2' = 2\(\times\)60" = 120" = 120\(\times\)4.85\(\times\)10^-6 rad
\(\theta\) = 5.82\(\times\)10^-4 rad;
d = 2 x r
d = 2 x 6400 = 12800 x 10³ m
The distance of heavenly body
\(D=\frac{d}{\theta}=\frac{12800\times10^3}{5.82\times10^{-4}}\)
D = 2.19\(\times\)10¹⁰m.

As F \(=G{m_1m_2\over r^2}\)
\(G={F.r^2\over m_1m_2}\)
\([G]={{MLT}^{-2}.L^2\over MM}={M}^{-1}L^3{T}^{-2}\)
\(\therefore\) a = -1, b ~ 3, c = -2

\(\therefore\) \(n_2=n_1{\left[ {M_1 \over M_2} \right]}^{a}{\left[ {L_1 \over L_2} \right]}^{b}{\left[ {T_1 \over T_2} \right]}^{c}\)
\(=6.67\times{10}^{-8}{\left[ {{1\over 1000}} \right]}^{-1}{\left[ {{1\over 100}} \right]}^{3}\left[ {1\over 1} \right]^{-2}=6.67\times{10}^{}-11\)
Hence in SI units,   G = 6.67\(\times\)10^-11Nm² kg^-2

Galileo's experiment with. the second plane (a) at same inclination angle Cisthe first (b) with increased smoothness (c) with reduced angle of inclination (d) with zero angle of inclination
When a ball rolls from the top of an inclined plane to its bottom, after reaching the ground it moves some distance and continues to move on to another inclined plane of same angle of inclination as shown in the Figure (a). By increasing the smoothness of both the inclined planes, the ball reach almost the same height (h) from where it was released (L1) in the second plane (L2) [figure (b)]. The motion of the ball is then observed by varying the angle of inclination of the second plane keeping the same smoothness. If the angle of inclination is reduced, the ball travels longer distance in the second plane to reach the same height [figure (c)). When the angle of inclination is made zero, the ball moves forever in the horizontal direction [figure (d)]. If the Aristotelian idea were true, the ball would not have moved in the second plane even if its smoothness is made maximum since no force acted on it in the horizontal direction.

Work done by a constant force:
(i) When a constant force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation,
dW= (F cos\(\theta\) ) dr
(ii) The total work done in producing a displacement from initial position r_i to final position r_f is,
\(W=\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } dw\)
\(W=\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } \left( F\cos { \theta } \right) dr=\left( F\cos { \theta } \right) \)\(\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } dr=\left( F\cos { \theta } \right) \left( { r }_{ f }-{ r }_{ i } \right) \)
(iii) The graphical representation of the work done by a constant force is shown in Figure. The area under the graph shows the work done by the constant force.
              
Work done by a variable force:
(i) When the component of a variable force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation
dw = (F cos\(\theta\)) dr
[F cos\(\theta\) is the component of the variable force F]
where, F and \(\theta\) are variables. The total work done for a displacement from initial position r_i to final position r_f is given by the relation, 
\(W=\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } dw=\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } F\cos { \theta } dr\)
(ii) A graphical representation of the work done by a variable force is shown in Figure. The area under the graph is the work done by the variable force.