Consider an object of mass m at a height h from the surface of the Earth. Acceleration experienced by the object due to Earth is
g' = \(\frac { GM }{ ({ R }_{ e }+h)^{ 2 } } \)       ..........(1)
g'= \(\frac { GM }{ { R }_{ e }^{ 2 }\left( 1+\frac { h }{ { R }_{ e } } \right) ^{ 2 } } \)
g'=\(\frac { GM }{ { R }_{ e }^{ 2 } } \left( 1+\frac { h }{ { R }_{ e } } \right) ^{ 2 }\)
If h << R_e
We can use Binomial expansion. Taking the terms up to first order
g'= \(\frac { GM }{ { R }_{ e }^{ 2 } } \left( 1-2\frac { h }{ { R }_{ e } } \right) \)
g' = \(g\left( 1-2\frac { h }{ { R }_{ e } } \right) \)  .....(2)
We find that g'< g. This means that as altitude h increases, the acceleration due to gravity g decreases.

\(\text {Centrifugal force experience by person } =\frac{m v^{2}}{r}=\frac{60 \mathrm{~kg} \times(50 \mathrm{~m} / \mathrm{s})^{2}}{10 \mathrm{~m}}  \)
\(=\frac{60 \times 2500}{10} \mathrm{~N}  \)
\(=15000 \mathrm{~N}\)

\(\overrightarrow { r } =3\hat { i } +2\hat { j } \)
\(\text { Unit vector } =\frac{\vec{r}}{|\vec{r}|}  \)
\(|\vec{r}| =\sqrt{3^{2}+2^{2}}=\sqrt{13}  \)
\(\therefore \text { Unit vector } =\frac{3 \hat{i}+2 \hat{j}}{\sqrt{13}}\)

F ∝ m^a v^b r^c ;
F = k m^a v^b r^c
where k is a dimensionless constant of proportionality. Rewriting above equation in terms of dimensions and taking k= 1, we have
[MLT^-2] = [M]^a [LT^-1]^b [L]^c
= [M^aL^bT^-bL^c]
[MLT^-2] = [M^aL^b+cT^-b]
Comparing the powers of M, L and T on both sides
 a = 1 ; b + c = 1; -b =-2
2 + c = 1 ; b = 2;
a = 1 b = 2 and c = -1
From the above equation we get
F = m^av^br^c
F =m¹v²r^-1
or F = \(\frac { m{ v }^{ 2 } }{ r } \)

As shown in the figure (ii), let R be the midpoint of arc PQ. Then \(\angle POR=45^0\) Magnitude of acceleration at R,

\(a=\frac{v^2}{r}=\frac{(2)^2}{4}=1\ cms^{-2}\)
The acceleration a acts along RO
Magnitude of component of along x-axis, ax = a cos45°
\(=1\times\frac{1}{\sqrt 2}=\frac{1}{\sqrt 2}\ cms^{-2}\)
\(\therefore\ \hat a_x=-\frac{1}{\sqrt 2}\hat x\)
Magnitude of component of a along y-axis, ay = 1\(\times\)sin45°
\(=\frac{1}{\sqrt 2}\ cms^{-2}\)
\(\therefore\hat a_y=-\frac{1}{\sqrt 2}\hat y\)
Hence \(\hat a=\hat a_x+\hat a_y=\frac{-1}{\sqrt 2}(\hat x+\hat y)\)

(i) If the object is at rest and no external force is applied on the object, the static friction acting on the object is zero (f_s = 0).
(ii) If the object is at rest, and there is an external force applied parallel to the surface, then the force of static friction acting on the object is exactly equal to the external force applied on the object (f_s = F_ext). But still the static friction Is is less than μ_sN.
(iii) When object begins to slide, the static friction (f_s) acting on the object attains maximum.

From law of conservation of momentum, along x-axis,
\(mu_1+0=mv_1\cos\theta_1+mv_2\cos\theta_2\)
\(u_1=v_1\cos\theta_1+v_2\cos\theta_2\)   ....(1)
along y-axis,  \(0=v_1\sin\theta_1-v_2\sin\theta_2\)   ....(2)
From energy of conservation,
\({1\over2}{mu}_{1}^{2}={1\over2}{mv}_{1}^{2}+{1\over 2}{mv}_{2}^{2}\)
\({u}_{1}^{2}={v}_{1}^{2}+{v}_{2}^{2}\)   ...(3)
By using equation (1) and (2) in equation (3) we get,
\(2v_1 v_2\cos(\theta_1+\theta_2)=0\)
\(\cos(\theta_1+\theta_2)=\cos{\pi \over 2}\)
\(\theta_1+\theta_2={\pi \over 2}\)
\(\therefore\) This shows that the particle move perpendicular to each other after collision.   

Mass of particle P m₁ = 1 kg.
Mass of particle Q m₂ = 3 kg
Velocity of particle P = velocity of particle Q but in opposite direction
Velocity of center of mass v_cm = ?
\({ vc }_{ m }=\frac { { m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
= 0.

Kinetic energy of rotation = \(\frac { 1 }{ 2 } I{ \omega }^{ 2 }\)
=\(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } { mr }^{ 3 } \right) { \omega }^{ 2 }\)
=\(\frac { 1 }{ 4 } \times 20\times (0.25)^{ 2 }\times 100\times 100\)
=
= 3125 J

Mass of the bullet (m) = 25 g
Initial velocity (m) = 400 ms^-1
Final velocity (v) = 300 ms^-1
Work done (w) =?
From work-energy theorem,
Work done = loss in kinetic energy,
\(W=\frac{1}{2}m(u^{2}-v^{2})\)
=\(\frac{1}{2}\times\frac{25}{100}kg\times(400^{2}-300^{2})\)
=\(\frac{1}{2}\times0.025\times(16\times10^{4}-9\times10^{4})\)
=\(\frac{1}{2}\times0.025\times(7\times10^{4})\)
= 0.0125\(\times(7\times10^{4})\)
Work done (W) = 875J

Errors in the difference of two quantities.
Let \(\triangle A\) and \(\triangle B\) be the absolute errors in the two quantities, A and B, respectively. Then,
Measured value of  \(A=A\pm\triangle A\)
Measured value of \(B=B\pm\triangle B\)
Consider the difference, Z =A - B
The error \(\triangle Z\) in Z is the given by
\(Z\pm \triangle Z=(A+\triangle A)-(B\pm \triangle B)\)
\(=(A-B)\pm(\triangle A+\triangle B)\)
\(=Z\pm(\triangle A+\triangle B)\)
(or) \(\triangle Z=\triangle A+\triangle B\) 
The maximum error in difference of two quantities is equal to the sum of the absolute errors in the individual quantities. Error in the division or quotient of two quantities
Let \(\triangle A\) and \(\triangle B\) be the absolute errors in the two quantities A and B respectively.
Consider the quotient, \(Z={{A}\over{B}}\)
The error \(\triangle Z\) in Z is given by
\(Z\pm Z={{A\pm \triangle A}\over{B+\triangle B}}={{A\left(1\pm{{{\triangle A}\over{A}}} \right)}\over{B\left( 1\pm{{\triangle B}\over{B}} \right)}}\)
\(={{A}\over{B}} \left( 1\pm{{\triangle A}\over{A}} \right)\left( 1\pm{{\triangle B}\over{B}} \right)^{-1}\)
or \(Z\pm \triangle Z=Z\left( 1\pm{{\triangle A}\over{A}} \right)\left( 1\mp{{\triangle B}\over{B}} \right)\)
[ using (1+x)n \(\approx\) 1 + nx, when x<<1]
Dividing both sides by Z, we get
\(1\pm{{\triangle Z}\over{Z}}=\left( 1\pm{{\triangle A}\over{A}} \right)\left( 1\mp {{\triangle B}\over{B}} \right)\)
\(=1\pm{{\triangle A}\over{A}}\mp{{\triangle B}\over{B}}\pm{{\triangle A}\over{A}}.{{\triangle B}\over{B}}\)
As the terms \(\triangle A/A\) and \(\triangle B/B\) are small, their product term can be neglected.
The maximum fractional error in Z is given by
\({{\triangle Z}\over{Z}}=\left( {{\triangle A}\over{A}} +{{\triangle B}\over{B}}\right)\)
The maximum fractional error in the quotient of two quantities is equal to the sum of their individual fractional errors.