Escape speed v_e = \(\sqrt 2g R\)
if g' = 4g 
then \(v'_e\) = \(\sqrt (4g) R\)
= \(\sqrt 2g R \) \(\times\)2
= 2v_e

\(\text { Escape speed } v_{e}=\sqrt{2 g R}\)
\(\text { If } g^{\prime}=4 \mathrm{~g}\)
\(\text { Then } v_{e}^{\prime}=\sqrt{2(4 g) R}\)
\(=\sqrt{2 g R} \times 2 \)
\(=2 v_{e} \)

g = \(\frac{GM_e}{R^2_e}\)
M_e = 2 M_e      R_e= 2 R_e then,
g' = \(\frac{G \times 2M_e}{(2R_e)^2}\) = \(2 \frac{Gm_e}{4R_e^2}\)
\(2 \frac{Gm_e}{4R_e^2}\) = \(\cfrac g2\)

T² = C (R + h)³
\(\therefore T \alpha (R_E)^\frac{3}{2}\)
R_E = 2R_E
\(\therefore T \alpha (2R_E)^\frac{3}{2}\)
Time period increases by \(2^\frac{2}{3}\)= 2\(\sqrt 2\) 
No. of days in a year = (365.4) \(\times\)2\(\sqrt 2\)
= 1032

\(\text { Gravitational force } F \propto m_{1} m_{2}\)
\(\text { If } m_{1}=2 m_{1} \text { and } m_{2}=2 m_{2} \text { then }\)
\(\text { Force } F \propto\left(2 m_{1}\right)\left(2 m_{2}\right)\)
\(\propto 4 m_{1} m_{2}\)

Mass per unit length = m
Velocity = v
(mass of water pumped in one second)
M = mv
\(\text { Kinetic energy }=\frac{1}{2} M v^{2}\)
\(=\frac{1}{2} m v^{3}\)

\(v_{0}=r \omega ; \quad \therefore v_{0} \alpha r\)
For a wheel (uniform ring) the distance of a point on the rim in level with the center 
\(\text { [i.e., radius] is } \sqrt{2} r\)
\(\therefore \text { The speed of the center is } \sqrt{2} v_{0}\)

Potential energy = Translational kinetic energy + Rotational kinetic energy
\(m g h=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2} \)
\(=\frac{1}{2} m v^{2}+\frac{1}{2} \times \frac{2}{5} M R^{2} \times \frac{v^{2}}{R^{2}}\left[\omega=\frac{v}{R}\right] \)
\(=\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2} \)
\(=\frac{5 m v^{2}+2 m v^{2}}{10}=\frac{7 m v^{2}}{10} \)
\(m g h=\frac{7 m v^{2}}{10} \)
\(g h=\frac{7 v^{2}}{10} \)
\(\therefore v^{2}=\frac{10 g h}{7} \)
\(\therefore v=\frac{\sqrt{10 g h}}{7} \)

Acceleration of the solid sphere while rolling down without slipping
\(a_{1}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}}\)
Acceleration developed while slipping down \(a_{2}=g \sin \theta\)
\(\text { Required ratio } \frac{a_{1}}{a_{2}}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}} / g \sin \theta\)
\(\frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{k^{2}}{r^{2}}}\)
\(\text { For a solid sphere } \frac{k^{2}}{r^{2}}=\frac{2}{5}\)
\(\therefore \text { Ratio of accelerations } \frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{2}{5}}\)
\(=\frac{1}{5+\frac{2}{5}}=\frac{1}{\frac{7}{5}}=\frac{5}{7}\)
\(\therefore a_{1}: a_{2}=5: 7 \)

The moments of inertia of two discs are I_a and I_b respectively The angular velocity of the disc A is \(\omega\).
The sum of kinetic energies of two discs before coming in contact is \(k_{1}=\frac{1}{2} I_{a} \omega_{1}^{2}+\frac{1}{2} I_{b} \omega_{2}^{2}\)
\(\text { But angular velocity of the disc be is } \omega_{2}=0 \ \text {(rest)}\)
\(\therefore k_{1}=\frac{1}{2} I_{a} \omega_{1}^{2}\)
The final kinetic energy of the two discs system \(k_{2}=\frac{1}{2} \frac{I_{a}^{2} \omega_{1}^{2}}{I_{a}+I_{b}}\)
The loss of kinetic energy is
\(k_{1}-k_{2} =\frac{1}{2} I_{a} \omega^{2}-\frac{1}{2}\left[\frac{I a^{2} \omega_{1}^{2}}{I_{a}+I_{2 b}}\right] \)
\(=\frac{1}{2} \frac{\left[I_{1}\left(I_{a}+I_{b}\right) \omega^{2}-I_{a}^{2} \omega^{2}\right]}{I_{a}+I_{b}} \)
\(k_{1}-k_{2} =\frac{1}{2} \frac{I_{a} b}{\left(I_{a}+I_{b}\right)} \omega^{2} \)

For any spring kl = constant
Length of the longer piece
\(=\frac{2 l}{3} \)
\(\therefore k^{1} \times \frac{2 l}{3} =k l \)
\(\therefore k^{1}=\frac{k l \times 3}{2 l}=\frac{3}{2} k \)
\(\therefore k^{1}=\frac{3}{2} k \)

For a conservative force
\(F=-\frac{d v}{d t} \)
\(\int_{0}^{u(x)} d v=-\int_{0}^{x} F d x=-\int_{0}^{x} k x d x \)
\(\text { As } v(0)=0\)
\(U(x)=-\frac{k x^{2}}{2}\)
Thus, the graph of U(x) versus (x) will be a parabola, symmetric about U - ax is bying below x - ax is with its vertex at the origin. Hence the correct answer is C

\(\text {Potential energy } P . E=\alpha-\frac{\beta}{2} x^{2}\)
P.E = Work = Fx
\(P=\alpha-\frac{\beta}{2} x^{2}\)
\(\text {Force }=\frac{d p}{d x}=\frac{d}{d x}\left(\alpha-\frac{\beta}{2} x^{2}\right)\)
\(=0-\frac{\beta}{2} \times 2 x =-\beta x \)

\(\text { Kinetic energy } E_{k}=\frac{p^{2}}{2 m}\)
\(\frac{\Delta E_{k}}{E_{k}}=\frac{2 \Delta p}{p}\)
\(\text {Given that } \frac{\Delta p}{p}=0.1\)
∴ Increase in kinetic energy
\(\frac{\Delta E_{k}}{E_{k}}=2 \frac{\Delta p}{p} \)
\(\frac{\Delta E_{k}}{E_{k}}=2 \times 0.1=0.2 \% \)

\(K \cdot E=\frac{1}{2} I \omega^{2} ; \quad L=I \omega ; \quad K \cdot E=\frac{2^{2}}{2^{2}} \)
\(\therefore K \cdot E \alpha L^{2} \quad E_{1} \alpha L_{1}^{2} \quad E_{2} \alpha L_{2}^{2}\)
\(\frac{E_{1}}{E_{2}}=\left(\frac{L_{1}}{L_{2}}\right)^{2} \)
\(\text { Here } E_{1}=E \quad E_{2}=\frac{E}{2} \)
\(L_{1}=L \quad \quad L_{2}=? \)
\(\frac{E}{\frac{E}{2}}=\left(\frac{L}{L_{2}}\right)^{2} \)
\(\frac{2 E}{E}=\left(\frac{L}{L_{2}}\right)^{2}\left(\frac{L_{1}}{L_{2}}\right)^{2}=2 \)
\(\therefore \frac{L}{L_{2}}=\sqrt{2} \)
\(L_{2}=\frac{L}{\sqrt{2}} \)

\(m=3 \mathrm{~kg} \quad r=40 \times 10^{-2} \mathrm{~m}=0.4 \mathrm{~m}\)
\(\text { Force }=30 N\)
Moment of inertia of a hollow-cylinder I= MR²
\(=3 \times\left(40 \times 10^{-2}\right)^{2} \)
\(=3 \times 0.4 \times 0.4=0.48 \mathrm{kgm}^{2} \)
\( F R =I \alpha \)
\(30 \times 40 \times 10^{-2}=0.48 d \)
\(d=\frac{12}{0.48}=\frac{1200}{48}=25 \mathrm{rad} \mathrm{s}^{-2} \)
\(\alpha=25 \mathrm{rad} \mathrm{s}^{-2} \)

Using law of conservation of momentum,
\(2 m v =\sqrt{m^{2} v^{2}+m^{2} v^{2}} \)
\(=\sqrt{2 m^{2} v^{2}} \)
\(v =\frac{\sqrt{2} m v}{2 m}=\frac{v}{\sqrt{2}} \)
Energy released in explosion = \(2 \times \frac{1}{2} m v^{2} +\frac{1}{2} \times 2 m \times\left(\frac{v^{2}}{\sqrt{2}}\right)^{2} \)
\(=m v^{2}+m \times \frac{v^{2}}{2} \)
\(=\frac{3}{2} m v^{2} \)

\(\text { Force } \overrightarrow{\mathbf{F}}=(2 i+\vec{j}) N\)
\(\text { Displacement } d=(5 \vec{i}+3 \vec{j})-(3 \vec{j}+\vec{k})\)
\(=(5 i-k) m\)
\(\text { Work done } W=F . d\)
\(=(2 \vec{i}+\vec{j})(5 i-k)\)
\(=10-0-0=10 J \)

\(\text {Time of flight }=\frac{2 u}{g}\)

It is a uniform circular motion. So the direction of velocity changes but not the magnitude. Therefore speed in considered constant. Again magnitude of acceleration does not change.

Distance travelled \(s=\frac{1}{2} g t^{2} ; s \alpha t^{2}\). The ratio of distances travelled by a freely falling body will be with ratio 1:4:9:...

\(\vec{v}=2 \hat{l}+t^{2} \hat{j}-9 \vec{k}\)
\(\vec{a}=\frac{d \vec{v}}{d t}=2 t \hat{j}\)
\(\text { When } t=0.5 \mathrm{~s}\)
\(a=1 \mathrm{~ms}^{-2}\)

Unit vector specifies only direction
\(\hat{A}=\frac{\vec{A}}{|\vec{A}|} \)
\(\therefore \hat{i}+\hat{j}=\frac{\hat{l}+\hat{j}}{|\hat{l}+\hat{j}|} \)
\(\hat{i} \text { and } \hat{j} \text { are orthogonal components of vectors. }\)
\(\therefore|\hat{i}+\hat{j}|=\sqrt{1^{2}+1^{2}}=\sqrt{1+1}=\sqrt{2}\)

\(\text {Dimension of permittivity } \varepsilon=A^{2} C^{2} N^{-1} m^{-2}\)
\(\text {Dimension of Boltzmann constant } k_{B}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]\)
\(\text {Dimension of Absolute Temperature }=T\)
No. of charged particles per unit
Volume = No. of dimensions = a
Dimension of charge = AT
Dimension of length = L
\(\text { Dimension of }=\frac{\sqrt{\varepsilon k_{B} T}}{n q^{2}}\)
\(=\frac{\sqrt{\left[\mathrm{A}^{2} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right]}\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right][\mathrm{T}]}{[\mathrm{AT}]^{2}}\)
\(=\sqrt{L^{2}}=L\)
\(\therefore \text { Dimension of length }=\mathrm{L}\)

\(\pi=3.14 \)
\(\pi^{2} =3.14 \times 3.14 \)
\(=9.8596=9.86 \)

\(\Delta l =0.01 \)
\(l =3.51 \)
\(\% \text { error } =\frac{\Delta l}{l} \times 100=\frac{0.01}{3.51} \times 100 \)
\(=0.00284 \times 100 \)
\(=0.284=0.28 \% \)