\(\mathrm{F}_{\mathrm{d}} \propto \mathrm{v} \)
\(\mathrm{F}_{\mathrm{d}}=-\mathrm{bv} ; \quad \mathrm{F}_{\mathrm{d}}=\mathrm{kv} \)
\(\therefore \mathrm{k}=\frac{F_{d}}{v} \)
\(\text { Units of } k=\frac{k g m s^{-2}}{m s^{-1}}\)
\(\mathrm{k}=\mathrm{kg} \mathrm{s}^{-1}\)

\(\mathrm{F} =-\mathrm{kx} \)
\(\therefore \mathrm{x} =\left|-\frac{F}{k^{\prime}}\right|=\frac{F}{k^{\prime}} \)
\(\mathrm{F} =\mathrm{Mg} \text { and } k^{\prime}=\frac{k}{2} \)
\(\therefore \mathrm{x} =\frac{M g}{\frac{k}{2}} \)
\(=\frac{2 M g}{k} \)

\(\mathrm{T}= \mathrm{T}_{1}=2 \pi \sqrt{\frac{l}{g}} \)
\(T_{1}^{2}=4 \pi^{2} \frac{l}{g} \)
\(\mathrm{y}=\mathrm{kt}^{2} \quad \mathrm{k}=1 \mathrm{~m} \mathrm{~s}^{-2} \)
\(\therefore \mathrm{y}=\mathrm{t}^{2} \quad \mathrm{y} \propto \mathrm{t}^{2} \)
\(\frac{y_{1}}{y_{2}}=\frac{t_{1}^{2}}{t_{2}^{2}} \)
\(\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{y_{1}}{y_{2}}=\frac{6}{5} \)

The sketch between cause (magnitude of acceleration) and effect (magnitude of displacement) is a straight line.

\(C_{p}-C_{v}=R\)
For diatomic gas (N₂) No of degrees of freedom = 5
\(\therefore S_{p}-S_{v}=R / 28\)

\(\gamma=\frac{27}{17}\)
Number of moles of helium
\(\mathrm{n}=\frac{8}{4}=2\)
Number of moles of oxygen
\(n^{\prime}=\frac{16}{32}=\frac{1}{2}\)
For mono atomic Helium gas
\(\mathrm{f} =3 \)
\(\mathrm{C}_{\mathrm{V}} =\frac{f}{2} R \)
\(=\frac{3}{2} R \)
For diatomic oxygen gas
f = 5
\(\mathrm{C}_{\mathrm{V}} =\frac{f}{2} R \)
\(=\frac{5}{2} R \)
\(\mathrm{C}_{\mathrm{V}} \text { mixture } =\frac{n c_{v}+n^{\prime} C_{v}^{\prime}}{n+n^{\prime}} \)
\(=\frac{2 \times \frac{3}{2} R+\frac{1}{2} \times \frac{5}{2} R}{2+\frac{1}{2}} \)
\(C_{V} =\frac{3 R+\frac{5}{4} R}{\frac{5}{2}} \)
\(=\frac{17 R}{10} \)
\(\gamma =\frac{C_{p}}{C_{v}} \)
\(=1+\frac{R}{C_{V}} \)
\(=1+\frac{R}{\frac{17 R} {10}}\)
\(=1+\frac{10}{17} \)
\(=\frac{27}{17} \)

\(\mathrm{T}_{2} =0^{\circ} \mathrm{C}+273=273 \mathrm{~K} \)
\(\mathrm{~T}_{1}=100^{\circ} \mathrm{C}=100+273=373 \mathrm{~K} \)
\(\eta =1-\frac{T_{2}}{T_{1}} \)
\(=1-\frac{273}{373} \)
\(=\frac{373-273}{373} \)
\(=\frac{100}{373}=0.26809 \times 100 \)
\(=26.809 \% \)

\(\Delta l \text { verses } \mathrm{E}\)
\(\text { Strain } \propto \text { Stress }\)
Hooke's law is followed

\(\frac{F}{A}=Y \frac{\Delta l}{l} \quad F=Y \frac{\Delta l}{l} \times A\)
\(A_{1}=\mathrm{A} \quad \mathrm{A}_{2}=2 \mathrm{~A}\)
\(\therefore F \propto \Delta l \times A\)
\(\mathrm{F} \propto \Delta l_{1} A_{1}\)
\(\mathrm{F}_{1} \propto 2 \Delta l_{2} \times A_{2}\)
\(\frac{F}{F_{1}}=\frac{\Delta l A}{2 \Delta l \times 2 A}\)
\(\therefore F_{1}=4 \mathrm{~F}\)

Rate of heat production
\(=\frac{\text { Work done }}{\text { Timetaken }}\)
Terminal velocity
\(v=\frac{2}{a} \frac{r^{2}(f-\sigma)}{\eta} g \)
\(v \propto r^{2} \)
\(\text { Work } \propto \text { Force } \times \text { distance }\)
\(\propto r^{2} \times r \Rightarrow 2^{5}\)

Moment of inertia of a disc passing through
\(\text { centre } I=\frac{1}{2} M R^{2}\)
\(\text { Energy of the first disc }=\frac{1}{4} I \omega_{1}^{2}\)
\(\text { Energy of the second disc }=\frac{1}{4} I \omega_{2}^{2}\)
\(\text { Loss of energy }=\frac{1}{4} I\left(\omega_{1}-\omega_{2}\right)^{2}\)

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

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 \(a_{2}=g \sin \theta\)
\(\text { Required ratio } \frac{a_{1}}{a_{2}}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}} / g \sin \theta\)
\(\frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{k^{2}}{r^{2}}}\)
\(\text { For a solid sphere } \frac{k^{2}}{r^{2}}=\frac{2}{5}\)
\(\therefore \text { Ratio of accelerations } \frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{2}{5}}\)
\(=\frac{1}{5+\frac{2}{5}}=\frac{1}{\frac{7}{5}}=\frac{5}{7}\)
\(\therefore a_{1}: a_{2}=5: 7 \)

Moment of inertia of a disc
\(\mathrm{I}_{1}=\frac{M R^{2}}{2}\)
\(\text { Mass of small disc }=\frac{M}{\pi R^{2}} \times \pi \times\left(\frac{R}{2}\right)^{2}\)
\(=\frac{M}{\pi R^{2}} \times \frac{\pi R^{2}}{4}=\frac{M}{4}\)
By the theorem of parallel axis, the moment of inertia of the small disc. About an axis passing through 0 is
\(I_{2} =\frac{1}{2} \times \frac{M}{4}\left(\frac{R}{2}\right)^{2}+\frac{M}{4}\left(\frac{R}{2}\right)^{2} \)
\(=\frac{M}{8} \times \frac{R^{2}}{4}+\frac{M}{4} \times \frac{R^{2}}{4} \)
\(=\frac{M R^{2}}{32}+\frac{M R^{2}}{16}=\frac{M R^{2}+2 M R^{2}}{32} \)
\(I_{2} =\frac{3 M R^{2}}{32} \)
Moment of inertia of the remaining part is I= I₁ - I₂
\(=\frac{M R^{2}}{2}-\frac{3 M R^{2}}{32} \)
\(=\frac{16 M R^{2}-3 M R^{2}}{32}=\frac{13 M R^{2}}{32}\)
\(I =\frac{13 M R^{2}}{32} \)

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
\(k_{1}-k_{2} =\frac{1}{2} I_{a} \omega^{2}-\frac{1}{2}\left[\frac{I a^{2} \omega_{1}^{2}}{I_{a}+I_{2 b}}\right] \)
\(=\frac{1}{2} \frac{\left[I_{1}\left(I_{a}+I_{b}\right) \omega^{2}-I_{a}^{2} \omega^{2}\right]}{I_{a}+I_{b}} \)
\(k_{1}-k_{2} =\frac{1}{2} \frac{I_{a} b}{\left(I_{a}+I_{b}\right)} \omega^{2} \)

\(F=-\frac{d u}{d x} \quad F(x) =k x+a x^{3} \)
\(d u =-F d x \)
\(u(x) =-\int_{0}^{x}\left(-k x+a x^{3}\right) d x \)
\(=\int_{0}^{x} k x d x-a \int_{0}^{x} x^{3} d x \)
\(=\frac{k x^{2}}{2}-\frac{a x^{4}}{2} \)
\(U(x) =\frac{x^{2}}{2}\left(k-\frac{a x^{2}}{2}\right) \)
\(u(x)=0 \text { at } x=0 \text { and }\)
\(U(x) =0 ; k-\frac{a x^{2}}{2}=0 \)
\(=\frac{a}{2} x^{2}=-k \)
\(x^{2} =\frac{2 k}{a} \)
\(\therefore x =\sqrt{\frac{2 k}{a}} \)
\(\text { Clearly } u(x)=0 \text { at } x=0 \text { and }\)
\(x=\sqrt{\frac{2 k}{a}}\)
\(\text { For } x>\sqrt{\frac{2 k}{a}} U(x) \text { will be negative. } \)
\(\text { At } x=0 ; F=\frac{-d u}{d x}=0\)
(i.e.,) Slope of V - x graph is zero at x = 0
Hence the most appropriate answer is d.

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

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

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 = \(2 \times \frac{1}{2} m v^{2} +\frac{1}{2} \times 2 m \times\left(\frac{v^{2}}{\sqrt{2}}\right)^{2} \)
\(=m v^{2}+m \times \frac{v^{2}}{2} \)
\(=\frac{3}{2} m v^{2} \)

Mass = 1kg
Velocity v = 20m/s
Height h = 18 m
\(g=10 \mathrm{~m} / \mathrm{s}^{2}\)
\(\text { Potential energy } P . E=m g h\)
\(=1 \times 10 \times 18 =180 \mathrm{~J} \)
\(\text { Kinetic energy } K . E=\frac{1}{2} m v^{2}\)
\(K . E=\frac{1}{2} \times 1(20)^{2}\)
\(=\frac{1}{2} \times 20 \times 20=200 J\)
\(\text { Loss of energy due to air friction }= K.E - P.E\)
\(=200-180=20 \mathrm{~J}\)

\(m_{1} =1, \quad m_{2}=2 \)
\(K . E . =m g(h-x) \)
\(\text { For both balls }(h-x)\)
\(=40 \text { i.e.) Same }\)
\(g=\text { constant }\)
\(\therefore K \cdot E_{1}=m_{1} g(h-x)=m_{1} g \times 40 \)
\(K \cdot E_{2}=m_{2} g(h-x)=m_{2} g \times 40 \)
\(\therefore \frac{K \cdot E_{1}}{K \cdot E_{2}}=\frac{m_{1} g \times 40}{m_{2} g \times 40}=\frac{m_{1}}{m_{2}} \)
\( \therefore K \cdot E_{1}: K \cdot E_{2}=1: 2 \)

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

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.

Initially velocity has maximum value and at maximum height velocity becomes zero. After that the velocity becomes negative

For freely falling body, velocity while the body, hit the ground \(v=\sqrt{2 g h}\)
\(v_{1}=\sqrt{2 g h_{1}} \text { and } v_{2}=\sqrt{2 g h_{2}} \)
\(\therefore \frac{m_{1} v_{1}}{m_{2} v_{2}}=\frac{m_{1} \sqrt{h_{1}}}{m_{2} \sqrt{h_{2}}}=\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)

Dimension of Planck's constant is \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\)
Dimension of Gravitational constant is \(\left[\mathrm{M}^{-1} \mathrm{~L}^{+3} \mathrm{~T}^{-1}\right]\)
Dimension of Velocity constant is LT^-1
Dimension of Length is L
\(\therefore \text { Dimension of } \frac{\sqrt{h G}}{C^{\frac{3}{2}}}\)
\(=\frac{\sqrt{\left(\mathrm{ML}^{2} \mathrm{~T}^{-1}\right)\left(\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right)}}{\left(\mathrm{LT}^{-1}\right)^{3 / 2}} \)
\(=\frac{\sqrt{\mathrm{L}^{5} \mathrm{~T}^{-3}}}{\mathrm{~L}^{3 / 2} \mathrm{~T}^{-3 / 2}} \)
\(=\frac{\mathrm{L}^{5 / 2} \mathrm{~T}^{-3 / 2}}{\mathrm{~L}^{3 / 2} \mathrm{~T}^{-3 / 2}} \)
\(=\mathrm{L}^{5 / 2-3 / 2} \mathrm{~T}^{3 / 2+3 / 2}=\mathrm{L}^{1} \mathrm{~T}^{0}=\mathrm{L}\)
Dimension of length = L

\(\text { Velocity of light } c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\)
\(c=\left(\mu_{0} \varepsilon_{0}\right)^{-\frac{1}{2}}\)
\(\text { Hence dimension }\left(\mu_{0} \varepsilon_{0}\right)^{-\frac{1}{2}} \text { is that of velocity. }\)

F = kv²
Dimensional of k
\(=\frac{\text { Dimension of } \mathrm{F}}{\text { Dimension of }(v)^{2}}\)
\(=\frac{\mathrm{MLT}^{-2}}{\left(\mathrm{LT}^{-1}\right)^{2}}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^{2} \mathrm{~T}^{-2}} \)
\(=\left[\mathrm{ML}^{-1-2} \mathrm{~T}^{-2+2}\right] \)
Dimension of proportionality constant \(=\left[\mathrm{ML}^{-1} \mathrm{~T}^{0}\right]\)

\(\text { Gravitational constant } G=\frac{F r^{2}}{m_{l} m_{2}}\)
\(\text { Dimensional formula of } \mathrm{G}=\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}{[\mathrm{M}][\mathrm{M}]}\)
\(=\frac{\mathrm{ML}^{3} \mathrm{~T}^{-2}}{\mathrm{M}^{2}}=\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\)

Dimensional formula of Planck's
\(\text { constant }=\frac{\text { Energy }}{\text { Frequency }}=\frac{\mathrm{ML}^{2} \mathrm{~T}^{-2}}{\mathrm{~T}^{-1}}\)
\(=M L^{2} T^{-2+1}=M L^{2} T^{-1}\)

\(T =2 \pi \sqrt{\frac{l}{g}} \)
\(g =4 \pi^{2} l / T^{2} \)
\(\frac{d g}{g} =\frac{d l}{l}-\frac{2 d T}{T} \)
\(\frac{d g}{g} \% =\left(\frac{d l}{l}\right) \%-2\left(\frac{d T}{T}\right) \% \)
\(=1 \%-2 \times(-3 \%) \)
\(=1+6=7 \% \)