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

Velocity of the Earth around Sun
y = 30 km/s
Kinetic energy \( \mathrm{K} . \mathrm{E}=\frac{1}{2} \frac{G M_{E} M_{\mathrm{s}}}{\left(\mathrm{R}_{E}+\mathrm{h}\right)}\)
Kinetic Energy \( \mathrm{K} . \mathrm{E} =\frac{1}{2} M_{s} V^{2}=\frac{1}{2} \times 1.9 \times 10^{30} \times\left(30 \times 10^{3}\right)^{2} \)
\(=\frac{1.9 \times 900 \times 10^{30+6}}{2} \)
\(\mathrm{~K} . \mathrm{E} =\mathrm{U}_{\mathrm{K}}=26.5 \times 10^{32} \mathrm{~J} \)
Total energy \(\mathrm{E} =-\frac{G M_{s} M_{E}}{2\left(\mathrm{R}_{E}\right)}  \)
\(\mathrm{E}_{\text {tot }} =-\frac{-6.67 \times 10^{-11} \times 1.9 \times 10^{36} \times 5.9 \times 10^{24}}{2 \times 6.4 \times 10^{6}}\)
\(\mathrm{E}_{\mathrm{tot}}=-23.29 \times 10^{32} \mathrm{~J}\)
Total energy has the negative sign. The negative sign in the total energy implies that the satellite is bound to the Earth and Earth cannot escape from the Sun i,e., -ve sign implies that Earth is bounded with Sun.

(i) Geostationary satellites are appeared to be stationary when seen from Earth. Its time period is same as that of the Earth.
(ii) A polar satellite orbits Earth from North pole to south pole. Its time period is nearly 100 minutes and it completes many revolution in a day.

The downward force on the elevator is :
F = mg + f = 22000 N
\(\therefore\) Power supplied by motor to balance this force is:
P = Fv = 44000 W =\(\frac { 44000 }{ 746 } \)= 59 hp.

If speed of an object increases with time, its acceleration is positive.
(Acceleration is in the direction of motion) and if speed of an object decreases with time its acceleration is negative (Acceleration is opposite to the direction of motion).

Minimum inaccuracy = Vernier constant
 = 1 MSD -1 VS.D
=1 MSD-\(\frac { 49 }{ 50 } \) MSD
= \(\frac { 1 }{ 50 } \)(0.5 mm)= 0.01 mm

I (first case) >1 (Second case)
\(\because \ \tau =I\alpha \)
\(\Rightarrow \alpha \left( first \ case \right) <\alpha \left( second \ case \right) \)

R₁ = 100 ± = 3\(\Omega\); R₂ = 150 ± 2\(\Omega\)
Equivalent resistance R =?
Equivalent resistance R = R₁+ R₂ = (100 ± 3) + (150 ± 2) = (100 + 150) ± (3 + 2)
R = (250 ± 5) \(\Omega\)

Jumping on a concrete cemented floor is more dangerous than jumping on the sand. Sand brings the body to rest slowly than the concrete floor, so that the average force experienced by the body will be lesser.

\(\sin\theta={5\over10}={1\over2}=\sin{\pi \over6}so,\theta={\pi \over 6}\)
Work done by frictional force, W \(=2\bar{f}.\bar{l}\)
W = \(2fil\ \cos\alpha(\alpha=\pi,\cos\alpha=-1)\)
W = \(-2\times0.15\times0.3\times\cos{\pi \over 6}\times10\times 9.8\)
= -7.6 J
kE at the end of trip = PE at top - work done against friction = 0.3\(\times\)9.8\(\times\)5 - 3.8 = 10.9J

When a raindrop gets detached from the cloud it experiences both downward gravitational force and upward air drag force. As it descends towards the Earth, the upward after drag force increases and after a certain time, the upward air drag force cancels the downward gravity. From then on the raindrop moves at constant velocity till it touches the surface of the Earth.

(i) Newton's second law gives the correct explanation for the experiment on the inclined plane.
(ii) In normal cases, where friction is not negligible, once the object reaches the bottom of the inclined plane, it travels some distance and stops.
(iii) Note that it stops because there is a frictional force acting in the direction opposite to its velocity.
(iv) It is this frictional force that reduces the velocity of the object to zero and brings it to rest.
(v) As per Aristotle's idea, as soon as the body reaches the bottom of the plane, it can travel only a small distance and stops because there is no force acting on the object.

(i) The average velocity is defined as ratio of the displacement vector to the corresponding time interval
\(\overrightarrow { { v }_{ avg } } =\frac { \Delta \overrightarrow { r } }{ \Delta t } \)
(ii) It is a vector quantity. The direction of average velocity is in the direction of the displacement vector (\(\Delta \)\(\overrightarrow { r } \)).

1. Equal mass balancing each side so reading shows Zero.
2. Mass of string acting downward direction in the inclined plane.
mg sin θ = T
T = 2kg x 9.8m/s² x sin(30^o)
T = 9.8N

\(\overrightarrow { { v }_{ 1 } } =5\hat { i } +6\hat { j } \)
\(\overrightarrow { { v }_{ 2 } } =2\hat { i } +3\hat { j } \)
\(\therefore \Delta \vec{r} =\vec{r}_{2}-\vec{r}_{1}=2 \hat{i}+3 \hat{j}-5 \hat{i}-6 \hat{j}  \)
\(=-3 \hat{i}-3 \hat{j}  \)
\(\Delta t =5 \mathrm{sec}  \)
\(\therefore \Delta v_{\text {avg }} s =\frac{\Delta \vec{r}}{\Delta t}  \)
\(=\frac{-3}{5}(\hat{i}+\hat{j})\)

Total distance to be covered = 1 km + 100 m = 1100 m (including both bridge and time)
Then, Speed=60 kmh^-1\(=60\times\frac{5}{18}ms^{-1}=\frac{50}{3}\ ms^{-1}\)
Then, time taken to cover this distance \(=\frac{1100}{\frac{150}{9}}s=66s\)

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.

Decreases.

\(\text {Distance of Jupiter } x =824.7 \times 10^{6} \mathrm{~km}  \)
\(\text {Angular diameter } Q =35.72^{\prime \prime}  \)
\(1^{\prime \prime} =4.85 \times 10^{-6} \mathrm{rad}  \)
\(Q =35.72 \times\left(4.85 \times 10^{-6}\right) \mathrm{rad}   \)
\(=173.242 \times 10^{-6} \mathrm{rad}\)
\(x =\frac{b}{Q} \)
Diameter of Jupiter b = x \(\times\) Q 
b = 824.7 x 10⁶ x 173.242 x 10^-6 
= 142,872.6 x 10^6-6 
= 1.428 x 10⁵ km

18.4

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.

Here R = 0.25m
M= 2kg
V= 2ms^-1
Rotational K.E = \(\frac { 1 }{ 2 } I\omega ^{ 2 }\) = \(\frac { 1 }{ 2 } MR ^{ 2 }\) \(\times \left( \frac { V }{ R } \right) ^{ 2 }\)
\(\frac { 1 }{ 2 } MV ^{ 2 }\) = \(\frac { 1 }{ 2 }\) ｘ2ｘ4 =4J

No, only that much amount of kinetic energy is lost as is necessary for the conservation of momentum.

No, because equations of motions are applicable as long as the acceleration is uniform.

As the disc is freely rotating, with the insect on it, the angular momentum of the system is conserved.
\(L=I\omega\)=constant

When the insect moves towards the center (from A to 0), the moment of inertia (I) increases. Thus, the angular velocity (co) increases. When it moves away from center (from 0 to B), the moment of inertia (I) decreases. Thus, the angular velocity decreases.

Let R be the radius of the ring and M be the total mass of the complete ring.
Let m be the mass of the section removed from the ring then, mass of the incomplete ring is M-m
Let us introduce a positive integer (n), such that, \(n\theta=360^{0} \), or \(n=\frac{360^{0}}{\theta}\)

mass of incomplete ring=M - m
\(m=\frac{M}{360}\times \theta\)
∴ Mass of complete ring = \(M-\frac{M}{360}\times \theta\)
Mass of incomplete ring = \(M-\frac{M}{n}=M(\frac{n-1}{n})\)
For example, (a) when \(\theta=60^{0}; n=\frac{360^{0}}{60^{0}}=6\)
∴ n-1 =5
Mass of incomplete ring = \(\frac{5}{6}M\)
(b)when \(\theta=30^{0}; n=\frac{360^{0}}{30^{0}}=12\)
n-1=11
Mass of incomplete ring = \(\frac{11}{12}M\)
The moment of inertia of the incomplete ring is, I=\(M \frac{(n-1)}{n}R^{2}.\)

Let G_SI be the gravitational constant in the SI system and G_cgs in the cgs system. Then 
G_SI = 6.6 10^-11 Nm² kg^-2;
G_cgs = ?
n₂ =\(n_1{ \left[ \frac { { M }_{ 1 } }{ { M }_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { L_{ 1 } }{ L_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { T_{ 1 } }{ T_{ 2 } } \right] }_{ }^{ c }\)
G_cgs = G_SI \({ \left[ \frac { { M }_{ 1 } }{ { M }_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { L_{ 1 } }{ L_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { T_{ 1 } }{ T_{ 2 } } \right] }_{ }^{ c }\)
M₁= 1 kg L₁ = 1 m T₁= 1s
M₂= 1 kg L₂ = 1 m T₂= 1s
The dimensional formula for G is M^-1L³ T^-2
a = -1 b = 3 and c =-2
G_cgs = 6.6\(\times\)10^-11 \(\left[ \frac { 1kg }{ 1g } \right] ^{ -1 }\left[ \frac { 1m }{ 1cm } \right] ^{ 3 }\left[ \frac { 1s }{ 1s } \right] ^{ -2 }\)
=  6.6\(\times\)10^-11 \(\left[ \frac { 1kg }{ { 10 }^{ -3 }kg } \right] ^{ -1 }\left[ \frac { 1m }{ { 10 }^{ -2 }m } \right] ^{ 3 }\left[ \frac { 1s }{ 1s } \right] ^{ -2 }\)
= 6.6\(\times\)10^-11\(\times\)10^-3\(\times\)10⁶\(\times\)1
G_cgs = 6.6\(\times\)10^-8 dyne cm² g^-2

Total initial energy = \({1 \over 2}{mu}^{2}+mgh\)
Energy after collision = 50% of \(\left( {1 \over 2}{mv}^{2}+mgh \right)\)
\(={1\over 2}\left({1 \over 2}{mv}^{2}+mgh \right)\)
As the ball rebounds to same height,
\({1\over 2}\left( {1 \over 2} {mv}^{2}+mgh \right)=mgh \)
\({1\over 4}{mv}^{2}={1 \over2}mgh\)
\(u=\sqrt{2gh}=\sqrt{2\times9.8\times10}\)
u = 14 m/s.