Correct Option is (C)

The kinetic energy of a body of mass $m$ and velocity $v$ is given by $K=\frac{1}{2}m{v}^2$. Since the bodies have the same linear momentum, we can write:

$p=mv$

where $p$ is the linear momentum of the bodies.

Let the masses of the two bodies be $m_1$ and $m_2$ and their kinetic energies be $K_1$ and $K_2$, respectively. Then, we have:

$\frac{K_1}{K_2}=\frac{16}{9}$

$\frac{1}{2}{m}_1{v}_1^2÷\frac{1}{2}{m}_2{v}_2^2=\frac{16}{9}$

Since $p=mv$, we have $v_1=\frac{p}{m_1}$ and $v_2=\frac{p}{m_2}$. Substituting these in the above equation, we get:

$\frac{m_2}{m_1}=\frac{9}{16}$

Therefore, the ratio of the masses of the two bodies is $\boxed{{\displaystyle 9:16}}$.

Correct Option is (B)

Statement I is correct: The kinetic energy of an object is given by $\frac{1}{2}m{v}^2$, where m is the mass of the object and v is its velocity. If a truck and a car are moving with the same kinetic energy and are brought to rest by applying brakes that provide equal retarding forces, both will come to rest in equal distances. This is because the distance required to stop an object depends on its initial kinetic energy and the force applied to bring it to rest. Since both the truck and car have the same initial kinetic energy and are subjected to the same retarding force, they will come to rest in the same distance.

Statement II is incorrect. When the car moves from east to north, even though its speed remains unchanged, its direction changes. Since velocity is a vector quantity that has both magnitude (speed) and direction, a change in direction implies a change in velocity. Acceleration is the rate of change of velocity, so when the velocity changes, there is acceleration. In this case, the car's acceleration is not zero as it turns from east to north.

Correct Option is (B)

First, we will use conservation of momentum to find the velocity of the bullet-block system just after the bullet gets embedded into the block.

The initial momentum of the system is given by the momentum of the bullet (as the block is initially at rest), and the final momentum of the system is the combined momentum of the bullet and the block.

Setting initial momentum equal to final momentum:

$m_{\text{bullet}}\cdot v_{\text{bullet}}=(m_{\text{bullet}}+m_{\text{block}})\cdot v_{\text{final}}$

Solving for ($v_{\text{final}}$):

$v_{\text{final}}=\frac{m_{\text{bullet}}\cdot v_{\text{bullet}}}{m_{\text{bullet}}+m_{\text{block}}}$

Substituting the given values:

$v_{\text{final}}=\frac{0.1\text{kg}\cdot 400\text{m/s}}{0.1\text{kg}+3.9\text{kg}}=10\text{m/s}$

Next, we know the block comes to rest after moving 20 m due to friction. The work done by the friction force is equal to the initial kinetic energy of the block (since it comes to rest, the final kinetic energy is 0). The work done by friction is given by the friction force times the distance, and the friction force is equal to the coefficient of friction times the normal force (which is equal to the weight of the block).

So, setting the work done by friction equal to the initial kinetic energy of the block:

$\mu \cdot (m_{\text{bullet}}+m_{\text{block}})\cdot g\cdot d=\frac{1}{2}\cdot (m_{\text{bullet}}+m_{\text{block}})\cdot v_{\text{final}}^2$

Solving for ($\mu$):

$\mu =\frac{\frac{1}{2}\cdot (m_{\text{bullet}}+m_{\text{block}})\cdot v_{\text{final}}^2}{(m_{\text{bullet}}+m_{\text{block}})\cdot g\cdot d}$

Substituting the given values:

$\mu =\frac{\frac{1}{2}\cdot (0.1\text{kg}+3.9\text{kg})\cdot (10\text{m/s}{)}^2}{(0.1\text{kg}+3.9\text{kg})\cdot 10{\text{m/s}}^2\cdot 20\text{m}}=0.25$

Correct Option is (C)

${\mathrm{KE}}_{\mathrm{in}}=\frac{1}{2}m{v}^2$

${\mathrm{KE}}_{\text{final }}=\frac{1}{2}m{v}^2{\cos }^2{30}^{\circ }=\frac{1}{2}m{v}^2{\left(\frac{\sqrt{3}}{2}\right)}^2$

$\frac{{\mathrm{KE}}_{\mathrm{in}}}{{\mathrm{KE}}_{\mathrm{f}}}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{v}^2\left(\frac{3}{4}\right)}=\frac{4}{3}$

Correct Option is (C)

$K.E.=E=\frac{1}{2}m{v}^2$

at highest point

$K.E^{\prime }=\frac{1}{2}m{v}^2{\cos }^2\theta$

$=\frac{1}{2}m{v}^2\left(\frac{1}{4}\right)$

$=\frac{E}{4}$