Correct answer option is (A)

The time period of a simple pendulum is given by the formula:

$T=2\pi \sqrt{\frac{l}{g}}$

where $T$ is the time period, $l$ is the length of the pendulum, and $g$ is the acceleration due to gravity at the location of the pendulum.

The acceleration due to gravity changes with height above the Earth's surface. The acceleration due to gravity at a height $h$ above the Earth's surface can be expressed as:

$g^{\prime }=g{\left(\frac{R}{R+h}\right)}^2$

where $g$ is the acceleration due to gravity at the surface of the Earth, $R$ is the radius of the Earth, and $h$ is the height above the Earth’s surface. Since the time period of the pendulum depends on the square root of the inverse of the acceleration due to gravity, any change in $g$ due to a change in height will affect the time period.

Given that the time period of the pendulum at a height $R$ above Earth's surface is $T_1$, and we're to find the time period $T_2$ at a height of $2R$, we can use the formula for acceleration due to gravity at different heights to express the relationship between $T_1$ and $T_2$.

For the initial case at height $R$:

$g_1=g{\left(\frac{R}{R+R}\right)}^2=g{\left(\frac{R}{2R}\right)}^2=\frac{g}{4}$

For the new case at height $2R$:

$g_2=g{\left(\frac{R}{R+2R}\right)}^2=g{\left(\frac{R}{3R}\right)}^2=\frac{g}{9}$

The time period is proportional to the square root of the inverse of $g$, so:

$\frac{T_1}{T_2}=\sqrt{\frac{g_2}{g_1}}=\sqrt{\frac{\frac{g}{9}}{\frac{g}{4}}}=\sqrt{\frac{4}{9}}=\frac{2}{3}$

Therefore:

$T_1=\frac{2}{3}{T}_2$

Rearranging this equation:

$3T_1=2T_2$

This corresponds to Option A.

Correct answer option is (A)

The initial momentum of the system (bullet + bob) is the momentum of the bullet because the bob is initially at rest. The momentum of the bullet is given by its mass times its velocity :

$p_{\text{initial}}=m_{\text{bullet}}\times v_{\text{bullet}}$

After the collision, the bullet and the bob move together with a common velocity. Let's denote this common velocity as $v^{\prime }$. The final momentum $p_{\text{final}}$ is the combined mass of the bullet and bob times the common velocity :

$p_{\text{final}}=(m_{\text{bullet}}+m_{\text{bob}})\times v^{\prime }$

According to the principle of conservation of linear momentum,

$p_{\text{initial}}=p_{\text{final}}$

$m_{\text{bullet}}\times v_{\text{bullet}}=(m_{\text{bullet}}+m_{\text{bob}})\times v^{\prime }$

Plugging in the values :

$({10}^{-2}\text{ kg})\times (2\times {10}^2\text{ m/s})=({10}^{-2}\text{ kg}+1\text{ kg})\times v^{\prime }$

Solving for $v^{\prime }$ :

$v^{\prime }=\frac{{10}^{-2}\times 2\times {10}^2}{{10}^{-2}+1}=\frac{2}{1.01}\approx 1.98\text{ m/s}$

After the collision, the system has some kinetic energy which will be completely converted to potential energy at the maximum height $h$ that the bob reaches. Using the principle of conservation of energy :

${\text{Kinetic Energy (KE)}}_{\text{initial}}={\text{Potential Energy (PE)}}_{\text{final}}$

$\frac{1}{2}(m_{\text{bullet}}+m_{\text{bob}})v^{\prime 2}=(m_{\text{bullet}}+m_{\text{bob}})gh$

Isolating $h$, we get :

$h=\frac{\frac{1}{2}(m_{\text{bullet}}+m_{\text{bob}})v^{\prime 2}}{(m_{\text{bullet}}+m_{\text{bob}})g}=\frac{v^{\prime 2}}{2g}$

Substituting the values for $v^{\prime }$ and $g$ :

$h=\frac{(1.98{)}^2}{2\times 10}=\frac{3.9204}{20}=0.19602\text{ m}$

Looking at the given options, the result most closely matches Option A :

$\boxed{{\displaystyle 0.20\text{ m}}}$

Therefore, the height to which the bob rises before swinging back is approximately $0.20\text{ m}$.

Correct answer option is (D)

$\ell =10\mathrm{m}$,

Initial energy $=\mathrm{mg}\ell$

$\begin{array}{rl}& \text{ So, }\frac{9}{10}\mathrm{mg}\ell =\frac{1}{2}{\mathrm{mv}}^2\\ & \Rightarrow \frac{9}{10}\times 10\times 10=\frac{1}{2}{\mathrm{v}}^2\\ & {\mathrm{v}}^2=180\\ & \mathrm{v}=\sqrt{180}=6\sqrt{5}\mathrm{m}/\mathrm{s}\end{array}$