B.

486 J

The linear momentum of the exploding part will remain conserved.

Applying coservation of linear momentum,

m₁ u₁ =m₂ u₂

Here, m₁ = 18kg, m₂ = 12kg

u₁ = 6ms^-1, u₂ = ?

${\mathrm{u}}_2 = \frac{18\times 6}{12} 9 {\mathrm{ms}}^{-1}$

Thus kinetic energy of 12 kg mass

$$\begin{gathered} {\mathrm{K}}_2 =\frac{1}{2}{\mathrm{m}}_1{\mathrm{u}}_2^2 \\ = \frac{1}{2}\times 12\times {\left(9\right)}^2 \\ = 6\times 81 \\ {\mathrm{K}}_2 = 486 \mathrm{J} \end{gathered}$$

A.

converts translational energy to rotational energy

When a body rolls down without slipping along an inclined plane of inclination θ, it rotates about a horizontal axis through its centre of mass and also its centre of mass moves. Therefore, rolling motion may be regarded as a rotational motion about an axis through its centre of mass plus a translational motion of the centre of mass. As it rolls down, it suffers loss in gravitational potential energy provided translational energy due to frictional force is converted into rotational energy.

A.

g' = 3g

The acceleration due to gravity on the planet can be found using the relation

$$\begin{gathered} \mathrm{g} = \frac{\mathrm{GM}}{{\mathrm{R}}^2} \\ \mathrm{but} \mathrm{M} = \frac{4}{3}{\mathrm{\pi R}}^3\mathrm{\rho } \\ \mathrm{\rho }- \mathrm{density} \\ \mathrm{g} = \frac{\mathrm{G} \times {\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3\mathrm{\rho }}{{\mathrm{R}}^2} \\ \mathrm{g} = \mathrm{G} \times \frac{4}{3}\mathrm{\pi R\rho } \end{gathered}$$

⇒ g ∝ R

$$\begin{gathered} \therefore \frac{\mathrm{g}\text{'}}{\mathrm{R}} = \frac{\mathrm{R}\text{'}}{\mathrm{R}} \\ \Rightarrow \frac{\mathrm{g}\text{'}}{\mathrm{g}} = \frac{3\mathrm{R}}{\mathrm{R}} \\ \Rightarrow \frac{\mathrm{g}\text{'}}{\mathrm{g}} = 3 \\ \Rightarrow \mathrm{g}\text{'} = 3\mathrm{g} \end{gathered}$$

C.

$-\frac{1}{2}$

Two vectors must be perpendicular if their dot product is zero.

$$\begin{gathered} \mathrm{let} \overrightarrow{\mathrm{a}} = 2\hat{\mathrm{i}} + 3\hat{\mathrm{j}} +8 \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{b}} = 4 \hat{\mathrm{j}} - 4 \hat{\mathrm{i}} + \mathrm{\alpha }\hat{\mathrm{k}} \\ = -4\hat{\mathrm{i}} + 4\hat{\mathrm{j}} + \mathrm{\alpha }\hat{\mathrm{k}} \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{above} \mathrm{hypothesis} \\ \overrightarrow{\mathrm{a}} \perp \overrightarrow{\mathrm{b}} \\ \Rightarrow \overrightarrow{\mathrm{a}} . \overrightarrow{\mathrm{b}} = 0 \\ \Rightarrow \left(2\widehat{\mathrm{i} } + 3\hat{\mathrm{j}} + 8\hat{\mathrm{k}}\right) . \left(-4\hat{\mathrm{i}} + 4\hat{\mathrm{j}} + \mathrm{\alpha }\hat{\mathrm{k}}\right) = 0 \\ \Rightarrow -8 + 12 +8\mathrm{\alpha } = 0 \\ \Rightarrow 8\mathrm{\alpha } = -4 \\ \mathrm{\alpha } =-\frac{4}{8} = -\frac{1}{2} \end{gathered}$$

$\overrightarrow{\mathrm{a}} . \overrightarrow{\mathrm{b}} = \mathrm{ab} \mathrm{cos\theta }.$ Here a and b are always positive as they are the magnitudes as $\overrightarrow{\mathrm{a}} \mathrm{and} \overrightarrow{\mathrm{b}}$.

B.

13.5 J

The work done will be the area of the F-x graph. Work done in moving the object from x = O to x = 6 m is given by

W = area of rectangle + area of triangle

    $= 3 \times 3 +\frac{1}{2} \times 3 \times 3$

W   = 9 + 4.5 = 13.5 J