CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 6 - MCQExams.com

Hint: Use formula of change in potential energy

$$Step - 1: Put\ formula\ of\ change\ in\ potential\ energy\ with\ height$$

Change in potential energy is given by the formula:

$$∆U = \dfrac{m\ g\ h}{1\ +\ \dfrac{h}{R}}$$

Now, if body is lifted to height three times to the radius of earth, then h = 3 R.

$$Step - 2: Put\ values\ in\ formula\ $$

On putting value in formula,

$$∆U = \dfrac{m\ g\ (3\ R)}{1\ +\ \dfrac{3\ R}{R}} = \dfrac{3}{4} m g R$$

$$Answer:$$

Hence, option C is the correct answer.

From mechanical energy conservation 0 + 0 = $$\displaystyle \frac{1}{2}mv^{2}-\frac{3GMm}{2R}\: \: \Rightarrow \: \: v=\sqrt{\frac{3GM}{R}}=\sqrt{1.5}v_{e}$$

$$\textbf{Hint: First calculate $\omega$ and Moment of inertia of planet is, $I=M{R}^{2}$.}$$

$$\textbf{Step 1: Calculating the angular velocity.}$$

Given that the planet has an aerial velocity of A, Aerial velocity is given by,

$$A=\dfrac{Area swept by radius vector}{Time taken}$$

$$\Rightarrow A=\dfrac{\pi {{R}^{2}}}{T}$$ (Where R is the radius of the planet)

We know that, $$T=\dfrac{2\pi }{\omega }$$

$$\Rightarrow A=\dfrac{\omega {{R}^{2}}}{2}$$

$$\Rightarrow \omega =\dfrac{2A}{{{R}^{2}}}$$

$$\textbf{Step 2: Calculating the angular momentum.}$$

Angular momentum of the planet will be, $$L=I\omega =M{{R}^{2}}\times \dfrac{2A}{{{R}^{2}}}$$

$$\Rightarrow L=2MA$$

$$\textbf{Correct option: Option (D).}$$

$$\textbf{Explanation:}$$

$$\bullet$$Gravitational force is the universal force of attraction experienced by all matters.

$$\bullet$$It is given by, $$F=\dfrac{GMm}{{{R}^{2}}}$$ (Where G is the universal gravitational constant whose value remains constant throughout the space, and M and m are the masses of the two objects and R be the distance between the bodies.)

Then, $$G=\dfrac{F{{R}^{2}}}{Mm}$$

For a force of 1 Newton between bodies of mass 1 kg and 1 meter apart, G will be

$$\bullet$$ S.I unit of G

$$\Rightarrow G=\dfrac{N\times {{m}^{2}}}{kg\times kg}$$

$$\Rightarrow G=N{{m}^{2}}/k{{g}^{2}}$$

$$\textbf{Hint: Formula for gravitational acceleration is g}$$ =$$\dfrac{GM}{{{R}^{2}}}$$

$$\textbf{Step 1: Weight of the body}$$

Weight of a body is (W)=mg.

Since, mass (m) is constant, ‘g’ is the variable force, also called gravitational force.

$$\textbf{Step 2: Utilising the formula for gravitational acceleration}$$

 The effect on gravitational force can be found by considering the formula

g = $$\dfrac{GM}{{{R}^{2}}}$$

where;

G = Gravitational constant ($$6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$$)

M = Mass of Earth

R = distance between centre of the earth and the body.

$$\textbf{Step 3: Finding the effect of ‘g’ above the surface}$$

Now, when the body goes away from the centre of the Earth, ‘R’ increases.

Since $$g\propto \dfrac{1}{R}$$, the gravitational force decreases.

Hence, gravitational force above the surface of the Earth is lesser than at the surface.

$$\textbf{Step 4: Finding the effect of ‘g’ below the surface}$$

When the body goes towards the centre of the Earth, although ‘R’ decreases; ‘M’ decreases as the mass of earth below the body is lesser/decreasing.

Since $$g\propto M$$, the gravitational force decreases.

Hence, gravitational force below the surface of the Earth is lesser than at the surface.

$$\textbf{Step 5: Finding the effect of ‘g’ at the centre}$$

When the body is at the centre of the Earth, M=0.

⸫ g = 0

Hence, gravitational force at centre of Earth is 0.

$$\textbf{Step 6: Conclusion}$$

Since ‘g’ is maximum for body at the surface of the earth, then maximum weight of the body (W) will also be at the surface of the Earth.

$$\textbf{Answer:}$$

$$\textbf{Hence, the correct option is (C) on the surface of the Earth}$$

Hint: using the formula of gravitional constant

Step1:We know that

$$\mathrm{F}=\mathrm{G} \times \dfrac{M_{1} M_{2}}{r^{2}} \\$$

$$\mathrm{~F} \times \mathrm{r}^{2}=\mathrm{G} \times \mathrm{M}_{1} \mathrm{M}_{2} \\$$

$$\mathrm{~F} \times \dfrac{r^{2}}{M_{1} M_{2}}=\mathrm{G} \\$$

$$\mathrm{G}=\mathrm{F} \times \dfrac{r^{2}}{M_{1} M_{2}}$$

SI Unit of

$$\text { Force }=\mathrm{F}=\text { Newton } \\$$

$$\text { Distance }=\mathrm{R}=\mathrm{m} \\$$

$$\text { Mass }=\mathrm{M}=\mathrm{kg}$$

Step2: Now,

$$\mathrm{G}=\mathrm{F} \times \dfrac{r^{2}}{M_{1} M_{2}}$$

$$\mathrm{G}=\dfrac{\text { Newton } \times \text { Meter }^{2}}{\text { Kilogram } \times \text { Kilogram }}$$

$$\mathrm{G}=\dfrac{\text { Newton Meter }^{2}}{\text { Kilogram }^{2}}$$

$$\mathrm{G}=\mathrm{Nm}^{2} / \mathrm{kg}^{2}$$

So, SI Unit of $$\mathrm{G}$$ is $$\mathrm{Nm}^{2} / \mathrm{kg}^{2}$$

$$\textbf{Thus, option (D) is correct.}$$