NEET Physics - MCQ Practice Questions

Understanding:

We need the orbital speed of a satellite at height $h$ above Earth's surface.

Formula:

For circular orbit, gravitational force provides centripetal force:

Step 1: Solve for orbital speed.

Step 2: Express $GM$ in terms of $g$ and $R$.

At Earth's surface: $g=\frac{GM}{R^2}$, so $GM=g{R}^2$.

Step 3: Substitute into the expression for $v$.

Answer:

The orbital speed of the satellite is:

Quick Tip:

For a satellite at the surface ($h=0$), this reduces to $v=\sqrt{gR}$, which is the first cosmic velocity — a value worth remembering.

Understanding:

We need the change in gravitational potential energy when a mass $m$ is raised by height $h$ above Earth's surface, for $h\ll R$.

Formula:

where $r$ is the distance from Earth's centre.

Step 1: Write $U_h$ and $U_s$.

Step 2: Compute $\Delta U$.

Step 3: Apply the approximation $h\ll R$, so $R+h\approx R$.

since $g=\frac{GM}{R^2}$.

Answer:

The approximate change in gravitational potential energy is:

Quick Tip:

This result is valid only for $h\ll R$. For large heights, always use the exact formula $\Delta U=\frac{GMmh}{R(R+h)}$.

Understanding:

We need the point on the line joining $m_1$ and $m_2$ where the net gravitational field is zero.

Formula:

At the null point, gravitational fields due to both masses are equal and opposite:

Step 1: Cancel $G$ and cross-multiply.

Step 2: Take square roots.

Step 3: Verify — distance from $m_2$ is $1-0.4=0.6\text{m}$, ratio of distances $=0.4:0.6=2:3=\sqrt{m_1}:\sqrt{m_2}$. ✓

Answer:

The gravitational field is zero at a distance of $0.4\text{m}$ from $m_1$.

Quick Tip:

The null point always lies closer to the lighter mass. The distances from each mass are in the ratio $\sqrt{m_1}:\sqrt{m_2}$.

Understanding:

We need the escape velocity from a planet with $M_p=4M_E$ and $R_p=2R_E$.

Formula:

Step 1: Write the ratio of escape velocities.

Step 2: Substitute the ratios $M_p=4M_E$ and $R_p=2R_E$.

Step 3: Compute the numerical value.

Answer:

The escape velocity from the planet's surface is approximately $15.8\text{km/s}$.

Quick Tip:

When both $M$ and $R$ change, use the ratio method: $v_{\text{esc}}\propto \sqrt{M/R}$. Here $M/R$ doubles, so $v_{\text{esc}}$ increases by $\sqrt{2}$.

Understanding:

Using Kepler's third law, find the period $T^{\prime }$ for a satellite at $r^{\prime }=10600\text{km}$, given that a geostationary satellite at $r=42400\text{km}$ has $T=24\text{h}$.

Formula:

Kepler's third law:

Step 1: Find the ratio $r^{\prime }/r$.

Step 2: Apply Kepler's third law.

Answer:

The time period of the satellite at radius $10600\text{km}$ is $3$ hours.

Quick Tip:

When the orbital radius is $1/n$ of a reference orbit, the period scales as $(1/n{)}^{3/2}$ of the reference period. Here $n=4$, giving $T^{\prime }=24/4^{3/2}=24/8=3\text{h}$.

Understanding:

We need the gravitational potential at the centre of a uniform thin spherical shell.

Formula:

Gravitational potential due to a mass element $dm$ at distance $d$ is:

Step 1: Identify the distance from every point on the shell to the centre.

For a thin spherical shell of radius $R$, every point on the shell is at the same distance $R$ from the centre. So $d=R$ for all mass elements.

Step 2: Integrate over the entire shell.

Step 3: Note the key result.

Inside a spherical shell, the gravitational field is zero (shell theorem), but the potential is constant and equal to the value at the surface, $-GM/R$. The centre is no exception.

Answer:

The gravitational potential at the centre of the spherical shell is:

Quick Tip:

Do not confuse field and potential inside a shell. The gravitational field $E=0$ inside, but the potential $V=-GM/R$ is uniform and non-zero throughout the interior.

Understanding:

Earth's mass $M$ remains constant but its radius becomes $R^{\prime }=R/2$. We need the new surface gravity $g^{\prime }$.

Formula:

Step 1: Write the new surface gravity.

Step 2: Express in terms of $g$.

Answer:

The acceleration due to gravity at the new surface is four times the original value.

Quick Tip:

$g\propto M/R^2$. With $M$ fixed, halving $R$ multiplies $g$ by $2^2=4$. This is a very common NEET question pattern.

Understanding:

Find the height $h$ at which the gravitational force (weight) reduces to $W/4$.

Formula:

Step 1: Set up the equation.

Step 2: Solve for $h$.

Answer:

The weight reduces to $W/4$ at a height equal to Earth's radius above the surface.

Quick Tip:

Weight $\propto 1/(R+h{)}^2$. For weight to become $1/n^2$ of the surface value, raise the body to a height $h=(n-1)R$. For $n=2$: $h=R$.

Understanding:

Using conservation of angular momentum to find the planet's speed at aphelion.

Formula:

By conservation of angular momentum (the velocity is perpendicular to the radius vector at both perihelion and aphelion):

Step 1: Solve for $v_2$.

Step 2: Physical check — since $r_2>r_1$ (aphelion is farther), we get $v_2<v_1$, consistent with Kepler's second law (the planet moves slowest at aphelion).

Answer:

The speed at aphelion is:

Quick Tip:

At perihelion and aphelion the velocity is perpendicular to the radius, so $L=mvr$ directly. At all other points you would need $L=mvr\sin \theta$.

Understanding:

The binding energy is the minimum energy needed to remove the satellite from its orbit to infinity (i.e., the negative of the total mechanical energy).

Formula:

Total mechanical energy of the satellite:

Step 1: Find kinetic energy using the orbital condition.

For circular orbit: $\frac{m{v}^2}{r}=\frac{GMm}{r^2}$, so $K=\frac{GMm}{2r}$.

Step 2: Find total energy.

Step 3: Binding energy is $E_b=-E$.

Answer:

The binding energy of the satellite is:

Quick Tip:

For a satellite: $K=-E_{\text{total}}$, $U=2E_{\text{total}}$, and $E_b=K=|E_{\text{total}}|$. These three relations are frequently tested in NEET.