The ratio of escape velocity of a planet to the escape velocity of earth will be:- Given: Mass of the planet is 16 times mass of earth and radius of the planet is 4 times the radius of earth.
- 2:1
- 1:√2
- 4:1
- 1:4
The Correct Option is A
Solution and Explanation
Top Questions on Speed and velocity
- Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s, the escape velocity in km/s from the planet will be:
- JEE Main - 2025
- Physics
- Speed and velocity
- A fluid flows through a pipe with varying cross-section. If the velocity at the narrow section is 3 m/s and the cross-sectional area is half of the wider section, what is the velocity in the wider section?
- BITSAT - 2025
- Physics
- Speed and velocity
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is:
- JEE Main - 2024
- Physics
- Speed and velocity
The velocity (v) - time (t) plot of the motion of a body is shown below :
The acceleration (a) - time(t) graph that best suits this motion is :
- NEET (UG) - 2024
- Physics
- Speed and velocity
A wheel of a bullock cart is rolling on a level road, as shown in the figure below. If its linear speed is v in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel, respectively) ?
- NEET (UG) - 2024
- Physics
- Speed and velocity
Questions Asked in JEE Main exam
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
- JEE Main - 2025
- Electromagnetic Field (EMF)
- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
- JEE Main - 2025
- Trigonometric Identities
- Ice at \( -5^\circ C \) is heated to become vapor with temperature of \( 110^\circ C \) at atmospheric pressure. The entropy change associated with this process can be obtained from:
- JEE Main - 2025
- Thermodynamics
- Let C be the circle of minimum area enclosing the ellipse E: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with eccentricity \( \frac{1}{2} \) and foci \( (\pm 2, 0) \). Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 29 is parallel to the major axis and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is:
- JEE Main - 2025
- Coordinate Geometry
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to: