The escape velocity is given by the formula: \[ v_{escape} = \sqrt{\frac{2GM}{R}} \]
Step 2: Given Mass and Radius RelationshipsWe are given the following mass and radius relationships: \[ \frac{M_{planet}}{M_{earth}} = \frac{1}{8}, \quad \frac{R_{planet}}{R_{earth}} = \frac{1}{2} \]
Step 3: Calculate the Ratio of Escape VelocitiesThe ratio of the escape velocities is: \[ \frac{v_{escape, planet}}{v_{escape, earth}} = \sqrt{\frac{M_{planet} R_{earth}}{M_{earth} R_{planet}}} = \frac{1}{2} \]
Step 4: Calculate the Escape Velocity from the PlanetTherefore, the escape velocity from the planet is: \[ v_{escape, planet} = \frac{1}{2} \times 11.2 = 5.6 \, \text{km/s} \]
Final Answer: \[ v_{escape, planet} = 5.6 \, \text{km/s} \]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:
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 :
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) ?

The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature \( R = 2 \, \text{m} \). Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the side view mirror is \( a \). The value of \( 100a \) is _____________ m/s\(^2\).
Let \( a_1, a_2, a_3, \ldots \) be in an A.P. such that \[ \sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1, \quad a_1 \neq 0. \] If \[ \sum_{k=1}^{n} a_k = 0, \] then \( n \) is: