Two satellitesP andQ go around the earth in circular orbits at heights ofh1 andh2 respectively from surface ofearth. Assuming the earth to be a uniform sphere of radiusre, the ratio of magnitudes of their orbital velocities will be(a)(b)(c)(d)6. A planet revolves in elliptical orbit around the sun. The linear speed of planet will be minimum at(a)A(b)D(c)C(d)B7. Wheel of an engine rotates with an angular speed of 120 rev/min. If radius of the wheel is 2 m, then linear speed ofany point on its rim is

The text you provided contains multiple questions. Here are the answers step by step:

1. The orbital velocity of a satellite is given by the formula v = sqrt(GM/r), where G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth. If we assume the Earth to be a uniform sphere, then the orbital velocities of the two satellites P and Q will be vp = sqrt(GM/(re+h1)) and vq = sqrt(GM/(re+h2)). The ratio of their magnitudes will be vp/vq = sqrt((re+h2)/(re+h1)).

2. In an elliptical orbit, the linear speed of a planet is minimum at the point farthest from the sun, which is the aphelion. In the options provided, without a diagram or additional context, it's impossible to definitively say which point represents the aphelion.

3. The angular speed of the wheel is given in revolutions per minute, so first we need to convert this to radians per second. 120 rev/min is equivalent to 1202π rad/min or 2π2 rad/sec. The linear speed v of any point on the rim of the wheel is given by the formula v = rω, where r is the radius of the wheel and ω is the angular speed. Substituting the given values, we get v = 2m * 2π*2 rad/sec = 8π m/sec.