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Physics NEET MCQ
Quiz 2
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Q.1
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small as compared to the mass of the earth. Which of the following statement is correct?
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The acceleration of S is always directed towards the center of the earth
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The angular momentum of S about the centete of the earth changes its direction but its magnitude remains constant
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The total mechanical energy of S remains constant
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The linear momentum of S remains constant in magnitude
Explanation
For ANY orbit governed purely by gravity — circular or elliptical — the gravitational force (and hence the acceleration) always points directly toward the central body (Earth's centre); this is the defining feature of motion under a central force, and it holds true throughout an elliptical orbit just as much as a circular one.
Q.2
Both earth and moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon
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will be elliptical
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will not be strictly elliptical because the total gravitational force on it is not central
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is not elliptical but will necessarily be a closed curve.
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deviates considerably from being elliptical due to influence of planets other than earth.
Explanation
The Moon is pulled by BOTH Earth's gravity and the Sun's gravity simultaneously. As seen from the Sun's frame, the total gravitational force acting on the Moon (Earth's pull plus the Sun's) is not purely directed at a single fixed centre — it isn't a simple central force — so the Moon's path around the Sun isn't a strictly, mathematically perfect ellipse, even though it's close to one.
Q.3
The radius vector, drawn from the sun to a planet sweeps out equal areas in equal lines. This is the statement of
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Kepler’s third law
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Kepler’s first law
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Newton’s third law
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Kepler’s second law
Explanation
The statement that a line from the Sun to a planet sweeps out equal areas in equal time intervals is Kepler's second law.
Q.4
Time period of pendulum, on a satellite orbiting the earth, is
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$\frac {1}{\pi}$
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0
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$\infty$
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$\pi$
Explanation
In an orbiting satellite, everything (including a pendulum's bob) is in continuous free fall together, so the effective gravity felt inside the satellite is zero. Since a pendulum's period T = 2π√(L/g), as g approaches zero, T approaches infinity — the pendulum simply doesn't swing back and forth at all in this weightless environment.
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