Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250N on the surface?
A.
125N
B.
250N
C.
0N
D.
120N
Answer: Option A
Explanation:
Here mg = 250N, d = R/2 Acceleration due to gravity at depth d = R/2, below the earth’s surface will bw gd=g(1-d/R)=g(1-(R/2)/R)=g/2 Therefore, new weight = mgd=mg/2=250/2=125N.
Calculate the force of attraction between two balls each of mass 1kg when their centres are 10cm apart. The value of gravitational constant G = 6.67×10-11 Nm2 kg(-2).
A.
6.67×10-11N
B.
6.67×10-10N
C.
6.67N
D.
6.67×1010 N
Answer: Option B
Explanation:
Force of attraction, F=G (m1 m2)/r2 = ((6.67×10-11)×1×1)/0.10=6.67×10-10 N.
If the kinetic energy of the satellite revolving in an orbit close to the earth’s surface happens to be doubled, what happens to the satellite?
A.
It revolves faster
B.
It revolves slower
C.
It remains unaltered
D.
It escapes
Answer: Option D
Explanation:
When the kinetic energy of the satellite is doubled, its orbital velocity increases √2 times and becomes equal to the escape velocity. So the satellite will escape.
The escape velocity in earth is 11.2km/s. What will be its value on a planet whose radius is double the radius of the earth and eight times the mass of the earth?
A spring balance is suspended from inside an artificial satellite revolving around the earth. If a body of mass 1kg is suspended from it, what would be its reading?
A.
1
B.
2
C.
Depends on the mass of the satellite
D.
0
Answer: Option D
Explanation:
If a body of mass 1 kg is suspended from a spring balance inside an artificial satellite revolving around the earth, its reading will be zero. This is because the satellite is in a state of free fall.
