Physics Test 155

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Physics Test 155

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Question 1
4 / -1

Determine the force that must be applied to end of the class 1 lever shown below to lift the 500 lb load.

A
500 lb

B
2500 lb

C
5 lb

D
100 lb

Solution

Balance the 2 opposite moments of inertia on the 2 sides

1 × 500 = F × 5

F = 100
Question 2
4 / -1

There are two blocks of mass 3 kg and 5 kg hanging from the ends of a rod of negligible mass. The rod is marked in eight equal parts as shown. At which of the points indicated should a string be attached if a rod is to remain horizontal when suspended from the string?

A
D

B
F

C
B

D
E

Solution

This is based on torque balance, That is F.d=const.

Therefore 3×d=5×(8-d)

by solving we get d=5.

and at point 5=E
Question 3
4 / -1

Four particles, each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction. (Take that gravitational force between two point masses m1 and m2 separated by is The speed of each particle is

A

B

C

D

Solution
Question 4
4 / -1

At what altitude will the acceleration due to gravity be 25% of that at the earth’s surface (given radius of earth is R)?

A
R/4

B
R

C
3R/8

D
R/2

Solution

Force on the body placed on Earth's surface is

F=GMm/R²

But, F=mg hence,

mg=GMm/R²

where, variables have their usual meanings.

gR²=GM

Now, force on the body at geo-potential height say h (altitude) where the acceleration due to gravity is 25% of that at the earth's surface i.e.

25g/100=g/4

Hence, we can write

g/4=GM/(R+h)²

g/4= gR² /(R+h)²

(R+h)²=4R²

Taking roots for both sides we get

R+h=2R

h=R
Question 5
4 / -1

A planet of mass M is revolving around sun in an elliptical orbit. If dA is the area swept in a time dt, angular momentum can be expressed as

A

B

C

D

Solution
Question 6
4 / -1

A straight rod of length L extends from x = a to x = L + a. Find the gravitational force exerted by it on a point mass m at x = 0 if the linear density of rod μ = A+ Bx²

A

B

C

D

Solution
Question 7
4 / -1

The work done in shifting a particle of mass m from centre of earth to the surface of earth is (0m at both places are fixed)

A
– m g R

B
+mgR/2

C
zero

D
none of these

Solution
Question 8
4 / -1

A particle is projected upward from the surface of earth (radius = R) with a speed equal to the orbital speed of a satellite near the earth’s surface. The height to which it would rise is

A
√2R

B
R/√2

C
R

D
2R

Solution

(orbital speed υ₀ of a satellite)

Near the earth’s surface is equal to 1/√2
times the escape velocity of a particle on earth’s surface) Now from conservation of mechanical energy: Decrease in kinetic energy = increase in potential energy
Question 9
4 / -1

Two concentric shells of uniform density of mass M₁ and M₂ are situated as shown in the figure. The forces experienced by a particle of mass m when placed at positions A, B and C respectively are (given OA = p, OB = q and OC = r)

A

B

C

D

Solution

We know that attraction at an external point due to spherical shell of mass M is GM/r² while at an internal point is zero. So, for particle at point C

OA = G (M1 + M2)m/p²

OB = G M₁m/q²

OC = 0
Question 10
4 / -1

If G is the universal gravitational constant and ρ is the uniform density of spherical planet.

Then shortest possible period of the planet can be

A

B

C

D

Solution

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. Let M be the mass of the planet R its radius and m the mass of a particle on its surface. Then