# National Science Olympiad Model Paper 3 Questions and Answers Part 23

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1. The earth (mass kg revolves around the sun with angular velocity 2 × rad/s in a circular orbit of radius km. The force exerted by the sun on the earth in Newton.

A) Zero

B) N

C) N

D)

Explanation:

m

ω

R

The force exerted by the sun on the earth

By substituting the value, we can get,

F

2. Two spheres of mass m and M are situated in air and the gravitational force between them is 2F. The space around the masses is now filled with a liquid of specific gravity 4. The gravitational force will now be

A) 2F

B) F

C)

D)

Explanation

Gravitational force does not depend on medium because gravity is unipolar.

Electric charges have two polarities, so an electric field polarizes the medium it goes through.

But gravity comes in only one polarity, attractive, so when it goes through a medium, there is no polarizing effect or polarizing forces.

3. The earth E moves in an elliptical orbit with the sun S at one of the foci as shown in the figure. Its speed of motion will be maximum at the point

A) 3

B) 1

C) 2

D) 4

Explanation

As gravitation force is acting towards the center so there is no change in angular momentum.

So angular momentum is constant

We know that Angular momentum L is equal to

L = mvr

So mvr

Mass is also constant so

So, v is inversely proportional to r

So, Where r is the distance between sun and earth and v is a velocity

So, if the distance is minimum velocity is maximum

at 1 position velocity is maximum

4. Where will it be profitable to purchase 1-kilogram salt.

At poles

At equator

At latitude

At latitude

Explanation:

We know that the value of acceleration due to gravity, g is defined as g = GM/R2 by the Law of Gravitational attraction.

According to this law, the value of, g depends on the mass and the radius of the Earth (G is a universal constant) .

However, as we know that the Earth is not perfectly spherical but bulged out (R is slightly more at the equator) at the equator on account of its rotation.

Thus, it is found that the value of g is 0.5 % more at the poles than it is at the equator. As a result, the weight (W = mg) of the same amount of salt must also be greater at the poles.

Weight of 1kg salt at equator measured by the weighing machine = 1x9.8 = 9.8N Mass measured (as per the calibrated weighing machine) will be = (9.8/9.8) = 1kg Weight of 1kg salt at poles = 1x 9.85 = 9.85N Mass measured (as per the calibrated weighing machine) will be = (9.85/9,8) = 1.005kg So at the poles, even 1kg salt will be read as 1.005 kg by the vendor and you will get lesser salt by him.

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