Problem 74 - A Complete Resource Book in Physics for JEE Main

Question

A particle of mass \(10 \mathrm{~g}\) is kept on the surface of a uniform sphere of mass \(100 \mathrm{~kg}\) and radius \(10 \mathrm{~cm}\). Find the work to the done against be gravitational force between them, to take the particle far away from the sphere. (you may take \(\left.G=6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{Kg}^{-2}\right)\) (A) \(13.34 \times 10^{-10} \mathrm{~J}\) (B) \(3.33 \times 10^{-10} \mathrm{~J}\) (C) \(6.67 \times 10^{-9} \mathrm{~J}\) (D) \(6.67 \times 10^{-10} \mathrm{~J}\)

Step-by-Step Solution

Answer

The work done against the gravitational force to take the particle far away from the sphere is (D) \(6.67 \times 10^{-10} \mathrm{~J}\).

Step 1: 1. Convert the given units

Before starting with the calculations, let's convert the given units into standard units (SI). 1. Mass of the particle: \(m_1 = 10 \mathrm{~g} = 10 \times 10^{-3} \mathrm{~kg}\) 2. Radius of the sphere: \(r = 10 \mathrm{~cm} = 10 \times 10^{-2} \mathrm{~m}\)

Step 2: 2. Calculate the gravitational potential energy

Now, let's calculate the gravitational potential energy (U) of the system, using the formula: $$U = -\frac{Gm_1 m_2}{r}$$ Plug in the values: $$U = -\frac{6.67 \times 10^{-11} \times (10 \times 10^{-3}) \times 100}{10 \times 10^{-2}}$$

Step 3: 3. Simplify and solve for U

We have: $$U = -\frac{6.67 \times 10^{-11} \times 10^{-2} \times 100}{10^{-1}}$$ Simplify and solve for U: $$U = -6.67 \times 10^{-10} \mathrm{~J}$$

Step 4: 4. Calculate the work done against the gravitational force

The work done against the gravitational force is equal to the absolute value of the potential energy of the system, which means we need to change the sign of the potential energy: Work done = \(|U| = 6.67 \times 10^{-10} \mathrm{~J}\) The correct answer is (D) \(6.67 \times 10^{-10} \mathrm{~J}\).

Key Concepts

Work Done in Physics

In the realm of physics, work is defined as the product of the force applied to an object and the displacement it causes in the direction of the force. If we're specifically discussing \textbf{gravitational work}, it's the work done against gravitational force to move an object from one point to another. For example, lifting a book off the floor requires work against the Earth's gravitational pull.

For an object held within a gravitational field, like the 10g particle near the sphere in our exercise, the work required to move it 'far away' (to infinity, where gravitational effects are negligible) is equal to the negative of the gravitational potential energy at the starting point. This is because work done by external forces to move an object from a point in a gravitational field to a point far away must overcome the gravitational pull; hence, it's equal to the gravitational potential energy of the system, but with the opposite sign.

The formulation for the work done against gravity is expressed in the equation: \[ W = -U \]where \( W \) is the work done and \( U \) is the gravitational potential energy, which brings us to our next crucial physical concept, gravitational force.

Gravitational Force

The gravitational force is an attractive force that acts between any two masses. Sir Isaac Newton quantified this force in his law of universal gravitation, which states that every point mass attracts every other point mass by a force pointed along the line intersecting both points. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula is given by:

\[ F = \frac{G \times m_1 \times m_2}{r^2} \]where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between the centers of the two masses. In our exercise, gravitational potential energy, which is related to this force, is used to find the work done to move the particle away from the sphere.

To move the particle far away from the gravitational influence of the sphere, the work done must be equivalent to the gravitational potential energy stored in the system while the particle is on the surface of the sphere. The formula for this potential energy is:\[ U = -\frac{G \times m_1 \times m_2}{r} \]This equation takes the negative sign because gravitational force is conservative, and the work done against gravity (to separate two masses) is opposite to the direction of the gravitational force which attracts them together.

Unit Conversion in Physics

Understanding unit conversion is vital for solving physics problems accurately. It involves transforming a measure of physical quantity from one unit to another. In physics, the International System of Units (SI) is the standard, and converting to these units allows for consistency and comparability of measurements. For instance, in our problem, the mass of the particle was given in grams (g) and needed to be converted to kilograms (kg) which is the SI unit for mass, and the radius was given in centimeters (cm) and converted to meters (m), the SI unit for length.

The conversion is done by multiplying by powers of 10 for each step up or down the SI prefix scale:

• To convert from grams to kilograms, you multiply by \( 10^{-3} \) because 1 gram is \( 10^{-3} \) kilograms.
• To convert from centimeters to meters, you multiply by \( 10^{-2} \) because 1 centimeter is \( 10^{-2} \) meters.

Proper unit conversion is crucial in our exercise, as it ensures that when we calculate the gravitational potential energy and the work done, our values are accurate and meaningful within the context of physical laws.