Question

Ten identical cells connected in series are needed to heat a wire of length one meter and radius 'r' by $10^\circ C$ in time 't'. How many cells will be required to heat the wire of length two meter of the same radius by the same temperature in time 't' ?

• A. 20
• B. 30
• C. 40
• D. 10

Answer 1

The Correct Option is A

To solve this problem, we must understand the relationship between the power supplied by the cells to the wire and how this affects the temperature change in the wire. The cells' power and the wire’s resistance are key elements here.

Step 1: Understanding Power Calculation

The power supplied to the wire can be calculated using Joule's law as:

P = I^2R, where I is the current and R is the resistance.

The resistance R of a uniform wire of length L and radius r is given by:

R = \rho \frac{L}{A}, where A = \pi r^2 is the cross-sectional area, and \rho is the resistivity.

Step 2: Initial Problem Setup

Given that ten identical cells are needed to increase the temperature by 10^\circ C for a 1-meter wire in time t, let's analyze the energy equation.

The energy required to change the temperature of the wire is:

Q = mc\Delta T,

where m is the mass of the wire, c is the specific heat capacity, and \Delta T is the temperature change.

The power supplied by 10 cells, P_{10}, for time t is:

P_{10} \times t = mc\Delta T

Step 3: Solving for 2-Meter Wire

Now, consider a wire of length 2 meters and the same radius. Its resistance will be twice, (R_{2L} = 2R).

The energy required for the 2-meter wire to have the same temperature change in the same time t is:

2P_{10} \times t = 2mc \Delta T

To supply this energy using identical cells in series, we need to increase the power by a factor of two. This can be achieved by using twice the number of cells.

Conclusion:

Hence, to heat a 2-meter wire to the same temperature in the same time, we need double the number of cells, which is 20 cells.

The correct answer is: 20