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Chapter 14: Problem 6

A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be (Assuming potential difference is same in both cases). (A) One-fourth (B) Halved (C) Doubled (D) Four times

Short Answer

Expert verified
The heat generated when only using one part of the coil is half of the original heat generation. Therefore, the correct answer is (B) Halved.

Step by step solution

01

Write the formula for heat generation

The heat generated in a resistor can be calculated using Joule's law, which states: \[H = I^2 R t\] Where H is the heat generated, I is the current, R is the resistance, and t is the time.
02

Write the formula for resistance

The resistance of a wire can be calculated using the formula: \[R = \frac{\rho L}{A}\] Where R is the resistance, 蟻 is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area.
03

Relate the resistance and heat generation

Substituting the expression for resistance into the heat generation formula, we get: \[H = I^2 \frac{\rho L}{A} t\]
04

Calculate the new resistance when the coil is cut in half

When the coil is cut into two equal parts, the length (L) of one part is half of the original length. So, the new resistance becomes: \[R' = \frac{\rho \frac{L}{2}}{A}\]
05

Calculate the new heat generated using one half of the coil

Using the new resistance value, we can calculate the new heat generation using Joule's law: \[H' = I^2 R' t = I^2 \frac{\rho \frac{L}{2}}{A} t\]
06

Compare the original heat generation and the new heat generation

Now, we will compare the heat generated in the original coil (H) with the heat generated in the half-coil (H'): \[\frac{H'}{H} = \frac{I^2 \frac{\rho \frac{L}{2}}{A} t}{I^2 \frac{\rho L}{A} t}\]
07

Simplify the equation and choose the correct answer

The current (I), resistivity (蟻), cross-sectional area (A), and time (t) cancel out, and we are left with: \[\frac{H'}{H} = \frac{\frac{1}{2}}{1}\] This means that the heat generated when only using one part of the coil is half of the original heat generation. Therefore, the correct answer is (B) Halved.

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Most popular questions from this chapter

A wire \(l=8 \mathrm{~m}\) long of uniform cross-sectional area \(A=8 \mathrm{~mm}^{2}\) has a conductance of \(G=2.45 \Omega^{-1}\). The resistivity of material of the wire will be (A) \(2.1 \times 10^{-7} \Omega \mathrm{m}\) (B) \(3.1 \times 10^{-7} \Omega \mathrm{m}\) (C) \(4.1 \times 10^{-7} \Omega \mathrm{m}\) (D) \(5.1 \times 10^{-7} \Omega \mathrm{m}\)

Figure \(14.47\) shows the circuit of a potentiometer. The length of the potentiometer wire \(A B\) is \(50 \mathrm{~cm}\). The EMF \(E_{1}\) of the battery is \(4 \mathrm{~V}\), having negligible internal resistance. Value of \(R_{1}\) and \(R_{2}\) are \(15 \Omega\) and \(5 \Omega\), respectively. When both the keys are open, the null point is obtained at a distance of \(31.25 \mathrm{~cm}\) from \(A\), but when both the keys are closed, the balance length reduces to \(5 \mathrm{~cm}\) only. Given \(R_{A B}=10 \Omega\) The balance length when key \(K_{2}\) is open and \(K_{1}\) is closed is given by (A) \(10.5 \mathrm{~cm}\) (B) \(11.5 \mathrm{~cm}\) (C) \(12.5 \mathrm{~cm}\) (D) \(13.5 \mathrm{~cm}\)

If \(\mathrm{EMF}\) in a thermocouple is \(\varepsilon=\alpha T+\beta T^{2}\), then the neutral temperature of the thermocouple is (A) \(-\beta /(2 \alpha)\) (B) \(-2 \beta / \alpha\) (C) \(-\alpha /(2 \beta)\) (D) \(-2 \alpha / \beta\)

In the circuit shown in Fig. \(14.27\), the heat produced in the \(5 \Omega\) resistor due to a current flowing in it is 10 calories per second. The heat produced in the \(4 \Omega\) resistor is (A) \(1 \mathrm{cal} \mathrm{s}^{-1}\) (B) \(2 \mathrm{cal} \mathrm{s}^{-1}\) (C) \(3 \mathrm{cal} \mathrm{s}^{-1}\) (D) \(4 \mathrm{cal} \mathrm{s}^{-1}\)

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