Q7. Two identical metal wires of thermal conductivities K1 and K2 respectively are connected in series. The effective thermal conductivity of the combination is: (1) 2 K1 K2 (2) K1+K2 K1+K2 2 K1 K2 (3) K1+K2 (4) K1 K2 K1 K2 K1+K2
What This Question Tests
This question directly tests the formula for equivalent thermal conductivity of two identical wires connected in series.
Concepts Tested
Formulas Used
K_eq = (L₁ + L₂)/(L₁/K₁ + L₂/K₂) for series
For identical wires, K_eq = 2K₁K₂/(K₁+K₂)
📚 NCERT Sections This Tests
11.3 — Zeroth Law Of Thermodynamics (A)
Physics Class 12 · Chapter 11
11.3 ZEROTH LAW OF THERMODYNAMICS (a) Imagine two systems A and B, separated by an adiabatic wall, while each is in contact with a third system C, via a conducting wall [Fig. 11.2(a)]. The states of the systems (i.e., their macroscopic variables) will change until both A and B come to thermal equilibrium with C. After this is achieved, suppose that the adiabatic wall between A and B is replaced by a conducting wall and C is insulated from A and B by an adiabatic wall [Fig.11.2(b)]. It is found that the states of A and B change no (b) further i.e. they are found to be in thermal Fig. 11.2 (a) Systems A and B are separated by an equilibrium with each other. This observation adiabatic wall, while each is in contact forms the basis of the Zeroth Law of with a third system C via a conducting Thermodynamics, which states that ‘two wall. (b) The adiabatic wall between A systems in thermal equilibrium with a third and B is replaced by a conducting wall, system separately are in thermal equilibrium while C is insulated from A and B by an adiabatic wall.with each other’. R.H. Fowler formulated this * Both the variables need not change. It depends on the constraints. For instance, if the gases are in containers of fixed volume, only the pressures of the gases would change to achieve thermal equilibrium. Reprint 2025-26 THERMODYNAMICS 229
2.14 — Combination Of Capacitors
Physics Class 11 · Chapter 2
2.14 COMBINATION OF CAPACITORS We can combine several capacitors of capacitance C1, C2,…, Cn to obtain a system with some effective capacitance C. The effective capacitance depends on the way the individual capacitors are combined. Two simple possibilities are discussed below. 2.14.1 Capacitors in series Figure 2.26 shows capacitors C1 and C2 FIGURE 2.26 Combination of two combined in series. capacitors in series. The left plate of C1 and the right plate of C2 are connected to two terminals of a battery and have charges Q and –Q , respectively. It then follows that the right plate of C1 has charge –Q and the left plate of C2 has charge Q. If this was not so, the net charge on each capacitor would not be zero. This would result in an electric field in the conductor connecting C1and C2. Charge would flow until the net charge on both C1 and C2 is zero and there is no electric field in the conductor connecting C1 and C2. Thus, in the series combination, charges on the two plates FIGURE 2.27 Combination of n (±Q) are the same on each capacitor. The total capacitors in series. 71 Reprint 2025-26 Physics potential drop V across the combination is the sum of the potential drops V1 and V2 across C1 and C2, respectively. Q Q + (2.55) V = V1 + V2 = C1 C 2 V 1 1 i.e., = + , (2.56) Q C1 C 2 Now we can regard the combination as an effective capacitor with charge Q and potential difference V. The effective capacitance of the combination is Q C = (2.57) V We compare Eq. (2.57) with Eq. (2.56), and obtain 1 1 1 = + (2.58) C C1 C 2 The proof clearly goes through for any number of capacitors arranged in a similar way. Equation (2.55), for n capacitors arranged in series, generalises to Q Q Q V = V1 + V 2 + ... + V n = + + ... + (2.59) C1 C 2 C n Following the same steps as for the case of two capacitors, we get the general formula for effective capacitance of a series combination of n capacitors: 1 1 1 1 1 = + + + ... + (2.60) C C1 C 2 C 3 C n 2.14.2 Capacitors in parallel Figure 2.28 (a) shows two capacitors arranged in parallel. In this case, the same potential difference is applied across both the capacitors. But the plate charges (±Q1) on capacitor 1 and the plate charges (±Q2) on the capacitor 2 are not necessarily the same: Q1 = C1V, Q2 = C2V (2.61) The equivalent capacitor is one with charge Q = Q1 + Q2 (2.62) and potential difference V. Q = CV = C1V + C2V (2.63) The effective capacitance C is, from Eq. (2.63), C = C1 + C2 (2.64) The general formula for effective capacitance C for parallel combination of n capacitors [Fig. 2.28 (b)] follows similarly, Q = Q1 + Q2 + ... + Qn (2.65) FIGURE 2.28 Parallel combination of i.e., CV = C1V + C2V + ... CnV(2.66) (a) two capacitors, (b) n capacitors. which gives C = C1 + C2 + ... Cn (2.67) 72 Reprint 2025-26 Electrostatic Potential and Capacitance Example 2.9 A network of four 10 mF capacitors is connected to a 500 V supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.) FIGURE 2.29 Solution (a) In the given network, C1, C2 and C3 are connected in series. The effective capacitance C¢ of these three capacitors is given by 1 1 1 1 = + + C ′ C1 C 2 C 3 For C1 = C2 = C3 = 10 mF, C¢ = (10/3) mF. The network has C¢ and C4 connected in parallel. Thus, the equivalent capacitance C of the network is 10 C = C¢ + C4 = 3 + 10 mF =13.3mF (b) Clearly, from the figure, the charge on each of the capacitors, C1, C2 and C3 is the same, say Q. Let the charge on C4 be Q¢. Now, since the potential difference across AB is Q/C1, across BC is Q/C2, across CD is Q/C3 , we have Q Q Q + + = 500 V . C1 C 2 C 3 Also, Q¢/C4 = 500 V. This gives for the given value of the capacitances, 10 −3 Q = 500 V × µ F = 1.7 × 10 C and EXAMPLE 3 Q ′ = 500 V × 10 µ F = 5.0 × 10 −3 C 2.9
11.2 — Thermal Equilibrium
Physics Class 12 · Chapter 11
11.2 THERMAL EQUILIBRIUM independent variables. Let the pressure and Equilibrium in mechanics means that the net volume of the gases be (PA, VA) and (PB, VB) external force and torque on a system are zero. respectively. Suppose first that the two systems The term ‘equilibrium’ in thermodynamics appears are put in proximity but are separated by an * Thermodynamics may also involve other variables that are not so obvious to our senses e.g. entropy, enthalpy, etc., and they are all macroscopic variables. However, a thermodynamic state is specified by five state variables viz., pressure, volume, temperature, internal energy and entropy. Entropy is a measure of disorderness in the system. Enthalpy is a measure of total heat content of the system. Reprint 2025-26 228 PHYSICS adiabatic wall – an insulating wall (can be law in 1931 long after the first and second Laws movable) that does not allow flow of energy (heat) of thermodynamics were stated and so numbered. from one to another. The systems are insulated The Zeroth Law clearly suggests that when two from the rest of the surroundings also by similar systems A and B, are in thermal equilibrium, adiabatic walls. The situation is shown there must be a physical quantity that has the schematically in Fig. 11.1 (a). In this case, it is same value for both. This thermodynamic found that any possible pair of values (PA, VA) will variable whose value is equal for two systems in be in equilibrium with any possible pair of values thermal equilibrium is called temperature (T ). (PB, VB). Next, suppose that the adiabatic wall is Thus, if A and B are separately in equilibrium replaced by a diathermic wall – a conducting wall with C, TA = TC and TB = TC. This implies that that allows energy flow (heat) from one to another. TA = TB i.e. the systems A and B are also in It is then found that the macroscopic variables of thermal equilibrium. the systems A and B change spontaneously until We have arrived at the concept of temperature both the systems attain equilibrium states. After formally via the Zeroth Law. The next question that there is no change in their states. The is : how to assign numerical values to situation is shown in Fig. 11.1(b). The pressure temperatures of different bodies ? In other words, and volume variables of the two gases change to how do we construct a scale of temperature ? (PB ′, VB ′) and (PA ′, VA ′) such that the new states Thermometry deals with this basic question to of A and B are in equilibrium with each other*. which we turn in the next section. There is no more energy flow from one to another. We then say that the system A is in thermal equilibrium with the system B. What characterises the situation of thermal equilibrium between two systems ? You can guess the answer from your experience. In thermal equilibrium, the temperatures of the two systems are equal. We shall see how does one arrive at the concept of temperature in thermodynamics? The Zeroth law of thermodynamics provides the clue.
📋 Question Details
- Chapter
- Thermodynamics & KTG
- Topic
- Thermal conductivity (series combination)
- Year
- 2021
- Shift
- 17 Mar Shift 1
- Q Number
- Q7
- Type
- MCQ
- NCERT Ref
- Class 11 Physics Ch 11: Thermal Properties of Matter
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