CUET UG Physics Booster Test – 2 Material Properties and Power

In electric power transmission, some power is inevitably lost as heat due to the resistance of transmission cables. According to NCERT, the power loss in a transmission line is given by: P_(loss)=I^2R First, calculate the current supplied through the cables. P=VI1000=100×II=10A Now calculate the power loss: P_(loss)=I^2R=(10)^2×10=100×10=1000W Thus, the power wasted in the cables is 1000 W.

According to NCERT, the power loss in transmission lines is: P_(loss)=I^2R To transmit a fixed amount of power, P=VI If the transmission voltage is increased significantly, the current required becomes smaller. Since power loss depends on the square of current, even a small reduction in current produces a large reduction in transmission losses. Therefore, Statement 2 is correct. After electricity reaches the destination, transformers are used to reduce the voltage to safe and practical values for domestic and industrial use. Hence Statement 4 is also correct. Statement 1 is incorrect because higher resistance increases energy loss. Statement 3 is incorrect because increasing current increases power dissipation in the cables. Thus, only Statements 2 and 4 are correct.

In an electrolytic cell or electrochemical cell, the positive and negative electrodes develop potentials relative to the electrolyte. When the circuit is open and no current flows, the total potential difference across the terminals is given by: ε=V_++V_- This quantity is called the electromotive force (EMF) of the cell. Despite its name, EMF is not a force. It is actually a potential difference measured when the cell supplies no current. According to NCERT, EMF represents the energy supplied by the cell per unit charge under open-circuit conditions. Once current begins to flow, the terminal voltage may differ from the EMF because of the internal resistance of the cell. Therefore, the quantity V_++V_-measured when no current flows is called the electromotive force.

In an electrolytic cell, the positive electrode develops a positive potential relative to the electrolyte, represented by V_+. Similarly, the negative electrode develops a potential -(V_-)relative to the electrolyte. When no current flows through the cell, the electrolyte is in equilibrium and remains at the same potential throughout. This condition corresponds to the open-circuit state. Under open-circuit conditions, current is zero and the cell exhibits its electromotive force. Therefore: • Positive electrode → develops V_+>0 • Negative electrode → develops -(V_-) • Electrolyte → same potential throughout • Open circuit → current zero Thus, the correct matching is 1-d, 2-c, 3-b and 4-a.

According to NCERT, the term electromotive force is somewhat misleading because it is not a force at all. It is actually a potential difference measured across the terminals of a cell when no current is flowing through the circuit. The term originated historically before the modern understanding of electric potential and has continued to be used. EMF represents the energy supplied by a source per unit charge. It is measured in volts and determines the maximum potential difference that the source can provide. When the circuit is open, no current flows and the terminal voltage equals the EMF. Once current begins to flow, the terminal voltage may differ from the EMF because of internal resistance effects. Thus, Statements 1, 2 and 3 are correct, while Statement 4 is incorrect because EMF is not a mechanical force.

Under open-circuit conditions, the external circuit is incomplete and therefore no current flows through the cell. Since the current is zero, the potential drop across the internal resistance of the cell is also zero. The relation between emf and terminal voltage is: V=ε-Ir For an open circuit, I=0 Therefore, V=ε This means the terminal voltage becomes exactly equal to the emf of the cell. The internal resistance does not produce any voltage drop because no current passes through it. Hence, the statement that the terminal voltage is strictly less than the emf is incorrect under open-circuit conditions.

Every practical cell possesses some resistance within itself due to its electrolyte, electrodes and internal construction. This resistance is known as internal resistance. When a current I flows through the cell, a voltage drop develops across the internal resistance according to Ohm's law: V_(drop)=Ir As a result, the terminal voltage of the cell becomes less than its emf: V=ε-Ir The larger the current, the greater the internal voltage drop. This is why cells supplying large currents show a noticeable reduction in terminal voltage. Thus, internal resistance is responsible for the loss of potential inside the cell and produces a voltage drop equal to Ir.

When a cell supplies current to an external circuit, part of its emf is consumed in overcoming its internal resistance. According to NCERT, the voltage drop across the internal resistance is: Ir Therefore, the terminal voltage available across the external resistance is: V=ε-Ir This equation shows that the terminal voltage is always less than the emf when the cell is discharging. The difference between emf and terminal voltage is exactly equal to the voltage lost inside the cell. The equation is widely used in numerical problems involving cells, batteries and electrical circuits. Hence, the correct expression for terminal voltage is ε-Ir.

When two cells are connected in series such that the negative terminal of one is connected to the positive terminal of the other, the cells assist each other. This arrangement is known as series aiding. If the emfs of the cells are ε_1 and ε_2, the equivalent emf becomes: ε_(eq)=ε_1+ε_2 Similarly, if their internal resistances are r_1 and r_2, then: r_(eq)=r_1+r_2 Such a combination is used when a larger voltage is required from multiple cells. Flashlights, battery packs and many electrical devices use this principle. Therefore, the equivalent emf of cells connected in series aiding is equal to the sum of their individual emfs.

When resistances are connected in series, their values add directly. The same principle applies to the internal resistances of cells connected in series. Given: r=0.5Ω Number of cells: n=3 Equivalent internal resistance: r_(eq)=nr=3×0.5=1.5Ω Thus, the equivalent internal resistance of the three-cell combination is 1.5 Ω. This result is important because increasing the number of cells in series increases both the total emf and the total internal resistance.

Kirchhoff's Junction Rule is based on the principle of conservation of charge. According to NCERT, in a steady-state circuit, charge cannot accumulate at any junction. Therefore, the total current entering a junction must equal the total current leaving it. For a junction in a parallel cell arrangement, I=I_1+I_2 where I is the current entering the junction and I_1 and I_2 are currents leaving through different branches. The rule states that as much charge enters a junction per second as leaves it per second. This condition ensures continuous current flow without charge accumulation. The validity of the junction rule does not depend on whether the cells are identical or different. Hence, Statements 1 and 3 are correct.

When two cells are connected in parallel, their internal resistances are also connected in parallel. According to the parallel resistance formula, 1/r_(eq)=1/r_1+1/r_2 Taking the reciprocal, r_(eq)=r_1r_2/r_1+r_2 This equivalent resistance is always smaller than either individual resistance. Parallel combination of cells is useful when a larger current output is required while keeping the emf unchanged. The result follows directly from the standard rule for combining resistances in parallel and is frequently used in NCERT-based numerical problems involving cell combinations.

Kirchhoff's Rules are based on two fundamental conservation laws. The junction rule follows from conservation of charge and states that the sum of currents entering a junction equals the sum of currents leaving it. This condition is valid only when there is no accumulation of charge at the junction. The loop rule follows from conservation of energy and states that the algebraic sum of all potential changes around a closed loop is zero. As a charge moves around a complete loop and returns to its starting point, its net energy change must be zero. Thus: • Steady currents → No accumulation of charges at any junction. • Junction rule → Sum of currents entering equals leaving. • Loop rule → Algebraic sum of potential changes is zero. • Total energy → Conserved around a closed loop. Hence, Option A gives the correct matching.

Kirchhoff's Junction Rule is based on the principle of conservation of charge. In a steady electrical circuit, charge cannot continuously accumulate at any point. Therefore, the total current entering a junction must equal the total current leaving it. Mathematically, ∑I_(in)=∑I_(out) The physical shape of the conducting wire has no effect on the validity of this law. Whether the wire is straight, bent or curved, the conservation of charge remains valid. Statement 4 is incorrect because charge accumulation would cause changing currents and violate steady-state conditions. Therefore, only Statements 1 and 2 are correct.

Kirchhoff's Loop Rule states that the algebraic sum of all potential rises and potential drops around any closed loop is zero. ∑ΔV=0 This principle follows from conservation of energy. As a charge moves around a complete loop and returns to its starting point, there can be no net gain or loss of energy. The loop may contain cells, resistors and other circuit elements. Potential rises occur across cells, while potential drops occur across resistors. Their algebraic sum must always equal zero. Option D is incorrect because the loop rule does not require the total potential change to be greater than the emf of any cell. In fact, the net change around the entire loop is zero.

Kirchhoff's Loop Rule states that the algebraic sum of all potential rises and potential drops around any closed loop is zero. ∑ΔV=0 This rule follows directly from the law of conservation of energy. As a charge moves through a complete loop and returns to its starting point, there can be no net gain or loss of electrical energy. The loop may contain cells, resistors and other circuit elements. Potential rises occur across cells, while potential drops occur across resistors. These changes exactly balance each other when summed algebraically. Therefore, Statement 1 is correct. Statement 4 is also correct because the rule is routinely applied to loops containing cells. Statements 2 and 3 are incorrect because the rule is based on energy conservation and does not require current to be zero.

The Wheatstone bridge is a network consisting of four resistors arranged in the form of a quadrilateral. According to NCERT, a source of emf is connected across one pair of diagonally opposite junctions, while a galvanometer is connected across the other pair. The galvanometer is used as a sensitive current detector. It helps determine whether the bridge is balanced or unbalanced. When the bridge reaches the balanced condition, the potential difference across the galvanometer becomes zero and no current flows through it. The arrangement of the battery and galvanometer across opposite diagonals is one of the defining features of the Wheatstone bridge. Therefore, the component connected across the second diagonal is the galvanometer.

A Wheatstone bridge is said to be balanced when the potential difference across the galvanometer is zero. Since both ends of the galvanometer are at the same potential, no current flows through it. As a result, the galvanometer shows a null deflection. This condition is called the null point or balance point of the bridge. The battery still supplies current through the resistor arms of the bridge. Also, the four resistors need not be equal. They only need to satisfy the balance condition: R_2/R_1=R_4/R_3 The balanced condition is extremely important because it allows accurate determination of unknown resistance values.

For a balanced Wheatstone bridge, R_2/R_1=R_4/R_3 Substituting the given values: 20/10=R_4/52=R_4/5 Multiplying both sides by 5: R_4=10Ω Thus, the unknown resistance is 10 Ω. The Wheatstone bridge balance condition is one of the most important applications of Kirchhoff's laws and is widely used in resistance measurements. At balance, no current flows through the galvanometer, making the calculation highly accurate.

In a Wheatstone bridge experiment, the known resistances R_1 and R_2 are generally kept fixed, while another resistance is adjusted until the galvanometer shows a null deflection. At this stage, the bridge is balanced. The balance condition is: R_2/R_1=R_4/R_3 Rearranging, R_4=R_3(R_2/R_1) This equation allows the unknown resistance R_4 to be determined accurately. The value of the galvanometer's internal resistance is not required because no current flows through it at balance. Moreover, Wheatstone bridges are normally operated using direct current sources rather than requiring alternating current. Therefore, Statements 1 and 2 are correct.