• Physics 5054 Doubts | Help Page 2

Question 2: [Turning effect of forces]

The diagram shows a plank of mass 5.0 kg and length 3.00 m resting horizontally on two trestles, P and Q, which are a distance of 2.50 m apart. When a student of mass 60.0 kg walks along the plank from one trestle to the other, the plank sags.

(a) Explain why the sag increases as the student walks towards the middle of the plank.

(b) Calculate the downward force exerted on each trestle when the student is at a distance of

(i) 0.50 m from trestle P,

(ii) 1.25 m from trestle P.

(g = 10.0 N/kg)

Reference: Past Exam Paper – N91 / P2 / QA2

Solution 2:

(a)

Assuming the plank is uniform, the weight of the plank acts at the midpoint of the plank causing the sag. As the bog walks from P towards the middle of the plank, the combined centre of gravity of the system of plank and boy acts in between the boy and the midpoint of the plank causing a larger sag. As the boy moves even nearer to the midpoint, the sag increases as the combined c.g. moves towards the centre of the plank.

(b)

(i) When x = 0.50 m from P,

By the Principle of Moments, moment about P

R_Q ×2.50 = (600×0.50) + (50×1.25)

R_Q = 145 N

For vertical equilibrium,

R_P + R_Q = 600 + 50 = 650 N

R_P = 650 – R_Q = 650 – 145 = 505 N

(ii) When x = 1.25 m from P, both the forces 600 N and 50 N act at the centre of mass of the plank.

By symmetry, R_P = R_Q            where  R_P + R_Q = 600 + 50 = 650 N

R_P = R_Q = 650/2 = 325 N

• Physics 5054 Doubts | Help Page 1

Question 1: [Electricity and Magnetism > D.C. Circuits]

A student sets up the circuit shown in Fig.

R is a fixed resistor in the circuit. Filament lamp is marked 12 V, 0.25 A.

Circuit is used to produce a current/voltage graph for the filament lamp. Ammeter and voltmeter needed are not shown.

To obtain different readings, the student changes position of the movable contact.

(a) On Fig, draw the symbols for ammeter and voltmeter in the correct positions.

(b) Explain why it is sensible to include resistor R in this circuit.

(c)

(i) On Fig, sketch current/voltage graph for the lamp.

(ii) State and explain how current/voltage graph for a fixed resistor is different from graph for a filament lamp.

(d) Fig shows position of the movable contact when the voltage across the lamp is 12 V and the current in the lamp is 0.25 A.

Determine

(i) current in the 50 Ω resistor

(ii) current in R

(iii) potential difference (p.d.) across R

(iv) resistance of R

Reference: Past Exam Paper – June 2014 Paper 22 Q11

Solution 1:

(a) The ammeter and the voltmeter should be drawn with the correct symbols. The ammeter is in series with the lamp and the voltmeter is in parallel with the lamp.

(b) The resistor R limits or reduces the current / voltage. Otherwise the lamp blows OR more of the 50 Ω can be used to adjust voltage / current.

(c)

(i) 12 V, 0.25 A correctly plotted (by eye). The graph is a curved line from the origin with a correct curvature – decreasing slope

(ii) The graph is a straight line (for a fixed resistor). The lamp has changing temperature or changing resistance OR the fixed resistor has constant temperature or constant

(d)

(i) {From Kirchhoff’s law, the p.d. across the 50 Ω resistor is the same as that across the filament lamp (since there are parallel in different loops), which is stated to be 12V.}

(Current I =) V / R in any algebraic or numerical for, e.g. 12/50

Current I = 0.24 A

(ii) {At a junction, the current splits and go into the different components. So, the current in the resistor R is the sum of currents passing through the components in parallel.}

Current (= 0.25 + 0.24) = 0.49 A

(iii) {The circuit consists of a potential divider between resistor R and the parallel combination of the lamp and the 50 Ω. From Kirchhoff’s law, the sum of p.d. across R and the parallel combination should be equal to the e.m.f of 18V. Since the p.d. across the parallel combination is 12V, the p.d. across R is 18 – 12 = 6V.}

p.d. = 6(.0) V

(iv) {The current through R is 0.49A and the p.d. across it is 6V. From Ohm’s law, R = V / I = 6 / 0.49 = 12.24Ω.}

Resistance = 12(.24) Ω