• Physics 9702 Doubts | Help Page 157

Question 777: [Dynamics > Equilibrium]

Diagram shows a crane supporting a load L.

A mass provides a balancing load W. Position of the load is such that the system is perfectly balanced with Wx = Ly. The ground provides a reaction force R. Distance x does not change.

If the load is moved further out so that distance y increases and the crane does not topple, which statement is correct?

Reference: Past Exam Paper – June 2012 Paper 11 Q15 & Paper 13 Q13

Solution 777:

Answer: D.

For equilibrium, the net torque on the system should be zero. That is, the clockwise moment should be equal to the anticlockwise moment.

The pivot in this system is in the middle of the vertical bar, at the top, where it crosses with the horizontal bar.

Wx = Ly

It is said that the distance x does not change.

Distance y has been increased. This increases the clockwise moment caused by load L. So we need a force that would cause an extra anticlockwise moment in order to achieve equilibrium again.

When the reaction force R moves to the right, it causes an anticlockwise moment about the pivot, thus achieving equilibrium again.

If the reaction force R moves to the lift, this could further increase the clockwise moment about the pivot. [C is incorrect]

Question 778: [Dynamics]

Brick weighing 20 N rests on an inclined plane. Weight of the brick has a component of 10 N parallel with the plane. The brick also experiences a frictional force of 4 N.

What is the acceleration of brick down the plane? Assume that the acceleration of free fall g is equal to 10 m s^–2.

A 0.3 m s^–2                  B 0.8 m s^–2                  C 3.0 m s^–2                  D 8.0 m s^–2

Reference: Past Exam Paper – June 2010 Paper 11 Q11 & Paper 12 Q14 & Paper 13 Q8

Solution 778:

Answer: C.

Weight = mg

Mass of brick, m = W / g = 20N / 10ms^-2 = 2kg

Resultant force F on brick = ma

Resultant force on brick = 10 – 4 N = 6N

Acceleration a = F / m = 6 / 2 = 3ms^-2

Question 779: [Current of Electricity > Resistance]

For June 2004 Paper 1 Q31:

Two wires made of same material and of the same length are connected in parallel to the same voltage supply. Wire P has a diameter of 2 mm. Wire Q has a diameter of 1 mm.

For November 2008 Paper 1 Q31 & June 2014 Paper 13 Q33:

Two wires P and Q made of same material and of the same length are connected in parallel to the same voltage supply. Wire P has diameter 2 mm and wire Q has diameter 1 mm.

What is the ratio current in P / current in Q?

A ¼                             B ½                             C 2                              D 4

Reference: Past Exam Paper – June 2004 Paper 1 Q31 & November 2008 Paper 1 Q31 & June 2014 Paper 13 Q33

Solution 779:

Answer: D.

Ohm’s law: V = IR

Current I = V / R

Resistance R of a wire = ρL / A

For the same material of the same length, resistivity ρ and length L are the same.

Cross-sectional area A = π(d/2)² = πd²/4

Resistance R is inversely proportional to area A (and thus to d²).

But current I is inversely proportional to resistance R, so current I is directly proportional to d².

Diameter of wire P = 2mm and diameter of wire Q is 1mm. 

Ratio of current in P to current in Q = 2² / 1² = 4 / 1

Question 780: [Current of Electricity > Charge]

Potential difference across a component in a circuit is 2.0 V.

How many electrons must flow through this component in order for it to be supplied with 4.8 J of energy?

A 2.6 × 10¹⁸                B 1.5 × 10¹⁹                 C 3.0 × 10¹⁹                 D 6.0 × 10¹⁹

Reference: Past Exam Paper – June 2014 Paper 11 Q35

Solution 780:

Answer: B.

Energy, E = Charge × p.d.

Energy, E = Q × V

Charge Q = E / V = 4.8 / 2.0 = 2.4C

Total charge Q = ne

where e is the charge of an electron and n is the number of electrons.

Number of electrons, n = Q / e = 2.4 / (1.6×10^-19) = 1.5×10¹⁹

Question 781: [Current of Electricity]

A student found two unmarked resistors. To determine resistance of the resistors, the circuit below was set up. The resistors were connected in turn between P and Q, noting the current readings. Voltage readings were noted without the resistors and with each resistor in turn.

Results were entered into a spreadsheet as shown.

The student forgot to enter column headings.

Which order of the headings would be correct?

Reference: Past Exam Paper – November 2014 Paper 13 Q34

Solution 781:

Answer: D.

The e.m.f. of the supply in the circuit is constant and independent of the other components in the circuit. Without the resistors in the circuit, the voltmeter will read the e.m.f. of the supply. Since the e.m.f is constant, the first column (which is the only column with constant values) would be values of the e.m.f. [B and C are incorrect]

Ohm’s law: V = IR

The supply may have some internal resistance r.

So, from Kirchhoff’s law,

p.d. across resistor + voltage drop due to internal resistance = e.m.f. of supply

p.d. across resistor = IR

Voltage drop due to internal resistance = Ir

IR + Ir = e.m.f.

I (R + r) = e.m.f.

To distinguish between the current I and resistance R columns, one method is to consider that the voltage drop due to internal resistance (= Ir) must be larger when the current is greater, so the first row of data is for a greater current than the second row.

The second column, which contain values close to 1.5, must be the p.d. V across the resistor. The values in this column increases (from 1.3 to 1.4) when the values of the last column increases (from 46 to 100). So, these must be the V and I columns respectively.

This can be verified as follows. From Ohm’s law, resistance R = V / I. The headings indicates that current is measured in mA – so we should multiply the value of the current column by a factor of 10^-3. So, the ratio of the V column (2^nd column) to that of the I column (3^rd column) should give the value of the resistance (4^th column).

Additionally, it can be seen that when the resistance R is greater, the current is smaller

Note that usually, when filling column, as in this example, the first column would be the values are remain constant (if any – here it is the e.m.f.) and the last column is a quantity that needs to be calculated (here the resistance). The middle columns usually contain the measured quantities (here, current and p.d.).