# Physics 9702 Doubts | Help Page 154

__Question 762: [Current of Electricity]__
In the circuit below, ammeter
reading is I and voltmeter reading is V.

When switch is closed, which row describes
what happens to I and V?

I
V

A decreases
decreases to zero

B increases
decreases to zero

C increases
stays the same

D stays
the same increases

**Reference:**

*Past Exam Paper – June 2012 Paper 11 Q36 & Paper 13 Q35*

__Solution 762:__**Answer: B.**

When the switch is opened, current
passes through both resistors in the circuit.

Let the resistance of one resistor
be R. Assume they are of the same resistance.

When switch is opened, Total resistance
R

_{T}in circuit = R + R = 2R
Ohm’s law: V = IR

This gives a current I (= E / R

_{T}where E is the e.m.f. of the supply). The total e.m.f. in the circuit is divided as p.d. across each of the resistor.
When the switched is closed, the
resistor connected in parallel to the wire is short circuited. That is, all
current would pass through the wire (of negligible resistance) instead of passing
through that resistor.

When switch is closed, Total resistance
R

_{T}in circuit = R + 0 = R
This gives a current I (= E / R

_{T}where E is the e.m.f. of the supply) which is now greater than before since R_{T}is less.
Ohm’s law for the resistor: V = IR

Since the current I through the
resistor (in parallel) is zero, the p.d. across it is also zero.

__Question 763: [Current of Electricity]__
A cell, two resistors of equal
resistance and ammeter are used to construct four circuits. The resistors are
the only parts of the circuits that have resistance.

In which circuit will ammeter show
the greatest reading?

**Reference:**

*Past Exam Paper – November 2011 Paper 11 Q37 & Paper 13 Q36*

__Solution 763:__**Answer: D.**

Ohms’ law: V = IR

Current I in the whole circuit = V /
R

_{T}where R_{T}is the combined resistance of the resistors.
Let the resistance of 1 resistor = R.
Let the e.m.f. of the supply = E

Circuit A consists of a series
connection of the 2 resistors. So, R

_{T}= R + R = 2R. Current I = E / 2R. Here, the current will be the lowest.
To measure the current flowing
through a resistor, the ammeter, which has a negligible resistance, must be
connected in series with it. If it is connected in parallel with a component,
the current would flows through it instead of the component (since the ammeter
has zero resistance) and that component would be short circuited.

In circuit B, R

_{T}= R + 0 = R. Current I = E / R.
Circuit C and D both consists of a
parallel combination of the 2 resistors.

Combined resistance R

_{T}in circuit = [1/R + 1/R]^{-1}= R / 2
Current I in circuit = E / (R/2) = 2
(E / R)

These 2 circuit contain the greatest
current of all the 4 choices. However, at a junction, the current splits. So,
the ammeter in circuit C gives only half the total current in the circuit.

Current flows from the positive terminal
of the supply to its negative terminal. In choice D, the ammeter is connected
directly to the positive terminal, so it reads the value of the current flowing
in the whole circuit.

__Question 764: [Dynamics > Collisions]__
Diagram shows a particle X, with
kinetic energy E

_{k}, about to collide with a stationary particle Y.
Both particles have the same mass.

After colliding, X and Y travel
onwards together as a single larger particle.

How much kinetic energy is lost in
collision?

A 0 B
E

_{k}/ 4 C E_{k}/ 2 D 3E_{k}/ 4**Reference:**

*Past Exam Paper – November 2014 Paper 13 Q16*

__Solution 764:__**Answer: C.**

In any closed system, momentum is always
conserved.

Before collision,

Let speed of X be v. Let mass of X =
mass of Y = m

Momentum of X = mu

Momentum of Y = m(0) = 0

After collision,

Both particles travel as a single
larger particle.

Mass = m + m = 2m

Let the velocity after collision be
v.

Total momentum = (2m)v = 2mv

From the conservation of momentum,

2mv = mu

Velocity v = u / 2

Kinetic energy before collision, E

_{k}= ½ mu^{2}
Kinetic energy after collision = ½ (2m)
(u/2)

^{2}= ¼ mu^{2}
Lost in Kinetic energy = ½ mu

^{2}– (¼ mu^{2}) = ¼ mu^{2}= E_{k}/ 2

__Question 765: [Current of Electricity > Potential divider]__
In circuit shown, XY is a length L
of uniform resistance wire. R

_{1}and R_{2}are unknown resistors. J is a sliding contact that joins the junction of R_{1}and R_{2}to points on XY through small signal lamp S.
To determine ratio V

_{1}/ V_{2}of the potential differences across R_{1}and R_{2}, a point is found on XY at which the lamp is off. This point is at a distance x from X.
What is value of the ratio V

_{1}/ V_{2}?
A L / x B x / L C
(L – x) / x D x
/ (L – x)

**Reference:**

*Past Exam Paper – June 2011 Paper 11 Q37 & Paper 13 Q36*

__Solution 765:__**Answer: D.**

For the lamp to be off, the
potential difference across it should be zero so that no current flows through
it. That is, the potential at the junction of R

_{1}and R_{2}should be equal to the potential at the sliding contact J
The p.d. across R

_{1}is V_{1}and the p.d. across R_{2}is V_{2}. So, V_{1}should be equal to the p.d. across XJ and V_{2}should be equal to the p.d. across JY.
Since point X is connected to the
negative terminal of the supply, it can be considered to be at a potential of
0. Point Y is at the maximum potential, equal to the e.m.f. of the supply.

The wire XY, together with the
sliding contact J, can be considered as a potential divider.

Resistance R of the wire = ρL / A

Since ρ and A are assumed to be constant,
the resistance of the wire is proportional to the length L.

Ohm’s law: V = IR

The p.d. across the wire is
proportional to R, which is itself proportional to L. So, the p.d. across the
wire is proportional to the length of wire being considered.

A length of L corresponds to a p.d.
equal to the e.m.f. E of the supply

A length of x corresponds to a p.d.
equal to (E / L) × x = (x / L) × E

Note that this is a p.d. (potential
difference – or in other words, a difference in potential). Since X is at a potential of zero, point J would
be at a potential of [(x / L) × E] – 0 = (x / L) × E.

V

_{1}= (x / L) × E
Now, the part JY of the wire
corresponds to a length (L – x) of the wire.

A length of (L – x) corresponds to a
p.d. equal to (E / L) × (L – x) = [(L – x) / L] × E

V

_{2}= [(L – x) / L] × E
Ratio = V

_{1}/ V_{2}= x / (L – x)

__Question 766: [Pressure]__
Atmospheric pressure at sea level
has value of 100 kPa.

Density of sea water is 1020 kg m

^{–3}.
At what depth in the sea would total
pressure be 110 kPa?

A 1.0 m B 9.8 m C
10 m D 11 m

**Reference:**

*Past Exam Paper – June 2010 Paper 11 Q17 & Paper 12 Q18 & Paper 13 Q15*

__Solution 766:__**Answer: A.**

The pressure at sea level is 100 kPa
due to the atmosphere (atmospheric pressure).

The pressure

**only**due to a liquid column of depth h and density ρ is hρg.
Total pressure at a depth h =
Atmospheric pressure + hρg

110 000 = 100 000 + (h × 1020 ×
9.81)

Depth h = 10 000 / (1020 × 9.81) =
0.999 = 1.0m

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