# Physics 9702 Doubts | Help Page 206

__Question 992: [Alternating Current > Transformer]__
A student is asked to design a
circuit by which a direct voltage of peak value 9.0 V is obtained from a 240 V
alternating supply.

The student uses a transformer that
may be considered to be ideal and a bridge rectifier incorporating four ideal
diodes.

The partially completed circuit
diagram is shown in Fig. 6.1.

**(a)**On Fig. 6.1, draw symbols for the four diodes so as to produce the polarity across the load as shown on the diagram.

**(b)**Calculate the ratio

number of turns on the secondary coil / number of turns on the
primary coil

**Reference:**

*Past Exam Paper – June 2010 Paper 41 Q6*

__Solution 992:__**(a)**

**(b)**

N

_{S}/ N_{P}= V_{S}/ V_{P}
Peak voltage V

_{0}= √2 × V_{rms}
{V

_{S}= 9.0 V. V_{P}= √2 × 240 V since 240V is the V_{rms}of the primary alternating voltage}
ratio = 9.0 / (√2 × 240) = 1/38 or 1/37 or 0.027

__Question 993: [Current of Electricity > Resistance]__
A wire RST is connected to another
wire XY as shown.

Each wire is 100 cm long with a
resistance per unit length of 10 Î© m

^{–1}.
What is the total resistance between
X and Y?

A 3.3 Î© B 5.0 Î© C
8.3 Î© D 13.3 Î©

**Reference:**

*Past Exam Paper – June 2015 Paper 11 Q38*

__Solution 993:__**Answer: C.**

Resistance per unit length of wire =
10 Î© m

^{–1}
Resistance of wire RST = 1m × 10 Î©
m

^{–1}= 10 Î©
Since the wire XY is 100cm, the sum
of XR + TY = 100 – 50 = 50cm

Sum of resistance of (XR+TY) = 0.5m × 10 Î©
m

^{–1}= 5 Î©
Resistance of wire RT = 0.5m × 10 Î©
m

^{–1}= 5 Î©
Hence the arrangement is effectively
a 10 Î© resistor RST in parallel with a 5 Î© resistor RT, and in series with another
5 Î© resistor (XR+TY).

Effective resistance of parallel
combination = [1/10 + 1/5]

^{-1}= 3.3 Î©
The total resistance is therefore 5
Î© + 3.3 Î© = 8.3 Î©.

__Question 994: [Electric field]__
A charged point mass is situated in
a vacuum. A proton travels directly towards the mass, as illustrated in Fig.
4.1.

When the separation of the mass and
the proton is r, the electric potential energy of the system is U

_{P}.
The variation with r of the
potential energy U

_{P}is shown in Fig. 4.2.**(a)**(i) Use Fig. 4.2 to state and explain whether the mass is charged positively or negatively.

(ii) The gradient at a point on the
graph of Fig. 4.2 is G.

Show that the electric field
strength E at this point due to the charged point mass is given by the
expression

Eq = G

where q is the charge at this point.

**(b)**Use the expression in (a)(ii) and Fig. 4.2 to determine the electric field strength at a distance of 4.0 cm from the charged point mass.

**Reference:**

*Past Exam Paper – June 2012 Paper 42 Q4*

__Solution 994:__**(a)**(i) As the separation r decreases, the electric potential energy U

_{p}decreases / work got out due to

__attraction__(attractive force) between the mass and the proton. So, the point mass is negatively charged.

(ii)

Electric potential energy is the
product of charge and electric potential. {U

_{p}= qV}
The electric field strength is the (electric)
potential gradient (rate of change of potential with respect to displacement). {E = dV / dr – may remove the ‘d’s}

So, the electric field strength is
equal to the gradient of the potential energy graph divided by charge.

{G = gradient = U

_{p }/ r
Replace V = U

_{p }/ q in E = dV / dr.
So, E = U

_{p}/ qdr = G/q. Hence, Eq = G}**(b)**Draw a tangent at the point (4.0, 14.5) obtaining a gradient of 3.6×10

^{-24}.

Field strength = (3.6×10

^{-24}) / (1.6×10^{-19}) = 2.3×10^{-5 }Vm^{-1}

__Question 995: [Dynamics > Resultant force]__
A wooden block rests on a rough
board. The end of the board is then raised until the block slides down the
plane of the board at constant velocity v.

Which row describes the forces
acting on the block when sliding with constant velocity?

frictional
force on block resultant force on
block

A down
the plane down the
plane

B down
the plane zero

C up
the plane down
the plane

D up
the plane zero

**Reference:**

*Past Exam Paper – November 2008 Paper 1 Q12 & June 2013 Paper 13 Q13*

__Solution 995:__**Answer: D.**

Since the block slides down with
constant velocity, the acceleration is zero. Thus, the resultant force on the
block is zero. [A and C are incorrect]

Frictional force opposes motion.
Since the block is moving downwards along the plane, the friction force would
be up the plane.

Please consider answering ALL of the following questions:

ReplyDelete4/O/N/02 Q.5(b),Q.6(c)(i)

6/O/N/02 Q.11(a)(b)

6/O/N/03 Q.9

04/M/J/04 Q.8(a),(b)(i),(ii)1.

06/M/J/04 Q.9(b)(iii),Q.11(b)

06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)

04/M/J/05 Q.7(a)

06/O/N/05 Q.8(b),Q.10(a)

04/M/J/06 Q.6(a),(c),Q.7(b)

06/M/J/06 Q.14(b)

04/O/N/06 Q.3(c)

06/O/N/06 Q.3(b)

05/M/J/07 Q.2(d)

04/O/N/07 Q.7(b)(i),(c),Q.10(c)

04/M/J/08 Q.5(b),Q.9(b),Q.11(a)(ii)

41/O/N/09 Q.6(a),(b)(i),Q.10

41/M/J/10 Q.7(a)

51/M/J/10 Q.2(d)

For 04/M/J/08 Q.11(a)(ii), see solution 997 at

Deletehttp://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-207.html

can u please answer june 2015 paper 11 question 15,18,23,25,26,32,33

ReplyDeleteFor Q15, see solution 1054 at

Deletehttp://physics-ref.blogspot.com/2015/11/physics-9702-doubts-help-page-221.html

This is so helpful!! Thanks a lot.

ReplyDeleteThank you for doing this bro really appreciate it

ReplyDelete