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Monday, July 23, 2018

The diagram shows an electric circuit in which the resistance of the external resistor is 2R and the internal resistance of the source is R.


Question 7
The diagram shows an electric circuit in which the resistance of the external resistor is 2R and the internal resistance of the source is R.


What is the ratio          power in external resistance / power in internal resistor ?
A ¼                 B ½                 C 2                  D 4





Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q33





Solution:
Answer: C.

P = I2R
where I is the current flowing through resistor R


In the circuit, the same current I flows as the resistor may be considered to be connected in series to the battery.

Pext = I2(2R) = 2 I2R

Pint = I2R


So, ratio = Pext / Pint = 2 I2R / I2R = 2

Friday, July 13, 2018

A uniform solid block has weight 500 N, width 0.4 m and height 0.6 m. The block rests on the edge of a step of depth 0.8 m, as shown.


Question 8
A uniform solid block has weight 500 N, width 0.4 m and height 0.6 m. The block rests on the edge of a step of depth 0.8 m, as shown.


The block is knocked over the edge of the step and rotates through 90° before coming to rest with the 0.6 m edge horizontal.

What is the change in gravitational potential energy of the block?
A 300 J                        B 400 J                        C 450 J                       D 550 J





Reference: Past Exam Paper – June 2015 Paper 12 Q15





Solution:
Answer: C.

Since the block is said to be uniform, the centre of mass can be considered to be at the centre.


At the top of the step,
Height of position of centre of mass above ground = 0.8 + (0.6/2) = 0.8 + 0.3 = 1.1 m


At the bottom of the step,
Height of position of centre of mass above ground = 0.4/2 = 0.2 m


Change in GPE = mgΔh = 500 × (1.1 – 0.2) = 450 J

Saturday, July 7, 2018

Three cells with e.m.f.s V1, V2 and V3, have negligible internal resistance. These cells are connected to three resistors with resistances R1, R2 and R3, as shown.


Question 6
Three cells with e.m.f.s V1, V2 and V3, have negligible internal resistance. These cells are
connected to three resistors with resistances R1, R2 and R3, as shown.


The current in the circuit is I.

Which equation is correct?
A V1 + V2 + V3 = I (R1 + R2 + R3)
B V1 + V2 V3 = I (R1 + R2 + R3)
C V1 V2 + V3 = I (R1 + R2 + R3)
D V1 V2 V3 = I (R1 + R2 + R3)





Reference: Past Exam Paper – March 2016 Paper 12 Q34





Solution:
Answer: D.

The sum of p.d. across the resistors in the circuit is equal to the overall e.m.f. in the circuit.

The current I is shown to flow from the positive terminal of cell V1. So, cell V1 is taken as the reference.


For cells to be connected correctly, the positive terminal of one cell should be connected to the negative terminal of the other. But the negative terminal of V3 is connected to the negative terminal of V1 and also, the positive terminal of V2 is connected to the positive terminal of V1. Thus, both V2 and V3 are wrongly connected to V1 and hence, the two cells are reducing.
 

Overall p.d. = V1 – V2 – V3 = I(R1 + R2 + R3)
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