Physics MCQs thread.

[Octahedral]

:%) :%)

 

[Hateexams93]

WHHY ITS A???how is I1 >I2 ?

Imagine

pipes and water

when the diameter of the pipes are small the ease of flow of water is less
whereas when the diameter of the pipes are bigger the ease of flow of water is much more

so when the value of resistor is high it means that the flow of current is reduced
and when value of resistor is small more current can flow

so, current through 3 ohm resistor is less than 2 ohm
I1<I2

now find the combined resistance of 3and6 ohm and 2and2 ohm resistors individually

comb resistance of 3and6=(3*6)/(3+6)=2-------------------(1)
comb resistance of 2and2=(2*2)/(2+2)=1-------------------(2)

as we know PD is proportional to R(1)>R(2)

so V1>V2 therefore answer C

THATS THE THING , the answer is A

 

[Hateexams93]

WHHY ITS A???how is I1 >I2 ?

IT's C as the others said . Current is divided equally in the Second one . And as you know Greater Resistance = Greater p.d

the answer is A -.-

 

[kakarocks]

It can't be , MS please!

 

[Hateexams93]

It can't be , MS please!

http://www.xtremepapers.me/CIE/Internat ... 4_ms_1.pdf
Q 37 , oct /nov 2004

 

[xHazeMx]

WHHY ITS A???how is I1 >I2 ?

IT's C as the others said . Current is divided equally in the Second one . And as you know Greater Resistance = Greater p.d

the answer is A -.-

A is correct .. the total current in loop 1 is equal to the total current in loop 2. the ratio of resistances in loop 1 is 1:2 so more current will be dissipated in the 3 ohms resistor in THIS loop as it is the best route which is 2/3 of the total current ( as I is inversely proportional to R ) . in the second loop, the ratio of resistances is 1:1 so either ways, 1/2 of the total current will be passing through the 2 ohms resistor which makes it
2/3 > 1/2
I1 > I2

 

[Arshiful]

WHHY ITS A???how is I1 >I2 ?

Imagine

pipes and water

when the diameter of the pipes are small the ease of flow of water is less
whereas when the diameter of the pipes are bigger the ease of flow of water is much more

so when the value of resistor is high it means that the flow of current is reduced
and when value of resistor is small more current can flow

so, current through 3 ohm resistor is less than 2 ohm
I1<I2

now find the combined resistance of 3and6 ohm and 2and2 ohm resistors individually

comb resistance of 3and6=(3*6)/(3+6)=2-------------------(1)
comb resistance of 2and2=(2*2)/(2+2)=1-------------------(2)

as we know PD is proportional to R(1)>R(2)

so V1>V2 therefore answer C

THATS THE THING , the answer is A

In my explanation I said I1<I2 and V1>V2 so C

 

[xHazeMx]

:%) :%)

B

 

[Hateexams93]

In my explanation I said I1<I2 and V1>V2 so C

BUT ITS wrong....the answer is A I1>I2

 

[xHazeMx]

In my explanation I said I1<I2 and V1>V2 so C

BUT ITS wrong....the answer is A I1>I2

check my answer for ur question in the above posts

 

[Hateexams93]

A is correct .. the total current in loop 1 is equal to the total current in loop 2. the ratio of resistances in loop 1 is 1:2 so more current will be dissipated in the 3 ohms resistor in THIS loop as it is the best route which is 2/3 of the total current ( as I is inversely proportional to R ) . in the second loop, the ratio of resistances is 1:1 so either ways, 1/2 of the total current will be passing through the 2 ohms resistor which makes it
2/3 > 1/2
I1 > I2

YAYYYYYaa thank u , i got it now

 

[xHazeMx]

A is correct .. the total current in loop 1 is equal to the total current in loop 2. the ratio of resistances in loop 1 is 1:2 so more current will be dissipated in the 3 ohms resistor in THIS loop as it is the best route which is 2/3 of the total current ( as I is inversely proportional to R ) . in the second loop, the ratio of resistances is 1:1 so either ways, 1/2 of the total current will be passing through the 2 ohms resistor which makes it
2/3 > 1/2
I1 > I2

YAYYYYYaa thank u , i got it now

you welcome

 

[hassam]

A is the answer for this hassam .. antinodes are formed at the open end!

its C...COS a doesnt shows reflected and incident wave

 

[xHazeMx]

A is the answer for this hassam .. antinodes are formed at the open end!

its C...COS a doesnt shows reflected and incident wave

:Yahoo!: and the second is D ??

 

[Arshiful]

A is correct .. the total current in loop 1 is equal to the total current in loop 2. the ratio of resistances in loop 1 is 1:2 so more current will be dissipated in the 3 ohms resistor in THIS loop as it is the best route which is 2/3 of the total current ( as I is inversely proportional to R ) . in the second loop, the ratio of resistances is 1:1 so either ways, 1/2 of the total current will be passing through the 2 ohms resistor which makes it
2/3 > 1/2
I1 > I2

YAYYYYYaa thank u , i got it now

you welcome

Thanks I got it too....

 

[xHazeMx]

Thanks I got it too....

you welcome too

 

[Banglarbagh]

ki netaji kecco bash ei khane ki koros..???

 

[Arshiful]

Summer 04 #9 pls some one explain this ques to me in details......

 

[hassam]

 

[xHazeMx]

Summer 04 #9 pls some one explain this ques to me in details......

confirm it first, if it's right. i will post my workings. is it D ?