Physics MCQs thread.

[Hateexams93]

can some1 explain this

 

[yubakkk]

i think this solution i in previous page u can look hatexam..........

 

[ShootingStar]

no, its B.
any1 plz explain...

ok..

E = mcT

W = fd
W = kgms^-2 x m
W = kgm^2s^-2

E = m x c x T
kgm^2s^-2 = kg x c x K

kgm^2s^-2
------------ = c
kg x K

m^2 x s^-2 x K^-1 = c

c = bT^3

m^2 x s^-2 x K^- 1
---------------------- = b
K^3

b = m^2s^-2k^-4

 

[Hateexams93]

i think this solution i in previous page u can look hatexam..........

which page???? i coudn't find it

 

[arlery]

It's on page 1.

 

[Hateexams93]

It's on page 1.

no its not..thats a different one

 

[yubakkk]

ok then see here
v1=at1 becoz u=o so
fer dist h
h=at1*(t2-t1) + 1/2*a*(t2-t1)^2
so u will solve 2 get a=option d
hope u have understand
if not i will xplain again clearly with full solution

 

[yubakkk]

have u understand hatexam????

 

[Hateexams93]

have u understand hatexam????

full working plz

 

[yubakkk]

ok
v1=at1 becoz u=o so
fer dist h
h=at1*(t2-t1) + 1/2*a*(t2-t1)^2
h=at1t2-at1^2+1/2a(t2^2-2*t1*t2+t2^2)
h=-1/2at1^2+1/2at2^2
2h=a(t2^2-t1^2)
a=2h/(t2^2-t1^2)
option d
hope now clear if not come on
http://www.facebook.com/yubak.ghimire i will make u more clear

 

[yubakkk]

have u underdtand now???hatexam??

 

[Zishi]

can some1 explain this

 

[arlery]

What about this one? How can I solve this??

 

[Zishi]

What about this one? How can I solve this??

We've to estimate the time in which a wave completes. This is 0.7 cm.

Multiply 0.7 by 10ms/cm and you'll get time period in milliseconds. To convert that is seconds, divide it by thousand.

Frequency = 1/Time period in seconds
Check which of the answers closely matches your calculation's result.
I get B as the answer.

 

[unique840]

What about this one? How can I solve this??

1 box is 10ms
n the wavelenght is about 0.7 box
to 0.7*10 = 7ms
7*10^-3 = time period
1/time= frequency.

 

[Hateexams93]

An area of land is an average of 2.0 m below sea level. To prevent flooding, pumps are used to lift
rainwater up to sea level.
What is the minimum pump output power required to deal with 1.3 × 109 kg of rain per day?
A 15 kW B 30 kW C 150 kW D 300kW

 

[Zishi]

An area of land is an average of 2.0 m below sea level. To prevent flooding, pumps are used to lift
rainwater up to sea level.
What is the minimum pump output power required to deal with 1.3 × 109 kg of rain per day?
A 15 kW B 30 kW C 150 kW D 300kW

Power = Work Done/Time

Work done = increase in potential energy

Increase in PE = mgh = 1.3x10^9 times 9.81 times 2 = 2.55 x 10^10

Seconds in a day = 24 x 3600 = 86400

Power = (2.55 x 10^10)/86400 = 300kW

D is the answer.

 

[Hateexams93]

lol i got 30 * 10^4 and was like but i don't have that option ..hahha ..so stupid of me , but yeah sometimes happens ..anyways thanx zishi

 

[Hateexams93]

can some1 plz explain this

 

[Zishi]

can some1 plz explain this

D, because increasing resistance of rheostat will decrease the potential drop across XY, as p.d across itself will increase. As the length of wire increases, the resistance by wire increases. To get null deflection, we'd have to increase p.d across XN and this can only be done by moving the slider towards Y.