Physics: Post your doubts here!

[amina1300]

Post the hardest MCQs !!

 

[Holmes]

Hope it helps

Thank you !Thelastmoment

 

[Holmes]

Post the hardest MCQs !!

Here you go


thanks in advance

 

[Holmes]

Post the hardest MCQs !!

 

[Holmes]

Post the hardest MCQs !!

 

[amina1300]

Here you go
View attachment 62310
thanks in advance

The first one was a bit tricky the others were easy.

Do u want me to show you how I got the answers?

Thanks for posting btw.

 

[amina1300]

Post the MCQs you found the hardest! I want to solve them for practice

 

[dumbledore.]

In one such pump the long approach pipe holds 500kg of water. A valve shuts when the speed of
this water reaches 2.0ms–1 and the kinetic energy of this water is used to lift a small quantity of
water by a height of 15m.
The efficiency of the pump is 10%.
Which mass of water could be lifted 15m?
A 0.15kg B 0.68kg C 1.5kg D 6.8kg

 

[Thelastmoment]

In one such pump the long approach pipe holds 500kg of water. A valve shuts when the speed of
this water reaches 2.0ms–1 and the kinetic energy of this water is used to lift a small quantity of
water by a height of 15m.
The efficiency of the pump is 10%.
Which mass of water could be lifted 15m?
A 0.15kg B 0.68kg C 1.5kg D 6.8kg

Answer:

I guess,
K . E= 1/2 x m x v2(squared)
= 1/2 x 500 x 2 (2)
=1000 J

then,
G.p.e = mgh
re-arrange to
m= G.p.e/( g x h)
=1000J / (9.81 x 15 )
= 6.8kg


6.8kg for 100%
m1 kg for 10%
cross multiply to get 0.68kg so answer is B.

 

[Bishnu Dev]

Solve this please... Answer is C. I get A

 

[Bishnu Dev]

Here you go
View attachment 62310

thanks in advance

answer is C for 13? Which year paper is this and what is the component code? I got 'C' for the question number 28 as well and 'B' for question number 10.

 

[amina1300]

answer is C for 13? Which year paper is this and what is the component code? I got 'C' for the question number 28 as well and 'B' for question number 10.

Your right, and this is March 2017.

 

[amina1300]

Solve this please... Answer is C. I get A

You get diameter 4.81mm so for radius d/2 and convet mm to m.
In m 4.81/2/1000
Next u find the volume using 4/3pi r^3
Density =mass(kg)/volume(m^3)
Convert g to kg 0.450/1000 ÷vol and you get C as the answer.

 

[amina1300]

que no 13 and 14 as well please

Q13 F1d1 + F2d2 =Ftdt
16000*0.6 +9000*2.1 =(16000+9000)*
28500÷2500=s
d=1.14m

 

[Bishnu Dev]

You get diameter 4.81mm so for radius d/2 and convet mm to m.
In m 4.81/2/1000
Next u find the volume using 4/3pi r^3
Density =mass(kg)/volume(m^3)
Convert g to kg 0.450/1000 ÷vol and you get C as the answer.

Thanks! i forgot to calculate radius and was recklessly using value of D in the formula (4/3)*pi*r^3

 

[amina1300]

Thanks! i forgot to calculate radius and recklessly using value of D in the formula (4/3)*pi*r^3

Did you solve Q14, march 2017 ?
Do Post it.

 

[Bishnu Dev]

Did you solve Q14, march 2017 ?
Do Post it.

Component Code?

 

[Shimmery woods]

Q13 F1d1 + F2d2 =Ftdt
16000*0.6 +9000*2.1 =(16000+9000)*
28500÷2500=s
d=1.14m

thankyou so much!

 

[Shimmery woods]

Component Code?

variant 12

 

[Bishnu Dev]

Well i am not sure if this is the question you're talking about (I could only find this component for 2017 March). If so, the answer is C. Three forces in equilibrium forms a closed triangle. Since the weight of the object is 10N and the angle is 90 degrees, by using Pythagoras theorem Weight = (F1^2 + F2^2)^(1/2). ''C' and 'D' only satisfies this equation. Now if you compare the angle between the vertical line and the rope of F1 attached to the object, you will notice that F1 makes greater angle than F2. With smaller angle, Value of Cos theta increases hence force also increases (F = R*Cos theta). So F2 has greater force