Apparently, the thicker spring has higher k, and the thinner spring has lower k. the spring with the lower k can be affected ( extended ) by small amount of force such as wind which gives an error in determining the mass. in that case, it should be kept in a rigid box to insulate it from any other external forces ( wind ). whereas in the thicker spring, the external forces are negligible because it needs much more force to be extended ( since it has a high k value). So A is the answer.arlery said:
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Physics MCQs thread.
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can anyone please explain me the relationship between upthrust, weight and air resistance of a falling body in a fluid pls i just dont seem to understand it.... please someone explain it in details...
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@hassam plz tell me the answers of above questions and plz explain that grating question
Arshiful said:
weight is the gravitational force acting on a body, it is a vector quantity which acts in the downward direction ALWAYS. it has a formula which is
weight = mass x gravity ( g = 9.81 ).
upthrust: is the upward force exerted on an object due to the pressure difference of the upper and lower surface of that object ( as they have different pressures due to different levels ) as it falls in the air or in a fluid ( higher in the fluids than in the air ). it acts upwards ALWAYS.
air resistance is the resistive force acting in the opposite direction of a falling/moving object, it increases as the velocity of the object increases. if air resistance is equal to the weight or acceleration, then the object will move/fall with steady speed.
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Zishi said:
how was ur chemistry paper??..which varient??
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What is the ans is it C?
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The answer is D
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xHazeMx said:
Ok, my reason is completely different! The one with the lower constant is kept in the box,cause : Picture this, as you add small amounts of weight, the spring with the lower k will stretch, instead of the higher k. So as we keep adding small masses, which keep increasing in weight, the spring with the lower K keeps stretching, until a point where it comes in contact with the box, preventing it from stretching further. So when larger masses ( the masses which are added after the spring with the lower K has fully stretched) are added, the spring with the HIGHER k , will start to stretch, which will result in lower sensitivity for the "large" masses.
The opposite would have happened with arrangement C, where the spring with the lower K will keep on stretching and eventually break!
Hope you got it! :friends:
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D is the answer for this.
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Since path difference for destructive interference is
= S2P -S1P
=(2n-1)lamda/2
S2P = (I1 + 12) & S1P = (I2 + I4)
we replace n with m and we get (I1 + 12) - (I2 + I4) = (2m+1)lamda/2
Hope you get it.
= S2P -S1P
=(2n-1)lamda/2
S2P = (I1 + 12) & S1P = (I2 + I4)
we replace n with m and we get (I1 + 12) - (I2 + I4) = (2m+1)lamda/2
Hope you get it.
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a b c are not correct because they do not give the path difference.. E is for constructive interference.. so its D!
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is the answer A? for revolution one?
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C is the answer for TRAIN question!