Thursday, March 26, 2015

Physics 9702 Doubts | Help Page 95

  • Physics 9702 Doubts | Help Page 95



Question 489: [Forces > Equilibrium]
Hinged trapdoor is held closed in the horizontal position by cable.
Three forces act on the trapdoor: weight W of the door, tension T in the cable and force H at the hinge.

Which list gives the three forces in increasing order of magnitude?
A H,T,W                     B T,H,W                     C W,H,T                     D W,T,H

Reference: Past Exam Paper – June 2013 Paper 11 Q16



Solution 489:
Answer: C.
For the trapdoor to stay horizontal (in equilibrium), the resultant force on it should be zero.

The weight W is purely vertical and acts downwards. The force H at the hinge can be resolved with a vertically component downwards and a horizontal component to the right.

The tension T can be resolved with a vertical component upwards and a horizontal component to the right.

Since the trapdoor is in equilibrium, the components of the forces should balance. The upward component of the tension should be equal to the sum of downward component of H + weight W. From here itself, it can be deduced that tension T is greater than weight W since just the vertically component of T is greater than W on its own. [A and B are incorrect]

So, the vertical component of T is also greater than the vertical component of H on its own. Now, for equilibrium, the horizontal components of T and H should be equal. Thus, greater T is greater than H. [D is incorrect]










Question 490: [Waves > Interference]
Teacher sets up apparatus shown to demonstrate a two-slit interference pattern on the screen.

Which change to the apparatus will increase fringe spacing?
A decreasing the distance p
B decreasing the distance q
C decreasing the distance r
D decreasing the wavelength of the light

Reference: Past Exam Paper – June 2013 Paper 12 Q28



Solution 490:
Answer: B.
For a double slit interference experiment: q sinθ = nλ

If the screen separation from the double slit, r is much greater than the slit separation, q (that is, r >> q), sinθ can be approximated to θ (sinθ ≈ θ)
qθ = nλ
Angle θ of nth fringe = nλ / q

Example:
Angle θ1 of 1st fringe = λ / q
Angle θ2 of 2nd fringe = 2λ / q
Fringe separation = θ2 – θ1 = λ / q

Angular spacing between successive fringes (fringe spacing) = λ/q










Question 491: [Kinematics > Linear motion]
(a) Sphere of radius R is moving through fluid with constant speed v. There is frictional force F acting on sphere, which is given by expression
F = 6πDRv
where D depends on the fluid
(i) Show that SI base units of quantity D are kgm-1s-1.
(ii) Raindrop of radius 1.5mm falls vertically in air at velocity of 3.7ms-1. Value of D for air = 6.6x10-4kgm-1s-1. Density of water, ρ is 1000kgm-3.
Calculate
1. Magnitude of frictional force F
2. Acceleration of raindrop

(b) Variation with time t of speed v of raindrop is shown in Fig.

(i) State variation with t of acceleration of raindrop:
(ii) A 2nd raindrop has radius that is smaller than in (a). On Fig, Sketch variation of speed with time for this 2nd raindrop:

Reference: Past Exam Paper – June 2011 Paper 22 Q2



Solution 491:
(a)
(i)
Base units of Force: kg ms-2
Radius: m                    Velocity: ms-1
Base units of D: [F/(R x v)]: kgms-2 / (m x ms-1) = kgm-1s-1

(ii)
1. F = 6πDRv = 6π x 6.6x10-4 x 1.5x10-3 x 3.7 = 6.9x10-5N

2.
Resultant force acting on raindrop: mg – F = ma       hence   a = g – [F/m]
R = 1.5x10-3m
m = ρV = ρ[(4/3)πR3] = (1.4x10-5kg)  ]
a = 9.81 – ([6.9x10-5] / ρ[(4/3)π(1.5x10-3)3])               (9.81 – 4.88)
a = 4.9(3)ms-2

(b)
(i) The acceleration is equal to the acceleration of free fall (a = g) at time t = 0. (since v = 0, friction which depends on v is also zero. So, a =g at t=0) The acceleration decreases as time increases (can be seen from graph in which gradient [which represents the acceleration] decreases. This due to frictional force.) until it goes to zero. (here, terminal velocity is reached since gradient is zero)

(ii) Correct shape below the original line. The sketch goes to terminal velocity earlier.



{For terminal velocity, frictional force = weight. The frictional force at any time depends on the velocity which changes according to the resultant force acting on it. So, let’s forget about it for now.

A smaller radius means that the volume and hence the mass of the raindrop is smaller.
Weight W = mg
Since mass m is now smaller, the weight is also smaller.
But remember that for terminal velocity, frictional force = weight.
Thus the frictional force required now for terminal velocity is smaller. Since F = 6πDRv, and for any particular raindrop only the velocity changes, we can conclude than as the frictional force F is smaller for terminal velocity the speed v at which terminal velocity occurs is also smaller. Hence, terminal velocity occurs at a smaller speed.



Now, the raindrop initially accelerates due to gravity until it reaches terminal velocity. But since the terminal velocity is now smaller, it would take less time to be reached.}








Question 492: [Kinematics > Linear motion]
Acceleration of free fall on Moon is one-sixth of that on Earth.
On Earth it takes time t for stone to fall from rest a distance of 2 m.
What is the time taken for stone to fall from rest a distance of 2 m on the Moon?
A 6t                             B t / 6                          C t √6                          D t / √6

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q5



Solution 492:
Answer: C.
Let the acceleration of free fall on Earth = g
Acceleration of free fall on Moon = g / 6

The stone is release from rest at a height of 2m.
Initial velocity, u = 0. Distance, s = 2m

Equation for uniformly accelerated motion: s = ut + ½ at2

On Earth, the time taken is t
2 = 0 + 0.5gt2
Acceleration, g = 4 / t2

Now, consider the stone falling on the Moon. Let the time taken be tM.
2 = 0 + 0.5 (g/6) tM2
Time tM = (4/g) = √(24/[4/t2]) = √(6/t2) = t√6  










Question 493: [Current of Electricity]
Power supply of electromotive force (e.m.f.) 12 V and internal resistance 2 Ω is connected in series with a load resistor. Value of the load resistor is varied from 0.5 Ω to 4 Ω.
Which graph shows how power P dissipated in the load resistor varies with the resistance of the load resistor?


Reference: Past Exam Paper – November 2011 Paper 12 Q35



Solution 493:
Answer: A.
e.m.f. of supply = 12V
Internal resistance, r = 2Ω

The resistance of the load resistor, R varies from 0.5 Ω to 4 Ω.

Power P dissipated in the load resistor = I2R

Total resistance in the circuit = R + r
From Ohm’s law: V = I (R+r)
So, current I = V / (R+r)
Power P dissipated = [V / (R+r)]2 R

Consider the point (r,P) [that is, internal resistance r on the x-axis and power P dissipated on y-axis].
Power P dissipated = [V / (R+r)]2 R
where V = 12V, r = 2 Ω and R varies from 0.5 Ω to 4 Ω

The following points are obtained (0.5, 11.52), (1, 16), (2, 18), (3, 17.28), (4, 16)










Question 494: [Forces > Moments]
A man holds 100 N load stationary in his hand. Combined weight of the forearm and hand is 20 N. Forearm is held horizontal, as shown.

What is vertical force F needed in the biceps?
A 750 N                      B 800 N                      C 850 N                      D 900 N

Reference: Past Exam Paper – June 2014 Paper 11 Q12



Solution 494:
Answer: C.
Equilibrium is reached since the load is stationary in his hand.

For equilibrium, the sum of clockwise moment is equal to the sum of anticlockwise moment.
Clockwise moment = 20(10) + 100(32)
Anticlockwise moment = F(4)

20(10) + 100(32) = F(4)
Vertical force F = [20(10) + 100(32)] / 4 = 850N

7 comments:

  1. thanks. these really help me a lot

    ReplyDelete
  2. What about the other condition for equilibrium? Shouldn't the upward forces be equal to the downward forces when the system is in equilibrium? In which case, Force in biceps = 100 + 20 = 120 N. How can this be different to the answer when the principle of moments is used? Puzzled.

    ReplyDelete
    Replies
    1. This question deals with moment since the forces are not along the same line.

      Even if the force of 120N acted on the biceps, it's still not on the load. i.e., the force on the load is sill only 100N.

      But this is not the case here. We want to know the force required to keep the forearm horizontal.

      Delete
  3. In solution 491 b ii) can you explain how you drew the curve??

    ReplyDelete
    Replies
    1. I have added an explanation above. See if it helps

      Delete
  4. In solution 489, the horizontal component of W has not been considered. I couldn't understand which had a greater horizontal component

    ReplyDelete
    Replies
    1. W is vertically downwards, so it has no horizontal components.

      Delete

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