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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 n^th fringe = nλ / q

Example:

Angle θ₁ of 1^st fringe = λ / q

Angle θ₂ of 2^nd 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 2^nd raindrop has radius that is smaller than in (a). On Fig, Sketch variation of speed with time for this 2^nd 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 t_M.

2 = 0 + 0.5 (g/6) t_M²

Time t_M = √(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:
Go to
A power supply of electromotive force (e.m.f.) 12 V and internal resistance 2 Ω is connected in series with a load resistor.









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