# 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 n

^{th}fringe = nλ / q
Example:

Angle θ

_{1}of 1^{st}fringe = λ / q
Angle θ

_{2}of 2^{nd}fringe = 2λ / q
Fringe separation = θ

_{2}– θ_{1}= 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

^{-1}s^{-1}.
(ii) Raindrop of radius 1.5mm falls
vertically in air at velocity of 3.7ms

^{-1}. Value of D for air = 6.6x10^{-4}kgm^{-1}s^{-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^{-1}s^{-1}
(ii)

1. F = 6πDRv = 6π x 6.6x10

^{-4}x 1.5x10^{-3}x 3.7 = 6.9x10^{-5}N
2.

Resultant force acting on raindrop:
mg – F = ma hence a = g – [F/m]

R = 1.5x10

^{-3}m
m = ρV
= ρ[(4/3)πR

^{3}] = (1.4x10^{-5}kg)**]**
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.

__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 at

^{2}
On Earth, the time taken is t

2 = 0 + 0.5gt

^{2}
Acceleration, g = 4 / t

^{2}
Now, consider the stone falling on
the Moon. Let the time taken be t

_{M}.
2 = 0 + 0.5 (g/6) t

_{M}^{2}
Time t

_{M}= √(4/g) = √(24/[4/t^{2}]) = √(6/t^{2}) = 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 = I

^{2}R

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

thanks. these really help me a lot

ReplyDeleteWhat 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.

ReplyDeleteThis question deals with moment since the forces are not along the same line.

DeleteEven 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.

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

ReplyDeleteI have added an explanation above. See if it helps

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

ReplyDeleteW is vertically downwards, so it has no horizontal components.

Delete